Closed form solution for $\int_0^1 \frac{\ln^2(x)\ln^2(1-x)}{x(1-x)} dx$ using Bose Integral

Question

I'm looking for the solution to the following integral, but by using the Bose integral: $$\int_0^1 \frac{\ln^2(x)\ln^2(1-x)}{x(1-x)} dx$$ I got to this form when looking for a solution to the following, $I=\int_0^{\pi/2} \frac{\ln^2(\sin x)\ln^2(\cos x)}{\sin x \cos x}dx$ , after some reworking. I know this specific integral has a post about it already but to be honest I can't find it. As a result, I do not recall the closed for solution, and wolfram doesn't seem to know either. As I mentioned before, I'm looking for a solution that utilizes the Bose Integral, which is as follows $$\Gamma(s)\zeta(s)=\int_0^\infty \frac{x^{s-1}}{e^x-1}dx$$ I used this integral to show the solution to $$J =\int_0^{\pi/2} \frac{\ln(\sin x)\ln(\cos x)}{\tan x}dx = \frac{1}{16}\int_0^\infty \frac{x^{2}}{e^x-1}dx = \frac{\zeta(3)}{8}$$ and am wondering if a similar argument can be made about the initial integral to reach a closed for solution like the one above due to the similarities between I and J.