Does $\int_0^{2\pi}\frac{d\phi}{2\pi} \,\ln\left(\frac{\cos^2\phi}{C^2}\right)\,\ln\left(1-\frac{\cos^2\phi}{C^2}\right)$ have a closed form? I am wondering if anyone has a nice way of approaching the following definite integral $\newcommand{\dilog}{\operatorname{Li}_2}$
$$\int_0^{2\pi}\frac{d\phi}{2\pi} \,\ln\left(\frac{\cos^2\phi}{C^2}\right)\,\ln\left(1-\frac{\cos^2\phi}{C^2}\right)\,.$$
Here $C$ is a positive, real constant that satisfies the constraint $C>1$. So far I have tried a simple $u$ substitution, $u = \cos\left(\phi\right)/C$. However, this doesn't get me anywhere. I have also tried performing a series expansion in small $1/C$ in the second log, performing the integration, and then summing in powers of $1/C$. However, the sum does not simplify nicely.
I have also tried relating the expression to $\dilog\left(1-\frac{\cos^2\phi}{C^2}\right)$ and $\dilog\left(\frac{\cos^2\phi}{C^2}\right)$ and performing the integration of these polylog functions using their series representation. However I have trouble performing the summation for the $\dilog\left(1-\frac{\cos^2\phi}{C^2}\right)$ term.

Claim. Let $H_k$ denote the $k$th harmonic number. Then for any $0<c<1$, $$\frac1{2\pi}\int_0^{2\pi}\log(c\cos^2\phi)\log(1-c\cos^2\phi)\,d\phi=\small2\log c\log\frac{1+\sqrt{1-c}}{2}-\sum_{k\ge1}\frac{(2k-1)!!(H_{k-1/2}-H_k)}{k\cdot2^k\cdot k!}c^k.$$ Proof: Let $\psi=c\cos^2\phi$ so that $d\psi=-2c\sin\phi\cos\phi\,d\phi$. Then \begin{align}\frac1{2\pi}\int_0^{2\pi}\log(c\cos^2\phi)\log(1-c\cos^2\phi)\,d\phi&=\frac2\pi\int_0^{\pi/2}\log(c\cos^2\phi)\log(1-c\cos^2\phi)\,d\phi\\&=\frac1\pi\int_0^c\frac{\log\psi\log(1-\psi)}{\sqrt\psi\sqrt{c-\psi}}\,d\psi\\&=-\frac1\pi\frac d{ds}\int_0^c\frac{\log(1-\psi)}{\psi^s(c-\psi)^{1/2}}\,d\psi\bigg\vert_{s=1/2}.\end{align} Using the Taylor expansion of $\log(1-\psi)$, we obtain \begin{align}\int_0^c\frac{\log(1-\psi)}{\psi^s(c-\psi)^{1/2}}\,d\psi&=\sum_{k\ge1}\frac1k\int_0^c\psi^{k-s}(c-\psi)^{-1/2}\,d\psi\\&=\sum_{k\ge1}\frac1k\cdot\frac{\Gamma(k-s+1)\Gamma(1/2)c^{k-s+1/2}}{\Gamma(k-s+3/2)}\end{align} where the last equality is due to a modification of the standard Beta function. This gives \begin{align}\frac1{2\pi}\int_0^{2\pi}\log(c\cos^2\phi)\log(1-c\cos^2\phi)\,d\phi&=-\sqrt{\frac c\pi}\sum_{k\ge1}\frac{c^k}k\frac d{ds}\frac{\Gamma(k-s+1)}{c^s\Gamma(k-s+3/2)}\bigg\vert_{s=1/2}\end{align} and since \begin{align}\small\frac d{ds}\frac{\Gamma(k-s+1)}{c^s\Gamma(k-s+3/2)}\bigg\vert_{s=1/2}&\small=-\frac{\Gamma(k-s+1)}{c^s\Gamma(k-s+3/2)}\left(\log c+\psi^{(0)}(k-s+1)-\psi^{(0)}\left(k-s+\frac32\right)\right)\bigg\vert_{s=1/2}\\&=-\frac{\Gamma(k+1/2)}{\sqrt c\cdot\Gamma(k+1)}\left(\log c+\psi^{(0)}\left(k+\frac12\right)-\psi^{(0)}(k+1)\right)\\&=-\sqrt{\frac\pi c}\frac{(2k-1)!!}{2^k\cdot k!}(\log c+H_{k-1/2}-H_k),\end{align} the integral simplifies to \begin{align}\frac1{2\pi}\int_0^{2\pi}\log(c\cos^2\phi)\log(1-c\cos^2\phi)\,d\phi&=-\sum_{k\ge1}\frac{(2k-1)!!(\log c+H_{k-1/2}-H_k)}{k\cdot2^k\cdot k!}c^k.\end{align} The result follows by observing the Taylor expansion identity $$\sum_{k\ge1}\frac{(2k-1)!!}{k\cdot2^k\cdot k!}c^k=-2\log\frac{1+\sqrt{1-c}}2.\tag*{$\square$}$$ Note the interchanges of summations, derivatives and integrals are permitted due to absolute convergence.
