I meet an integral, but it is beyond my ability. $$ {\rm I}\left(a\right) = \int_{a}^{1}{\arcsin\left(\,\sqrt{\,{1 - x^{2} \over 1 - a^{2}}\,}\,\right) \over x + 1}\,{\rm d}x, 0\le a <1. $$
I can work it out when $a = 0$, but failed otherwise. I don't intend to find a closed-form currently, an approximation expression will be OK. Any Suggestion?
For $0\le a<1$, $$\small{\mathcal{I}{\left(a\right)}:=\int_{a}^{1}\frac{\arcsin{\left(\sqrt{\frac{1-x^2}{1-a^2}}\right)}}{x+1}\,\mathrm{d}x=-\frac{\pi}{2}\,\ln{\left(\frac{1+a}{2}\right)}-\frac{\pi\left(1-a^2\right)}{16}\,{_4F_3}{\left(1,1,\frac32,\frac32;2,2,2;1-a^2\right)}.}$$
Proof:
Let $a\in\mathbb{R}\land0\le a<1$, and for each such $a$ define $a^{\prime}:=\sqrt{1-a^2}$ to be the complement of $a$. Changing variables in the integral so that the argument of the arcsine function is as simple as possible, we arrive at in integral that starts to more closely resemble an elliptic integral.
$$\begin{align} \mathcal{I}{\left(a\right)} &=\int_{a}^{1}\frac{\arcsin{\left(\sqrt{\frac{1-x^2}{1-a^2}}\right)}}{x+1}\,\mathrm{d}x\\ &=\int_{1}^{0}\frac{\arcsin{\left(y\right)}}{1+\sqrt{1-\left(1-a^2\right)y^2}}\cdot\frac{-\left(1-a^2\right)y}{\sqrt{1-\left(1-a^2\right)y^2}}\,\mathrm{d}y;~~~\small{\left[\sqrt{\frac{1-x^2}{1-a^2}}=y\right]}\\ &=\int_{0}^{1}\frac{\arcsin{\left(y\right)}}{1+\sqrt{1-a^{\prime\,2}y^2}}\cdot\frac{a^{\prime\,2}y}{\sqrt{1-a^{\prime\,2}y^2}}\,\mathrm{d}y\\ &=\int_{0}^{1}\left(\frac{1-\sqrt{1-a^{\prime\,2}y^2}}{y\sqrt{1-a^{\prime\,2}y^2}}\right)\arcsin{\left(y\right)}\,\mathrm{d}y.\\ \end{align}$$
Integrating by parts next,
$$\begin{align} \mathcal{I}{\left(a\right)} &=\int_{0}^{1}\left(\frac{1-\sqrt{1-a^{\prime\,2}y^2}}{y\sqrt{1-a^{\prime\,2}y^2}}\right)\arcsin{\left(y\right)}\,\mathrm{d}y\\ &=\left[-\arcsin{\left(y\right)}\,\ln{\left(1+\sqrt{1-a^{\prime\,2}y^2}\right)}\right]_{0}^{1}+\int_{0}^{1}\frac{\ln{\left(1+\sqrt{1-a^{\prime\,2}y^2}\right)}}{\sqrt{1-y^2}}\,\mathrm{d}y\\ &=-\frac{\pi}{2}\,\ln{\left(1+\sqrt{1-a^{\prime\,2}}\right)}+\int_{0}^{1}\frac{\ln{\left(1+\sqrt{1-a^{\prime\,2}y^2}\right)}}{\sqrt{1-y^2}}\,\mathrm{d}y\\ &=-\frac{\pi}{2}\,\ln{\left(1+a\right)}+\int_{0}^{1}\frac{\ln{\left(1+\sqrt{1-a^{\prime\,2}y^2}\right)}}{\sqrt{1-y^2}}\,\mathrm{d}y.\\ \end{align}$$
Using trigonometric substitution,
$$\begin{align} \mathcal{I}{\left(a\right)} &=-\frac{\pi}{2}\,\ln{\left(1+a\right)}+\int_{0}^{\frac{\pi}{2}}\ln{\left(1+\sqrt{1-a^{\prime\,2}\sin^2{\left(\theta\right)}}\right)}\,\mathrm{d}\theta;~~~\small{\left[y=\sin{\theta}\right]}.\\ \end{align}$$
At this point I recognized the integral as essentially equivalent to this one. It seems that the best one can hope for is a closed form in terms of generalized hypergeometric functions. The closed form in terms of hypergeometric functions can be found via differentiating under the integral sign. Define the function $\mathcal{J}{\left(\alpha\right)}$ for $0<\alpha\le1$ as
$$\mathcal{J}{\left(\alpha\right)}:=\int_{0}^{1}\frac{\ln{\left(1+\sqrt{1-\alpha^2y^2}\right)}}{\sqrt{1-y^2}}\,\mathrm{d}y.$$
Then,
$$\begin{align} \mathcal{J}{\left(a^{\prime}\right)} &=\int_{0}^{1}\frac{\ln{\left(1+\sqrt{1-a^{\prime\,2}y^2}\right)}}{\sqrt{1-y^2}}\,\mathrm{d}y\\ &=\mathcal{J}{\left(0\right)}+\int_{0}^{a^{\prime}}\mathrm{d}\alpha\,\frac{\partial}{\partial\alpha}\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1+\sqrt{1-\alpha^2y^2}\right)}}{\sqrt{1-y^2}}\\ &=\int_{0}^{1}\frac{\ln{\left(2\right)}}{\sqrt{1-y^2}}\,\mathrm{d}y+\int_{0}^{a^{\prime}}\mathrm{d}\alpha\int_{0}^{1}\mathrm{d}y\,\frac{\partial}{\partial\alpha}\frac{\ln{\left(1+\sqrt{1-\alpha^2y^2}\right)}}{\sqrt{1-y^2}}\\ &=\ln{(2)}\int_{0}^{1}\frac{\mathrm{d}y}{\sqrt{1-y^2}}+\int_{0}^{a^{\prime}}\mathrm{d}\alpha\int_{0}^{1}\mathrm{d}y\,\frac{\left(-\alpha\,y^2\right)}{\sqrt{1-y^2}\sqrt{1-\alpha^2y^2}\left(1+\sqrt{1-\alpha^2y^2}\right)}\\ &=\frac{\pi}{2}\ln{(2)}+\int_{0}^{a^{\prime}}\mathrm{d}\alpha\int_{0}^{1}\mathrm{d}y\,\frac{\sqrt{1-\alpha^2y^2}-1}{\alpha\sqrt{1-y^2}\sqrt{1-\alpha^2y^2}}\\ &=\frac{\pi}{2}\ln{(2)}+\int_{0}^{a^{\prime}}\mathrm{d}\alpha\,\frac{1}{\alpha}\int_{0}^{1}\mathrm{d}y\,\frac{\sqrt{1-\alpha^2y^2}-1}{\sqrt{1-y^2}\sqrt{1-\alpha^2y^2}}\\ &=\frac{\pi}{2}\ln{(2)}+\int_{0}^{a^{\prime}}\mathrm{d}\alpha\,\frac{1}{\alpha}\left[\frac{\pi}{2}-K{\left(\alpha\right)}\right],\\ \end{align}$$
where $K{\left(k\right)}$ is the complete elliptic integral of the first kind, defined here with the argument convention,
$$K{\left(k\right)}:=\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1-t^2}\sqrt{1-k^2t^2}};~~~\small{\left[0<k^2<1\right]}.$$
In terms of hypergeometric functions, the complete elliptic integral of the first kind may be written (see eq.3 here) as
$$K{\left(k\right)}=\frac{\pi}{2}\,{_2F_1}{\left(\frac12,\frac12;1;k^2\right)}.$$
Thus,
$$\begin{align} \mathcal{I}{\left(a\right)} &=-\frac{\pi}{2}\,\ln{\left(1+a\right)}+\mathcal{J}{\left(a^{\prime}\right)}\\ &=-\frac{\pi}{2}\,\ln{\left(1+a\right)}+\frac{\pi}{2}\ln{(2)}+\int_{0}^{a^{\prime}}\mathrm{d}\alpha\,\frac{\left[\frac{\pi}{2}-K{\left(\alpha\right)}\right]}{\alpha}\\ &=-\frac{\pi}{2}\,\ln{\left(\frac{1+a}{2}\right)}+\int_{0}^{a^{\prime}}\mathrm{d}\alpha\,\frac{\left[\frac{\pi}{2}-\frac{\pi}{2}\,{_2F_1}{\left(\frac12,\frac12;1;\alpha^2\right)}\right]}{\alpha}\\ &=-\frac{\pi}{2}\,\ln{\left(\frac{1+a}{2}\right)}-\frac{\pi}{2}\int_{0}^{a^{\prime}}\mathrm{d}\alpha\,\frac{\left[{_2F_1}{\left(\frac12,\frac12;1;\alpha^2\right)}-1\right]}{\alpha}.\\ \end{align}$$
The following reduction identity will be useful:
$${_3F_2}{\left(1,b,c;2,d;z\right)}=\frac{d-1}{\left(b-1\right)\left(c-1\right)z}\left[{_2F_1}{\left(b-1,c-1;d-1;z\right)}-1\right].$$
Setting $b=c=\frac32\land d=2\land z=\alpha^2$ yields
$${_3F_2}{\left(1,\frac32,\frac32;2,2;\alpha^2\right)}=\frac{4}{\alpha^2}\left[{_2F_1}{\left(\frac12,\frac12;1;\alpha^2\right)}-1\right].$$
Thus,
$$\begin{align} \mathcal{I}{\left(a\right)} &=-\frac{\pi}{2}\,\ln{\left(\frac{1+a}{2}\right)}-\frac{\pi}{2}\int_{0}^{a^{\prime}}\mathrm{d}\alpha\,\frac{\left[{_2F_1}{\left(\frac12,\frac12;1;\alpha^2\right)}-1\right]}{\alpha}\\ &=-\frac{\pi}{2}\,\ln{\left(\frac{1+a}{2}\right)}-\frac{\pi}{2}\int_{0}^{a^{\prime}}\mathrm{d}\alpha\,\frac{\alpha}{4}\,{_3F_2}{\left(1,\frac32,\frac32;2,2;\alpha^2\right)}\\ &=-\frac{\pi}{2}\,\ln{\left(\frac{1+a}{2}\right)}-\frac{\pi}{8}\int_{0}^{a^{\prime}}\mathrm{d}\alpha\,\alpha\,{_3F_2}{\left(1,\frac32,\frac32;2,2;\alpha^2\right)}\\ &=-\frac{\pi}{2}\,\ln{\left(\frac{1+a}{2}\right)}-\frac{\pi\,a^{\prime\,2}}{8}\int_{0}^{1}\mathrm{d}u\,u\,{_3F_2}{\left(1,\frac32,\frac32;2,2;a^{\prime\,2}u^2\right)};~~~\small{\left[\frac{\alpha}{a^{\prime}}=u\right]}\\ &=-\frac{\pi}{2}\,\ln{\left(\frac{1+a}{2}\right)}-\frac{\pi\,a^{\prime\,2}}{16}\int_{0}^{1}\mathrm{d}t\,{_3F_2}{\left(1,\frac32,\frac32;2,2;a^{\prime\,2}t\right)};~~~\small{\left[u^2=t\right]}.\\ \end{align}$$
For $\Re{\left(r\right)}>\Re{\left(d\right)}>0$, we have the integral representation,
$${_4F_3}{\left(a,b,c,d;p,q,r;z\right)}=\frac{\Gamma{\left(r\right)}}{\Gamma{\left(d\right)}\,\Gamma{\left(r-d\right)}}\int_{0}^{1}\mathrm{d}t\,t^{d-1}\left(1-t\right)^{r-d-1}\,{_3F_2}{\left(a,b,c;p,q;zt\right)}.$$
Setting $d=1\land r=2$, we then have the integral representation,
$${_4F_3}{\left(a,b,c,1;p,q,2;z\right)}=\int_{0}^{1}\mathrm{d}t\,{_3F_2}{\left(a,b,c;p,q;zt\right)}.$$
Finally,
$$\begin{align} \mathcal{I}{\left(a\right)} &=-\frac{\pi}{2}\,\ln{\left(\frac{1+a}{2}\right)}-\frac{\pi\,a^{\prime\,2}}{16}\int_{0}^{1}\mathrm{d}t\,{_3F_2}{\left(1,\frac32,\frac32;2,2;a^{\prime\,2}t\right)}\\ &=-\frac{\pi}{2}\,\ln{\left(\frac{1+a}{2}\right)}-\frac{\pi\,a^{\prime\,2}}{16}\,{_4F_3}{\left(1,\frac32,\frac32,1;2,2,2;a^{\prime\,2}\right)}\\ &=-\frac{\pi}{2}\,\ln{\left(\frac{1+a}{2}\right)}-\frac{\pi\left(1-a^2\right)}{16}\,{_4F_3}{\left(1,\frac32,\frac32,1;2,2,2;1-a^2\right)}\\ &=-\frac{\pi}{2}\,\ln{\left(\frac{1+a}{2}\right)}-\frac{\pi\left(1-a^2\right)}{16}\,{_4F_3}{\left(1,1,\frac32,\frac32;2,2,2;1-a^2\right)}.\\ \end{align}$$
