Another Ramanujan's formula dealing with $\coth^{2}(5\pi)$

This is just an idea. I write the series as:$$\sum_{n = 1}^{\infty}\frac{n^{5}q^{2n}}{1 - q^{2n}}\cdot\frac{1}{2500 + n^{4}}= \sum_{n = 1}^{\infty}\frac{n^{4}q^{2n}}{1 - q^{2n}}\cdot\frac{n}{2500 + n^{4}} $$ As you noticed: $$n^{4} + 2500 = (n^{2} + 50)^{2} - 100n^{2} = (n^{2} - 10n + 50)(n^{2} + 10n + 50)$$ Then $$\begin{aligned}S &= \sum_{n = 1}^{\infty}\left( \frac{1}{n^{2} - 10n + 50}-\frac{1}{n^{2} + 10n + 50} \right)\\ &= \sum_{n = 1}^{\infty} \left( \frac{1}{n^{2} - 10n + 50}-\frac{1}{(n+10)^{2} - 10(n+10) + 50} \right)\\ &= \sum_{n = 1}^{\infty} \left( \frac{1}{n^{2} - 10n + 50} \right)-\sum_{n = 10}^{\infty} \left( \frac{1}{(n-10)^{2} - 10(n-10) + 50} \right)\\ &=\sum_{n = 1}^{10} \left( \frac{1}{n^{2} - 10n + 50} \right)\\ &=\frac{4118807}{13138450}\end{aligned}$$ And hence $$ \sum_{n = 1}^{\infty}\frac{n}{2500 + n^{4}}=\frac{4118807}{262769000}$$ The fraction $$ \frac{4118807}{262769000}=\frac{3\cdot4118807}{500\cdot1576614}=\frac{12356421}{500\cdot1576614}$$ Thus $$ \frac{123826979}{6306456}=\frac{123826979}{4\cdot1576614}=\frac{15478372375}{12356421}\cdot\sum_{n = 1}^{\infty}\frac{n}{2500 + n^{4}}$$ We can write: $$ \sum_{n = 1}^{\infty}\frac{n}{2500 + n^{4}} \left( \frac{15478372375}{12356421}-\frac{n^{4}}{e^{2 \pi n}-1}\right)=\frac{25}{4} \pi \coth^{2}(5 \pi)$$