$ \int_{-\infty}^{\infty} \frac{e^{2x}}{ae^{3x}+b} dx,$ where $a,b \gt 0$ Evaluate
$$ \int_{-\infty}^{\infty} \frac{e^{2x}}{ae^{3x}+b} dx,$$ where $a,b \gt 0$ 
I tried using $y=e^x$, but I still can't solve it.
I get $\displaystyle\int_0^\infty \frac y{ay^3+b} \, dy.$
Is there any different method to solve it?

$$
\begin{align}
\int_{-\infty}^\infty\frac{e^{2x}}{ae^{3x}+b}\,\mathrm{d}x
&=\frac1{3b}\left(\frac ba\right)^{2/3}\int_0^\infty\frac{u^{-1/3}}{u+1}\,\mathrm{d}u\tag{1}\\
&=\frac13\left(\frac1{ba^2}\right)^{1/3}\pi\csc\left(\frac23\pi\right)\tag{2}\\
&=\frac{2\pi}{3\sqrt3}\left(\frac1{ba^2}\right)^{1/3}\tag{3}
\end{align}
$$
Explanation:
$(1)$: $u=\frac abe^{3x}$
$(2)$: result from this answer
$(3)$: $\csc\left(\frac23\pi\right)=\frac2{\sqrt3}$
