Integrate $\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx$ Please help me to solve this integral:
$$\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx.$$
I managed to calculate an indefinite integral of the left part:
$$\int\frac{\cos x}{\sin x}x\ \mathrm dx=\ x\log(2\sin x)+\frac{1}{2} \Im\ \text{Li}_2(e^{2\ x\ i}),$$
where $\Im\ \text{Li}_2(z)$ denotes the imaginary part of the dilogarithm. The corresponding definite integral $$\int_0^\pi\frac{\cos x}{\sin x}x\ \mathrm dx$$ diverges. So, it looks like in the original integral summands compensate each other's singularities to avoid divergence.
I tried a numerical integration and it looks plausible that
$$\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx\stackrel{?}{=}\pi \log 54,$$
but I have no idea how to prove it.

Here's one way to go. 
First, note that 
$$\begin{eqnarray*}
\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx
&=& \int_0^\pi\frac{3x(1+\cos x)}{\sin x} \mathrm dx 
+\int_0^\pi\frac{3x}{\sin x} \left(-1+\sqrt{1-\frac{\sin^2x}{9}}\right)\ \mathrm dx.
\end{eqnarray*}$$
For now I'll simply claim that 
\begin{equation*}
\int_0^\pi\frac{3x(1+\cos x)}{\sin x} \mathrm dx = \pi\log 64.\tag{1}
\end{equation*}
(I would be surprised if this integral has not been handled somewhere on this site.)
But
$$\begin{eqnarray*}
\int_0^\pi\frac{3x}{\sin x} \left(-1+\sqrt{1-\frac{\sin^2x}{9}}\right)\ \mathrm dx
&=& \int_0^\pi\frac{3x}{\sin x} \sum_{k=1}^\infty {1/2\choose k} \frac{(-1)^k}{3^{2k}} \sin^{2k}x \ \mathrm dx \\
&=& \sum_{k=1}^\infty {1/2\choose k} \frac{(-1)^k}{3^{2k-1}}
\int_0^\pi x  \sin^{2k-1}x \ \mathrm dx \\
&=& \sum_{k=1}^\infty {1/2\choose k} \frac{(-1)^k}{3^{2k-1}}
\frac{\pi^{3/2}\Gamma(k)}{2\Gamma(k+1/2)} \\
&=& -\pi \sum_{k=1}^\infty \frac{1}{3^{2k-1}2k(2k-1)} \\
&=& -\pi \log \frac{32}{27}.
\end{eqnarray*}$$
(The last sum can be found by standard methods.
Schematically, $\sum \frac{a^{2k-1}}{2k(2k-1)} = \sum \int {\mathrm da} \frac{a^{2k-2}}{2k}$.) 
Thus, the integral is $\pi \log 54$ as claimed. 

Proof of (1): 
We have 
$$\begin{eqnarray*}
\int_0^\pi \frac{3x(1+\cos x)}{\sin x} \ \mathrm dx 
&=& \int_{0^+}^\pi \frac{3x(1+\cos x)}{\sin x} \ \mathrm dx \\
&=& 3\int_{0^+}^\pi x \cot\frac{x}{2} \ \mathrm dx
\hspace{5ex}\textrm{(double angle formulas)} \\
&=& 12 \int_{0^+}^{\pi/2} t\cot t \ \mathrm dt
\hspace{5ex} (t = x/2) \\
&=& -12\int_{0^+}^{\pi/2} \log\sin t \ \mathrm dt
\hspace{5ex}\textrm{(integrate by parts)} \\
&=& -6\int_{0^+}^{\pi/2} \log\sin^2 t \ \mathrm dt \\
&=& -6\int_{0^+}^{\pi/2} \log(1-\cos^2 t) \ \mathrm dt \\
&=& 6 \int_{0^+}^{\pi/2} \sum_{k=1}^\infty \frac{1}{k}\cos^{2k}t \ \mathrm dt
\hspace{5ex}\textrm{(series for log)} \\
&=& 6\sum_{k=1}^\infty \frac{1}{k} \int_{0^+}^{\pi/2} \cos^{2k}t \ \mathrm dt 
\hspace{5ex} \textrm{(Tonelli's theorem)}\\
&=& 6\sum_{k=1}^\infty \frac{1}{k}
    \frac{\sqrt{\pi}\Gamma(k+1/2)}{2\Gamma(k+1)}  \\
&=& 3\pi \sum_{k=1}^\infty {1/2 \choose k}(-1)^{k+1}\frac{2k-1}{k} \\
&=& \pi \log 64.
\end{eqnarray*}$$
Note that 
$$\begin{eqnarray*}
6\pi \sum_{k=1}^\infty {1/2 \choose k}(-1)^{k+1} 
&=& -6\pi \left[\sum_{k=0}^\infty {1/2 \choose k}(-1)^{k} - 1\right] \\
&=& -6\pi[(1-1)^{1/2} - 1] \\
&=& 6\pi 
\end{eqnarray*}$$
and 
$$\begin{eqnarray*}
-3\pi \sum_{k=1}^\infty {1/2 \choose k}(-1)^{k+1} \frac{1}{k}
&=& 3\pi \sum_{k=1}^\infty {1/2\choose k}(-1)^k  \int_0^1 x^{k-1} \ \mathrm dx \\
&=& 3\pi \int_0^1 \frac{1}{x}  \left[
\sum_{k=0}^\infty {1/2\choose k}(-1)^k x^{k} -1
\right] \ \mathrm dx \\
&=& 3\pi \int_0^1 \frac{1}{x}  \left(
\sqrt{1-x} -1
\right) \ \mathrm dx \\
&=& 3\pi(-2+\log 4) \\
&=& -6\pi + \pi\log 64.
\end{eqnarray*}$$
