Request for help to find a closed-form expression for $\sum_{n=0}^{\infty}\frac{(2n+15)^2}{(2n+11)^3\sqrt{(2n+13)}}$

Not a closed form but a rather good approximation of it.

Rewriting as $$a_n=\frac{(2n+15)^2}{(2n+11)^3\sqrt{2n+13}}=\frac{(2n+15)^2}{(2n+11)^3\sqrt{2n+11+2}}=\frac{(2n+15)^2}{(2n+11)^{\frac 72}\sqrt{1+\frac 2{2n+11}}}$$ we can write $$\frac 1{\sqrt{1+\frac 2{2n+11}}}=\sum_{k=0}^\infty \binom{-\frac{1}{2}}{k}\frac{2^k}{(2n+11)^k}$$ and then face summations of terms $$S_k=\sum_{n=0}^\infty\frac{(2n+15)^2}{(2n+11)^{k+\frac 72}}=2^{-k-\frac{3}{2}} \left(\zeta \left(k+\frac{3}{2},\frac{11}{2}\right)+4 \zeta \left(k+\frac{5}{2},\frac{11}{2}\right)+4 \zeta \left(k+\frac{7}{2},\frac{11}{2}\right)\right)$$ and then for the considered summation $$\Sigma=\sum_{n=0}^\infty \frac{(2n+15)^2}{(2n+11)^3\sqrt{2n+13}}$$ So, the partial sums $$\Sigma_p=\frac 1{2\sqrt 2}\sum_{k=0}^p \binom{-\frac{1}{2}}{k} \left(\zeta \left(k+\frac{3}{2},\frac{11}{2}\right)+4 \zeta \left(k+\frac{5}{2},\frac{11}{2}\right)+4 \zeta \left(k+\frac{7}{2},\frac{11}{2}\right)\right)$$ Computing $$\left( \begin{array}{cc} p & \Sigma_p \\ 5 & 0.39483148307398830445048206504782536496698034745960346801382376024 \\ 10 & 0.39483198676616324735777911451436673248113391244073226133701558847 \\ 15 & 0.39483198670639658527502248284985700122310258168712958250026452734 \\ 20 & 0.39483198670640571076206255275333399438974388443990146144695977886 \\ 25 & 0.39483198670640570922643131344152894240169346679460070946980052257 \\ 30 & 0.39483198670640570922670214360763624644791582367286943018650562365 \\ 35 & 0.39483198670640570922670209457967764179263574295643280410747243813 \\ 40 & 0.39483198670640570922670209458869824722904384200792304484978282169 \\ 45 & 0.39483198670640570922670209458869656914069678422236918330533995116 \\ 50 & 0.39483198670640570922670209458869656945538601634396101492730552390 \\ 55 & 0.39483198670640570922670209458869656945532663821679612114844784572 \\ 60 & 0.39483198670640570922670209458869656945532664947597620995355367822 \\ 65 & 0.39483198670640570922670209458869656945532664947383263347402134364 \\ 70 & 0.39483198670640570922670209458869656945532664947383304295989988132 \\ 75 & 0.39483198670640570922670209458869656945532664947383304288145065669 \\ 80 & 0.39483198670640570922670209458869656945532664947383304288146572333 \\ 85 & 0.39483198670640570922670209458869656945532664947383304288146572043 \\ 90 & 0.39483198670640570922670209458869656945532664947383304288146572044 \\ 95 & 0.39483198670640570922670209458869656945532664947383304288146572044 \\ 100 & 0.39483198670640570922670209458869656945532664947383304288146572044 \end{array} \right)$$

Edit

In order to know how many terms have to be added for $p$ significant digits, since we face an alternatin series, consider $$a_k=\frac 1{2\sqrt 2}\binom{-\frac{1}{2}}{k} \left(\zeta \left(k+\frac{3}{2},\frac{11}{2}\right)+4 \zeta \left(k+\frac{5}{2},\frac{11}{2}\right)+4 \zeta \left(k+\frac{7}{2},\frac{11}{2}\right)\right)$$ A quick and dirty linear regression (built for $10 \leq k \leq 1000$ by steps of $10$) shows that (and this is an almost perfect fit) $$\log_{10}(|a_k|)=-0.740989\, k-2.51445$$ So, for $p$ exact digits, we need to sum up $\lceil 1.35\, p -3.40 \rceil$ terms (just as reflected by the values in the above table). For $74$ digits as given in the post, then $k=97$ which has been verified. Using the original summation, I suppose that billions of terms have been added (summing from $n=0$ to $n=10^9$ leading to $0.39481$). This look normal since, for large values of $n$, $\frac{a_{n+1}}{a_n}=1-\frac{3}{2 n}+O\left(\frac{1}{n^2}\right)$ which is extremely slow.