What special functions do we need to get a closed form for $\sum\limits_{n=0}^\infty \frac{2^{n+1}}{\binom{2 n}{n}} \cdot \frac{1}{(2n+1)^p}$? This question generalizes the considerations in Is there a closed form for $\sum_{n=0}^{\infty}{2^{n+1}\over {2n \choose n}}\cdot\left({2n-1\over 2n+1}\right)^2?$ . 
Let $p\ge 3$ be an integer. We consider the following sum:
\begin{equation}
S_{p} := \sum\limits_{n=0}^\infty \frac{2^{n+1}}{\binom{2 n}{n}} \cdot \frac{1}{(2n+1)^p}
\end{equation}
By pushing through the approach developed in the aforementioned question we got the following result:
\begin{eqnarray}
S_3 = -2 \imath \sqrt{2} \left(\right.\\
\left.
(\pi-\imath \log(2))\cdot \frac{2 \pi ^2+2 \pi  \left(\psi ^{(0)}\left(\frac{3}{8}\right)-\psi ^{(0)}\left(\frac{1}{8}\right)\right)-\psi ^{(1)}\left(\frac{1}{8}\right)-\psi
   ^{(1)}\left(\frac{3}{8}\right)}{32 \sqrt{2}}+ \right.\\
\left.\frac{\sqrt{2}}{256}\cdot(448 \sqrt{2} \zeta (3)+28 \pi ^3+4 \pi ^2 (\psi ^{(0)}(\frac{3}{8})-\psi ^{(0)}(\frac{1}{8}))-4 \pi  (\psi^{(1)}(\frac{1}{8})+\psi ^{(1)}(\frac{3}{8}))+\psi ^{(2)}(\frac{1}{8})-\psi^{(2)}(\frac{3}{8}))+\right.\\
\left.
-\int\limits_1^{\exp(\imath \frac{\pi}{4})} \log(z) \left(\log(z-1)+\log(z+1)\right) \cdot \left( \frac{1}{z-1} - \frac{1}{z+1}\right)dz \right.\\
\left.
\right)
\end{eqnarray}
Here $\psi^{(m)}(z)$ is the poly-gamma function.
Now, the question is twofold. Firstly can we evaluate the remaining integral in the right hand side and express the whole thing a a "closed form". Secondly, can we come up with closed forms for arbitrary values of $p$. If yes what kind of special functions will be involved?
 A: We provide a partial answer to question number one. It is not hard to see that one can construct anti-derivatives of the integrand by using poly-logarithms.
As a matter of fact we have:
\begin{eqnarray}
\int \log(z) \log(z+1) \frac{1}{z+1} dz&=& \frac{1}{2} \imath \pi [\log(1+z)]^2- \log(1+z) Li_2(1+z)+Li_3(1+z)\\
\int \log(z) \log(z-1) \frac{1}{z-1} dz&=& - \log(-1+z) Li_2(1-z)+Li_3(1-z)
\end{eqnarray}
Now using the above after tedious but elementary simplifications we got the following:
\begin{eqnarray}
&&\int\limits_1^{\exp(\imath \frac{\pi}{4})} \log(z) \log(z+1) \frac{1}{z+1} dz=-\frac{7}{8} \zeta(3)+\frac{1}{512}(-\pi+2 \imath(-2 \log(-1+\sqrt{2})+\log(2)))\cdot\\
&&\left(16 C+\left(4 \sqrt{2}+13 i\right) \pi ^2-4 \pi  \left(\log (2)-2 \log \left(\sqrt{2}-1\right)\right)-\sqrt{2} \left(\psi ^{(1)}\left(\frac{1}{8}\right)+\psi
   ^{(1)}\left(\frac{3}{8}\right)\right)\right)+\\
&&+Li_3[1+\exp(\imath \frac{\pi}{4})]
\end{eqnarray}
Likewise
\begin{eqnarray}
&&\int\limits_1^{\exp(\imath \frac{\pi}{4})} \log(z) \log(z-1) \frac{1}{z-1} dz=\frac{1}{512}(5\pi+2 \imath(-2\log(-1+\sqrt{2})-\log(2)))\cdot\\
&&\left(-16 C+\left(4 \sqrt{2}-i\right) \pi ^2-4 \pi  \left(\log (2)+2 \log \left(\sqrt{2}-1\right)\right)-\sqrt{2} \left(\psi ^{(1)}\left(\frac{1}{8}\right)+\psi
   ^{(1)}\left(\frac{3}{8}\right)\right)\right)+\\
&&+Li_3[1-\exp(\imath \frac{\pi}{4})]
\end{eqnarray}
where $C$ is the Catalan constant.
The remaining tri-logarithms are expressed in terms of the Hurwitz zeta functions $\Psi$ as follows:
\begin{eqnarray}
Li_3[1+\exp(\imath \frac{\pi}{4})]&=&\frac{1}{8^3} \sum\limits_{\xi=1}^8 \exp(\imath \xi \frac{\pi}{8})(2 \cos(\frac{\pi}{8}))^\xi \cdot \Psi(-(2 \cos(\frac{\pi}{8}))^8,3,\frac{\xi}{8})\\
Li_3[1-\exp(\imath \frac{\pi}{4})]&=&\frac{1}{8^3} \sum\limits_{\xi=1}^8 \exp(-\imath \xi \frac{3\pi}{8})(2 \sin(\frac{\pi}{8}))^\xi \cdot \Psi(-(2 \sin(\frac{\pi}{8}))^8,3,\frac{\xi}{8})\\
\end{eqnarray}
