<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] Ket us use $$\int \frac{dt}{(m^2+t^2)^{1]2}}=\ln (t+\sqrt{m^2+t^2}]+C_1~~~)1)$$
D. w.r.t $a$ we get $$\int \frac{dt}{(m^2+t^2)^{3/2}}=\frac{t}{m^2\sqrt{m^2+t^2}}+C_2~~~~(2) $$
Then $$I= \int \frac{\cosh x}{[(a^2+b^2) \sinh^2 x+b^2]^{3/2}} dx= \frac{1}{(a^2+b^2)^{3/2}}\int \frac{dt} {(m^2+t^2)^{3/2}}, m=b/\sqrt{a^2+b^2}.$$
Now (2) can be used.
</s>