Evaluate $\int_0^1 \frac{\log^3 x}{1+x}\arcsin^2\left(\frac{\sqrt{x}}{2}\right)\mathrm{d}x.$ I'm looking for evaluation of the integral
$$\int_0^1 \frac{\log^3 x}{1+x}\arcsin^2\left(\dfrac{\sqrt{x}}{2}\right)\mathrm{d}x.$$
I've tried some trivial substitutions and series expansions so far but it didn't get me anything satisfactory. I'm wondering if there's a closed form for this integral in terms of classical special functions and constants?
Any help would be highly appreciated. Thanks!
 A: Just a beginning.
$$\arcsin^2\left(\frac{\sqrt{x}}{2}\right) = \frac{1}{2}\sum_{n\geq 1}\frac{x^n}{n^2\binom{2n}{n}} $$
and
$$ \int_{0}^{1}\frac{x^n \log^3(x)}{1+x}\,dx = \sum_{m\geq 0}(-1)^m\int_{0}^{1} x^{n+m}\log^3(x)\,dx = -6\sum_{m\geq 0}\frac{(-1)^m}{(m+n+1)^4} $$
so our integral equals
$$ 3\sum_{n\geq 1}\frac{(-1)^n}{n^2\binom{2n}{n}}\sum_{m > n}\frac{(-1)^m}{(m+1)^4}=-\frac{7\pi^4}{120}\log^2\left(\frac{1+\sqrt{5}}{2}\right)-3\sum_{n\geq 1}\frac{(-1)^n}{n^2\binom{2n}{n}}\sum_{m=0}^{n}\frac{(-1)^m}{(m+1)^4} $$
which should be related to Euler sums with weigth $6$.
A: If to take a closer look at the integrand then it turns out that the main line segment that contributes to the integral is concentrated in a small zone near zero.
Extreme point of the integrand lies at $x=0.044$ and after that the integrand quickly approaches zero.
That means we can try to compute (approximately) the integral using the first few terms of the Taylor expansion of $\arcsin^2(x)$
$$\arcsin^2(x)=x^2+\frac{x^4}{3}+...$$
We get
$$I_{approx}=\int_{0}^{1}\frac{\frac{x}{4}+\frac{x^2}{48}}{1+x}\log^3(x)\,dx =\frac{77\pi^4-7965}{5760}$$
The approximation error is about $0.00005$
