\[\Large\displaystyle \sum_{n=1}^\infty\frac{\Gamma\left(n+\dfrac{1}{2}\right)}{(2n+1)^4\,4^n\,n!}
\]
\(\Large\mathbf{Solution:}\)
First, in view of Legrende's duplication formula,
\[\begin{align*}
S&=\sum_{n=1}^\infty\frac{\Gamma\left(n+\dfrac{1}{2}\right)}{(2n+1)^4\,4^n\,n!}\\&=2\sqrt{\pi}\sum_{n=1}^\infty\frac{\Gamma(2n)}{\Gamma(n)\,n!\,(2n+1)^4\, 16^n}
\\&
=-\frac{\sqrt{\pi}}{3}\int_0^1 \ln^3(x)\sum_{n=1}^{\infty}\frac{\Gamma(2n)}{\Gamma(n)\,n!}\left(\frac{x^2}{16}\right)^n\mathrm{d}x\\
&=-\sqrt{\pi}-\frac{\sqrt{\pi}}{6}\int_0^1\frac{\ln^3(x)}{\sqrt{1-\dfrac{x^2}{4}}}\mathrm{d}x\\&=-\sqrt{\pi}-\frac{\sqrt{\pi}}{3}\int_0^{\frac{\pi}{6}}\ln^3(2\sin x)\mathrm{d}x
\end{align*}\]
Claim: for \(0<a\leq \dfrac{\pi}{2}\),
\[\begin{align*}
\int_0^a \ln^3\left(\frac{\sin x}{\sin a}\right)\mathrm{d}x&=\frac{4a-3\pi}{2}a^2\ln(2\sin a)-\frac{3\pi}{4}\zeta(3)\\
&+3\left(\frac{\pi}{2}-a\right)\Re\left(\frac12 \operatorname{Li}_3(e^{2ia})+\operatorname{Li}_3(1-e^{2ia})\right)\\
&+3\Im\left(\frac14\operatorname{Li}_4(e^{2ia})+\operatorname{Li}_4(1-e^{2ia})\right)
\end{align*}\]
Proof: The idea is exactly identical to the proof displayed in this question.
The proof is rather tedious (and obviously inefficient), and ends with a somewhat of a cancellation (implying the existence of a shortcut) , so I omit the boring algebra and outline the main ideas, which can be repeated systematically to obtain closed forms for even higher powers of logsine.
Things to know:
\[\begin{align*}
\ln(2\sin x)=\ln(1-e^{2ix})+i\left(\frac{\pi}{2}-x\right) \tag{1}
\end{align*}\]
\[\begin{align*}
\int\frac{\ln^3(1-x)}{x}\mathrm{d}x&=\ln^3(1-x)\ln(x)+3\ln^2(1-x)\text{Li}_2(1-x)\\
&-6\ln(1-x)\text{Li}_3(1-x)+6\text{Li}_4(1-x) \tag{2}
\end{align*}\]
\[\begin{align*}
\int_0^a x\ln(2\sin x)\mathrm{d}x=-\frac{a}{2}\text{Cl}_2(2a)-\frac14\Re\text{Li}_3(e^{2ia})+\frac{\zeta(3)}{4}\tag{3}
\end{align*}\]
\[\begin{align*}
\int_0^a x^2\ln(2\sin x)\mathrm{d}x=-\frac{a^2}{2}\text{Cl}_2(2a)-\frac{a}{2}\Re\text{Li}_3(e^{2ia})+\frac14\Im\text{Li}_4(e^{2ia})\tag{4}
\end{align*}\]
\[\begin{align*}
\int_0^a \ln(\sin x)\mathrm{d}x=-a\ln2-\frac12 \text{Cl}_2(2a)\tag{5}
\end{align*}\]
\[\begin{align*}
\int_0^a \ln^2(\sin x)\mathrm{d}x=\frac{a^3}{3}+a\ln^2 2-a\ln^2(2\sin a)-\ln(\sin a)\text{Cl}_2(2a)-\Im\text{Li}_3(1-e^{2ia})\tag{6}
\end{align*}\]
(1) is trivial, (2) is not too hard to find, (5) and (6) are shown in the linked answer, and (3),(4) are easily found using \(\displaystyle \ln(2\sin x)=-\sum_{n\geq1}\frac{\cos(2xn)}{n}\).
It is obvious that since we have (5) and (6), the claim (0) depends on a closed form for \(\displaystyle\int_0^a \ln^3(\sin x)\mathrm{d}x\), and the latter may be evaluated in terms of \(\displaystyle\int_0^a \ln^3(2\sin x)\mathrm{d}x\).
But, with the help of (1),
\[\begin{align*}
\int_0^a \ln^3(2\sin x)\mathrm{d}x&=\Re\int_0^a \ln^3(1-e^{2ix})\mathrm{d}x+3\int_0^a \ln(2\sin x)\left(\frac{\pi}{2}-x\right)^2\mathrm{d}x\\
&=\frac12\Im\int_1^{e^{2ia}}\frac{\ln^3(1-x)}{x} \mathrm{d}x+3\int_0^a \ln(2\sin x)\left(\frac{\pi}{2}-x\right)^2\mathrm{d}x
\end{align*}\]
(Same idea RandomVariable had in this answer.)
Now we employ (2),(3),(4), and (5). Some expressions cancel and claim follows.
This result, together with the fact that \(e^{i\pi/3}\) and \(1-e^{i\pi/3}\) are conjugates, yields
\[\displaystyle \int_0^{\frac{\pi}{6}} \ln^3(2\sin x)\mathrm{d}x=-\frac{\pi}{4}\zeta(3)-\frac94\Im\text{Li}_4(e^{i\pi/3})
\]
and use
\[\frac{2}{\sqrt{3}}\Im\text{Li}_4(e^{i\pi/3})=\sum_{n\geq 0}\frac{(-1)^n}{(3n+1)^4}+\sum_{n\geq 0}\frac{(-1)^n}{(3n+2)^4} =\frac{\psi^{(3)}\left(\dfrac13\right)}{216}-\frac{\pi^4}{81}
\]
Hence we have
\[\Large\boxed{\displaystyle \sum_{n=1}^\infty\frac{\Gamma\left(n+\dfrac{1}{2}\right)}{(2n+1)^4\,4^n\,n!}= \color{Blue}{\sqrt{\pi}\left(\frac{\pi}{12}\zeta(3)+\frac{\psi^{(3)}\left(\dfrac13\right)}{192\sqrt3}-\frac{\pi^4}{72\sqrt3}-1\right)}}
\]
It can also be expressed in generalized hypergeometric function with the help of WolframAlpha:
\[\Large\boxed{\displaystyle \sum_{n=1}^\infty\frac{\Gamma\left(n+\dfrac{1}{2}\right)}{(2n+1)^4\,4^n\,n!}=\color{blue}{\frac{\sqrt{\pi}}{648} {_6F_5}\left(\begin{array}c\ 1,\dfrac32,\dfrac32,\dfrac32,\dfrac32,\dfrac32\\2,\dfrac52,\dfrac52,\dfrac52,\dfrac52\end{array}\middle|\,\frac14\right)}}
\]
