The problem is to show $$\int_{0}^{1} \frac{\operatorname{arsinh}x}{x\sqrt{1-x^4}} \,\mathrm{d}x = \frac{\pi^2}{8}$$ which is a typical integral emerges in many specific value, yet I have no more insight how to prove this one directly (any possible method other than the example below is acceptable).
Here is an example when the integral pops out in Legendre-chi function $\chi_2$, like (1) in this post, we have $$ \int_{0}^{\pi/2} {\arctan(\sin x) \,\mathrm{d}x} = 2\chi_2(\sqrt2-1) = \frac{\pi^2}{8} - \frac{\ln^{2}(\sqrt{2}+1)}{2} $$ which is a very typical integral, like in this post, from where we already know that $$ \int_{0}^{\pi/2} {\arctan(\sin x) \,\mathrm{d}x} = \int_{0}^{1} {\frac{\arctan x}{\sqrt{1-x^2}} \,\mathrm{d}x} = \int_{0}^{1} {\frac{\operatorname{arsinh}x}{x\sqrt{1+x^{2}}} \,\mathrm{d}x} $$ or $$ \int_{0}^{\pi/4} {\arcsin(\tan x) \,\mathrm{d}x} = \frac{\pi^2}{8} - \int_{0}^{\pi/2} {\arctan(\sin x) \,\mathrm{d}x} = \frac{\ln^{2}(\sqrt{2}+1)}{2} $$ Now, for $a<1$ consider the following parameterization $$ I(a) = \int_{0}^{\pi/4} {\arcsin(a\tan x) \,\mathrm{d}x} $$ let $u=\tan x$ and $y^2=(1-a^2u^2)/(1+u^2)$ in its derivative $$ \begin{aligned} I'(a) &= \int_{0}^{\pi/4} {\frac{\tan x}{\sqrt{1-a^2\tan^2x}} \,\mathrm{d}x} = \int_{0}^{1} {\frac{u}{(1+u^2)\sqrt{1-a^2u^2}} \,\mathrm{d}u} \\ &= \frac1{\sqrt{1+a^2}}\int_{\sqrt{(1-a^2)/2}}^{1} {\frac{\mathrm{d}y}{\sqrt{a^2+y^2}}} = \frac1{\sqrt{1+a^2}}\operatorname{arsinh}\left(\frac{y}{a}\right)\biggr|_{y=\sqrt{(1-a^2)/2}}^{1} \\ &= \frac{1}{\sqrt{1+a^{2}}} \operatorname{arsinh}\frac1{a} - \frac{1}{\sqrt{1+a^{2}}} \operatorname{arsinh}\sqrt{\frac{1-a^2}{2a^2}} \end{aligned} $$ Integrtation by parts gives $$ \begin{aligned} \int_{0}^{\pi/4} {\arcsin(\tan x) \,\mathrm{d}x} &= \int_{0}^{1} \frac{1}{\sqrt{1+a^{2}}} \operatorname{arsinh}\frac1{a} \,\mathrm{d}x - \int_{0}^{1} \frac{1}{\sqrt{1+a^{2}}} \operatorname{arsinh}\sqrt{\frac{1-a^2}{2a^2}} \,\mathrm{d}x \\ &= \operatorname{arsinh}^2(1) + \int_{0}^{1} {\frac{\operatorname{arsinh}a}{a\sqrt{1+a^{2}}}\,\mathrm{d}a} - \int_{0}^{1} \frac{\operatorname{arsinh}a}{a\sqrt{1-a^4}} \,\mathrm{d}a \end{aligned} $$ Combining all the special case of $\chi_2(\sqrt2-1)$ from above, we will find the required integral. And there are other equivalent form of required integral, for example $$ \int_{0}^{1} \frac{\operatorname{arsinh}x}{x\sqrt{1-x^4}} \,\mathrm{d}x = \frac1{2} \int_{0}^{1} \frac{1}{\sqrt{1+x^2}} \operatorname{arcosh}\frac1{x^2} \,\mathrm{d}x = \frac1{4} \int_{0}^{1} \frac{1}{\sqrt{x(1+x)}} \operatorname{arcosh}\frac1{x} \,\mathrm{d}x $$ Yet, I still can not find any convenience for further calculation.
Thanks in advance for any help.
Integrate[ArcSinh[x]/(x*Sqrt[1 - x^4]), {x, 0, 1}] // FullSimplify
Mathematica shows the result is $\frac{3 \Gamma \left(\frac{1}{4}\right)^2 \, _3F_2\left(\frac{1}{4},\frac{1}{4},\frac{1}{4};\frac{1}{2},\frac{5}{4};1\right)-2 \Gamma \left(\frac{3}{4}\right)^2 \, _3F_2\left(\frac{3}{4},\frac{3}{4},\frac{3}{4};\frac{3}{2},\frac{7}{4};1\right)}{12 \sqrt{2 \pi }}$ $\endgroup$