3 vs 5

We'll use $\displaystyle \sum_{k=1}^{n}k^5=\frac{n^2(n+1)^2(2n^2+2n-1)}{12}$ & $\displaystyle \sum_{k=1}^{n}k^3 = \frac{n^2(n+1)^2}{4}$ ,

So our sum equals $\displaystyle S=\sum_{n=1}^{\infty} \frac{3}{2n^2+2n-1}=\frac{3}{2} \sum_{n=1}^{\infty} \frac{1}{(n+\frac{\sqrt{3}+1}{2})(n+\frac{1-\sqrt{3}}{2}) }$

By partial fraction decomposition we have, $\displaystyle S = \frac{\sqrt{3}}{2} \sum_{n=1}^{\infty} \frac{1}{n+\frac{1-\sqrt{3}}{2}} -\frac{1}{n+\frac{\sqrt{3}+1}{2}}$ and by Digamma reflection formula,

$\displaystyle S= \frac{\sqrt{3}}{2} (\psi(\frac{\sqrt{3}+1}{2})-\psi(\frac{1-\sqrt{3}}{2}) +\frac{2}{1-\sqrt{3}}-\frac{2}{1+\sqrt{3}}) = \frac{\sqrt{3}\pi}{2}\cot(\frac{\pi}{2}-\frac{\sqrt{3}}{2}) + 3 = \frac{\sqrt{3}\pi}{2}\tan(\pi\sin(\frac{\pi}{3}))+3 = \frac{6+\sqrt{3}\pi\tan(\pi\sin(\frac{\pi}{3}))}{2}$ and thus making the answer $\boxed{2+3+6=11}$