TITLE: Closed-form of $\int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx $
QUESTION [46 upvotes]: I'm looking for a closed form of this integral.
$$I = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx ,$$
where $\operatorname{Li}_2$ is the dilogarithm function.
A numerical approximation of it is
$$ I \approx 1.39130720750676668181096483812551383015419528634319581297153...$$
As Lucian said $I$ has the following equivalent forms:
$$I = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx = \int_0^1 \frac{\operatorname{Li}_2\left( \sqrt{x} \right)}{2 \, \sqrt{x} \, \sqrt{1-x}} \,dx = \int_0^{\frac{\pi}{2}} \operatorname{Li}_2(\sin x) \, dx = \int_0^{\frac{\pi}{2}} \operatorname{Li}_2(\cos x) \, dx$$
According to Mathematica it has a closed-form in terms of generalized hypergeometric function, Claude Leibovici has given us this form.
With Maple using Anastasiya-Romanova's form I could get a closed-form in term of Meijer G function. It was similar to Juan Ospina's answer, but it wasn't exactly that form. I also don't know that his form is correct, or not, because the numerical approximation has just $6$ correct digits. 
I'm looking for a closed form of $I$ without using generalized hypergeometric function, Meijer G function or $\operatorname{Li}_2$ or $\operatorname{Li}_3$.
I hope it exists. Similar integrals are the following.
$$\begin{align}
J_1 & = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{1+x} \,dx = \frac{\pi^2}{6} \ln 2 - \frac58 \zeta(3) \\
J_2 & = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x}} \,dx = \pi^2 - 8 \end{align}$$
Related techniques are in this or in this paper. This one also could be useful.

REPLY [3 votes]: From writing $$\operatorname{Li}_2(x)=-\int_0^1\frac{x\ln u}{1-xu}du$$
It follows that 
$$-I=-\int_0^1\frac{\operatorname{Li}_2(x)}{\sqrt{1-x^2}}dx=\int_0^1\ln u\left[\int_0^1\frac{x}{(1-ux)\sqrt{1-x^2}}dx\right]du$$
$$=\int_0^1\ln u\left[\frac{\pi}{2}\cdot\left(\frac{1}{u\sqrt{1-u^2}}-\frac1u\right)+\frac{\sin^{-1}(u)}{u\sqrt{1-u^2}}\right]du$$
$$=\frac{\pi}2\int_0^1\frac{\ln u}{u}\left(\frac1{\sqrt{1-u^2}}-1\right)du+\int_0^1\frac{\ln u\sin^{-1}(u)}{u\sqrt{1-u^2}}du$$
For the first integral, let $u^2\to u$ first then apply integration by parts, we obtain
$$\frac{\pi}{2}\int_0^1\frac{\ln u}{u}\left(\frac{1}{\sqrt{1-u^2}}-1\right)\ du=\frac{\pi}{8}\int_0^1\ln^2u\ du\left(\frac{1}{\sqrt{1-u}}-1\right)du\\=-\frac{\pi}{32}\int_0^1\ln^2u (1-u)^{-3/2}du=-\frac{\pi}{32}\frac{\partial^2}{\partial\alpha^2}\lim_{\alpha\ \mapsto1}\text{B}\left(\alpha,-\frac12\right)\\=-\frac{\pi}{32}\left(\frac23\pi^2-8\ln^22\right)=\boxed{\frac{\pi}4\ln^2(2)-\frac{\pi^3}{48}}\, .$$
The second integral is already calculated here
$$\int_0^1\frac{\ln(x) \sin^{-1}(x)}{x\sqrt{1-x^2}}dx=\boxed{4 \operatorname{Im} \operatorname{Li}_3(1+\mathrm{i}) -\frac{3 \pi^3}{16} -\frac{\pi}{4} \ln^2(2)}  \, .$$
Collecting the boxed results we get

$$I= \frac{5 \pi^3}{24}-4 \operatorname{Im} \operatorname{Li}_3(1+\mathrm{i})  \, .$$