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)?

We will outline of a way forward leaving some of the work to the reader.
Denote the integral of interest by $I$ where  
$$I=\int \arcsin(x) \log^3(x)\,dx \tag 1$$
Integrating $(1)$ by parts by letting $u=\arcsin(x)$ and $v=x\left(\log^3(x)-3\log^2(x)+6\log(x)-6\right)$, we find that
$$\begin{align}
I&=x\arcsin(x)\left(\log^3(x)-3\log^2(x)+6\log(x)-6\right)\\\\&-\int \left(\frac{\log^3(x)-3\log^2(x)+6\log(x)-6}{\sqrt{1-x^2}}\right)\,x\,dx \tag 2
\end{align}$$

Next, denote the integral on the right-hand side of $(2)$ by $J$.  Enforcing the substitution $x=\sqrt{1-y^2}$ yields
$$\begin{align}
J&=-\int \left(\frac{\log^3(x)-3\log^2(x)+6\log(x)-6}{\sqrt{1-x^2}}\right)\,x\,dx\\\\
&=J_3+J_2+J_1+J_0
\end{align}$$
where 
$$\begin{align}
J_3&=\int \log^3(\sqrt{1-y^2})\,dy \tag 3\\\\
J_2&=-3\int \log^2(\sqrt{1-y^2})\,dy \tag 4\\\\
J_1&=6\int \log(\sqrt{1-y^2})\,dy \tag 5\\\\
J_0&=-6\int 1\,dy \tag 6
\end{align}$$

The integrals in $(5)$ and $(6)$ can be evaluated in terms of elementary functions with 
$$J_0=-6y$$
and 
$$J_1=3y\log(1-y^2)-6y-3\log(1-y)+3\log(1+y)$$

The integrals in $(3)$ and $(4)$ can be expressed in terms of polylogarithm functions.  For $J_2$ we can write
$$\begin{align}
J_2&=-3\int \log^2(\sqrt{1-y^2})\,dy\\\\
&=-\frac34 \left(K_1+K_2+K_3\right)
\end{align}$$
where
$$\begin{align}
K_1&=\int \log^2(1-y)\,dy \tag 7\\\\
K_2&=\int \log^2(1+y)\,dy \tag 8\\\\
K_3&=2\int \log(1-y)\log(1+y)\,dy \tag 9
\end{align}$$
The integrals $K_1$ and $K_2$ can be written in closed form with 
$$\begin{align}
K_1&=(y-1)\left(\log^2(1-y)-2\log(1-y)+2\right)\\\\
\end{align}$$
and
$$\begin{align}
K_2&=(y+1)\left(\log^2(1+y)-2\log(1+y)+2\right)\\\\
\end{align}$$

For $K_3$ we integrate by parts with $u=\log(1-y)$ and $v=(y+1)\log(y+1)-y$ and obtain
$$\begin{align}
K_3&=2(y+1)\log(1-y^2)-2y\log(1-y)+2\int \frac{(y+1)\log(y+1)-y}{1-y}\,dy\\\\
&=2(y+1)\log(1-y^2)-2y\log(1-y)+2y+2\log(1-y)+2\int \frac{(y+1)\log(y+1)}{1-y}\,dy\\\\
&=2(y+1)\log(1-y)+2y\left(1-\log(1-y)\right)+4\int \frac{\log(1+y)}{1-y}\,dy \tag{10}
\end{align}$$
To evaluate the integral in $(10)$, we make the substitution $y=1-2z$.  Then, 
$$\begin{align}
\int \frac{\log(1+y)}{1-y}\,dy&=-\log(2)\log(w)-\int \frac{\log(1-w)}{w}\,dw\\\\
&=-\log(2)\log\left(\frac{1-y}{2}\right)+\text{Li}_2\left(\frac{1-y}{2}\right)
\end{align}$$ 

The integral $J_3$ can be evaluated in terms of the dilogarithm function $\text{Li}_2$ and trilogarithm function $\text{Li}_3$ using a similar approach to the one used herein to evaluate $K_2$.  We will leave that very tedious analysis to the reader.
