Closed-form of $\int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx $ 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.
 A: According to a CAS, $$I = \int_0^1 \frac{\operatorname{Li}_2\left( \sqrt{t} \right)}{2 \, \sqrt{t} \, \sqrt{1-t}} \,dt =\,
   _4F_3\left(\frac{1}{2},\frac{1}{2},1,1;\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right
   )+\frac{\pi ^3}{48}-\frac{1}{4} \pi  \log ^2(2)$$
Enjoy !
A: My attempt. This is by no means closer to the answer, but I want to address several equivalent forms that might be helpful for future calculations.
First, from Landen's identity of the following form
$$ \mathrm{Li}_2(z) = -\mathrm{Li}_2\left(-\frac{z}{1-z}\right) - \frac{1}{2}\log^{2}(1-z), \quad z \notin [1, \infty)$$
we observe that
\begin{align*}
I
&= -\int_{0}^{1} \frac{1}{\sqrt{1-x^2}} \left\{ \mathrm{Li}_2\left(-\frac{x}{1-x}\right) - \frac{1}{2}\log^{2}(1-x) \right\} \, dx \\
&= -\int_{-\infty}^{0} \frac{2\mathrm{Li}_2 (t) + \log^{2}(1-t)}{2\sqrt{1-2t}(1-t)} \, dt
\tag{1}
\end{align*}
By noting that
$$ \frac{d}{dt} \arctan\left(\frac{1}{\sqrt{1-2t}}\right) = \frac{1}{2\sqrt{1-2t}(1-t)}, $$
integration by parts and the substitution $x = (1-2t)^{-1/2}$ shows that (1) is equal to
\begin{align*}
I
&= \int_{-\infty}^{0} \arctan\left(\frac{1}{\sqrt{1-2t}}\right) \frac{2\log(1-t)}{t(t-1)}  \, dt \\
&= \int_{0}^{1} \frac{8x \arctan x}{1 - x^4} \log \left( \frac{1+x^2}{2x^2} \right) \, dx \tag{2}
\end{align*}
The following observation
$$ \Re \log \left(\frac{1+ix}{\sqrt{2}} \right) = \frac{1}{2}\log \left( \frac{1+x^2}{2} \right)
\quad \text{and} \quad
\Im \log \left(\frac{1+ix}{\sqrt{2}} \right) = \arctan x $$
somehow seems to suggest complex-analytic approach, but I have not been successful with such approaches so far. Next, from the following simple formula
$$ \log \left( \frac{1+x^2}{2x^2} \right) \, dx = \int_{0}^{1} \frac{d}{dy} \log \left( \frac{y^2+x^2}{y^2 + 1} \right) \, dy $$
the integral (2) can be further decomposed into the following form
$$ I = \int_{0}^{1}\int_{0}^{1} \frac{16xy \arctan x}{(1+x^2)(1+y^2)(x^2+y^2)} \,dxdy. \tag{3} $$
Simple calculation shows that
$$ \int_{0}^{1}\int_{0}^{1} \frac{16xy}{(1+x^2)(1+y^2)(x^2+y^2)} \,dxdy = 2\zeta(2), $$
so I suspect that the situation in (3) is not that bad.
A: Following Anastasiya-Romanova's approach, we have:
$$ I = \frac{1}{2}\sum_{n\geq 1}\frac{1}{n^2}\int_{0}^{\pi/2}\sin^n x\,dx =\frac{\pi}{16}\sum_{n\geq 1}\frac{\binom{2n}{n}}{n^2 4^n}+\frac{1}{4}\sum_{n\geq 1}\frac{4^n}{\binom{2n}{n}n(2n-1)^2}\tag{1}$$
where:
$$ S_1 = \sum_{n\geq 1}\frac{\binom{2n}{n}}{n^2 4^n} = \zeta(2)-2\log^2 2 \tag{2}$$
and the second sum is the problematic one, leading to a value for a hypergeometric function $\phantom{}_4 F_3$:
$$\begin{eqnarray*}S_2 = \sum_{n\geq 1}\frac{4^n}{\binom{2n}{n}n(2n-1)^2} &=& -\int_{0}^{1}\frac{2\arcsin x}{x\sqrt{1-x^2}}\log x\,dx\\&=&-2\int_{0}^{\pi/2}\frac{\theta}{\sin\theta}\log\sin\theta\,d\theta. \tag{3}\end{eqnarray*}$$
However, since the Fourier cosine series of $\log\sin\theta$ is well-known:
$$\log\sin\theta = -\log 2-\sum_{n\geq 1}\frac{\cos(2n\theta)}{n}\tag{4} $$
we just need to compute the Fourier cosine series of $\frac{\theta}{\sin\theta}$. Since the Fourier sine series of the triangle wave is given by:
$$ \theta = \sum_{n\geq 1}\frac{(-1)^{n+1}}{n}\sin(2n\theta) \tag{5}$$
by exploiting $\frac{\sin(2n\theta)}{\sin\theta}=2\left(\cos\theta+\cos(3\theta)+\ldots+\cos((2n-1)\theta)\right)$ we have:
$$\frac{\theta}{\sin\theta}=2\sum_{k=1}^{+\infty}\left(\sum_{n\geq k}\frac{(-1)^{n+1}}{n}\right)\cos((2k-1)\theta)\tag{6}$$
so, at least in principle, $S_2$ is computable through a Fourier-analytic approach, by exploiting:
$$ \int_{0}^{\pi/2}\cos((2n-1)\theta)\cos(2m\theta)\,d\theta = \frac{(2n-1)(-1)^m}{4m^2-(2n-1)^2}.\tag{7}$$
It is also interesting to notice that the last integral appearing in $(3)$ is very similar to the one appearing in this related question, but the latter is way easier to compute since in the Fourier cosine series of $\frac{\theta}{\sin\theta}\,\cos^2\theta$ there are only "even cosines".
A: 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

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

A: Using Maple I am obtaining
$$1+\frac{\pi }{16}{\ _4F_3(1,1,1,3/2;\,2,2,2;\,1)}+\frac{\sqrt {\pi }}{8}
G^{4, 1}_{4, 4}\left(-1\, \Big\vert\,^{1, 5/2, 5/2, 5/2}_{2, 3/2, 3/2, 1}\right)
$$
and a numerical approximation is
$$1.3913063720392030337$$
A: An alternative approach. As shown by nospoon here,
\begin{equation}\label{shalev} \int_{0}^{1}\frac{\log^2(x)}{\sqrt{x(1- x\sin^2\theta )}}\,dx = \frac{8}{\sin\theta}\left[\frac{\theta^3}{3}-\text{Im}\,\text{Li}_3\left(1-e^{2i\theta}\right)\right]\tag{1}\end{equation}
holds for any $\theta\in\left(0,\frac{\pi}{2}\right)$. This is an istance of a very nice principle, according to which every hypergeometric function of the $\phantom{}_{p+1}F_{p}\left(\frac{1}{2},\frac{1}{2},\ldots;\frac{3}{2},\frac{3}{2},\ldots;z\right)$ kind has a closed form in terms of polylogarithms. We need to evaluate $\phantom{}_4 F_3\left(\frac{1}{2},\frac{1}{2},1,1;\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right)$, hence our arrival point is already pretty close to the statement of $(1)$.
The evaluation of $(1)$ at $\theta=\frac{\pi}{4}$ leads to
\begin{equation}\label{shalev2} \int_{0}^{1}\frac{\log^2(x)}{\sqrt{x(2- x )}}\,dx = 8\left[\frac{\pi^3}{192}-\text{Im}\,\text{Li}_3\left(1-i\right)\right]\tag{2}\end{equation}
and the functional relations for $\text{Li}_2$ reduce the original problem to $(2)$. In particular
$$\begin{eqnarray*} \int_{0}^{\pi/2}\text{Li}_2(\sin\theta)\,d\theta &=& \int_{0}^{1}\frac{2}{1+t^2}\text{Li}_2\left(\frac{1-t^2}{1+t^2}\right)\,dt\\
\int_{0}^{\pi/2}\text{Li}_2(\cos\theta)\,d\theta&=&\int_{0}^{1}\frac{2}{1+t^2}\text{Li}_2\left(\frac{2t}{1+t^2}\right)\,dt\\ 
&=&\frac{5\pi^3}{24}+4\,\text{Im}\,\text{Li}_3(1-i)\tag{3}\end{eqnarray*}$$
as already shown by Reshetnikov.
A: Another one... replacing the Meijer G in Juan's answer
$$
{\mbox{$_4$F$_3$}(1/2,1/2,1,1;\,3/2,3/2,3/2;\,1)}+
\frac{\pi \,
{\mbox{$_4$F$_3$}(1,1,1,3/2;\,2,2,2;\,1)}}{16}
\\
\approx 1.3913072075067666818109648381255138301541952863
$$
and with user's comment
$$
{\mbox{$_4$F$_3$}(1/2,1/2,1,1;\,3/2,3/2,3/2;\,1)}+\frac{\pi^3}{48}-\frac{\pi (\log 2)^2}{4}
$$
agreeing with Claude.
