On $\sum_{n=0}^\infty \frac{1}{n^2+bn+c} = \frac{\pi \tan\big(\frac{\pi}2\sqrt{b^2-4c}\big)}{\sqrt{b^2-4c}}+x$ and solvable Galois groups?

Well, if a quadratic polynomial $p(x)$ has a negative discriminant, $\sum_{n\in\mathbb{Z}}\frac{1}{p(n)}$ can be found through the Poisson summation formula or just by considering $\frac{d}{dx}\log(\cdot)$ applied to the Weierstrass product for the cosine function. For instance

$$\sum_{n\in\mathbb{Z}}\frac{1}{n^2+9n+c}=4\sum_{n\in\mathbb{Z}}\frac{1}{(2n+9)^2+(4c-49)}=4\sum_{n\in\mathbb{Z}}\frac{1}{(2n+9)^2+\sqrt{4c-81}^2}$$ and assuming $d>0$ $$ \sum_{n\in\mathbb{Z}}\frac{1}{(2n+1)^2+d^2}=\frac{\pi}{2d}\,\tan\frac{\pi d}{2} $$ so by assuming $c>\frac{81}{4}$ we have $$\sum_{n\in\mathbb{Z}}\frac{1}{n^2+9n+c}=\frac{2\pi\tan\left(\pi\sqrt{c-\tfrac{81}{4}}\right)}{\sqrt{4c-81}} $$ and $p(n)=n(n+9)+c$ fulfills $p(n)=p(-n-9)$, so the LHS of the previous line can be written as $$ 2\sum_{n\geq 0}\frac{1}{n^2+9n+c}+\sum_{k=1}^{8}\frac{1}{p(-k)}$$ such that $$\sum_{n\geq 0}\frac{1}{n^2+9n+c}=\frac{\pi\tan\left(\pi\sqrt{c-\tfrac{81}{4}}\right)}{\sqrt{4c-81}}-\frac{4 \left(c^3-45c^2+654c-3044\right)}{(c-8)(c-14)(c-18)(c-20)}.$$ $P_2(c)$ has a pretty clear structure, I am not so confident about the Galois group of $P_1(c)$ over $\mathbb{Q}$, but the same manipulation applies in the general case.

It looks like the correct thing to look at is $$\sum_{n=-\infty}^\infty\frac1{n^2+bn+c}.$$ This is surely expressible in terms of the digamma function.

Your rational fraction is surely an artefact of insisting on taking a one-sided sum. As an example, $$\sum_{n=0}^\infty\frac1{n^2+9n+c} =\sum_{m=-\infty}^{-9}\frac1{(-9-m)^2+9(-9-m)+c} =\sum_{m=-\infty}^{-9}\frac1{m^2+9m+c}$$ and so

\begin{align} \sum_{n=0}^\infty\frac1{n^2+9n+c} &=\frac12\sum_{n=-\infty}^\infty\frac1{n^2+9n+c}-\frac12 \sum_{n=-8}^{-1}\frac1{n^2+9n+c}\\ &=\frac12\sum_{n=-\infty}^\infty\frac1{n^2+9n+c}- \left(\frac1{c-8}+\frac1{c-14}+\frac1{c-18}+\frac1{c-20}\right) \end{align} etc.