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