evaluation of a definite integral involving inverse hyperbolic functions The following integral appears during the analysis of statistical mechanical models.
$$\int\limits_0^{\pi /2}{\operatorname{arctanh} \left[ {\sqrt {1 - \sin (x)\sin (x){a^2}} } \right]dx} $$
Is there a closed form solution for the integral?
 A: Consider
$$I(b)=\int_0^{\frac \pi 2} \tanh ^{-1}\left(\sqrt{1-b \sin ^2(x)}\right)\,dx$$
$$I'(b)=-\frac 1{2b}\int_0^{\frac \pi 2} \frac{dx}{\sqrt{1-b \sin ^2(x)}}=-\frac{K\left(\frac{b}{b-1}\right)}{2 b\sqrt{1-b}}$$ and $I(1)=2 C$. So
$$I(a^2)=2C+\frac 12 \int_{a^2}^1\frac{1}{ b\sqrt{1-b}}K\left(\frac{b}{b-1}\right)\,db$$
You could use the series
$$I'(b)=\frac{\pi }{4 b}+ \pi \sum_{n=0}^\infty \frac {\alpha_n}{2^{\beta_n}} b^n $$
The $\alpha_n$ correspond to sequence $A038534$ in $OEIS$ and the $\beta_n$ form the sequence
$$\{4,8,10,16,18,22,24,32,34,38,40,46,48,52,54,64,66,70,72,78,80,84,86
   ,94,96,100,\cdots\}$$
Integrating termwise leads to more than decent approximations.
$$J=\int_0^{\frac \pi 2} \tanh ^{-1}\left(\sqrt{1-a^2 \sin ^2(x)}\right)\,dx=2C-\frac{1}{2} \pi  \log (a)+\pi \sum_{n=0}^\infty \frac {\alpha_n}{2^{\beta_n}\,(n+1)}(1-a^{2(n+1)})$$
Edit
Reworking
$$J=\frac 12 \int_{a^2}^1\frac{1}{ b\sqrt{1-b}}K\left(\frac{b}{b-1}\right)\,db$$
let $b=\sin^2(t)$ to make
$$J=\int_{\sin ^{-1}(a)}^{\frac \pi 2} \csc (t) \,K\left(-\tan ^2(t)\right)\,dt$$
Using
$$K\left(-z^2\right)=\pi \sum_{n=0}^\infty (-1)^n\, \frac {\alpha_n}{2^{\gamma_n}}\, z^{2n} $$ the $\gamma_n$ forming the sequence
$$\{1,3,7,9,15,17,21,23,31,33,37,39,45,47,51,53,63,65,69,71,77,79,8
   3,85,93,\cdots\}$$
$$J=-\frac{1}{2} \pi  \log \left(\tan \left(\frac{\sin
   ^{-1}(a)}{2} \right)\right)+$$ $$\pi \sum_{n=0}^\infty (-1)^n\, \frac {\alpha_n}{2^{\gamma_n}}\,2^{2 n-1} \left(B_1(n,1-2 n)-B_{\tan ^2\left(\frac{\sin
   ^{-1}(a)}{2} \right)}(n,1-2 n)\right)$$
A: 
Define the function $\mathcal{I}:\left(0,1\right]\rightarrow\mathbb{R}$ via the definite integral
$$\mathcal{I}{\left(b\right)}:=\int_{0}^{\pi/2}\mathrm{d}\theta\,\operatorname{artanh}{\left(\sqrt{1-b\sin^{2}{\left(\theta\right)}}\right)},$$
where here the real inverse hyperbolic tangent is defined by
$$\operatorname{artanh}{\left(x\right)}:=\int_{0}^{x}\mathrm{d}t\,\frac{1}{1-t^{2}}=\frac12\ln{\left(\frac{1+x}{1-x}\right)};~~~\small{x\in\left(-1,1\right)}.$$
Consider the following specific value for the generalized hypergeometric function $_3F_2$:
$${_3F_2}{\left(1,1,\frac32;2,2;z\right)}=-\frac{4}{z}\ln{\left(\frac{1+\sqrt{1-z}}{2}\right)};~~~\small{z\in(-\infty,1]}.$$
We can use this fact to rewrite the most complicated part of the integrand of $\mathcal{I}$ as a hypergeometric function.

For any $b\in\left(0,1\right]$,
$$\begin{align}
\mathcal{I}{\left(b\right)}
&=\int_{0}^{\pi/2}\mathrm{d}\theta\,\operatorname{artanh}{\left(\sqrt{1-b\sin^{2}{\left(\theta\right)}}\right)}\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{artanh}{\left(\sqrt{1-bx^{2}}\right)}}{\sqrt{1-x^{2}}};~~~\small{\left[\theta=\arcsin{\left(x\right)}\right]}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{\operatorname{artanh}{\left(\sqrt{1-bt}\right)}}{2\sqrt{t}\sqrt{1-t}};~~~\small{\left[x=\sqrt{t}\right]}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(\frac{1+\sqrt{1-bt}}{1-\sqrt{1-bt}}\right)}}{4\sqrt{t}\sqrt{1-t}}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(\frac{(1+\sqrt{1-bt})^{2}}{bt}\right)}}{4\sqrt{t}\sqrt{1-t}}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{2\ln{\left(1+\sqrt{1-bt}\right)}-\ln{\left(bt\right)}}{4\sqrt{t}\sqrt{1-t}}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{2\ln{\left(1+\sqrt{1-bt}\right)}-2\ln{\left(2\right)}+2\ln{\left(2\right)}-\ln{\left(b\right)}-\ln{\left(t\right)}}{4\sqrt{t}\sqrt{1-t}}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{2\ln{\left(\frac{1+\sqrt{1-bt}}{2}\right)}-\ln{\left(\frac{b}{4}\right)}-\ln{\left(t\right)}}{4\sqrt{t}\sqrt{1-t}}\\
&=-\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(\frac{b}{4}\right)}}{4\sqrt{t}\sqrt{1-t}}-\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(t\right)}}{4\sqrt{t}\sqrt{1-t}}+\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(\frac{1+\sqrt{1-bt}}{2}\right)}}{2\sqrt{t}\sqrt{1-t}}\\
&=\pi\ln{\left(2\right)}-\frac{\pi}{4}\ln{\left(b\right)}+\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(\frac{1+\sqrt{1-bt}}{2}\right)}}{\sqrt{t}\sqrt{1-t}}\\
&=\pi\ln{\left(2\right)}-\frac{\pi}{4}\ln{\left(b\right)}-\frac18\int_{0}^{1}\mathrm{d}t\,\frac{bt}{\sqrt{t}\sqrt{1-t}}\left[-\frac{4}{bt}\ln{\left(\frac{1+\sqrt{1-bt}}{2}\right)}\right]\\
&=\pi\ln{\left(2\right)}-\frac{\pi}{4}\ln{\left(b\right)}-\frac{b}{8}\int_{0}^{1}\mathrm{d}t\,t^{1/2}\left(1-t\right)^{-1/2}\,{_3F_2}{\left(1,1,\frac32;2,2;bt\right)}\\
&=\pi\ln{\left(2\right)}-\frac{\pi}{4}\ln{\left(b\right)}-\frac{b}{8}\operatorname{B}{\left(\frac32,\frac12\right)}\,{_4F_3}{\left(1,1,\frac32,\frac32;2,2,2;b\right)}\\
&=\pi\ln{\left(2\right)}-\frac{\pi}{4}\ln{\left(b\right)}-\frac{\pi}{16}b\,{_4F_3}{\left(1,1,\frac32,\frac32;2,2,2;b\right)},\\
\end{align}$$
where the final integration comes from Euler's integral formula for higher-order hypergeometric functions:
$$\int_{0}^{1}\mathrm{d}t\,t^{d-1}\left(1-t\right)^{r-d-1}\,{_3F_2}{\left(a,b,c;p,q;zt\right)}=\operatorname{B}{\left(d,r-d\right)}\,{_4F_3}{\left(a,b,c,d;p,q,r;z\right)};~~~\small{0<d<r}.$$

A: Given:
$$I(a)=\int\limits_0^{\pi /2}{\operatorname{arctanh} ( {\sqrt {1 - a^2\sin^2 x}})  \;dx}$$
The inverse hyperbolic tangent is given by
$$\operatorname{arctanh}z=\frac{1}{2}\ln\frac{1+z}{1-z}$$
So we can after some transformations present the integrand in the form
$$\ln(1+\sqrt {1 - a^2\sin^2 x})-\ln a -\ln \sin x $$
Using well known result
$$\int\limits_0^{\pi /2}\ln \sin x\;dx=-\frac{\pi}{2}\ln 2$$
we get after integrating:
$$I(a)=\frac{\pi}{2}\ln\frac{2}{a}+\int\limits_0^{\pi /2}\ln(1+\sqrt {1 - a^2\sin^2 x})\;dx$$
We can already get useful information from this result.
Asymptotic behavior of $I(a)$ as $a$ tends to zero:
$$I(a)\to \frac{\pi}{2}\ln\frac{2}{a}$$
At $a=1$ we get:
$$I(1)=\frac{\pi}{2}\ln 2+\int\limits_0^{\pi /2}\ln(1+\cos x)\;dx=2G$$
where $G$ is  Catalan's constant.
Well known result
$$\int\limits_0^{\pi /2}\ln(1+\cos x)\;dx=2G-\frac{\pi}{2}\ln 2$$
is used.
For values around $a=\frac{1}{2}$ simply expand $I(a)$ in Taylor's series about $a$ as given by Mariusz Iwaniuk.
Some of the first terms are already giving an excellent accuracy.
