I encountered an integral identity:
$$\int_{0}^{1} \frac{K(x)K\left ( \sqrt{2} \sqrt{\frac{1-x^2}{2-x^2} } \right ) }{2-x^2}\,dx =\frac{\pi^3}{8\sqrt{2}} {}_6F_5\left ( \frac{1}{4},\frac{1}{4},\frac{1}{4}, \frac{3}{4},\frac{3}{4},\frac{3}{4}; \frac{1}{2},\frac{1}{2},1,1,1;1 \right ).$$ Where $K(x)=\int_{0}^{1} \frac{1}{\sqrt{1-t^2}} \frac{1}{\sqrt{1-x^2t^2} }\text{d}t$ and ${}_6F_5$ is the Generalized Hypergeometric function.\
Here is my attempt: $$ \begin{aligned} &\int_{0}^{1} \frac{K(x)K\left ( \sqrt{2} \sqrt{\frac{1-x^2}{2-x^2} } \right ) }{2-x^2}\text{d}x\\ \\ &=\frac{1}{2} \int_{0}^{1} \frac{K(\sqrt{x} )K\left ( \sqrt{2} \sqrt{\frac{1-x}{2-x}}\right) }{\sqrt{x} (2-x)}\text{d}x\\ \\ &=\frac{1}{2}\int_{0}^{1} \frac{K(\sqrt{1-x} )K\left ( \sqrt{2} \sqrt{\frac{x}{1+x} } \right ) }{\sqrt{1-x}(1+x) }\text{d}x\\ \\ &=\;... \end{aligned} $$
Thanks to the user who answered this question. I had found these closed-forms: $$\int_{0}^{1} \frac{K^2\left ( \sqrt{\frac{1}{2}-\frac{1}{\sqrt{2} } \sqrt{\frac{1-x^2}{2-x^2} } } \right ) K(\sqrt{1-x^2} )}{2-x^2}\text{d}x = \frac{\pi ^{7/2} \,\Gamma \left(\frac{1}{4}\right) \, _8F_7\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{3}{4},\frac{3}{4},\frac{3}{4};\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},1,1,1;1\right)}{32 \sqrt{2}\, \Gamma \left(\frac{3}{4}\right)}+\frac{\pi ^{7/2} \,\Gamma \left(\frac{3}{4}\right) \, _8F_7\left(\frac{3}{4},\frac{3}{4},\frac{3}{4},\frac{3}{4},\frac{3}{4},\frac{5}{4},\frac{5}{4},\frac{5}{4};1,1,1,\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right)}{128 \sqrt{2} \,\Gamma \left(\frac{1}{4}\right)}$$ $$\int_{0}^{1} \frac{\operatorname{Li}_4\left ( \frac{x}{\sqrt{2-x^2} } \right ) K(\sqrt{1-x^2}) }{2-x^2}\text{d}x =\frac{\pi}{4 \sqrt{2}}\, _7F_6\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{2},\frac{1}{2},1;\frac{3}{4},\frac{5}{4},\frac{5}{4},\frac{5}{4},\frac{5}{4},\frac{5}{4};1\right) +\frac{1}{243 \sqrt{2}}\, _7F_6\left(\frac{3}{4},\frac{3}{4},\frac{3}{4},\frac{3}{4},1,1,1;\frac{5}{4},\frac{7}{4},\frac{7}{4},\frac{7}{4},\frac{7}{4},\frac{7}{4};1\right) +\frac{\pi ^{3/2} \,\Gamma \left(\frac{3}{4}\right)}{64 \sqrt{2} \,\Gamma \left(\frac{1}{4}\right)} \, _6F_5\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{3}{4},\frac{3}{4};\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right) +\frac{\pi ^{3/2} \, \Gamma \left(\frac{1}{4}\right)}{16384 \sqrt{2}\,\Gamma \left(\frac{3}{4}\right)}\, _7F_6\left(1,1,1,1,1,\frac{5}{4},\frac{5}{4};\frac{3}{2},2,2,2,2,2;1\right)$$
Our objective is to verify the conjectured identity
$$\int_{0}^{1}\mathrm{d}x\,\frac{K{\left(x\right)}K{\left(\sqrt{\frac{2-2x^{2}}{2-x^{2}}}\right)}}{2-x^{2}}=\frac{\pi^{3}}{8\sqrt{2}}\,{_6F_5}{\left(\frac14,\frac14,\frac14,\frac34,\frac34,\frac34;\frac12,\frac12,1,1,1;1\right)},\tag{1}$$
where here $K$ denotes the complete elliptic integral of the first kind and is defined via the usual integral representation
$$K{\left(k\right)}:=\int_{0}^{1}\mathrm{d}x\,\frac{1}{\sqrt{\left(1-x^{2}\right)\left(1-k^{2}x^{2}\right)}};~~~\small{k\in(-1,1)},$$
and for positive real parameters $\left(a_{1},a_{2},a_{3},a_{4},a_{5},a_{6}\right)\in\mathbb{R}_{>0}^{6}$ and $\left(b_{1},b_{2},b_{3},b_{4},b_{5}\right)\in\mathbb{R}_{>0}^{5}$ the generalized hypergeometric function ${_6F_5}$ is given by the infinite series
$${_6F_5}{\left(a_{1},a_{2},a_{3},a_{4},a_{5},a_{6};b_{1},b_{2},b_{3},b_{4},b_{5};z\right)}:=\sum_{n=0}^{\infty}\frac{\left(a_{1}\right)_{n}\,\left(a_{2}\right)_{n}\,\left(a_{3}\right)_{n}\,\left(a_{4}\right)_{n}\,\left(a_{5}\right)_{n}\,\left(a_{6}\right)_{n}}{\left(b_{1}\right)_{n}\,\left(b_{2}\right)_{n}\,\left(b_{3}\right)_{n}\,\left(b_{4}\right)_{n}\,\left(b_{5}\right)_{n}}\,\frac{z^{n}}{n!},$$
which converges absolutely for $z\in\mathbb{C}\land|z|\le1$ provided that $0<\sum_{j=1}^{5}b_{j}-\sum_{j=1}^{6}a_{j}$.
The Pochhammer symbol $(a)_{n}$ is defined for positive real $a$ and nonnegative integer index $n$ in terms of the gamma function as
$$\left(a\right)_{n}:=\frac{\Gamma{\left(a+n\right)}}{\Gamma{\left(a\right)}};~~~\small{a\in\mathbb{R}_{>0}\land n\in\mathbb{Z}_{\ge0}}.$$
Let $\mathcal{H}$ denote the following generalized hypergeometric series:
$$\mathcal{H}:={_6F_5}{\left(\frac14,\frac14,\frac14,\frac34,\frac34,\frac34;\frac12,\frac12,1,1,1;1\right)}.$$
The beta function will be our friend in the evaluation of $\mathcal{H}$:
$$\operatorname{B}{\left(x,y\right)}=\frac{\Gamma{\left(x\right)}\,\Gamma{\left(y\right)}}{\Gamma{\left(x+y\right)}}=\int_{0}^{1}\mathrm{d}t\,t^{x-1}\left(1-t\right)^{y-1};~~~\small{(x,y)\in\mathbb{R}_{>0}^{2}}.$$
Consider the following result which allows us to express a certain ratio of Pochhammer symbols as a beta function: for any $n\in\mathbb{Z}_{\ge0}$,
$$\begin{align} \frac{\left(\frac14\right)_{n}\,\left(\frac34\right)_{n}}{\left(\frac12\right)_{n}\,\left(1\right)_{n}} &=\frac{\Gamma{\left(\frac12\right)}\,\Gamma{\left(1\right)}\,\Gamma{\left(\frac14+n\right)}\,\Gamma{\left(\frac34+n\right)}}{\Gamma{\left(\frac14\right)}\,\Gamma{\left(\frac34\right)}\,\Gamma{\left(\frac12+n\right)}\,\Gamma{\left(1+n\right)}}\\ &=\frac{\Gamma{\left(\frac14+n\right)}\,\Gamma{\left(\frac34+n\right)}}{\sqrt{2\pi}\,\Gamma{\left(\frac12+n\right)}\,\Gamma{\left(1+n\right)}}\\ &=\frac{\Gamma{\left(\frac12\right)}\,\Gamma{\left(2n+\frac12\right)}}{\pi\,\Gamma{\left(2n+1\right)}}\\ &=\frac{1}{\pi}\operatorname{B}{\left(2n+\frac12,\frac12\right)}.\\ \end{align}$$
Then, the hypergeometric series $\mathcal{H}$ can be rewritten as a multiple integral as follows:
$$\begin{align} \mathcal{H} &={_6F_5}{\left(\frac14,\frac14,\frac14,\frac34,\frac34,\frac34;\frac12,\frac12,1,1,1;1\right)}\\ &=\sum_{n=0}^{\infty}\frac{\left(\frac14\right)_{n}\,\left(\frac14\right)_{n}\,\left(\frac14\right)_{n}\,\left(\frac34\right)_{n}\,\left(\frac34\right)_{n}\,\left(\frac34\right)_{n}}{\left(\frac12\right)_{n}\,\left(\frac12\right)_{n}\,\left(1\right)_{n}\,\left(1\right)_{n}\,\left(1\right)_{n}}\cdot\frac{1}{n!}\\ &=\sum_{n=0}^{\infty}\frac{\left(\frac14\right)_{n}\,\left(\frac14\right)_{n}\,\left(\frac14\right)_{n}\,\left(\frac34\right)_{n}\,\left(\frac34\right)_{n}\,\left(\frac34\right)_{n}}{\left(\frac12\right)_{n}\,\left(\frac12\right)_{n}\,\left(\frac12\right)_{n}\,\left(1\right)_{n}\,\left(1\right)_{n}\,\left(1\right)_{n}}\cdot\frac{\left(\frac12\right)_{n}}{\Gamma{\left(n+1\right)}}\\ &=\sum_{n=0}^{\infty}\left[\frac{\left(\frac14\right)_{n}\,\left(\frac34\right)_{n}}{\left(\frac12\right)_{n}\,\left(1\right)_{n}}\right]^{3}\frac{\Gamma{\left(\frac12+n\right)}}{\Gamma{\left(\frac12\right)}\,\Gamma{\left(n+1\right)}}\\ &=\sum_{n=0}^{\infty}\left[\frac{\left(\frac14\right)_{n}\,\left(\frac34\right)_{n}}{\left(\frac12\right)_{n}\,\left(1\right)_{n}}\right]^{3}\frac{\Gamma{\left(n+\frac12\right)}\,\Gamma{\left(\frac12\right)}}{\left[\Gamma{\left(\frac12\right)}\right]^{2}\,\Gamma{\left(n+1\right)}}\\ &=\frac{1}{\pi^{4}}\sum_{n=0}^{\infty}\left[\operatorname{B}{\left(2n+\frac12,\frac12\right)}\right]^{3}\operatorname{B}{\left(n+\frac12,\frac12\right)}\\ &=\frac{1}{\pi^{4}}\sum_{n=0}^{\infty}\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}u\int_{0}^{1}\mathrm{d}v\int_{0}^{1}\mathrm{d}w\,\frac{t^{2n-1/2}}{\sqrt{1-t}}\,\frac{u^{2n-1/2}}{\sqrt{1-u}}\,\frac{v^{2n-1/2}}{\sqrt{1-v}}\,\frac{w^{n-1/2}}{\sqrt{1-w}}\\ &=\frac{1}{\pi^{4}}\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}u\int_{0}^{1}\mathrm{d}v\int_{0}^{1}\mathrm{d}w\,\frac{\sum_{n=0}^{\infty}\left(t^{2}u^{2}v^{2}w\right)^{n}}{\sqrt{t\left(1-t\right)u\left(1-u\right)v\left(1-v\right)w\left(1-w\right)}}\\ &=\frac{1}{\pi^{4}}\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}u\int_{0}^{1}\mathrm{d}v\int_{0}^{1}\mathrm{d}w\,\frac{\left(1-t^{2}u^{2}v^{2}w\right)^{-1}}{\sqrt{t\left(1-t\right)u\left(1-u\right)v\left(1-v\right)w\left(1-w\right)}}.\\ \end{align}$$
It can be shown that
$$\frac{1}{\pi}\int_{0}^{1}\mathrm{d}w\,\frac{1}{\left(1-zw\right)\sqrt{w\left(1-w\right)}}={_2F_1}{\left(1,\frac12;1;z\right)}=\frac{1}{\sqrt{1-z}};~~~\small{z<1}.$$
Hence,,
$$\begin{align} \mathcal{H} &=\frac{1}{\pi^{4}}\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}u\int_{0}^{1}\mathrm{d}v\int_{0}^{1}\mathrm{d}w\,\frac{\left(1-t^{2}u^{2}v^{2}w\right)^{-1}}{\sqrt{t\left(1-t\right)u\left(1-u\right)v\left(1-v\right)w\left(1-w\right)}}\\ &=\frac{1}{\pi^{3}}\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}u\int_{0}^{1}\mathrm{d}v\,\frac{1}{\sqrt{t\left(1-t\right)u\left(1-u\right)v\left(1-v\right)}}\\ &~~~~~\times\frac{1}{\pi}\int_{0}^{1}\mathrm{d}w\,\frac{1}{\left(1-t^{2}u^{2}v^{2}w\right)\sqrt{w\left(1-w\right)}}\\ &=\frac{1}{\pi^{3}}\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}u\int_{0}^{1}\mathrm{d}v\,\frac{1}{\sqrt{t\left(1-t\right)u\left(1-u\right)v\left(1-v\right)}}\cdot\frac{1}{\sqrt{1-t^{2}u^{2}v^{2}}}\\ &=\frac{1}{\pi^{3}}\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}u\,\frac{1}{\sqrt{t\left(1-t\right)u\left(1-u\right)}}\int_{0}^{1}\mathrm{d}v\,\frac{1}{\sqrt{v\left(1-v\right)\left(1-t^{2}u^{2}v^{2}\right)}}\\ &=\frac{1}{\pi^{3}}\int_{0}^{1}\mathrm{d}t\int_{0}^{t}\mathrm{d}x\,\frac{1}{\sqrt{t\left(1-t\right)x\left(t-x\right)}}\int_{0}^{1}\mathrm{d}v\,\frac{1}{\sqrt{v\left(1-v\right)\left(1-x^{2}v^{2}\right)}};~~~\small{\left[u=t^{-1}x\right]}\\ &=\frac{1}{\pi^{3}}\int_{0}^{1}\mathrm{d}t\int_{0}^{t}\mathrm{d}x\,\frac{1}{\sqrt{t\left(1-t\right)x\left(t-x\right)}}\int_{0}^{x}\mathrm{d}y\,\frac{1}{\sqrt{\left(x-y\right)y\left(1-y^{2}\right)}};~~~\small{\left[v=x^{-1}y\right]}\\ &=\frac{1}{\pi^{3}}\int_{0}^{1}\mathrm{d}t\int_{0}^{t}\mathrm{d}x\,\frac{1}{\sqrt{t\left(1-t\right)x\left(t-x\right)}}\\ &~~~~~\times\frac{1}{\sqrt{1+x}}\int_{0}^{\frac{x}{1+x}}\mathrm{d}u\,\frac{1}{\sqrt{u\left(\frac{x}{1+x}-u\right)\left(1-2u\right)}};~~~\small{\left[y=\frac{u}{1-u}\right]}\\ &=\frac{1}{\pi^{3}}\int_{0}^{1}\mathrm{d}t\int_{0}^{t}\mathrm{d}x\,\frac{1}{\sqrt{t\left(1-t\right)x\left(t-x\right)}}\\ &~~~~~\times\frac{2}{\sqrt{1+x}}\int_{0}^{1}\mathrm{d}v\,\frac{1}{\sqrt{\left(1-v^{2}\right)\left(1-\frac{2x}{1+x}v^{2}\right)}};~~~\small{\left[u=\frac{x}{1+x}v^{2}\right]}\\ &=\frac{1}{\pi^{3}}\int_{0}^{1}\mathrm{d}t\int_{0}^{t}\mathrm{d}x\,\frac{1}{\sqrt{t\left(1-t\right)x\left(t-x\right)}}\cdot\frac{2}{\sqrt{1+x}}K{\left(\sqrt{\frac{2x}{1+x}}\right)}\\ &=\frac{2}{\pi^{3}}\int_{0}^{1}\mathrm{d}x\,\frac{K{\left(\sqrt{\frac{2x}{1+x}}\right)}}{\sqrt{x\left(1+x\right)}}\int_{x}^{1}\mathrm{d}t\,\frac{1}{\sqrt{\left(1-t\right)\left(t-x\right)t}}\\ &=\frac{2}{\pi^{3}}\int_{0}^{1}\mathrm{d}x\,\frac{K{\left(\sqrt{\frac{2x}{1+x}}\right)}}{\sqrt{x\left(1+x\right)}}\int_{0}^{1-x}\mathrm{d}u\,\frac{1}{\sqrt{u\left(1-x-u\right)\left(1-u\right)}};~~~\small{\left[t=1-u\right]}\\ &=\frac{4}{\pi^{3}}\int_{0}^{1}\mathrm{d}x\,\frac{K{\left(\sqrt{\frac{2x}{1+x}}\right)}}{\sqrt{x\left(1+x\right)}}\int_{0}^{1}\mathrm{d}v\,\frac{1}{\sqrt{\left(1-v^{2}\right)\left(1-(1-x)v^{2}\right)}};~~~\small{\left[u=(1-x)v^{2}\right]}\\ &=\frac{4}{\pi^{3}}\int_{0}^{1}\mathrm{d}x\,\frac{K{\left(\sqrt{\frac{2x}{1+x}}\right)}K{\left(\sqrt{1-x}\right)}}{\sqrt{x\left(1+x\right)}}\\ &=\frac{4}{\pi^{3}}\int_{0}^{1}\mathrm{d}y\,\frac{K{\left(\sqrt{\frac{2-2y}{2-y}}\right)}K{\left(\sqrt{y}\right)}}{\sqrt{\left(1-y\right)\left(2-y\right)}};~~~\small{\left[x=1-y\right]}\\ &=\frac{8\sqrt{2}}{\pi^{3}}\int_{0}^{1}\mathrm{d}y\,\frac{K{\left(\sqrt{\frac{2-2y}{2-y}}\right)}K{\left(\sqrt{y}\right)}}{2\left(2-y\right)\sqrt{2\left(\frac{1-y}{2-y}\right)}}\\ &=\frac{8\sqrt{2}}{\pi^{3}}\int_{0}^{1}\mathrm{d}t\,\frac{K{\left(\sqrt{t}\right)}K{\left(\sqrt{\frac{2-2t}{2-t}}\right)}}{2\left(2-t\right)\sqrt{t}};~~~\small{\left[\frac{2-2y}{2-y}=t\right]}\\ &=\frac{8\sqrt{2}}{\pi^{3}}\int_{0}^{1}\mathrm{d}x\,\frac{K{\left(x\right)}K{\left(\sqrt{\frac{2-2x^{2}}{2-x^{2}}}\right)}}{2-x^{2}};~~~\small{\left[t=x^{2}\right]},\\ \end{align}$$
thus proving the validity of identity $(1)$, so we're done. Cheers!