So, following the procedure I outlined here, I get for the transformed integral:
$$I(r,s) = \int_0^{\infty} dv \frac{4 s \left(v^2-1\right) \left(v^4-(4 r-6) v^2+1\right)}{v^8+4 \left(2 r-s^2-1\right) v^6 +2 \left(8 r^2-8 r-4 s^2+3\right) v^4 +4 \left(2 r-s^2-1\right) v^2 +1} \log{v} $$
Note that this reduces to the integral in the original problem when $r=3$ and $s=2$. Then we see that the roots of the denominator satisfy the same symmetries as before, so we need only find one root of the form $\rho e^{i \theta}$ where
$$\rho = \sqrt{\frac{r+\sqrt{r^2-s^2}}{2}} + \sqrt{\frac{r+\sqrt{r^2-s^2}}{2}-1}$$
and
$$\theta = \arctan{\sqrt{\frac{2 \left (r+\sqrt{r^2-s^2}\right )}{s^2}-1}}$$
Using the same methodology I derived, I am able to confirm your conjecture.
First I want to define with the Stirling numbers of the first kind $\left[ \begin{array}{c} n \\ k \end{array} \right]$ a special generalization of the Riemann Zeta function :
$$\zeta_n(m):=\sum\limits_{k=1}^\infty \frac{1}{k^m}\left(\frac{n!}{(k-1)!}\left[\begin{array}{c} k \\ n+1 \end{array} \right]\right)$$
and
$$\eta_n(m):=\sum\limits_{k=1}^\infty \frac{(-1)^{k-1}}{k^m}\left(\frac{n!}{(k-1)!}\left[\begin{array}{c} k \\ n+1 \end{array} \right]\right)$$
which are convergent for the integer values $\enspace m\geq 2$ .
For $\enspace n=0\enspace$ we have $\enspace\zeta_0(m)=\zeta(m)\enspace$ and $\enspace\eta_0(m)=\eta(m)\enspace$ .
Note: Obviously (because of the other results) these series can be expressed by sums of the polylogarithm function and modifications of that.
Please also see here, part Expansion by harmonic numbers, with $\enspace\displaystyle w(n,m):=\frac{m!}{(n-1)!}\left[ \begin{array}{c} n \\ {m+1} \end{array} \right]\enspace$ and it's recursion formula.
Secondly, an extension of an integral as a series, $n\in\mathbb{N}_0$ and $z\in\mathbb{R}\setminus \{2\mathbb{N}\}$ and $nz>-1$:
$ \displaystyle \int\limits_0^\pi x^n \left(2\sin\frac{x}{2}\right)^z dx=i^{-z} \int\limits_0^\pi x^n e^{i\frac{xz}{2}}(1- e^{-ix})^z dx= e^{-i\frac{\pi z}{2}} \int\limits_0^\pi x^n \sum\limits_{k=0}^\infty\binom{z}{k}(-1)^k e^{-ix(\frac{z}{2}-k)} dx$
$\displaystyle =\int\limits_0^\pi x^n e^{i(x-\pi)\frac{z}{2}} dx+ \sum\limits_{v=0}^n \frac{(-1)^v\pi^{n-v} n!}{i^{v+1}(n-v)!} \sum\limits_{k=1}^\infty \binom{z}{k}\frac{1}{(\frac{z}{2}-k)^{v+1}} $
$\displaystyle \hspace{3.5cm} -i^{n-1}n!e^{-i\frac{\pi z}{2}} \sum\limits_{k=1}^\infty \binom{z}{k}\frac{ (-1)^k}{(\frac{z}{2}-k)^{n+1}}$
using the main branch of the logarithm and therefore $\displaystyle i=e^{i\frac{\pi}{2}}$ .
The Stirling numbers of the first kind are usually defined by $\enspace \displaystyle \sum\limits_{k=0}^n \left[ \begin{array}{c} n \\ k \end{array} \right] x^k := x(x+1)…(x+n-1) $ .
Because of $\enspace \displaystyle (\sum\limits_{v=0}^\infty x^v \frac{d^k}{dz^k}\binom{z}{v}) |_{z=0} =\frac{d^k}{dz^k}(1+x)^z |_{z=0} =(\ln(1+x))^k=k!\sum\limits_{v=k}^\infty (-1)^{v-k} \left[\begin{array}{c} v \\ k \end{array} \right] \frac{x^v}{v!}$
we get $\enspace \displaystyle \binom{z}{k}|_{z=0}=0^k\enspace$ , $\enspace \displaystyle \frac{d}{dz} \binom{z}{k} |_{z=0} = (-1)^{k-1} \left[\begin{array}{c} k \\ 1 \end{array} \right] \frac{1}{k!}= \frac{(-1)^{k-1}}{k} \enspace$ , $\enspace \displaystyle \frac{d^2}{dz^2} \binom{z}{k} |_{z=0} = (-1)^{k-2} \left[\begin{array}{c} k \\ 2 \end{array} \right] \frac{2!}{k!}= \frac{(-1)^k 2}{k}\sum\limits_{j=1}^{k-1}\frac{1}{j} \enspace$ and $\enspace \displaystyle \frac{d^3}{dz^3} \binom{z}{k} |_{z=0} = (-1)^{k-3} \left[\begin{array}{c} k \\ 3 \end{array} \right] \frac{3!}{k!}= \frac{(-1)^{k-1} 3}{k}( (\sum\limits_{j=1}^{k-1}\frac{1}{j})^2 - \sum\limits_{j=1}^{k-1}\frac{1}{j^2} ) $ .
For $(n;k):=(3;3)$ follows
$\displaystyle \int\limits_0^\pi x^3 \left(\ln\left(2\sin\frac{x}{2} \right)\right)^3 dx =$
$\hspace{2cm}\displaystyle =\frac{9\pi^2}{2}\left(\zeta(5)+3\eta(5)-4\eta_1(4)+2\eta_2(3)\right) $
$\hspace{2.5cm}\displaystyle - 90\left(\zeta(7)+\eta(7)\right) +72\left(\zeta_1(6)+\eta_1(6)\right) - 18\left(\zeta_2(5)+\eta_2(5)\right) $
Note:
For the calculations I have used $\enspace\displaystyle\int\limits_0^\pi x^n e^{iax}dx = \frac{(-1)^{n+1} n!}{(ia)^{n+1}}+e^{i\pi a}\sum\limits_{v=0}^n\frac{(-1)^v \pi^{n-v}n!}{(ia)^{v+1}(n-v)!}$
with $\enspace\displaystyle a=-(\frac{z}{2}-k)$ .
And it was necessary to calculate $\enspace\displaystyle\frac{d^m}{dz^m} \binom{z}{k}\frac{1}{(\frac{z}{2}-k)^{v+1}}|_{z=0}\enspace$ and $\enspace\displaystyle\frac{d^m}{dz^m} e^{-i\frac{\pi z}{2}}\binom{z}{k}\frac{1}{(\frac{z}{2}-k)^{n+1}}|_{z=0}\enspace$ for $\enspace m\in\{0,1,2,3\}$ .
Best Answer
The given problem is equivalent to the evaluation of $$ \int_{0}^{1}\frac{\arcsin(x)}{\sqrt{1-x^2}}\cdot\frac{\log(1-x)}{x}\,dx =\sum_{n\geq 1}\frac{4^n}{2n\binom{2n}{n}}\int_{0}^{1}x^{2n-2}\log(1-x)\,dx=\sum_{n\geq 1}\frac{4^n H_{2n-1}}{2n\binom{2n}{n}(1-2n)}$$ which is a twisted hypergeometric series. On the other hand $$ \mathcal{J}= 2\int_{0}^{\pi/4}\frac{2x \log(1-\sin(2x))}{\sin(2x)}\,dx=2\int_{0}^{1}\frac{\arctan(t)}{t}\log\left(\frac{(1-t)^2}{1+t^2}\right)\,dt $$ appears to be manageable through the polylogarithms machinery.
Indeed $\arctan t=\text{Im}\log(1+it)$ and the integrals $$ \int \frac{\log(1+it)\log(1\pm it)}{t}\,dt, \qquad \int \frac{\log(1+it)\log(1-t)}{t}\,dt $$ have closed forms in terms of $\text{Li}_2$ and $\text{Li}_3$. However the simplest way to recover $\mathcal{J}=-\frac{\pi^3}{8}$ might be to exploit complex analysis and contour integration, as it often happens when integrating multiples of $\frac{x}{\sin x}$.
Through the Fourier series of $\log\sin$ we have $$ \log(1-\cos x)=-\log(2)-2\sum_{n\geq 1}\frac{\cos(nx)}{n} $$ pointwise on $(0,\pi/2)$. We have that $\int_{0}^{\pi/2}\frac{x}{\sin x}\,dx $ equals $2K$, with $K$ being Catalan's constant, and by induction
$$ \int_{0}^{\pi/2}\frac{x}{\sin x}\cos\left[n\left(\frac{\pi}{2}-x\right)\right]\,dx $$ up to the sign, equals $\sum_{m>n/2}\frac{2(-1)^m}{(2m+1)^2}$ or $\sum_{m> n/2}\frac{1}{(2m+1)^2}$, according to the parity of $n$. This allows to write the original twisted sum in terms of standard Euler sums. $K$ disappears from the outcome after some simplification and $$ \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^3} = \frac{\pi^3}{32} $$ is well-known.