On the summation $\sum \limits_{n=1}^{\infty} \arctan \left ( \frac{1}{n^3+n^2+n+1} \right )$

We have $$S = \sum_{n=1}^{\infty}\arctan\left(\frac{1}{n^3 + n^2 + n + 1}\right) \\ = \Im\left(\ln\left(\prod_{n=1}^{\infty} \left(1 + \frac{i}{n^3+n^2+n+1}\right)\right)\right) \\ = \Im\left(\ln\left(\prod_{n=1}^{\infty} \frac{n^3+n^2+n+1+i}{n^3+n^2+n+1}\right)\right) \\ = \Im\left(\ln\left(\frac{\pi\operatorname{csch}(\pi)}{1+i}\prod_{k=1}^{3}\frac{1}{\Gamma(r_k)}\right)\right)$$ where $r_k$ is the $k$th root of $x^3-x^2+x-1-i$ (order doesn't matter since the product iterates over all of them). The last step comes from employing equation 19 on here (which comes directly from the Weierstrass factorization product formula for $\Gamma(x)$), and the fact that $\Gamma(i)\Gamma(-i) = \pi\operatorname{csch}(\pi)$.

Something tells me this might be difficult to do by hand. Plugging in your product formula into Mathematica and then Simplifying, this is what comes out: