How to solve $\int\tfrac{\sqrt{x^2+1}}{(x^2+a)^{3/2}}dx$? I am trying to find a solution for the following integration for a physics problem. Could anyone give hint on how to do this?
$$
\int\frac{\sqrt{x^2+1}}{\left(x^2+a\right)^\frac{3}{2}}dx
$$
Thanks for your time.

Allow us to use generalized binomial expansion theorem on the $\sqrt{x^2+1}$ and the $u$ substitution $x^2+a=u$.
$$\sqrt{x^2+1}=\sum_{n=1}^\infty\binom{1/2}{n}x^{1-2n}=x+\frac12x^{-1}-\frac18x^{-2}+\dots$$
$$\begin{align}
\int\frac{\sqrt{x^2+1}}{(x^2+a)^{3/2}}dx & = \int\frac{\sum_{n=1}^\infty\binom{1/2}{n}x^{1-2n}}{(x^2+a)^{3/2}}dx \\
& = \frac12\sum_{n=1}^\infty\binom{1/2}{n}\int\frac{x^{-2n}}{(x^2+a)^{3/2}}(2xdx) \\
& = \frac12\sum_{n=1}^\infty\binom{1/2}{n}\int u^{-3/2}(u-a)^{-n}du \\
\end{align}$$
Again, binomial expand the $(u-a)^{-n}$
$$\begin{align}
& = \frac12\sum_{n=1}^\infty\sum_{k=0}^\infty\binom{1/2}{n}\binom nk(-a)^k\int u^{-3/2}u^{-1-k}du \\
& = C+\frac12\sum_{n=1}^\infty\sum_{k=0}^\infty\binom{1/2}{n}\binom nk(-a)^k\frac{(x^2+a)^{-(3/2)-k}}{-(3/2)-k} \\
\end{align}$$
I guess you could say its no longer an integral problem and more of a summation problem.
Also, don't expect anything simple, because according to Wolframalpha, the integral involves elliptic integrals.
