How to evaluate
$$\int_0^1\frac{\ln x\operatorname{Li}_2(x)\ln(1-\frac{x}{2})}{x}\ dx\ ?$$
I came across this integral while I was trying to calculate $\int_0^{1/2}\frac{\operatorname{Li}_2^2(x)}{x}\ dx$ in a different way and as follows,
Using the identity
$$\int_0^1\frac{\ln x\operatorname{Li}_2(x)}{1-ax}\ dx=\frac{\operatorname{Li}_2^2(a)}{2a}+3\frac{\operatorname{Li}_4(a)}{a}-2\zeta(2)\frac{\operatorname{Li}_2(a)}{a}$$
Integrate both sides from $a=0$ to $a=1/2$ we get
$$\frac12\int_0^{\frac{1}{2}}\frac{\operatorname{Li}_2^2(a)}{a}\ da+3\operatorname{Li}_5\left(\frac12\right)-2\zeta(2)\operatorname{Li}_3\left(\frac12\right)$$ $$=\int_0^1\ln x\operatorname{Li}_2(x)\left(\int_0^{\frac{1}{2}}\frac{da}{1-ax}\right)\ dx=-\int_0^1\frac{\ln x\operatorname{Li}_2(x)\ln(1-\frac{x}{2})}{x}\ dx$$
its ok to use harmonic series but not the advanced ones like $\displaystyle\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^32^n}$, $\displaystyle\sum_{n=1}^\infty \frac{H_n^{(3)}}{n^22^n}$, $\displaystyle\sum_{n=1}^\infty \frac{H_n^{(4)}}{n2^n}$ and their alternating versions.
