Evaluation of integral involving $ \tanh(ax) $ Is it possible to evaluate in closed form the integral
$$ \int_{-\sqrt{x}}^{\sqrt{x}}\frac{r\tanh(ar)}{\sqrt{x-r^{2}}}dr=2\int_{0}^{\sqrt{x}}\frac{r\tanh(ar)}{\sqrt{x-r^{2}}}dr$$
here $a$ is a positive constant, $ \tanh(x) $ means the Hyperbolic tangent and $x$ is also positive.
If I apprximate $ \tanh(x) \sim 1+ \frac{1}{x} $ I know how to compute the integral of course :)
 A: A way forward, albeit not very pretty, is to express the $\tanh$ term as exponentials:
$$\begin{align} \tanh{a r} &= \frac{1-e^{-2 a r}}{1+e^{-2 a r}}\\ &= (1-e^{-2 a r}) \sum_{k=0}^{\infty} (-1)^k e^{-2 k a r} \\ &= \sum_{k=0}^{\infty} (-1)^k (e^{-2 k a r}-e^{-2 (k+1)a r}) \end{align}$$
so that the integral becomes
$$\begin{align}\int_{0}^{\sqrt{x}}dr \:\frac{r}{\sqrt{x-r^{2}}} \sum_{k=0}^{\infty} (-1)^k (e^{-2 k a r}-e^{-2 (k+1)a r}) &= \sum_{k=0}^{\infty} (-1)^k \int_{0}^{\sqrt{x}}dr \:\frac{r}{\sqrt{x-r^{2}}} (e^{-2 k a r}-e^{-2 (k+1)a r}) \\ \end{align}$$
It turns out that the integrals may be evaluated in terms of Bessel and Struve functions:
$$\int_{0}^{\sqrt{x}}dr \:\frac{r}{\sqrt{x-r^{2}}} (e^{-2 k a r}-e^{-2 (k+1)a r}) = \frac{1}{2} \pi  \sqrt{x} (\pmb{L}_{-1}(2 a \sqrt{x} k)-\pmb{L}_{-1}(2 a \sqrt{x} (k+1))-I_1(2 a \sqrt{x} k)+I_1(2 a \sqrt{x} (k+1)))$$
This is about as far as I am able to go for now.  I cannot determine if the series is summable in closed form; Mathematica is not able to do it, but that does not mean it is so.  Also I do not have a feel for the rate of convergence of the series, so I am not sure how useful it is compared to a simple numerical evaluation or, as you imply, a power series approximation for small $r$.
