breqn: specifying the characters at which equations can be broken

Question

Is there a way to force breqn to break at a certain character in favor of another. For instance, in the following example, one might prefer that the line be broken at a + sign rather than at a \cdot. Can one instruct breqn the exact point at which an equation should be broken?

\documentclass{article}
\usepackage{amsmath}
\usepackage{breqn}
\begin{document}
\begin{dmath*}
x = \left( y + \frac{1}{2}y^2z^3 \int_{-\infty}^{+\infty} xy^2\exp(z+xy)\,\mathrm{d}y  + \frac{9\pi}{7} \int_{-\infty}^{+\infty} xy^3\cdot\ln z \,  \mathrm{d}z \cdot \int_{-\infty}^{+\infty} \frac{x}{(1-y)^2} \cdot \mathrm{e}^{-3z} \,\mathrm{d}y \right)
\end{dmath*}
\end{document}

Answer 1

You can force a line break with \\

I think breqn by default gives the same breaking penalty to all binary relations, with no user-level customisation of that.

On the other hand I think that it is safe to do this

 \let\oldcdot\cdot
 \usepackage{breqn}
 \let\cdot\oldcdot

which restores \cdot to its normal non-breqn definition so breaking won't occur at the dot. On your example, this causes the break to happen before the + \frac{9\pi}

Answer 2

There is no way to force breqn to break at a certain character in favor of another. However, you can influence its default behavior by enclosing sections in {} or use an explicit break with \\. If you use the latter you will not get proper alignment.

Eq(1a) was printed using \exp({z+xy}), whereas eq(1c) was printed with the default behavior. Obviously breqn's algorithm needs some improvements here as breaking at the plus sign of the exponential is a poor choice.

Eq(1d) was obtained with a forced break using \\.

Here is the MWE for anyone wanting to experiment further.

\documentclass{article}
\usepackage{amsmath}
\let\oldcdot\cdot
\usepackage{breqn}
\let\cdot\oldcdot
\begin{document}
\begin{dgroup}[compact]
\begin{dmath}
x  = \left( y + \frac{1}{2}y^2z^3 \int_{-\infty}^{+\infty} xy^2\exp({z+xy})\,\mathrm{d}y  + \frac{9\pi}{7} \int_{-\infty}^{+\infty} xy^3\cdot\ln z \,  \mathrm{d}z \cdot \int_{-\infty}^{+\infty} \frac{x}{(1-y)^2} \cdot \mathrm{e}^{-3z} \,\mathrm{d}y \right)
\end{dmath}
\begin{dmath}
f=0
\end{dmath}
\begin{dmath}
x = \left( y + \frac{1}{2}y^2z^3 \int_{-\infty}^{+\infty} xy^2\exp(z+xy)\,\mathrm{d}y + \frac{9\pi}{7} \int_{-\infty}^{+\infty} xy^3\cdot\ln z \,  \mathrm{d}z \cdot \int_{-\infty}^{+\infty} \frac{x}{(1-y)^2} \cdot \mathrm{e}^{-3z} \,\mathrm{d}y \right)
\end{dmath}
\begin{dmath}
x = \left( y + \frac{1}{2}y^2z^3 \int_{-\infty}^{+\infty} xy^2\exp(z+xy)\,\mathrm{d}y + \frac{9\pi}{7}\\ \int_{-\infty}^{+\infty} xy^3\cdot\ln z \,  \mathrm{d}z \cdot \int_{-\infty}^{+\infty} \frac{x}{(1-y)^2} \cdot \mathrm{e}^{-3z} \,\mathrm{d}y \right)
\end{dmath}
\begin{dmath}
f=0
\end{dmath}
\end{dgroup}
\end{document}