Calculating $\int_0^1 \frac{\log (x) \log \left(\frac{1}{2} \left(1+\sqrt{1-x^2}\right)\right)}{x} \, dx$ How would you like to calculate this one? Do you see a fast, neat way here? Ideas?
$$\int_0^1 \frac{\log (x) \log \left(\frac{1}{2} \left(1+\sqrt{1-x^2}\right)\right)}{x} \, dx$$
Sharing solutions is only optional.
The closed form revealed is 
$$\frac{1}{4} \left(\frac{2 }{3}\log ^3(2)-\zeta(2) \log (2)+\zeta (3)\right).$$

Here is an alternative solution.
\begin{align}
\int^1_0\frac{\ln{x}\ln\left(\frac{1+\sqrt{1-x^2}}{2}\right)}{x}\ {\rm d}x
&=\frac{1}{4}\int^1_0\frac{\ln{x}\ln\left(\frac{1+\sqrt{1-x}}{2}\right)}{x}\ {\rm d}x\tag1\\
&=\frac{1}{4}\int^1_0\frac{\ln(1-x)}{1-x}\ln\left(\frac{1+\sqrt{x}}{2}\right)\ {\rm d}x\tag2\\
&=\frac{1}{16}\int^1_0\frac{\ln^2(1-x)}{\sqrt{x}(1+\sqrt{x})}\frac{1-\sqrt{x}}{1-\sqrt{x}}\ {\rm d}x\tag3\\
&=-\frac{1}{96}\int^1_0x^{-3/2}\ln^3(1-x)\ {\rm d}x\tag4\\
&=\frac{1}{48}\lim_{q\to 1}\frac{\partial^3}{\partial q^3}\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(q\right)}{\Gamma\left(q-\frac{1}{2}\right)}\tag5
\end{align}

Explanation:
$(1)$: Substituted $x\mapsto\sqrt{x}$. 
$(2)$: Substituted $x\mapsto 1-x$. 
$(3)$: Integrated by parts. 
$(4)$: Integrated by parts. 
$(5)$: Used the integral representation of the Beta function.

Using Wolfram Alpha (or differentiating by hand),

Setting $q=1$ gives us the required result.
$$\int^1_0\frac{\ln{x}\ln\left(\frac{1+\sqrt{1-x^2}}{2}\right)}{x}\ {\rm d}x=\frac{\zeta(3)}{4}-\frac{\pi^2}{24}\ln{2}+\frac{\ln^3{2}}{6}$$
