About the integral $\int_{0}^{1}\text{arctanh}(x)\arcsin(x)\frac{dx}{x}$

Question

I am receiving a lot of feedback from the recent article I published on Arxiv, whose first part was originally presented here as an answer to a question of John Campbell. About that, today Steven Finch asked me if it is possible to evaluate in terms of a nice closed form

$$\begin{eqnarray*} \phantom{}_4 F_3\left(\frac{1}{2},\frac{1}{2},1,1;\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right)&=&\sum_{n\geq 0}\frac{4^n}{(2n+1)^3 {\binom{2n}{n}}}\\&=&\sum_{n\geq 0}\frac{4^n B(n+1,n+1)}{(2n+1)^2}\\&=&\int_{0}^{1}\sum_{n\geq 0}\frac{(1-x^2)^n}{(2n+1)^2}\,dx\\&=&\int_{0}^{\pi/2}\frac{\sin^{2n+1}\theta}{(2n+1)^2}\,d\theta\\&=&\int_{0}^{1}\frac{\text{Li}_2(x)-\text{Li}_2(-x)}{2\sqrt{1-x^2}}\,dx\\&=&\frac{\pi^3}{16}-\color{blue}{\int_{0}^{1}\text{arctanh}(x)\arcsin(x)\frac{dx}{x}}.\end{eqnarray*}$$

I would really like to help him, and I invoke your help too. I am still struggling to find an efficient approach for dealing with the blue integral. Fourier-Chebyshev or Fourier-Legendre series expansions seem promising, and the Lemma

$$ \int_{0}^{1}\text{arctanh}^s(x)\,dx = \frac{2\zeta(s)(2^s-2)\Gamma(s+1)}{4^s} \tag{Lemma}$$ (proved at page 81 here) might be relevant too. It is important to mention that John Campbell himself recently dealt with the similar integrals $$\int_{0}^{1}\arcsin(x)\log(x)\,dx =2-\frac{\pi}{2}-\log 2,\\ \int_{0}^{1}\arcsin(x)\log(x)\frac{dx}{x}=-\frac{\pi^3}{48}-\frac{\pi}{4}\log^2(2),\tag{SimInt} $$ but something seems to suddenly stop working if $\log(x)$ is replaced by $\log(1\pm x)$.

Addendum (thanks to Tolaso J Kos): the integral $\int_{0}^{1}\frac{\text{Li}_2(x)}{\sqrt{1-x^2}}\,dx$ has been proved to be depending on the imaginary part of a trilogarithm by Vladimir Reshetnikov, here. By following his technique verbatim, I got

$$ \phantom{}_4 F_3\left(\frac{1}{2},\frac{1}{2},1,1;\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right)=\color{blue}{\frac{3\pi^3}{16}+\frac{\pi}{4}\log^2(2)-4\,\text{Im}\,\text{Li}_3(1+i)}$$

and I guess that settles the question.
In terms of an absolutely convergent series, the RHS of the last line equals

$$ -\frac{\pi^3}{32}-\frac{\pi}{8}\log^2(2)+4\sum_{n\geq 1}\frac{\sin(\pi n/4)}{n^3 \sqrt{2}^n}. $$

This can probably be seen as an instance of an acceleration technique for the series $\sum_{n\geq 0}\frac{4^n}{(2n+1)^3\binom{2n}{n}}$, whose general term roughly behaves like $\frac{\sqrt{\pi}}{8}\cdot\frac{1}{n^{5/2}}$.

Answer 1

This is not an answer to the main question, which has been solved through the pointers provided by Tolaso and others. Instead, let us prove James Arathoon conjecture in the comments. I have the strong feeling this thread might be an interesting starting point for studying "twisted" Euler sums and the interplay between $\text{Li}_{\color{red}{3}}$ and $\phantom{}_4 F_3$.
It is well-know (and not difficult to prove) that for any $x\in(-1,1)$ $$\arcsin(x)=\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n(2n+1)}x^{2n+1}\tag{ArcSin}$$ and the following identity holds for any $n\in\mathbb{N}$: $$ \int_{0}^{1} x^{2n}\text{arctanh}(x)\,dx = \frac{H_n+2\log 2}{4n+2}\tag{ArcTanh}$$ as a simple consequence of integration by parts:

$$\begin{eqnarray*} \int_{0}^{1} x^{2n}\text{arctanh}(x)\,dx &=& \left[\frac{x^{2n+1}-1}{2n+1}\text{arctanh}(x)\right]_{0}^{1}+\frac{1}{2n+1}\int_{0}^{1}\frac{x^{2n+1}-1}{x^2-1}\,dx\\&=&\frac{1}{2n+1}\int_{0}^{1}\left(\frac{x^{2n+1}-x}{x^2-1}+\frac{1}{x+1}\right)\,dx\\&=&\frac{1}{2n+1}\left(\log 2+\frac{1}{2}\int_{0}^{1}\frac{z^n-1}{z-1}\,dz\right)=\frac{\log 2+\frac{H_n}{2}}{2n+1}.\end{eqnarray*}$$ By combining $(\text{ArcSin})$ and $(\text{ArcTanh})$ we have $$ \int_{0}^{1}\arcsin(x)\text{arctanh}(x)\frac{dx}{x}=\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n(2n+1)^2}\left(\log 2+\frac{H_n}{2}\right)$$ and James Arathoon's conjecture is proved by computing $$\begin{eqnarray*} \sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n(2n+1)^2}&=&\int_{0}^{1}\frac{\arcsin(x)}{x}\,dx=\int_{0}^{\pi/2}\theta\cot\theta\,d\theta\\&\stackrel{\text{IBP}}{=}&\int_{0}^{\pi/2}\log\cos\theta\,d\theta = \frac{\pi\log 2}{2}.\end{eqnarray*}$$ This proves:

$$\begin{eqnarray*}\sum_{n\geq 0}\frac{\binom{2n}{n}H_n}{4^n(2n+1)^2}&=&2\int_{0}^{1}\arcsin(x)\text{arctanh}(x)\frac{dx}{x}-\pi\log^2(2)\\&=&\frac{\pi^3}{8}-\pi\log^2(2)-2\cdot\phantom{}_4 F_3\left(\frac{1}{2},\frac{1}{2},1,1;\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right)\\&=&-\frac{\pi^3}{4}-\frac{3\pi}{2}\log^2(2)+8\,\text{Im}\,\text{Li}_3(1+i)\\&=&\frac{3\pi^3}{16}-\frac{3\pi}{4}\log^2(2)-8\sum_{n\geq 1}\frac{\sin(\pi n/4)}{n^3\sqrt{2}^n}.\end{eqnarray*}$$

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Just in order to keep collecting interesting material, I would like to mention that in this thread by Markus Scheuer it is shown that $\frac{1}{2k+1}$ is the binomial transform of $\frac{4^k}{(2k+1)\binom{2k}{k}}$ and $\frac{1}{2k+3}$ is the binomial transform of $\frac{4^k}{(2k+1)(2k+3)\binom{2k}{k}}$. That might be a useful lemma for dealing with values of $\phantom{}_4 F_3$ whose associated series contains terms of such form or their squares.

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This thread contains other interesting informations about the interplay between $\text{Li}_3$ and the value of $\phantom{}_4 F_3$ mentioned in the above question.