<s>[INST] Integrate $I=\int_0^1\frac{\arcsin{(x)}\arcsin{(x\sqrt\frac{1}{2})}}{\sqrt{2-x^2}}dx$ How to prove 
\begin{align}
 I &= \int_0^1\frac{\arcsin{(x)}\arcsin{(x\sqrt\frac{1}{2})}}{\sqrt{2-x^2}}dx \\
   &= \frac{\pi}{256}\left[ \frac{11\pi^4}{120}+2{\pi^2}\ln^2{2}-2\ln^4{2}-12\zeta{(3)}\ln{2} \right]
\end{align}
 By asking  $$x=\sqrt{2}y$$ then using integration by parts, we have
 $$I=\frac{\pi^5}{2048}-\frac{1}{4}\int_0^1{\arcsin^4\left( \frac{z}{\sqrt{2}}\right) }\frac{dz}{\sqrt{1-x^2}}$$
But how to calculate this integral? I would appreciate your help
 [/INST] This integral can be done by recognizing that
$$\frac{\arcsin{\frac{x}{\sqrt{2}}}}{\sqrt{2-x^2}} = \sum_{n=0}^{\infty} \frac{2^n x^{2 n+1}}{(2 n+1) \binom{2 n}{n}}$$
and that
$$\int_0^1 dx \, x^{2 n+1} \arcsin{x} = \frac{\pi}{4 (n+1)} \left [1- \frac1{2^{2 n+2}} \binom{2 n+2}{n+1}\right ] $$
To see this, integrate by parts and see this answer.
With a bit of algebra, we find that the integral is equal to
$$\frac{\pi}{2} \sum_{n=0}^{\infty} \frac{2^n}{(2 n+1)(2 n+2) \binom{2 n}{n}} - \frac{\pi}{16} \sum_{n=0}^{\infty} \frac1{2^n (n+1)^2}$$
The first sum may be evaluated by recognizing that it is 
$$\int_0^1 dx \frac{\arcsin{\frac{x}{\sqrt{2}}}}{\sqrt{2-x^2}} = \frac{\pi^2}{32}$$
The second sum is recognized as $\operatorname{Li_2}{\left ( \frac12 \right )} = \frac{\pi^2}{6}-\log^2{2} $
Putting all of this together, we find that the integral is equal to

$$\int_0^1 dx \, \frac{\arcsin{x} \arcsin{\frac{x}{\sqrt{2}}}}{\sqrt{2-x^2}} = \frac{\pi^3}{192} + \frac{\pi}{16} \log^2{2} $$

Numerical evaluation in Mathematica confirms the result, which differs from that asserted by the OP.
</s>