Evaluating $\int_{0}^{2\pi}x^2\ln^2(1-\cos x)dx$

After spending a lot of time I've reached the answer (not without help of "MathStackExchangians"). I'll continue the derivation by Larry starting from $I_2$.

I will use the following integral:

$$\int_0^\pi x^2\cos(2kx)~dx=\frac\pi{2k^2}$$

We have

$$\small\begin{align} I_2 &= 32\int_{0}^{\pi}x^2\ln^2(\sin x)~dx\\ &= 32\int_{0}^{\pi}x^2\left(\ln(2)+\sum_{n=1}^{\infty}\frac{\cos (2nx)}{n}\right)^2~dx\\ &= 32\int_{0}^{\pi}x^2\ln^2(2)~dx+64\ln(2)\int_{0}^{\pi}x^2\sum_{n=1}^{\infty}\frac{\cos (2nx)}{n}~dx+32\int_{0}^{\pi}x^2\left(\sum_{n=1}^{\infty}\frac{\cos (2nx)}{n}\right)^2~dx\\ &=\frac{32}{3}\pi^3\ln^2(2)+\sum_{n=1}^{\infty}\frac{64\ln(2)}{n}\int_{0}^{\pi}x^2\cos(2nx)~dx+32\underbrace{\int_{0}^{\pi}x^2\left(\sum_{n=1}^{\infty}\frac{\cos (2nx)}{n}\right)^2~dx}_{J} \end{align}$$

$$\small\begin{align} J &= \int_{0}^{\pi}x^2\left(\sum_{n=1}^{\infty}\frac{\cos (2nx)}{n}\right)^2~dx\\ &=\int_{0}^{\pi}x^2\left(\sum_{n=1}^{\infty}\frac{\cos^2 (2nx)}{n^2}+\sum_{m,n=1;m\neq n}^{\infty}\frac{\cos (2mx)\cos (2nx)}{mn}\right)~dx\\ &=\sum_{n=1}^{\infty}\frac1{n^2}\int_{0}^{\pi}x^2\cos^2 (2nx)~dx+\sum_{m,n=1;m\neq n}^{\infty}\frac1{mn}\int_{0}^{\pi}x^2\cos (2mx)\cos (2nx)~dx\\ &=\sum_{n=1}^{\infty}\frac1{2n^2}\int_{0}^{\pi}x^2(1+\cos (4nx))~dx+\sum_{m,n=1;m\neq n}^{\infty}\frac1{2mn}\int_{0}^{\pi}x^2(\cos (2(m+n)x)+\cos (2(m-n)x))~dx\\ &=\sum_{n=1}^{\infty}\frac1{2n^2}\left(\int_{0}^{\pi}x^2~dx+\int_{0}^{\pi}x^2\cos (4nx)~dx\right)+\sum_{m,n=1;m\neq n}^{\infty}\frac1{2mn}\left(\int_{0}^{\pi}x^2\cos (2(m+n)x)~dx+\int_{0}^{\pi}x^2\cos (2(m-n)x)~dx\right)\\ &=\sum_{n=1}^{\infty}\frac1{2n^2}\left(\frac{\pi^3}3+\frac\pi{2(2n)^2}\right)+\sum_{m,n=1;m\neq n}^{\infty}\frac1{2mn}\left(\frac\pi{2(m+n)^2}+\frac\pi{2(m-n)^2}\right)\\ &=\frac{\pi^3}6\sum_{n=1}^{\infty}\frac1{n^2}+\frac\pi{16}\sum_{n=1}^{\infty}\frac1{n^4}+\frac\pi2\sum_{m,n=1;m\neq n}^{\infty}\frac{m^2+n^2}{mn(m^2-n^2)^2}\\ &=\frac{\pi^3}6\frac{\pi^2}6+\frac{\pi}{16}\frac{\pi^4}{90}+\frac\pi2\frac{11\pi^4}{720}=\frac{13\pi^5}{360} \end{align} $$

The last sum is evaluated (my thanks to Robert Z and Zvi) in this question

Finally we have

$$\small\begin{align} I_2 &= \frac{32}{3}\pi^3\ln^2(2)+\sum_{n=1}^{\infty}\frac{64\ln(2)}{n}\int_{0}^{\pi}x^2\cos(2nx)~dx+32\frac{13\pi^5}{360}\\ &= \frac{32}{3}\pi^3\ln^2(2)+\sum_{n=1}^{\infty}\frac{64\ln(2)}{n}\frac\pi{2n^2}+\frac{52\pi^5}{45}\\ &=\frac{32}{3}\pi^3\ln^2(2)+32\ln(2)\pi\sum_{n=1}^{\infty}\frac1{n^3}+\frac{52\pi^5}{45}\\ &= \pi^3\ln^2(2)+32\pi\ln(2)\zeta(3)+\frac{52\pi^5}{45} \end{align}$$

And thus

$$\begin{align} I&=I_1+I_2\\ &= -8\pi^3\ln^2(2)-16\pi\ln(2)\zeta(3)+\frac{32}{3}\pi^3\ln^2(2)+32\pi\ln(2)\zeta(3)+\frac{52\pi^5}{45}\\ &= 16\pi\zeta(3)\ln(2)+\frac{8\pi^3\ln^2(2)}{3}+\frac{52\pi^5}{45} \end{align}$$