How to calculate $ \sum_{r=1}^{\infty}\left(r+1\right)^{2}2^{-\left(2r-1\right)^{2}} $

Is there a way to calculate $ \sum_{r=1}^{\infty}\left(r+1\right)^{2}2^{-\left(2r-1\right)^{2}} $ ?

I tried to differentiate term by term, but it did not lead me anywhere. Basically, I want to know if its possible to calculate this sum precisly, or the best we can do is find an upper bound for the mistake.

Thanks in advance

If $$a_r=\left(r+1\right)^{2}\,\, 2^{-\left(2r-1\right)^{2}}$$ the partial sums should convere very fast since $$\frac{a_{r+1}}{a_r}=\left(\frac{r+2}{r+1}\right)^2\,2^{-8 r}$$ If we compute them $$S_p=\sum_{r=0}^p a_r$$ we have $$\left( \begin{array}{cc} p & S_p\\ 0 & 0.5000000000000000000000000000000000000000000000000 \\ 1 & 2.5000000000000000000000000000000000000000000000000 \\ 2 & 2.5175781250000000000000000000000000000000000000000 \\ 3 & 2.5175786018371582031250000000000000000000000000000 \\ 4 & 2.5175786018372026120459850062616169452667236328125 \\ 5 & 2.5175786018372026120459998955126428997648713181690 \\ 6 & 2.5175786018372026120459998955126429181966227390624 \\ 7 & 2.5175786018372026120459998955126429181966227390625 \end{array} \right)$$

To give you an idea, $a_{10}=2.576\times 10^{-107}$ and $a_{100}=8.314\times 10^{-11918}$.

Just for the fun, if you accept an error of $6.0\times 10^{-10}$, an inverse symbolic calculator of mine proposes for the infinite sum $$10 \sin \left(\frac{40 \pi }{381}\right) \cos \left(\frac{81 \pi }{374}\right)=\color{red}{2.51757860}78$$

Now, identifying the corresponding theta function is another story (I also would like to know the formal answer).