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.

It seems that a series solution is possible
$$I=\int_0^{2\pi} \log \left(\frac{\cos ^2(x)}{c^2}\right) \log \left(1-\frac{\cos ^2(x)}{c^2}\right)\,dx=4\int_0^{\frac \pi 2} \log \left(\frac{\cos ^2(x)}{c^2}\right) \log \left(1-\frac{\cos ^2(x)}{c^2}\right)\,dx$$ Expanding the second logarithms
$$I= -4\sum_{n=1}^\infty \frac 1 {n\,c^{2n}}\int_0^{\frac \pi 2} \log \left(\frac{\cos ^2(x)}{c^2}\right)\cos ^{2n}(x) \,dx$$ The integrals
$$J_n=\int\log \left(\frac{\cos ^2(x)}{c^2}\right)\cos ^{2n}(x) \,dx$$ are relatively simple. One integration by parts gives
$$J_n=-\frac{\cos ^{2 n+1}(x) \log \left(\frac{\cos ^2(x)}{c^2}\right) \,
   _2F_1\left(\frac{1}{2},n+\frac{1}{2};n+\frac{3}{2};\cos ^2(x)\right)}{2 n+1}-$$
$$\frac{1}{2} \Gamma \left(n+\frac{1}{2}\right)^2 \cos ^{2 n+1}(x) \,
   _3\tilde{F}_2\left(\frac{1}{2},n+\frac{1}{2},n+\frac{1}{2};n+\frac{3}{2},n+\frac {3}{2};\cos ^2(x)\right)$$
Using the bounds
$$K_n=\int_0^{\frac \pi 2}\log \left(\frac{\cos ^2(x)}{c^2}\right)\cos ^{2n}(x) \,dx$$
$$K_n=\frac{\sqrt{\pi }\, \Gamma \left(n+\frac{1}{2}\right) \left(H_{n-\frac{1}{2}}-H_n-2 \log
   (c)\right)}{2 \Gamma (n+1)}$$ Then
$$I=2 \sqrt{\pi }\sum_{n=1}^\infty \frac{  \Gamma \left(n+\frac{1}{2}\right) }{n \,c^{2 n}\,\Gamma (n+1)}\left(2 \log
   (c)+H_n-H_{n-\frac{1}{2}}\right)$$
$$\color{blue}{I=8 \pi  \log (c) \left(\log (2 c)-\cosh ^{-1}(c)\right)+2 \sqrt{\pi }\sum_{n=1}^\infty \frac{  \Gamma \left(n+\frac{1}{2}\right) }{n \,c^{2 n}\,\Gamma (n+1)}\left(H_n-H_{n-\frac{1}{2}}\right)}$$
If we denote the summand by $a_n$ we have, for large values of $n$
$$\frac{a_{n+1}}{a_n}=\frac 1{c^2}\left(1-\frac{5}{2 n} \right)+O\left(\frac{1}{n^2}\right)$$ which leads to a quite fast convergence.
For $c=2$, the partial sums (from $n=0$ to $n=p$)
$$\left(
\begin{array}{cc}
p & S_p \\
 0 & 1.20789 \\
 1 & 1.39219 \\
 2 & 1.51043 \\
 3 & 1.52617 \\
 4 & 1.52870 \\
 5 & 1.52914 \\
 6 & 1.52923 \\
 7 & 1.52925
\end{array}
\right)$$
We ca have save calculation time using for large enough $n$ the expansion
$$a_n=\frac{\sqrt \pi} {c^{2n}\, n^{\frac 52}}\left(1-\frac{3}{8 n}+\frac{5}{128 n^2}+\frac{35}{1024
   n^3}+O\left(\frac{1}{n^4}\right)\right)$$
