TITLE: Integral involving Clausen function ${\large\int}_0^{2\pi}\operatorname{Cl}_2(x)^2\,x^p\,dx$
QUESTION [25 upvotes]: Consider the Clausen function $\operatorname{Cl}_2(x)$ that can be defined for $0<x<2\pi$ in several equivalent ways:
$$\begin{align}\operatorname{Cl}_2(x)&=-\int_0^x\ln\left(2\sin\left(\tfrac t2\right)\right)dt\\&=\sum_{n=1}^\infty\frac{\sin\,(n x)}{n^2}\\&=\Im\operatorname{Li}_2\left(e^{i x}\right)\\&=i\left(\frac{\pi^2}6+\frac{x^2}4-\frac{\pi x}2-\operatorname{Li}_2\left(e^{i x}\right)\right).\end{align}\tag1$$
I'm interested in integrals of the form
$$I(p)=\int_0^{2\pi}\operatorname{Cl}_2(x)^2\,x^p\,dx.\tag2$$
I found that $$I(0)=\frac{\pi^5}{90}\tag3$$ and conjectured next several values based on numeric evidence:
$$I(1)\stackrel?=\frac{\pi^6}{90},\ \ I(2)\stackrel?=\frac{44\,\pi^7}{2835},\ \ I(3)\stackrel?=\frac{23\,\pi^8}{945}.\tag4$$
One might expect that $I(4)$ is a rational multiple of $\pi^9$, but apparently it is not (unless the denominator is huge).
I'm asking for your help in proving conjectured values $(4)$, finding a closed form of $I(4)$, and, if possible, a general formula for $I(p)$.
Update: Values $I(-1)$ and $I(-2)$ are also interesting.

REPLY [2 votes]: We can use the identity 
$$\mathrm{cl}_2(x)^2 = \mathrm{cl}_2(2\pi -x)^2 $$
Hence we have 
$$I(5)=\int_0^{2\pi}\operatorname{Cl}_2(x)^2\,(2\pi -x)^5\,dx$$
Expand 
$$I(5) = 32 \pi^5I(0)-80 \pi^4 I(1)+80 \pi^3 I(2)-40 \pi^2 I(3)+10 \pi I(4)-I(5)$$
Solve for $I(5)$
$$I(5) = 16\pi^5I(0)-40\pi^4 I(1)+40\pi^3 I(2)-20\pi^2 I(3)+5\pi I(4)$$
This simplifies to 
$$I(5) = \frac{544}{2835}π^{10}-960 π^2 \zeta(3)\zeta(5)+480 π^2 \zeta(6,2)$$
This method can be generalized for $I(2q+1)$ given we know all the previous terms
$$I(2q+1) = \int^{2\pi}_0 (2\pi-x)^{2q+1} \mathrm{cl}_2(x)^2\,dx$$
Use the binomial theorem 
$$I(2q+1) = \sum _{ k=1 }^{ 2q+1 }{ { \left( -1 \right)  }^{ k }\binom{2q+1}{k} { 2\pi }^{ 2q+1-k } }\int^{2\pi}_0 x^k \mathrm{cl}_2(x)^2\,dx$$
$$I(2q+1) = \frac{1}{2}\sum _{ k=1 }^{ 2q }{ { \left( -1 \right)  }^{ k }\binom{2q+1}{k} { 2\pi }^{ 2q+1-k } }I(k)$$