How to automatically make integral size fit the integrand?

Question

The integral sign is too small in this MWE

\documentclass[12pt]{article}
\usepackage{amsmath}
%\usepackage{wasysym}  provides \varint but did not work
\begin{document}  
\begin{align*}
z_1(x) &= e^{ \int \omega \,dx} \\ 
       &= \int \frac{7 x -3+\sqrt{x^{2}+\left(\left(x -1\right)^{2} x^{3}\right)^{\frac{1}{3}}-x}+\sqrt{-\frac{2 \left(\left(-x^{2}+x +\frac{\left(\left(x -1\right)^{2} x^{3}\right)^{\frac{1}{3}}}{2}\right) \sqrt{x^{2}+\left(\left(x -1\right)^{2} x^{3}\right)^{\frac{1}{3}}-x}+x^{2} \left(x -1\right)\right)}{\sqrt{x^{2}+\left(\left(x -1\right)^{2} x^{3}\right)^{\frac{1}{3}}-x}}}}{12 x \left(x -1\right)}dx\\
       &= {\mathrm e}^{\int \frac{7 x -3+\sqrt{x^{2}+\left(\left(x -1\right)^{2} x^{3}\right)^{\frac{1}{3}}-x}+\sqrt{-\frac{2 \left(\left(-x^{2}+x +\frac{\left(\left(x -1\right)^{2} x^{3}\right)^{\frac{1}{3}}}{2}\right) \sqrt{x^{2}+\left(\left(x -1\right)^{2} x^{3}\right)^{\frac{1}{3}}-x}+x^{2} \left(x -1\right)\right)}{\sqrt{x^{2}+\left(\left(x -1\right)^{2} x^{3}\right)^{\frac{1}{3}}-x}}}}{12 x \left(x -1\right)}dx}                  
\end{align*}
\end{document}

Compiled with lualatex (which is what I only use) gives

I know there is bigint package which provides different sizes. But this is not what asking for as I do not want to try different sizes each time. I need the integral size to automatically adjust to fit the size of the integrand. Similar to how \left( and \right) automatically adjust to the size of what is inside.

Is there a package for this? I tried \varint but it did work. the integral sign remained the same.

Answer 1

Rather than increase the size of the integral symbol, I'd fix the way the integrand \omega(x) gets typeset. Once you fix the integrand, it turns out that you no longer have to worry about getting the size of integral symbol right.

I suggest you do something like the following.

% !TEX TS-program = lualatex
\documentclass[12pt]{article}
\usepackage{mathtools}
\begin{document} 
\noindent
For $x\notin\{0,1\}$, put 
\begin{align*}
S(x)&=x(x-1)\\
T(x)&={[xS(x)^2]}^{1/3} \\
U(x)&=\sqrt{S(x)+T(x)} \\
V(x)&=-S(x) +\tfrac{1}{2}T(x)\\
\shortintertext{and}
\omega(x)&=\frac{7x-3 +U(x)+\sqrt{-2 [U(x)V(x)+xS(x)]/U(x)}}{12 S(x)}\,.
\end{align*}
Then,
\[
z_1(x) = \mathrm{e}^{ \int \! \omega(x) \,dx} 
       = \exp\biggl(\int \omega(x) \,dx \!\biggr) \,.
\]
\end{document}