Integral $\frac{1}{\pi}\int_0^{\pi/3}\log\big( \mu(\theta)+\sqrt{\mu^2(\theta)-1} \big)\ d\theta, \quad \mu(\theta)=\frac{1+2\cos\theta}{2}.$ Hi I am trying to calculate this integral:
$$
I=\frac{1}{\pi}\int_0^{\pi/3}\log\left( \frac{1+2\cos\theta}{2}+\sqrt{\bigg( \frac{1+2\cos\theta}{2}   \bigg)^2-1}         \right)\ d\theta.
$$
The integral evaluation is related to Mahler measures.  You may also recognize that it is related to the evaluation of log-sine integrals at $\dfrac{\pi}{3}$.
This integral $I$ is somewhat related to
$$
\int_0^1\log\big|2a+2b\cos (2\pi \theta)\big|\ d\theta=\log \big( |a|+\sqrt{a^2-b^2} \big),
$$
for $a,b\in\mathbb{R}$ with $|a|\geq|b|> 0$. This can be seen in Gradstein and Ryzhik's tables of integrals, but I am not sure how to use that to help me to solve $I$.  
Thanks!
 A: The integral can be expressed in terms of a series of Gauss hypergeometric functions. It is doubtfull that it would be possible to go further on this way.
$I = \frac{1}{\pi}\int_0^{\pi/3}\ln\left(\mu(\theta)+\sqrt{\mu^2(\theta)-1}\right)d\theta = \frac{1}{\pi}\int_0^{\pi/3}\cosh^{-1}(\mu(\theta))$, where $\mu(\theta) = \frac{1}{2}+\cos(\theta)$.
$$\cosh^{-1}(\mu)=\sum_{k=0}^\infty \frac{(-1)^k\Gamma(k+1/2)}{2^{k-1/2}(2k+1)k!\sqrt{\pi}}(\mu-1)^{k+1/2}$$
so
$$\mu(\theta)=\frac{1}{2}+\cos(\theta)\to I = \frac{1}{\pi^{3/2}}\sum_{k=0}^\infty \frac{(-1)^k\Gamma(k+1/2)}{2^{k-1/2}(2k+1)k!}(\cos(\theta)-1/2)^{k+1/2}$$
Let $I_k = \int(cos(\theta) - 1/2)^{k+1/2}d\theta$. Then
$$I_k = -\frac{1}{(2k+3)2^k}\sqrt{\frac{2}{3}}(2\cos(\theta)-1)^{k+3/2}F_1\left(k+\frac{3}{2};\frac{1}{2},\frac{1}{2};k+\frac{5}{2};\frac{1}{3}(1-2\cos(\theta)),2\cos(\theta)-1\right)+C$$
where $C$ is some constant and $F_1$ is the Appell hypergeometric function of two variables.
$2\cos(0)-1=1$, $2\cos(\pi/3)-1=0$, and $F_1(a;b,b;c;-1/3,1)=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}_2F_1(a,b;c-b;-1/3)$, so
$$I_k=\frac{1}{(2k+3)2^k}\sqrt{\frac{2}{3}}\frac{\Gamma(k+5/2)\Gamma(1/2)}{\Gamma(1)\Gamma(k+2)} {}_2F_1\left(k+\frac{3}{2},\frac{1}{2};k+2;-\frac{1}{3}\right)$$
$$I_k=\sqrt{\frac{2\pi}{3}}\frac{(2k+1)\Gamma(k+1/2)}{2^{k+2}(k+1)!} {}_2F_1\left(k+\frac{3}{2},\frac{1}{2};k+2;-\frac{1}{3}\right)$$
Finally, from $I = \frac{1}{\pi^{3/2}}\sum_{k=0}^\infty \frac{(-1)^k\Gamma(k+1/2)}{2^{k-1/2}(2k+1)k!}I_k$, we have
$$\frac{1}{\pi}\int_0^{\pi/3}\ln\left(\mu(\theta)+\sqrt{\mu^2(\theta)-1}\right)d\theta=\boxed{\frac{1}{\pi\sqrt{3}}\sum_{k=0}^\infty \frac{(-1)^k(\Gamma(k+1/2))^2}{2^{2k+1}(2k+1)k!(k+1)!} {}_2F_1\left(k+\frac{3}{2},\frac{1}{2};k+2;-\frac{1}{3}\right)}$$
