A surprising dilogarithm integral identity arising from a generalised point enclosure problem
Under Construction: This isn't a complete answer, but I wanted to go ahead and point out some symmetries and simplifications to your integral that'll make it much friendlier to work with...
Define the function $f:\left[0,1\right]\times\left[0,\pi\right]\setminus\{\left(1,0\right)\}\rightarrow\left[0,\frac{\pi}{2}\right]$ via the inverse trigonometric expression,
$$f{\left(x,\alpha\right)}:=\arcsin{\left(\frac{x\sin{\left(\alpha\right)}}{\sqrt{1-2x\cos{\left(\alpha\right)}+x^{2}}}\right)},$$
and then define the function $\mathcal{P}:\left[0,1\right]\rightarrow\mathbb{R}$ via the double integral,
$$\begin{align} \mathcal{P}{\left(x\right)} &:=\int_{0}^{\pi}\mathrm{d}\alpha\int_{0}^{\alpha}\mathrm{d}\beta\,\left[f{\left(x,\beta\right)}-f{\left(x,\alpha\right)}\right].\\ \end{align}$$
It so happens that any double integral of this form can be reduced to a single-variable integral without even specifying the function $f$ inside the integrand, using basic symmetries.
For any $x\in\left[0,1\right]$, we obtain
$$\begin{align} \mathcal{P}{\left(x\right)} &=\int_{0}^{\pi}\mathrm{d}\alpha\int_{0}^{\alpha}\mathrm{d}\beta\,\left[f{\left(x,\beta\right)}-f{\left(x,\alpha\right)}\right]\\ &=\int_{0}^{\pi}\mathrm{d}\alpha\int_{0}^{\alpha}\mathrm{d}\beta\,f{\left(x,\beta\right)}-\int_{0}^{\pi}\mathrm{d}\alpha\int_{0}^{\alpha}\mathrm{d}\beta\,f{\left(x,\alpha\right)}\\ &=\int_{0}^{\pi}\mathrm{d}\beta\int_{0}^{\beta}\mathrm{d}\alpha\,f{\left(x,\alpha\right)}-\int_{0}^{\pi}\mathrm{d}\alpha\int_{0}^{\alpha}\mathrm{d}\beta\,f{\left(x,\alpha\right)}\\ &=\int_{0}^{\pi}\mathrm{d}\alpha\int_{\alpha}^{\pi}\mathrm{d}\beta\,f{\left(x,\alpha\right)}-\int_{0}^{\pi}\mathrm{d}\alpha\int_{0}^{\alpha}\mathrm{d}\beta\,f{\left(x,\alpha\right)}\\ &=\int_{0}^{\pi}\mathrm{d}\alpha\,\left[\int_{\alpha}^{\pi}\mathrm{d}\beta\,f{\left(x,\alpha\right)}-\int_{0}^{\alpha}\mathrm{d}\beta\,f{\left(x,\alpha\right)}\right]\\ &=\int_{0}^{\pi}\mathrm{d}\alpha\,\left[\left(\pi-\alpha\right)f{\left(x,\alpha\right)}-\alpha\,f{\left(x,\alpha\right)}\right]\\ &=\int_{0}^{\pi}\mathrm{d}\alpha\,\left(\pi-2\alpha\right)f{\left(x,\alpha\right)}.\\ \end{align}$$
Given $x\in\left(0,1\right)$, we then find using the trigonometric identity
$$\arcsin{\left(x\right)}=\arctan{\left(\frac{x}{\sqrt{1-x^{2}}}\right)};~~~\small{x^{2}<1},$$
that $\mathcal{P}$ reduces to:
$$\begin{align} \mathcal{P}{\left(x\right)} &=\int_{0}^{\pi}\mathrm{d}\alpha\int_{0}^{\alpha}\mathrm{d}\beta\,\left[f{\left(x,\beta\right)}-f{\left(x,\alpha\right)}\right]\\ &=\int_{0}^{\pi}\mathrm{d}\alpha\,\left(\pi-2\alpha\right)f{\left(x,\alpha\right)}\\ &=\int_{0}^{\pi}\mathrm{d}\alpha\,\left(\pi-2\alpha\right)\arcsin{\left(\frac{x\sin{\left(\alpha\right)}}{\sqrt{1-2x\cos{\left(\alpha\right)}+x^{2}}}\right)}\\ &=\int_{0}^{\pi}\mathrm{d}\alpha\,\left(\pi-2\alpha\right)\arctan{\left(\frac{x\sin{\left(\alpha\right)}}{1-x\cos{\left(\alpha\right)}}\right)}\\ &=\int_{0}^{\pi}\mathrm{d}\alpha\,\left(\pi-\alpha\right)\arctan{\left(\frac{x\sin{\left(\alpha\right)}}{1-x\cos{\left(\alpha\right)}}\right)}\\ &~~~~~-\int_{0}^{\pi}\mathrm{d}\alpha\,\alpha\arctan{\left(\frac{x\sin{\left(\alpha\right)}}{1-x\cos{\left(\alpha\right)}}\right)}\\ &=\int_{0}^{\pi}\mathrm{d}\alpha\,\left(\pi-\alpha\right)\arctan{\left(\frac{x\sin{\left(\alpha\right)}}{1-x\cos{\left(\alpha\right)}}\right)}\\ &~~~~~-\int_{0}^{\pi}\mathrm{d}\alpha\,\left(\pi-\alpha\right)\arctan{\left(\frac{x\sin{\left(\alpha\right)}}{1+x\cos{\left(\alpha\right)}}\right)};~~~\small{\left[\alpha\mapsto\pi-\alpha\right]}\\ &=\int_{0}^{\pi}\mathrm{d}\alpha\,\left(\pi-\alpha\right)\left[\arctan{\left(\frac{x\sin{\left(\alpha\right)}}{1-x\cos{\left(\alpha\right)}}\right)}-\arctan{\left(\frac{x\sin{\left(\alpha\right)}}{1+x\cos{\left(\alpha\right)}}\right)}\right]\\ &=\int_{0}^{\pi}\mathrm{d}\alpha\,\left(\pi-\alpha\right)\arctan{\left(\frac{x^{2}\sin{\left(2\alpha\right)}}{1-x^{2}\cos{\left(2\alpha\right)}}\right)}\\ &=\frac14\int_{0}^{2\pi}\mathrm{d}\varphi\,\left(2\pi-\varphi\right)\arctan{\left(\frac{x^{2}\sin{\left(\varphi\right)}}{1-x^{2}\cos{\left(\varphi\right)}}\right)};~~~\small{\left[2\alpha=\varphi\right]}\\ &=\frac14\int_{0}^{2\pi}\mathrm{d}\varphi\,\left(-\varphi\right)\arctan{\left(\frac{x^{2}\sin{\left(\varphi\right)}}{1-x^{2}\cos{\left(\varphi\right)}}\right)};~~~\small{\left[\varphi\mapsto2\pi-\varphi\right]}\\ &=\frac14\int_{0}^{2\pi}\mathrm{d}\varphi\,\left(\pi-\varphi\right)\arctan{\left(\frac{x^{2}\sin{\left(\varphi\right)}}{1-x^{2}\cos{\left(\varphi\right)}}\right)}\\ &=\frac12\int_{0}^{\pi}\mathrm{d}\varphi\,\left(\pi-\varphi\right)\arctan{\left(\frac{x^{2}\sin{\left(\varphi\right)}}{1-x^{2}\cos{\left(\varphi\right)}}\right)}\\ &=\frac12\int_{0}^{\pi}\mathrm{d}\varphi\int_{\varphi}^{\pi}\mathrm{d}\omega\,\arctan{\left(\frac{x^{2}\sin{\left(\varphi\right)}}{1-x^{2}\cos{\left(\varphi\right)}}\right)}\\ &=\frac12\int_{0}^{\pi}\mathrm{d}\omega\int_{0}^{\omega}\mathrm{d}\varphi\,\arctan{\left(\frac{x^{2}\sin{\left(\varphi\right)}}{1-x^{2}\cos{\left(\varphi\right)}}\right)}...\\ \end{align}$$
...TBC
Continuing from the last expression for the original double integral in David's answer: $$\mathcal P(x)=\int_0^\pi\int_0^a(\dots)\,db\,da=\frac12\int_0^\pi\int_0^\omega\tan^{-1}\frac{x^2\sin\varphi}{1-x^2\cos\varphi}\,d\varphi\,d\omega$$ Using the logarithmic form of $\tan^{-1}$ we can rewrite the integrand of the RHS as $$\begin{align}\tan^{-1}\frac{x^2\sin\varphi}{1-x^2\cos\varphi}&=\frac i2\ln\frac{1-\frac{ix^2\sin\varphi}{1-x^2\cos\varphi}}{1+\frac{ix^2\sin\varphi}{1-x^2\cos\varphi}}\\ &=\frac i2\ln\frac{\frac{1-x^2\cos\varphi-ix^2\sin\varphi}{1-x^2\cos\varphi}}{\frac{1-x^2\cos\varphi+ix^2\sin\varphi}{1-x^2\cos\varphi}}\\ &=\frac i2\ln\frac{1-x^2(\cos\varphi+i\sin\varphi)}{1-x^2(\cos\varphi-i\sin\varphi)}\\ &=\frac i2\ln\frac{1-x^2e^{i\varphi}}{1-x^2e^{-i\varphi}}\\ &=\frac i2(\ln(1-x^2e^{i\varphi})-\ln(1-x^2e^{-i\varphi})) \end{align}$$ Using the Maclaurin series of $\ln(1-z)$, which is valid since $|x^2e^{i\varphi}|=x^2\le1$: $$\begin{align} \frac i2(\ln(1-x^2e^{i\varphi})-\ln(1-x^2e^{-i\varphi}))&=\frac i2\left(\sum_{n=1}^\infty-\frac{(x^2e^{i\varphi})^n}n-\sum_{n=1}^\infty-\frac{(x^2e^{-i\varphi})^n}n\right)\\ &=-\frac i2\sum_{n=1}^\infty\frac{x^{2n}}n(e^{in\varphi}-e^{-in\varphi})\\ &=-\frac i2\sum_{n=1}^\infty\frac{x^{2n}}n\cdot2i\sin n\varphi\\ &=\sum_{n=1}^\infty\frac{x^{2n}}n\sin n\varphi \end{align}$$ Therefore $$\frac12\int_0^\pi\int_0^\omega\tan^{-1}\frac{x^2\sin\varphi}{1-x^2\cos\varphi}\,d\varphi\,d\omega=\frac12\int_0^\pi\int_0^\omega\sum_{n=1}^\infty\frac{x^{2n}}n\sin n\varphi\,d\varphi\,d\omega$$ We can move the summation and $\frac{x^{2n}}n$ factor to the outside, then perform simple termwise integration to get the desired result: $$\begin{align} \frac12\int_0^\pi\int_0^\omega\sum_{n=1}^\infty\frac{x^{2n}}n\sin n\varphi\,d\varphi\,d\omega&=\frac12\sum_{n=1}^\infty\frac{x^{2n}}n\int_0^\pi\int_0^\omega\sin n\varphi\,d\varphi\,d\omega\\ &=\frac12\sum_{n=1}^\infty\frac{x^{2n}}n\int_0^\pi\left[-\frac1n\cos n\varphi\right]_0^\omega\,d\omega\\ &=\frac12\sum_{n=1}^\infty\frac{x^{2n}}n\int_0^\pi\frac1n(1-\cos n\omega)\,d\omega\\ &=\frac12\sum_{n=1}^\infty\frac{x^{2n}}{n^2}\int_0^\pi(1-\cos n\omega)\,d\omega\\ &=\frac12\sum_{n=1}^\infty\frac{x^{2n}}{n^2}\left[\omega-\frac1n\sin n\omega\right]_0^\pi\\ &=\frac12\sum_{n=1}^\infty\frac{x^{2n}}{n^2}\cdot\pi\\ &=\frac\pi2\sum_{n=1}^\infty\frac{x^{2n}}{n^2}\\ \mathcal P(x)&=\frac\pi2\operatorname{Li}_2(x^2) \end{align}$$
More generally, we have the following identity for $|z|\le1$: $$\int_0^\pi\int_0^\omega\tan^{-1}\frac{z\sin\varphi}{1-z\cos\varphi}\,d\varphi\,d\omega=\pi\operatorname{Li}_2(z)$$