Seeking closed-form solution to $\sum_{n=1}^{\infty}\frac{\log{(1+n)}}{(1+n)^{\alpha}-1}$ I'm looking for a closed-form solution to this infinite series:

$$S(\alpha):=\sum_{n=1}^{\infty}\frac{\log{(1+n)}}{(1+n)^{\alpha}-1},~~~\Re(\alpha)>1.$$


My attempt
All I've really been able to do is confirm this series converges for the specified values of $\alpha$ via the integral test:
$$\begin{align}
S(\alpha)\leq I(\alpha)&=\int_{0}^{\infty}\frac{\log{(1+x)}}{(1+x)^{\alpha}-1}\mathrm{d}x\\
&=\int_{0}^{\infty}\frac{u\,e^u}{e^{\alpha u}-1}\mathrm{d}u\\
&=\int_{0}^{\infty}ue^u\sum_{n=1}^{\infty}e^{-n\alpha u}\mathrm{d}u\\
&=\sum_{n=1}^{\infty}\int_{0}^{\infty}ue^ue^{-n\alpha u}\mathrm{d}u\\
&=\sum_{n=1}^{\infty}\frac{1}{(n\alpha-1)^2},~~~\text{for }\Re(\alpha)>1\\
&=\frac{1}{\alpha^2}\psi^{(1)}{\left(1-\frac{1}{\alpha}\right)}
\end{align}$$
where $\psi^{(1)}(x)$ is the first derivative of the digamma function.
I thought about expanding the logarithmic part in terms of its power series to write $S(\alpha)$ as a double-sum, and then try switching the order of summation, e.g.:
$$\begin{align}
S(\alpha)&=\sum_{n=1}^{\infty}\frac{\log{(1+n)}}{(1+n)^{\alpha}-1}\\
&=-\sum_{n=1}^{\infty}\frac{1}{(1+n)^{\alpha}-1}\sum_{k=1}^{\infty}\frac{(-1)^kn^k}{k}\\
&=-\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{(-1)^k}{k}\frac{n^k}{(1+n)^{\alpha}-1}\\
&=-\sum_{k=1}^{\infty}\frac{(-1)^k}{k}\sum_{n=1}^{\infty}\frac{n^k}{(1+n)^{\alpha}-1}.
\end{align}$$
But I'm not sure how to proceed from there. Suggestions?
 A: 
I'd be surprised if $S(2)$ had a closed form.

$S(2)$ admits a closed form in terms of the poly-Stieltjes constants unveiled here. 
Moreover, $S(3),S(4),S(5),\cdots$, admit a closed form in terms of these special functions. 
We have the following result.

Proposition. Let $\alpha=2,3,4,5,\ldots$. 
Then
  $$
\bbox[15px,border:1px solid orange]{S(\alpha)=\frac1\alpha\sum_{k=0}^{\alpha-1}e^{2k\pi i/\alpha}\:\gamma_1\!\!\left(1,1-e^{2k\pi i/\alpha}\right)} \tag1
$$ where $$
\gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\frac12 \log^2 \!N\right). \tag2
$$

Proof. For $\alpha=2,3,\ldots $, we have
$$
\begin{align}
S(\alpha):&=\lim_{N \to \infty}\sum_{n=1}^N\frac{\log{(1+n)}}{(1+n)^{\alpha}-1}\\
&=\frac1\alpha\lim_{N \to \infty}\left(\sum_{k=0}^{\alpha-1}e^{2k\pi i/\alpha}\sum_{n=1}^N\frac{\log (n+1)}{n+1-e^{2k\pi i/\alpha}}\right)\\
&=\frac1\alpha\lim_{N \to \infty}\left(\sum_{k=0}^{\alpha-1}e^{2k\pi i/\alpha}\left(\sum_{n=1}^N\frac{\log (n+1)}{n+1-e^{2k\pi i/\alpha}}-\frac12\log^2 N\right)+\frac12\log^2 N\sum_{k=0}^{\alpha-1}e^{2k\pi i/\alpha}\right)\\
&=\frac1\alpha\lim_{N \to \infty}\left(\sum_{k=0}^{\alpha-1}e^{2k\pi i/\alpha}\left(\sum_{n=1}^N\frac{\log (n+1)}{n+1-e^{2k\pi i/\alpha}}-\frac12\log^2 N\right)+0\right)\\
&=\frac1\alpha\sum_{k=0}^{\alpha-1}e^{2k\pi i/\alpha}\lim_{N \to \infty}\left(\sum_{n=1}^N\frac{\log (n+1)}{n+1-e^{2k\pi i/\alpha}}-\frac12\log^2 N\right)\\
&=\frac1\alpha\sum_{k=0}^{\alpha-1}e^{2k\pi i/\alpha}\gamma_1(1,1-e^{2k\pi i/\alpha})
\end{align}
$$ where we have used a partial fraction decomposition over the complex numbers,
$$
\frac{1}{X^\alpha-1}=\frac1\alpha\sum_{k=0}^{\alpha-1}\frac{e^{2k\pi i/\alpha}}{X-e^{2k\pi i/\alpha}},\quad \alpha=1,2,3,\ldots, \tag3
$$ giving
$$
\frac{\log (n+1)}{(n+1)^\alpha-1}=\frac1\alpha\sum_{k=0}^{\alpha-1}e^{2k\pi i/\alpha}\frac{\log (n+1)}{n+1-e^{2k\pi i/\alpha}},\quad \alpha=1,2,3,\ldots, 
$$ and we have used the standard fact that $\displaystyle \sum_{k=0}^{\alpha-1}e^{2k\pi i/\alpha}=0$.
