1

I am trying to type this extremely long equation without success. I got this result out of Mathematica and copied it. For some reason, the parentheses are not changing their shape according to the fraction's height.

I tried using an automatic line brake with \usepackage{breqn} and \begin{dmath} without success.

enter image description here

-\frac{2 u_g \cosh \left(\frac{h \pi }{D}\right) \sin \left(\frac{h \pi }{D}\right) \sin \left(\frac{\pi  z}{D}\right) \sinh \left(\frac{\pi  z}{D}\right) \cos ^2\left(\frac{h \pi }{D}\right)}{\left(\cos \left(\frac{2 h \pi }{D}\right)+\cosh \left(\frac{2 h \pi }{D}\right)\right) \left(\cos ^2\left(\frac{h \pi }{D}\right) \cosh ^2\left(\frac{h \pi }{D}\right)+\sin ^2\left(\frac{h \pi }{D}\right) \sinh ^2\left(\frac{h \pi }{D}\right)\right) \left(\frac{\cosh \left(\frac{h \pi }{D}\right) \sinh \left(\frac{h \pi }{D}\right)}{\cos ^2\left(\frac{h \pi }{D}\right) \cosh ^2\left(\frac{h \pi }{D}\right)+\sin ^2\left(\frac{h \pi }{D}\right) \sinh ^2\left(\frac{h \pi }{D}\right)}-\frac{\cos \left(\frac{h \pi }{D}\right) \sin \left(\frac{h \pi }{D}\right)}{\cos ^2\left(\frac{h \pi }{D}\right) \cosh ^2\left(\frac{h \pi }{D}\right)+\sin ^2\left(\frac{h \pi }{D}\right) \sinh ^2\left(\frac{h \pi }{D}\right)}\right)}-\frac{4 \pi  \tau_y \cosh ^2\left(\frac{h \pi }{D}\right) \sin \left(\frac{\pi  z}{D}\right) \sinh \left(\frac{\pi  z}{D}\right) \cos ^2\left(\frac{h \pi }{D}\right)}{D f \rho_0 \left(\cos \left(\frac{2 h \pi }{D}\right)+\cosh \left(\frac{2 h \pi }{D}\right)\right) \left(\cos ^2\left(\frac{h \pi }{D}\right) \cosh ^2\left(\frac{h \pi }{D}\right)+\sin ^2\left(\frac{h \pi }{D}\right) \sinh ^2\left(\frac{h \pi }{D}\right)\right) \left(\frac{\cosh \left(\frac{h \pi }{D}\right) \sinh \left(\frac{h \pi }{D}\right)}{\cos ^2\left(\frac{h \pi }{D}\right) \cosh ^2\left(\frac{h \pi }{D}\right)+\sin ^2\left(\frac{h \pi }{D}\right) \sinh ^2\left(\frac{h \pi }{D}\right)}-\frac{\cos \left(\frac{h \pi }{D}\right) \sin \left(\frac{h \pi }{D}\right)}{\cos ^2\left(\frac{h \pi }{D}\right) \cosh ^2\left(\frac{h \pi }{D}\right)+\sin ^2\left(\frac{h \pi }{D}\right) \sinh ^2\left(\frac{h \pi }{D}\right)}\right)}+\frac{4 h \pi  u_g \cosh \left(\frac{h \pi }{D}\right) \sin \left(\frac{\pi  z}{D}\right) \sinh \left(\frac{\pi  z}{D}\right) \cos \left(\frac{h \pi }{D}\right)}{D \left(\cos \left(\frac{2 h \pi }{D}\right)+\cosh \left(\frac{2 h \pi }{D}\right)\right) \left(\frac{\cosh \left(\frac{h \pi }{D}\right) \sinh \left(\frac{h \pi }{D}\right)}{\cos ^2\left(\frac{h \pi }{D}\right) \cosh ^2\left(\frac{h \pi }{D}\right)+\sin ^2\left(\frac{h \pi }{D}\right) \sinh ^2\left(\frac{h \pi }{D}\right)}-\frac{\cos \left(\frac{h \pi }{D}\right) \sin \left(\frac{h \pi }{D}\right)}{\cos ^2\left(\frac{h \pi }{D}\right) \cosh ^2\left(\frac{h \pi }{D}\right)+\sin ^2\left(\frac{h \pi }{D}\right) \sinh ^2\left(\frac{h \pi }{D}\right)}\right)}+\frac{4 \pi  \tau_y \cosh \left(\frac{h \pi }{D}\right) \sin \left(\frac{\pi  z}{D}\right) \sinh \left(\frac{\pi  z}{D}\right) \cos \left(\frac{h \pi }{D}\right)}{D f \rho_0 \left(\cos \left(\frac{2 h \pi }{D}\right)+\cosh \left(\frac{2 h \pi }{D}\right)\right) \left(\frac{\cosh \left(\frac{h \pi }{D}\right) \sinh \left(\frac{h \pi }{D}\right)}{\cos ^2\left(\frac{h \pi }{D}\right) \cosh ^2\left(\frac{h \pi }{D}\right)+\sin ^2\left(\frac{h \pi }{D}\right) \sinh ^2\left(\frac{h \pi }{D}\right)}-\frac{\cos \left(\frac{h \pi }{D}\right) \sin \left(\frac{h \pi }{D}\right)}{\cos ^2\left(\frac{h \pi }{D}\right) \cosh ^2\left(\frac{h \pi }{D}\right)+\sin ^2\left(\frac{h \pi }{D}\right) \sinh ^2\left(\frac{h \pi }{D}\right)}\right)}+\frac{2 \pi  \tau_y \cos \left(\frac{\pi  (h+z)}{D}\right) \cosh \left(\frac{h \pi }{D}\right) \sinh \left(\frac{\pi  (h+z)}{D}\right) \cos \left(\frac{h \pi }{D}\right)}{D f \rho_0 \left(\cos \left(\frac{2 h \pi }{D}\right)+\cosh \left(\frac{2 h \pi }{D}\right)\right)}-\frac{2 u_g \cos \left(\frac{\pi  z}{D}\right) \cosh \left(\frac{h \pi }{D}\right) \cosh \left(\frac{\pi  z}{D}\right) \cos \left(\frac{h \pi }{D}\right)}{\cos \left(\frac{2 h \pi }{D}\right)+\cosh \left(\frac{2 h \pi }{D}\right)}-\frac{2 \pi  \tau_y \cosh \left(\frac{h \pi }{D}\right) \cosh \left(\frac{\pi  (h+z)}{D}\right) \sin \left(\frac{\pi  (h+z)}{D}\right) \cos \left(\frac{h \pi }{D}\right)}{D f \rho_0 \left(\cos \left(\frac{2 h \pi }{D}\right)+\cosh \left(\frac{2 h \pi }{D}\right)\right)}+\frac{2 u_g \cos \left(\frac{\pi  z}{D}\right) \cosh \left(\frac{\pi  z}{D}\right) \sin ^2\left(\frac{h \pi }{D}\right) \sinh \left(\frac{h \pi }{D}\right) \cos \left(\frac{h \pi }{D}\right)}{\left(\cos \left(\frac{2 h \pi }{D}\right)+\cosh \left(\frac{2 h \pi }{D}\right)\right) \left(\cos ^2\left(\frac{h \pi }{D}\right) \cosh ^2\left(\frac{h \pi }{D}\right)+\sin ^2\left(\frac{h \pi }{D}\right) \sinh ^2\left(\frac{h \pi }{D}\right)\right) \left(\frac{\cosh \left(\frac{h \pi }{D}\right) \sinh \left(\frac{h \pi }{D}\right)}{\cos ^2\left(\frac{h \pi }{D}\right) \cosh ^2\left(\frac{h \pi }{D}\right)+\sin ^2\left(\frac{h \pi }{D}\right) \sinh ^2\left(\frac{h \pi }{D}\right)}-\frac{\cos \left(\frac{h \pi }{D}\right) \sin \left(\frac{h \pi }{D}\right)}{\cos ^2\left(\frac{h \pi }{D}\right) \cosh ^2\left(\frac{h \pi }{D}\right)+\sin ^2\left(\frac{h \pi }{D}\right) \sinh ^2\left(\frac{h \pi }{D}\right)}\right)}-\frac{2 u_g \cosh ^2\left(\frac{h \pi }{D}\right) \sin \left(\frac{\pi  z}{D}\right) \sinh \left(\frac{h \pi }{D}\right) \sinh \left(\frac{\pi  z}{D}\right) \cos \left(\frac{h \pi }{D}\right)}{\left(\cos \left(\frac{2 h \pi }{D}\right)+\cosh \left(\frac{2 h \pi }{D}\right)\right) \left(\cos ^2\left(\frac{h \pi }{D}\right) \cosh ^2\left(\frac{h \pi }{D}\right)+\sin ^2\left(\frac{h \pi }{D}\right) \sinh ^2\left(\frac{h \pi }{D}\right)\right) \left(\frac{\cosh \left(\frac{h \pi }{D}\right) \sinh \left(\frac{h \pi }{D}\right)}{\cos ^2\left(\frac{h \pi }{D}\right) \cosh ^2\left(\frac{h \pi }{D}\right)+\sin ^2\left(\frac{h \pi }{D}\right) \sinh ^2\left(\frac{h \pi }{D}\right)}-\frac{\cos \left(\frac{h \pi }{D}\right) \sin \left(\frac{h \pi }{D}\right)}{\cos ^2\left(\frac{h \pi }{D}\right) \cosh ^2\left(\frac{h \pi }{D}\right)+\sin ^2\left(\frac{h \pi }{D}\right) \sinh ^2\left(\frac{h \pi }{D}\right)}\right)}+\frac{4 \pi  \tau_y \cos \left(\frac{\pi  z}{D}\right) \cosh \left(\frac{h \pi }{D}\right) \cosh \left(\frac{\pi  z}{D}\right) \sin \left(\frac{h \pi }{D}\right) \sinh \left(\frac{h \pi }{D}\right) \cos \left(\frac{h \pi }{D}\right)}{D f \rho_0 \left(\cos \left(\frac{2 h \pi }{D}\right)+\cosh \left(\frac{2 h \pi }{D}\right)\right) \left(\cos ^2\left(\frac{h \pi }{D}\right) \cosh ^2\left(\frac{h \pi }{D}\right)+\sin ^2\left(\frac{h \pi }{D}\right) \sinh ^2\left(\frac{h \pi }{D}\right)\right) \left(\frac{\cosh \left(\frac{h \pi }{D}\right) \sinh \left(\frac{h \pi }{D}\right)}{\cos ^2\left(\frac{h \pi }{D}\right) \cosh ^2\left(\frac{h \pi }{D}\right)+\sin ^2\left(\frac{h \pi }{D}\right) \sinh ^2\left(\frac{h \pi }{D}\right)}-\frac{\cos \left(\frac{h \pi }{D}\right) \sin \left(\frac{h \pi }{D}\right)}{\cos ^2\left(\frac{h \pi }{D}\right) \cosh ^2\left(\frac{h \pi }{D}\right)+\sin ^2\left(\frac{h \pi }{D}\right) \sinh ^2\left(\frac{h \pi }{D}\right)}\right)}+u_g

Any suggestions?

  • 3
    The size of the brackets might be related to the used font. But as someone who is somewhat accustomed to long, horrible equations, my only suggestion is to rewrite this equation. Introduce some abbreviation for the two quantities h\pi/D and \pi z/D and for other recurring combinations; factorize long denominators. I fear that no splitting will make this equation readable. – campa May 22 at 16:17
  • 1
    @campa yes, that was my fear, there's also some common expressions as denominators so that in fact may be the only solution. Thanks! – ASPVL May 22 at 16:21
  • 1
    I would remove all the left and right to allow breaking and add \renewcommand\frac[2]{(#1)/(#2)}` to inline fractions, then set it in inline math and allow it to wrap over lines, there are some examples on site I'll find a link – David Carlisle May 22 at 17:51
  • @campa - Good piece of advice to introduce abbreviations and to factorize the denominator terms. :-) – Mico May 23 at 5:45
3

Something like this?

enter image description here

\documentclass{article}
\usepackage[letterpaper,margin=1in]{geometry} % set page parameters appropriately
\usepackage{amsmath} % for 'align*' env.
\begin{document}
Put $\lambda=h\pi/D$, $\mu=\pi z/D$, and $\nu=\lambda+\mu$. Put 
$P=\cos(2\lambda)+\cosh(2\lambda)$, 
$Q=\cos^2\lambda \cosh^2\lambda + \sin^2\lambda \sinh^2\lambda$, and
$R=\cosh\lambda \sinh\lambda - \cos\lambda \sin\lambda$. Then
\begin{align*}
u_g
&-\frac{2 u_g \cosh\lambda \sin\lambda \sin\mu \sinh\mu \cos^2\lambda}{PR}
 -\frac{4\pi \tau_y \cosh^2\lambda \sin\mu \sinh\mu \cos^2\lambda}{f\rho_0 DPR} \\
&+\frac{4 h\pi u_g \cosh\lambda \sin\mu \sinh\mu \cos\lambda}{DPR/Q}
 +\frac{4\pi \tau_y \cosh\lambda \sin\mu \sinh\mu \cos\lambda}{f\rho_0 DPR/Q}\\
&+\frac{2\pi \tau_y \cos\nu \cosh\lambda \sinh\nu \cos\lambda}{f\rho_0 DP}
 -\frac{2 u_g \cos\mu \cosh\lambda \cosh\mu \cos\lambda}{P}\\
&-\frac{2\pi \tau_y \cosh\lambda \cosh\nu \sin\nu \cos\lambda}{f\rho_0 DP}
 +\frac{2 u_g \cos\mu \cosh\mu \sin^2\lambda \sinh\lambda \cos\lambda}{PR} \\
&-\frac{2 u_g \cosh^2\lambda \sin\mu \sinh\lambda \sinh\mu \cos\lambda}{PR}
 +\frac{4\pi \tau_y \cos\mu \cosh\lambda \cosh\mu \sin\lambda \sinh\lambda \cos\lambda}{f\rho_0 DPR}\,.
\end{align*}
\end{document}

Addendum, stimulated by a follow-up comment by @Thev: Once one has shown that Mathematica's big honking formula can be displayed as the sum of 10 \frac expressions (plus a lone u_g term), one can (should??) look for further ways to make the formula more accessible. For instance, one could note that 5 of the 10 \frac expressions are multiples of 2u_g, whereas the other 5 are multiples of \frac{2\pi\tau_y}{f\rho_0 D}. One could also organize the numerators some more; for instance, one could impose the ordering \lambda-terms before \mu-terms before \nu-terms, along with a secondary ordering of \cos, \cos^2, \cosh, \sin, \sin^2, \sinh. Collecting these thoughts, and increasing the line spacing as per @Thev's suggestion, one might end up with the following result (the horizontal line in the screenshot is there to indicate the width of the text block):

enter image description here

%% (compile with the same preamble as above)
\begin{align*}
u_g+2u_g \smash{\biggl\{}
&{-}\frac{\cos^2\lambda \cosh\lambda \sin\lambda \sin\mu \sinh\mu}{PR} 
 +\frac{2\pi h \cos\lambda \cosh\lambda \sin\mu \sinh\mu}{DPR/Q}
 -\frac{\cos\lambda \cosh\lambda \cos\mu \cosh\mu}{P}\\[0.75ex]
&\quad+\frac{\cos\lambda \sin^2\lambda \sinh\lambda \cos\mu \cosh\mu}{PR} 
 -\frac{\cos\lambda \cosh^2\lambda \sinh\lambda \sin\mu \sinh\mu}{PR}
 \smash{\biggr\}} \\[1.5ex]
{}+\frac{2\pi\tau_y}{f\rho_0 D} \smash{\biggl\{}
&{-}\frac{2 \cos^2\lambda \cosh^2\lambda \sin\mu \sinh\mu}{PR} 
 +\frac{2\pi \cos\lambda \cosh\lambda \sin\mu \sinh\mu}{PR/Q} 
 +\frac{\cos\lambda \cosh\lambda \cos\nu \sinh\nu}{P}\\[0.75ex]
&\quad-\frac{\cos\lambda \cosh\lambda \cosh\nu \sin\nu}{P}
 +\frac{2 \cos\lambda \cosh\lambda \sin\lambda \sinh\lambda \cos\mu \cosh\mu}{PR}
 \smash{\biggr\}}\,.
\end{align*}

I have no doubt whatsoever that further tweaks could be applied...

| improve this answer | |
  • +1, and I personally add \\[0.07in] for some (imo) much-needed whitespace between the lines. – Thev May 23 at 1:28
  • @Thev - Thanks. Your comment prompted me to provide an addendum that applies further streamlining to the code -- and, per your suggestion, provides for more whitespace between the lines. – Mico May 23 at 5:32
  • @Mico Thank you a lot for your input. It's very helpful for a beginner like me to have such type of information! – ASPVL 2 days ago
  • @ASPVL - You're most welcome. Just to satisfy my own curiosity: Was there an aspect in my answer that was paritcularly helpful to you? For instance, was most of the value added achieved with the first formula, or was there meaningful value added by showing that the full 10-fractional-term expression could be split into two groups with 5 terms each? – Mico 2 days ago
  • 1
    @Mico The use of whitespace for fine-tuning the overall appearance of the equation and, of course, the substitutions. Thanks again, Mico – ASPVL 2 days ago
1

enter image description here

Letting tex do some inline substitutions, and inline fractions.

\documentclass{article}

\begin{document}
\begin{flushleft}

$\displaystyle
\alpha=\frac{h \pi }{D},
\beta=\frac{\pi z}{D}
\gamma=\frac{2 h \pi }{D}
$

\def\za{h \pi}
\def\zb{D}
\def\zc{\pi z}
\def\zd{2 h \pi }

In

$\displaystyle
\let\left\relax
\let\right\relax
\def\frac#1#2{%
\def\zz{#1}\def\zzz{#2}%
\ifx\zzz\zb
  \ifx\zz\za
     \alpha
   \else
  \ifx\zz\zc
      \beta
     \else
  \ifx\zz\zd
      \gamma
     \else
    (#1)/D
   \fi
   \fi
    \fi
\else
\penalty-1000(#1)/(#2)%
\fi}
 -\frac{2 u_g \cosh \left(\frac{h \pi }{D}\right) \sin
  \left(\frac{h \pi }{D}\right) \sin \left(\frac{\pi z}{D}\right)
  \sinh \left(\frac{\pi z}{D}\right) \cos ^2\left(\frac{h \pi
    }{D}\right)}{\left(\cos \left(\frac{2 h \pi }{D}\right)+\cosh
    \left(\frac{2 h \pi }{D}\right)\right) \left(\cos ^2\left(\frac{h
        \pi }{D}\right) \cosh ^2\left(\frac{h \pi }{D}\right)+\sin
    ^2\left(\frac{h \pi }{D}\right) \sinh ^2\left(\frac{h \pi
      }{D}\right)\right) \left(\frac{\cosh \left(\frac{h \pi
        }{D}\right) \sinh \left(\frac{h \pi }{D}\right)}{\cos
      ^2\left(\frac{h \pi }{D}\right) \cosh ^2\left(\frac{h \pi
        }{D}\right)+\sin ^2\left(\frac{h \pi }{D}\right) \sinh
      ^2\left(\frac{h \pi }{D}\right)}-\frac{\cos \left(\frac{h \pi
        }{D}\right) \sin \left(\frac{h \pi }{D}\right)}{\cos
      ^2\left(\frac{h \pi }{D}\right) \cosh ^2\left(\frac{h \pi
        }{D}\right)+\sin ^2\left(\frac{h \pi }{D}\right) \sinh
      ^2\left(\frac{h \pi }{D}\right)}\right)}-\frac{4 \pi \tau_y
  \cosh ^2\left(\frac{h \pi }{D}\right) \sin \left(\frac{\pi
      z}{D}\right) \sinh \left(\frac{\pi z}{D}\right) \cos
  ^2\left(\frac{h \pi }{D}\right)}{D f \rho_0 \left(\cos \left(\frac{2
        h \pi }{D}\right)+\cosh \left(\frac{2 h \pi }{D}\right)\right)
  \left(\cos ^2\left(\frac{h \pi }{D}\right) \cosh ^2\left(\frac{h \pi
      }{D}\right)+\sin ^2\left(\frac{h \pi }{D}\right) \sinh
    ^2\left(\frac{h \pi }{D}\right)\right) \left(\frac{\cosh
      \left(\frac{h \pi }{D}\right) \sinh \left(\frac{h \pi
        }{D}\right)}{\cos ^2\left(\frac{h \pi }{D}\right) \cosh
      ^2\left(\frac{h \pi }{D}\right)+\sin ^2\left(\frac{h \pi
        }{D}\right) \sinh ^2\left(\frac{h \pi }{D}\right)}-\frac{\cos
      \left(\frac{h \pi }{D}\right) \sin \left(\frac{h \pi
        }{D}\right)}{\cos ^2\left(\frac{h \pi }{D}\right) \cosh
      ^2\left(\frac{h \pi }{D}\right)+\sin ^2\left(\frac{h \pi
        }{D}\right) \sinh ^2\left(\frac{h \pi
        }{D}\right)}\right)}+\frac{4 h \pi u_g \cosh \left(\frac{h \pi
    }{D}\right) \sin \left(\frac{\pi z}{D}\right) \sinh
  \left(\frac{\pi z}{D}\right) \cos \left(\frac{h \pi }{D}\right)}{D
  \left(\cos \left(\frac{2 h \pi }{D}\right)+\cosh \left(\frac{2 h \pi
      }{D}\right)\right) \left(\frac{\cosh \left(\frac{h \pi
        }{D}\right) \sinh \left(\frac{h \pi }{D}\right)}{\cos
      ^2\left(\frac{h \pi }{D}\right) \cosh ^2\left(\frac{h \pi
        }{D}\right)+\sin ^2\left(\frac{h \pi }{D}\right) \sinh
      ^2\left(\frac{h \pi }{D}\right)}-\frac{\cos \left(\frac{h \pi
        }{D}\right) \sin \left(\frac{h \pi }{D}\right)}{\cos
      ^2\left(\frac{h \pi }{D}\right) \cosh ^2\left(\frac{h \pi
        }{D}\right)+\sin ^2\left(\frac{h \pi }{D}\right) \sinh
      ^2\left(\frac{h \pi }{D}\right)}\right)}+\frac{4 \pi \tau_y
  \cosh \left(\frac{h \pi }{D}\right) \sin \left(\frac{\pi
      z}{D}\right) \sinh \left(\frac{\pi z}{D}\right) \cos
  \left(\frac{h \pi }{D}\right)}{D f \rho_0 \left(\cos \left(\frac{2 h
        \pi }{D}\right)+\cosh \left(\frac{2 h \pi }{D}\right)\right)
  \left(\frac{\cosh \left(\frac{h \pi }{D}\right) \sinh \left(\frac{h
          \pi }{D}\right)}{\cos ^2\left(\frac{h \pi }{D}\right) \cosh
      ^2\left(\frac{h \pi }{D}\right)+\sin ^2\left(\frac{h \pi
        }{D}\right) \sinh ^2\left(\frac{h \pi }{D}\right)}-\frac{\cos
      \left(\frac{h \pi }{D}\right) \sin \left(\frac{h \pi
        }{D}\right)}{\cos ^2\left(\frac{h \pi }{D}\right) \cosh
      ^2\left(\frac{h \pi }{D}\right)+\sin ^2\left(\frac{h \pi
        }{D}\right) \sinh ^2\left(\frac{h \pi
        }{D}\right)}\right)}+\frac{2 \pi \tau_y \cos \left(\frac{\pi
      (h+z)}{D}\right) \cosh \left(\frac{h \pi }{D}\right) \sinh
  \left(\frac{\pi (h+z)}{D}\right) \cos \left(\frac{h \pi
    }{D}\right)}{D f \rho_0 \left(\cos \left(\frac{2 h \pi
      }{D}\right)+\cosh \left(\frac{2 h \pi
      }{D}\right)\right)}-\frac{2 u_g \cos \left(\frac{\pi
      z}{D}\right) \cosh \left(\frac{h \pi }{D}\right) \cosh
  \left(\frac{\pi z}{D}\right) \cos \left(\frac{h \pi
    }{D}\right)}{\cos \left(\frac{2 h \pi }{D}\right)+\cosh
  \left(\frac{2 h \pi }{D}\right)}-\frac{2 \pi \tau_y \cosh
  \left(\frac{h \pi }{D}\right) \cosh \left(\frac{\pi (h+z)}{D}\right)
  \sin \left(\frac{\pi (h+z)}{D}\right) \cos \left(\frac{h \pi
    }{D}\right)}{D f \rho_0 \left(\cos \left(\frac{2 h \pi
      }{D}\right)+\cosh \left(\frac{2 h \pi
      }{D}\right)\right)}+\frac{2 u_g \cos \left(\frac{\pi
      z}{D}\right) \cosh \left(\frac{\pi z}{D}\right) \sin
  ^2\left(\frac{h \pi }{D}\right) \sinh \left(\frac{h \pi }{D}\right)
  \cos \left(\frac{h \pi }{D}\right)}{\left(\cos \left(\frac{2 h \pi
      }{D}\right)+\cosh \left(\frac{2 h \pi }{D}\right)\right)
  \left(\cos ^2\left(\frac{h \pi }{D}\right) \cosh ^2\left(\frac{h \pi
      }{D}\right)+\sin ^2\left(\frac{h \pi }{D}\right) \sinh
    ^2\left(\frac{h \pi }{D}\right)\right) \left(\frac{\cosh
      \left(\frac{h \pi }{D}\right) \sinh \left(\frac{h \pi
        }{D}\right)}{\cos ^2\left(\frac{h \pi }{D}\right) \cosh
      ^2\left(\frac{h \pi }{D}\right)+\sin ^2\left(\frac{h \pi
        }{D}\right) \sinh ^2\left(\frac{h \pi }{D}\right)}-\frac{\cos
      \left(\frac{h \pi }{D}\right) \sin \left(\frac{h \pi
        }{D}\right)}{\cos ^2\left(\frac{h \pi }{D}\right) \cosh
      ^2\left(\frac{h \pi }{D}\right)+\sin ^2\left(\frac{h \pi
        }{D}\right) \sinh ^2\left(\frac{h \pi
        }{D}\right)}\right)}-\frac{2 u_g \cosh ^2\left(\frac{h \pi
    }{D}\right) \sin \left(\frac{\pi z}{D}\right) \sinh \left(\frac{h
      \pi }{D}\right) \sinh \left(\frac{\pi z}{D}\right) \cos
  \left(\frac{h \pi }{D}\right)}{\left(\cos \left(\frac{2 h \pi
      }{D}\right)+\cosh \left(\frac{2 h \pi }{D}\right)\right)
  \left(\cos ^2\left(\frac{h \pi }{D}\right) \cosh ^2\left(\frac{h \pi
      }{D}\right)+\sin ^2\left(\frac{h \pi }{D}\right) \sinh
    ^2\left(\frac{h \pi }{D}\right)\right) \left(\frac{\cosh
      \left(\frac{h \pi }{D}\right) \sinh \left(\frac{h \pi
        }{D}\right)}{\cos ^2\left(\frac{h \pi }{D}\right) \cosh
      ^2\left(\frac{h \pi }{D}\right)+\sin ^2\left(\frac{h \pi
        }{D}\right) \sinh ^2\left(\frac{h \pi }{D}\right)}-\frac{\cos
      \left(\frac{h \pi }{D}\right) \sin \left(\frac{h \pi
        }{D}\right)}{\cos ^2\left(\frac{h \pi }{D}\right) \cosh
      ^2\left(\frac{h \pi }{D}\right)+\sin ^2\left(\frac{h \pi
        }{D}\right) \sinh ^2\left(\frac{h \pi
        }{D}\right)}\right)}+\frac{4 \pi \tau_y \cos \left(\frac{\pi
      z}{D}\right) \cosh \left(\frac{h \pi }{D}\right) \cosh
  \left(\frac{\pi z}{D}\right) \sin \left(\frac{h \pi }{D}\right)
  \sinh \left(\frac{h \pi }{D}\right) \cos \left(\frac{h \pi
    }{D}\right)}{D f \rho_0 \left(\cos \left(\frac{2 h \pi
      }{D}\right)+\cosh \left(\frac{2 h \pi }{D}\right)\right)
  \left(\cos ^2\left(\frac{h \pi }{D}\right) \cosh ^2\left(\frac{h \pi
      }{D}\right)+\sin ^2\left(\frac{h \pi }{D}\right) \sinh
    ^2\left(\frac{h \pi }{D}\right)\right) \left(\frac{\cosh
      \left(\frac{h \pi }{D}\right) \sinh \left(\frac{h \pi
        }{D}\right)}{\cos ^2\left(\frac{h \pi }{D}\right) \cosh
      ^2\left(\frac{h \pi }{D}\right)+\sin ^2\left(\frac{h \pi
        }{D}\right) \sinh ^2\left(\frac{h \pi }{D}\right)}-\frac{\cos
      \left(\frac{h \pi }{D}\right) \sin \left(\frac{h \pi
        }{D}\right)}{\cos ^2\left(\frac{h \pi }{D}\right) \cosh
      ^2\left(\frac{h \pi }{D}\right)+\sin ^2\left(\frac{h \pi
        }{D}\right) \sinh ^2\left(\frac{h \pi }{D}\right)}\right)}+u_g
$

\end{flushleft}

\end{document}
| improve this answer | |
  • +1. Another substitution possibility: \frac{\pi(h+z)}{D}=\alpha+\beta. – Mico May 22 at 22:12

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