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 
$$\frac{1}{4} \left(\frac{2 }{3}\log ^3(2)-\zeta(2) \log (2)+\zeta (3)\right).$$
 A: We want to compute:
$$ I=\int_{0}^{\frac{\pi}{2}}\log(\sin\theta)\log\left(\cos^2\frac{\theta}{2}\right)\cot(\theta)\,d\theta \tag{1}$$
and by integration by parts, the problem boils down to computing:
$$ J = \int_{0}^{\pi/2}\log^2(\sin\theta)\tan\frac{\theta}{2}\,d\theta =2\int_{0}^{1}\log^2\left(\frac{t}{1+t^2}\right)\frac{t\,dt}{1+t^2}.\tag{2}$$
On the other hand, we have:
$$ \int_{0}^{1}\frac{t\log^2(t)}{1+t^2}\,dt=\frac{3\zeta(3)}{16},\qquad \int_{0}^{1}\frac{t\log^2(1+t^2)}{1+t^2}\,dt=\frac{\log^3(2)}{6},\tag{3}$$
where the first integral can be computed through differentiation under the integral sign, by exploiting the Euler beta function, while the second integral is elementary. In the same way we get:
$$ \int_{0}^{1}\frac{t\log(t)\log(1+t^2)}{1+t^2}\,dt = -\frac{\zeta(3)}{32} \tag{4}$$
so it is straightforward to compute $(2)$, then $(1)$.
A: 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.
$$\int^1_0\frac{\ln{x}\ln\left(\frac{1+\sqrt{1-x^2}}{2}\right)}{x}\ {\rm d}x=\frac{\zeta(3)}{4}-\frac{\pi^2}{24}\ln{2}+\frac{\ln^3{2}}{6}$$
