Verify that $\int_{0}^{\pi/2}\tan x\ln(\tan x)\ln(\cos x)\ln^3(\sin x)\mathrm dx=-\frac{3\zeta(6)}{2^8}$ $$\int_{0}^{\pi/2}\tan x\ln(\tan x)\ln(\cos x)\ln^3(\sin x)\mathrm dx=-\frac{3\zeta(6)}{2^8}$$
Any hints how to simplify this integral to  more manageable to be solve.

Considering $$
\begin{aligned}
J(a, b, c) &=\int_0^{\frac{\pi}{2}} \tan ^a x \cos ^b x \sin ^c x d x \\
&=\int_0^{\frac{\pi}{2}} \sin ^{a+c} x \cos ^{b-a} x d x \\
&=\frac{1}{2} B\left(\frac{a+c+1}{2}, \frac{b-a+1}{2}\right),
\end{aligned}
$$
then $$\boxed{I=\left.\frac{1}{2} \frac{\partial^3}{\partial a \partial b \partial c^3} B\left(\frac{a+c+1}{2}, \frac{b-a+1}{2}\right)\right|_{a=1, b=0=c}}$$
