<s>[INST] Evaluate $\int_0^1\arcsin^2(\frac{\sqrt{-x}}{2}) (\log^3 x) (\frac{8}{1+x}+\frac{1}{x}) \, dx$ Here is an interesting integral, which is equivalent to the title
$$\tag{1}\int_0^1 \log ^2\left(\sqrt{\frac{x}{4}+1}-\sqrt{\frac{x}{4}}\right) (\log ^3x) \left(\frac{8}{1+x}+\frac{1}{x}\right) \, dx = \frac{5 \pi ^6}{1134}-\frac{22 \zeta (3)^2}{5}$$

Question: how to prove $(1)$?


Using power expansion of $\log^2(\sqrt{x+1}-\sqrt{x})$, we can derive an equivalent form of $(1)$:
$$\tag{2}\sum _{n=1}^{\infty } \frac{1}{n^2 \binom{2 n}{n}} \left(8 \sum _{j=1}^n \frac{(-1)^j}{j^4}+\frac{(-1)^n}{n^4}\right)=-\frac{22 \zeta (3)^2}{15}-\frac{97 \pi ^6}{34020}$$
Letting $\sqrt{\frac{x}{4}+1}-\sqrt{\frac{x}{4}}=\sqrt{u+1}$ gives
$$\tag{3}\int_0^{\phi} \log^2 (1+u) \log^3\left(\frac{u^2}{1+u}\right)  \frac{(u+2)(9 u^2+u+1)}{u (u+1) (u^2+u+1)} du = \frac{10 \pi ^6}{567}-\frac{88 \zeta (3)^2}{5}$$
with $\phi = (\sqrt{5}+1)/2$. But all these variations look equally difficult.
Any idea is welcomed.
 [/INST] Too long for comment: Let
$$I_1 = \int_0^1\log^2\left(\sqrt{\frac x4+1}+\sqrt{\frac x4}\right) \log^3(x) \, \frac{dx}x \\
I_2 = \int_0^1 \log^2\left(\sqrt{\frac x4+1}+\sqrt{\frac x4}\right) \log^3(x) \, \frac{dx}{x+1}$$
so the desired integral is $I=I_1+8I_2$.
Taking a cue from Ali Shadhar's solution to a related problem, we may be able to handle $I_1$ via the chain of substitutions
$$u = \log\left(\sqrt{\frac x4}+\sqrt{\frac x4+1}\right) =\operatorname{arsinh}\left(\sqrt{\frac x4}\right) \to v=e^u \to w=\sqrt{v+1}$$
Then
$$\begin{align*}
I_1 &= 16\int_0^{\log(\phi)}u^2\log^3(2\sinh(u))\coth(u)\,du \\[1ex]
&= 16\int_1^\phi\log^2(v)\log^3\left(\frac{v^2-1}v\right)\frac{v^2+1}{v^2-1}\,\frac{dv}v \\[1ex]
&= 2\int_0^\phi \log^2(w+1) \left[\log^3(w)-\frac32\log^2(w)\log(w+1)\right.\\
&\qquad\qquad\qquad\qquad\qquad\left.+\frac34\log(w)\log^2(w+1)-\frac18\log^3(w+1)\right]\left(\frac2w-\frac1{w+1}\right) \, dw \\[1ex]
&= 4\int_0^\phi \frac{\log^3(w)\log^2(w+1)}w \, dw - 4\int_0^\phi \frac{\log^2(w)\log^3(w+1)}w \, dw \\
&\qquad + \frac32 \int_0^\phi \frac{\log(w)\log^4(w+1)}w\,dw - \frac15 \int_0^\phi \frac{\log^5(w+1)}w \, dw + K
\end{align*}$$
where
$$K = - \frac23 \log^3(\phi) \log^3(\phi+1) + \frac34 \log^2(\phi) \log^4(\phi+1) - \frac3{10} \log(\phi) \log^5(\phi+1) + \frac1{24}\log^6(\phi+1)$$
Fully expanding the penultimate integrand produces some other integrals with denominators $y+1$, but they can each be done by parts to get absorbed into the integrals and constant $K$ shown here.
With $f(x)=\frac{\log^a(x) \log^b(x+1)}x$, let
$$\mathcal K(a,b) = \int_0^\phi f(x) \, dx, \quad \mathcal L(a,b) = \int_0^1 \cdots ,\quad \mathcal M(a,b) = \int_{\frac1\phi}^1 \cdots, \quad \mathcal N(a,b) = \int_0^{\frac1\phi} \cdots$$
We find
$$\begin{align*}
\mathcal K(a,b) &= \int_0^\phi \frac{\log^a(x) \log^b(x+1)}x \, dx \\[1ex]
&= \int_0^1 \frac{\log^a(x)\log^b(x+1)}x \, dx + \int_{\frac1\phi}^1 \frac{\log^a(x) \left(\log(x+1)-\log(x)\right)^b}x \, dx \\[1ex]
&= \mathcal L(a,b) + \sum_{\beta=0}^b (-1)^{a+\beta} \binom b\beta \mathcal M(a+\beta,b-\beta)
\end{align*}$$
so we can rewrite $I_1$ as several "simpler" integrals,
$$\begin{align*}
I_1 &= 4 \mathcal K(3,2) - 4 \mathcal K(2,3) + \frac32 \mathcal K(1,4) - \frac15 \mathcal K(0,5) + K \\[1ex]
&= -4 \mathcal L(2,3) + 2 \mathcal M(2,3) + 3 \mathcal M(3,2) + 2 \mathcal M(4,1) - \frac32 \mathcal M(5,0) + \frac32 \mathcal N(1,4) + 4 \mathcal N(3,2) - \frac15 \mathcal K(0,5) + K
\end{align*}$$
If Ali's answer is any indication, this should resolve into something involving polylogarithms of $\phi$ and zetas. I suppose zeta-free terms vanish in the end by virtue of a polylog ladder. Not sure if $I_2$ can be approached in the same way.
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