Computing a double gamma-digamma-trigamma series What are your thoughts on this series?
$$\sum _{k=1}^{\infty } \sum _{n=1}^{\infty } \frac{\Gamma (k)^2 \Gamma (n) }{\Gamma (2 k+n)}((\psi ^{(0)}(n)-\psi ^{(0)}(2 k+n)) (\psi ^{(0)}(k)-\psi ^{(0)}(2 k+n))-\psi ^{(1)}(2 k+n)).$$
EDIT: Noting the interest for this series, I wanna add that the series gets reduced to the calculation of $$\int_0^1 \left(\frac{\text{Li}_3(x)}{x^2-2 x+2}+\frac{\text{Li}_3\left(x-x^2\right)}{x^2-2 x+2}-\frac{\text{Li}_3\left(\frac{x}{x-1}\right)}{x^2-2 x+2}-\frac{\text{Li}_3\left(\frac{(x-1) x}{x^2-x+1}\right)}{x^2-2 x+2}-\frac{\text{Li}_2\left(\frac{x}{x-1}\right) \log (1-x)}{x^2-2 x+2}-\frac{\text{Li}_2(x) \log (1-x)}{x^2-2 x+2}+\frac{\text{Li}_2\left(\frac{x}{x-1}\right) \log (x)}{x^2-2 x+2}+\frac{\text{Li}_2\left(\frac{(x-1) x}{x^2-x+1}\right) \log (x)}{x^2-2 x+2}-\frac{\text{Li}_2(x) \log (x)}{x^2-2 x+2}-\frac{\text{Li}_2\left(x-x^2\right) \log (x)}{x^2-2 x+2}-\frac{\text{Li}_2\left(x-x^2\right) \log \left(x^2-x+1\right)}{x^2-2 x+2}-\frac{\text{Li}_2\left(\frac{(x-1) x}{x^2-x+1}\right) \log \left(x^2-x+1\right)}{x^2-2 x+2}-\frac{\log ^3(1-x)}{3 \left(x^2-2 x+2\right)}-\frac{\log ^3\left(x^2-x+1\right)}{3 \left(x^2-2 x+2\right)}-\frac{\log ^2(x) \log (1-x)}{x^2-2 x+2}-\frac{\log ^2(x) \log \left(x^2-x+1\right)}{x^2-2 x+2}+\frac{\pi ^2 \log (1-x)}{6 \left(x^2-2 x+2\right)}+\frac{\pi ^2 \log \left(x^2-x+1\right)}{6 \left(x^2-2 x+2\right)}\right) \, dx$$
where we are pretty familiar with all the stuff in here. One can find the closed form by calculating the integral, a bit long, but it's a nice journey to go.
A 300 points bounty moment: After 2 years and 10 months since the problem has been posed no full solution has been provided. Is it possible to find a slick solution?
 A: I'd try to use: $$\displaystyle \int\limits_{0<x+y<1} x^{u-1}y^{u-1}(1-x-y)^{v-1} \,dx\,dy= \frac{\Gamma(u)^2\Gamma(v)}{\Gamma(2u+v)}$$
for reals $u,v \in \mathbb{R}$.
Hence, $$\displaystyle \begin{align}\sum\limits_{n,k=1}^{\infty} \frac{\Gamma(k+u)^2\Gamma(n+v)}{\Gamma(2k+2u+n+v)} &= \int\limits_{0<x+y<1}  \frac{(xy)^{u}(1-x-y)^v}{(1-xy)(x+y)}\,dx\,dy\end{align}$$
Since, $\displaystyle \frac{\partial^2}{\partial u\partial v} \left(\frac{\Gamma(u)^2\Gamma(v)}{\Gamma(2u+v)}\right) = \frac{2\Gamma(u)^2\Gamma(v)}{\Gamma(2u+v)}\left((\psi^{(0)}(u) - \psi^{(0)}(2u+v))(\psi^{(0)}(v) - \psi^{(0)}(2u+v)) - \psi^{(1)}(2u+v)\right)$
Hence, $$\begin{align}&2\sum _{k=1}^{\infty } \sum _{n=1}^{\infty } \frac{\Gamma (k)^2 \Gamma (n) }{\Gamma (2 k+n)}((\psi ^{(0)}(n)-\psi ^{(0)}(2 k+n)) (\psi ^{(0)}(k)-\psi ^{(0)}(2 k+n))-\psi ^{(1)}(2 k+n))\\&=  \lim\limits_{(u,v) \to (0,0)} \frac{\partial^2}{\partial u\partial v}\int\limits_{0<x+y<1}  \frac{(xy)^{u}(1-x-y)^v}{(1-xy)(x+y)}\,dx\,dy\\&= \lim\limits_{(u,v) \to (0,0)} \int\limits_{0<x+y<1}  \frac{(xy)^{u}(1-x-y)^v\log (xy)\log(1-x-y)}{(1-xy)(x+y)}\,dx\,dy\end{align}$$
We need to compute the integral and take the limit after that. This is only a partial attempt.
