<s>[INST] $ \int \frac{\cosh(x)}{ (a^2 \sinh^2(x) + b^2 \cosh^2(x))^{\frac{3}{2}} } dx$ I am tring to obtain following formula [actually I obtained this resut via mathematica]
\begin{align}
 \int  \frac{\cosh(x)}{ (a^2 \sinh^2(x) + b^2 \cosh^2(x))^{\frac{3}{2}} } dx 
= \frac{\sqrt{2} \sinh(x)}{a^2 \sqrt{a^2-b^2 + (a^2+b^2) \cosh(2x)}} + C
\end{align}
Since the answer contains $\cosh(2x)$ 
My first trial was
\begin{align}
a^2 \sinh^2(x) + b^2 \cosh^2(x) = a^2 \cosh(2x) + (b^2 -a^2) \cosh^2(x) 
\end{align}
Seems not good.. 
My second trial is 
\begin{align}
a^2 \sinh^2(x) + b^2 \cosh^2(x)  = (a^2+b^2) \sinh^2(x) + b^2 
\end{align}
and this also does not good for integrand.... 
How one can obtain this integral? 
 [/INST] Using substitution: Let $u=\sinh(x)$. Then $\mathrm du$ corresponds to $\cosh(x)\,\mathrm dx$ and thus (since $a^2\sinh^2(x)+b^2\cosh^2(x)=(a^2+b^2)\sinh^2(x)+b^2$) $$I=\int \big(u^2\cdot(a^2+b^2)+b^2\big)^{-\frac32}\,\mathrm du.$$
Now we want to substitute $$u^2\cdot(a^2+b^2)=b^2\tan^2(s).$$
Then $$\mathrm du\sim \frac{b \sec^2(s)}{\sqrt{a^2+b^2}} \,\mathrm ds$$
and $$\big(u^2\cdot(a^2+b^2)+b^2\big)^{-\frac32}=\big(b^2\tan^2(s)+b^2\big)^{-\frac32}=\frac{1}{b^3\sec^3(s)}.$$ 
Hence, $$I=\frac{1}{b^2\sqrt{a^2+b^2}}\int \cos(s)\,\mathrm ds.$$
I'll let you do the rest. 

You can use that $$s=\arctan\left(u\frac{\sqrt{a^2+b^2}}b\right)$$ and $$\sin(y)=\frac{\tan(y)}{\sqrt{1+\tan^2(y)}}$$ for all $y\in]-\frac\pi2,\frac\pi2[$.
Also, note that $$\cosh(2y)=\sinh^2(y)+\cosh^2(y).$$
</s>