Integral $\int_0^{1/2}\arcsin x\cdot\ln^3x\,dx$ It's a follow-up to my previous question.
Can we find an anti-derivative
$$\int\arcsin x\cdot\ln^3x\,dx$$
or, at least, evaluate the definite integral
$$\int_0^{1/2}\arcsin x\cdot\ln^3x\,dx$$
in a closed form (ideally, as a combination of elementary functions and polylogarithms)?

The indefinite integral can be expressed in terms of hypergeometrics. 
We start with integration by parts with $dv=\sin^{-1}(x)$ and $u=\log^3x$. 
$$\begin{align}
I&=\int \arcsin x \cdot \ln^3x\,dx\\
&=\ln^3x\cdot\left(\sqrt{1-x^2}+x\arcsin x\right)-\int \left(\sqrt{1-x^2}+x\arcsin x\right)\cdot \frac{3\ln^2x}{x}\,dx
\end{align}$$
Expanding the integrand we get
$$\frac{3\sqrt{1-x^2}\cdot\ln^2x}{x}+3\arcsin(x)\ln^2(x)$$
The integration of the second term is addressed in your previous question so I will focus on just the first term. 
$$K=\int\frac{3\sqrt{1-x^2}\ln^2x}{x}\,dx$$
Mathematica gives the result $K=$
$$\frac{3\sqrt{1-x^2}}{x\sqrt{1-\frac{1}{x^2}}} \left[2 x \, _4F_3\left(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2};\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{1}{x^2}\right)-2 x \log (x) \, _3F_2\left(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2};\frac{1}{2},\frac{1}{2};\frac{1}{x^2}\right)+x \sqrt{1-\frac{1}{x^2}} \log ^2(x)+\log ^2(x) \csc ^{-1}(x)\right]$$
Therefore 
$$I=\ln^3x\cdot\left(\sqrt{1-x^2}+x\arcsin x\right)-K-3\int\arcsin x\ln^2x\,dx$$
Here are the sum representations of the hypergeometrics
$$\, _3F_2\left(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2};\frac{1}{2},\frac{1}{2};\frac{1}{x^2}\right)=-\frac{1}{\sqrt \pi}\sum_{k=0}^\infty\frac{x^{-2k}}{(2k-1)^3}\cdot\frac{\Gamma\left(k+\frac 1 2\right)}{\Gamma(k+1)}$$
$$\, _4F_3\left(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2};\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{1}{x^2}\right)=\frac{1}{\sqrt \pi}\sum_{k=0}^\infty \frac{x^{-2k}}{(2k-1)^4}\cdot\frac{\Gamma\left(k+\frac 1 2\right)}{\Gamma(k+1)}$$
