Given $|x|\leq 1$, has the series
$$\sum_{n=0}^{\infty}\frac{x^n}{(n-a)^2+b^2}$$
a closed form expression in simple functions?
It is known that for $x=1$ from Closed form for $\sum_{n=-\infty}^{\infty}\frac{1}{(n-a)^2+b^2}$. this sums up to
$$\frac{\pi}{b} \frac{\sinh(2\pi b)}{\cosh(2\pi b)-\cos(2\pi a)}$$
Also for $x=-1$ exists simple result
Can it be generalized as mentioned above?
EDIT:
Closed form above for $x=1$ relates to a modified series $$\sum_{n=-\infty}^{\infty}\frac{x^{|n|}}{(n-a)^2+b^2}$$
which looks more promising for candidate for close-term expression via Poisson's summation formula as @N74 suggests.
As Felix Marin commented, there is a solution in terms of the incomplete Beta function.
Starting with $$\frac{1}{(n-a)^2+b^2}=\frac i{2b}\left(\frac{1}{n-(a-i b)}-\frac{1}{n-(a+i b)}\right)$$
$$\sum_{n=0}^{\infty}\frac{x^n}{(n-a)^2+b^2}=\frac i{2b}\left(x^{(a-i b)} B_x(-(a-i b),0)-x^{(a+i b)} B_x(-(a+i b),0) \right)$$ What I suspect is that this could also write using Hurwitz-Lerch transcendent function since $$\sum_{n=0}^{\infty}\frac{x^n}{n+k}=\Phi (x,1,k)$$