First, in view of Legrende's duplication formula,
$$S=\sum_{n=1}^\infty\frac{\Gamma\left(n+\frac{1}{2}\right)}{(2n+1)^4\,4^n\,n!}=2\sqrt{\pi}\sum_{n=1}^\infty\frac{\Gamma(2n)}{\Gamma(n)\,n!\,(2n+1)^4\, 16^n}
\\=-\frac{\sqrt{\pi}}{3}\int_0^1 \ln^3(x)\sum_{n=1}^{\infty}\frac{\Gamma(2n)}{\Gamma(n)\,n!}\left(\frac{x^2}{16}\right)^ndx\\
=-\sqrt{\pi}-\frac{\sqrt{\pi}}{6}\int_0^1\frac{\ln^3(x)}{\sqrt{1-x^2/4}}dx=-\sqrt{\pi}-\frac{\sqrt{\pi}}{3}\int_0^{\frac{\pi}{6}}\ln^3(2\sin x)dx$$
Claim: for $0<a\leq \frac{\pi}{2}$,
$$\int_0^a \ln^3\left(\frac{\sin x}{\sin a}\right)dx\tag{0}=\frac{4a-3\pi}{2}a^2\ln(2\sin a)-\frac{3\pi}{4}\zeta(3)+3\left(\frac{\pi}{2}-a\right)\Re\left(\frac12 \operatorname{Li}_3(e^{2ia})+\operatorname{Li}_3(1-e^{2ia})\right)+3\Im\left(\frac14\operatorname{Li}_4(e^{2ia})+\operatorname{Li}_4(1-e^{2ia})\right)
$$
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:
$$\ln(2\sin x)=\ln(1-e^{2ix})+i\left(\frac{\pi}{2}-x\right) \tag{1}$$
$$\small\int\frac{\ln^3(1-x)}{x}dx=\ln^3(1-x)\ln(x)+3\ln^2(1-x)\text{Li}_2(1-x)-6\ln(1-x)\text{Li}_3(1-x)+6\text{Li}_4(1-x) \tag{2}$$
$$\int_0^a x\ln(2\sin x)dx=-\frac{a}{2}\text{Cl}_2(2a)-\frac14\Re\text{Li}_3(e^{2ia})+\frac{\zeta(3)}{4}\tag{3}$$
$$\int_0^a x^2\ln(2\sin x)dx=-\frac{a^2}{2}\text{Cl}_2(2a)-\frac{a}{2}\Re\text{Li}_3(e^{2ia})+\frac14\Im\text{Li}_4(e^{2ia})\tag{4}$$
$$\int_0^a \ln(\sin x)dx=-a\ln2-\frac12 \text{Cl}_2(2a)\tag{5}$$
$$\int_0^a \ln^2(\sin x)dx=\frac{a^3}{3}+a\ln^2 2-a\ln^2(2\sin a)-\ln(\sin a)\text{Cl}_2(2a)-\Im\text{Li}_3(1-e^{2ia})\tag{6}$$
$(1)$ 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 $\,\,\ln(2\sin x)=-\sum_{n\geq1}\frac{\cos(2xn)}{n}$.
It is obvious that since we have $(5)$ and $(6)$, the claim $(0)$ depends on a closed form for $\displaystyle\int_0^a \ln^3(\sin x)dx$, and the latter may be evaluated in terms of $\displaystyle\int_0^a \ln^3(2\sin x)dx$.
But, with the help of $(1)$,
$$\int_0^a \ln^3(2\sin x)dx=\Re\int_0^a \ln^3(1-e^{2ix})dx+3\int_0^a \ln(2\sin x)\left(\frac{\pi}{2}-x\right)^2dx\\
=\frac12\Im\int_1^{e^{2ia}}\frac{\ln^3(1-x)}{x} dx+3\int_0^a \ln(2\sin x)\left(\frac{\pi}{2}-x\right)^2dx$$
(Same idea @RandomVariable had in this answer.)
Now we employ $(2),(3),(4),$ and $(5)$. Some expressions cancel and claim follows.$\square $
This result, together with the fact that $e^{i\pi/3}$ and $1-e^{i\pi/3}$ are conjugates, yields $\displaystyle \int_0^{\frac{\pi}{6}} \ln^3(2\sin x)dx=-\frac{\pi}{4}\zeta(3)-\frac94\Im\text{Li}_4(e^{i\pi/3})$,
and
$$S=\sqrt{\pi}\left(\frac{\pi}{12}\zeta(3)+\frac{9}{12}\Im\text{Li}_4(e^{i\pi/3})-1\right)$$
This form is equivalent to @user153012's form, as
$$\frac{2}{\sqrt{3}}\Im\text{Li}_4(e^{i\pi/3})=\sum_{n\geq 0}\frac{(-1)^n}{(3n+1)^4}+\sum_{n\geq 0}\frac{(-1)^n}{(3n+2)^4} \\=\frac{\psi^{(3)}\left(\frac13\right)}{216}-\frac{\pi^4}{81}$$
Also, as noted in the comments in the linked question, this may be used to write a closed form for a certain hypergeometric function.
This serves as a generalisation for the series, because $\displaystyle \sum_{n=1}^{\infty} \frac{\Gamma(n+1/2)}{(2n+1)^4 n!}a^{2n}=-\sqrt{\pi}\left(1+\frac1{6a}\int_0^{\sin^{-1} a}\ln^3\left(\frac{\sin x}{a}\right)dx\right)$
As an example, using closed forms for trilogarithms displayed in this post, we have
$$\int_0^{\frac{\pi}{4}}\ln^3(\sqrt{2}\sin x)dx=-\frac{\pi^3}{128}\ln2-\frac{3\pi}{8}\zeta(3)+\frac34\beta(4)+3\Im\text{Li}_4(1-i)$$
where $\beta(4)=\Im\text{Li}_4(i)$ is a value of Dirichlet's beta function.
Or equivalently,
$$\sum_{n=1}^{\infty} \frac{\Gamma\left(n+\frac12\right)}{(2n+1)^4\,2^n\,n!}=-\sqrt{\pi}-\frac{\sqrt{2\pi}}{6}\left(-\frac{\pi^3}{128}\ln2-\frac{3\pi}{8}\zeta(3)+\frac34\beta(4)+3\Im\text{Li}_4(1-i)\right)$$
Best Answer
Let $\mathcal{S}$ denote the sum of the following infinite series:
$$\mathcal{S}:=\sum_{n=1}^{\infty}\frac{a_{n}}{\left(n+1\right)4^{n}},$$
where
$$a_{n}:=\sum_{k=1}^{n}\binom{2n}{n-k}\frac{1+(-1)^{k+1}}{k^{2}};~~~\small{n\in\mathbb{N}}.$$
We first show that the series can be reduced to an integral as
$$\mathcal{S}=\frac43\zeta{\left(2\right)}+4\int_{0}^{1}\mathrm{d}x\,\frac{2x\ln{\left(1-x+x^{2}\right)}}{1+x^{2}}.$$
We then evaluate the integral using Feynman's trick to obtain the conjectured value of $\mathcal{S}$.
Consider the following basic integration formula
$$\int_{0}^{y}\mathrm{d}z\,z^{k-1}=\frac{y^{k}}{k};~~~\small{k\in\mathbb{N}\land y\in\mathbb{R}}.$$
It follows that
$$\frac{x^{k}}{k^{2}}=\int_{0}^{x}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,\frac{z^{k}}{yz};~~~\small{k\in\mathbb{N}\land x\in\mathbb{R}}.$$
For each $n\in\mathbb{N}$, we can then express $a_{n}$ as a double integral as follows:
$$\begin{align} a_{n} &=\sum_{k=1}^{n}\binom{2n}{n-k}\frac{1+(-1)^{k+1}}{k^{2}}\\ &=\sum_{k=1}^{n}\binom{2n}{n-k}\left[1+(-1)^{k+1}\right]\int_{0}^{1}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,\frac{z^{k}}{yz}\\ &=\int_{0}^{1}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,\sum_{k=1}^{n}\binom{2n}{n-k}\left[1+(-1)^{k+1}\right]y^{-1}z^{k-1}\\ &=\int_{0}^{1}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,y^{-1}\sum_{k=1}^{n}\binom{2n}{n-k}\left[z^{k-1}+\left(-z\right)^{k-1}\right]\\ &=\int_{0}^{1}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,y^{-1}\left[\sum_{k=1}^{n}\binom{2n}{n-k}z^{k-1}+\sum_{k=1}^{n}\binom{2n}{n-k}\left(-z\right)^{k-1}\right].\\ \end{align}$$
Recall the basic definition of the Gauss hypergeometric function ${_2F_1}$ for real parameters $(a,b,c)\in\mathbb{R}^{3}\land c\notin\mathbb{Z}_{\le0}$ and complex argument $z\in\mathbb{C}\land|z|\le1$ by the power series
$${_2F_1}{\left(a,b;c;z\right)}:=\sum_{n=0}^{\infty}\frac{\left(a\right)_{n}\,\left(b\right)_{n}\,z^{n}}{\left(c\right)_{n}\,n!};~~~\small{|z|<1\lor\left(|z|=1\land c-a-b>0\right)},$$
where the Pochhammer symbol $\left(r\right)_{n}$ is defined for $r\in\mathbb{R}$ and $n\in\mathbb{Z}_{\ge0}$ by
$$\left(r\right)_{n}= \begin{cases} 1;&n=0,\\ \prod_{k=1}^{n}\left(r+k-1\right);&n>0.\\ \end{cases}$$
The series terminates if $a$ is a nonpositive integer, in which case the Gauss hypergeometric function reduces to a polynomial:
$${_2F_1}{\left(-m,b;c;z\right)}:=\sum_{n=0}^{m}\left(-1\right)^{n}\binom{m}{n}\frac{\left(b\right)_{n}\,z^{n}}{\left(c\right)_{n}};~~~\small{m\in\mathbb{Z}_{\ge0}}.$$
For $(a,b,c)\in\mathbb{R}^{3}\land0<b<c$ and $z\in\mathbb{C}\setminus(1,\infty)$, the Gauss hypergeometric function can be expressed as an integral through Euler's integration formula:
$$\int_{0}^{1}\mathrm{d}t\,\frac{t^{b-1}\left(1-t\right)^{c-b-1}}{\left(1-zt\right)^{a}}=\operatorname{B}{\left(b,c-b\right)}\,{_2F_1}{\left(a,b;c;z\right)};~~~\small{z\neq1\lor\left(z=1\land c-a-b>0\right)}.$$
Using the machinery of hypergeometric functions, we can then show that for $n\in\mathbb{N}\land z\in\mathbb{C}\land|z|\le1$,
$$\begin{align} \sum_{k=1}^{n}\binom{2n}{n-k}z^{k-1} &=\sum_{k=1}^{n}\frac{(2n)!}{(n-k)!\,(n+k)!}\,z^{k-1}\\ &=\frac{(2n)!}{(n-1)!\,(n+1)!}\sum_{k=1}^{n}\frac{(n-1)!}{(k-1)!\,(n-k)!}\cdot\frac{(n+1)!\,(k-1)!\,z^{k-1}}{(n+k)!}\\ &=\binom{2n}{n-1}\sum_{k=1}^{n}\binom{n-1}{k-1}\frac{\left(1\right)_{k-1}\,z^{k-1}}{\left(n+2\right)_{k-1}}\\ &=\binom{2n}{n-1}\sum_{k=0}^{n-1}\binom{n-1}{k}\frac{\left(1\right)_{k}\,z^{k}}{\left(n+2\right)_{k}}\\ &=\binom{2n}{n-1}\,{_2F_1}{\left(1-n,1;n+2;-z\right)}\\ &=\left(n+1\right)\binom{2n}{n-1}\int_{0}^{1}\mathrm{d}t\,\left(1-t\right)^{n}\left(1+zt\right)^{n-1}\\ &=n\binom{2n}{n}\int_{0}^{1}\mathrm{d}t\,\left(1-t\right)^{n}\left(1+zt\right)^{n-1}.\\ \end{align}$$
For $n\in\mathbb{N}$, we can then write $a_{n}$ as
$$\begin{align} a_{n} &=\int_{0}^{1}\mathrm{d}y\int_{0}^{y}\mathrm{d}z\,y^{-1}\left[\sum_{k=1}^{n}\binom{2n}{n-k}z^{k-1}+\sum_{k=1}^{n}\binom{2n}{n-k}\left(-z\right)^{k-1}\right]\\ &=\int_{0}^{1}\mathrm{d}z\int_{z}^{1}\mathrm{d}y\,y^{-1}\left[\sum_{k=1}^{n}\binom{2n}{n-k}z^{k-1}+\sum_{k=1}^{n}\binom{2n}{n-k}\left(-z\right)^{k-1}\right]\\ &=\int_{0}^{1}\mathrm{d}z\,\left[-\ln{\left(z\right)}\right]\left[\sum_{k=1}^{n}\binom{2n}{n-k}z^{k-1}+\sum_{k=1}^{n}\binom{2n}{n-k}\left(-z\right)^{k-1}\right]\\ &=\int_{0}^{1}\mathrm{d}z\,\left[-\ln{\left(z\right)}\right]\\ &~~~~~\times\left[n\binom{2n}{n}\int_{0}^{1}\mathrm{d}t\,\left(1-t\right)^{n}\left(1+zt\right)^{n-1}+n\binom{2n}{n}\int_{0}^{1}\mathrm{d}t\,\left(1-t\right)^{n}\left(1-zt\right)^{n-1}\right]\\ &=n\binom{2n}{n}\int_{0}^{1}\mathrm{d}z\,\left[-\ln{\left(z\right)}\right]\int_{0}^{1}\mathrm{d}t\,\left(1-t\right)^{n}\left[\left(1+zt\right)^{n-1}+\left(1-zt\right)^{n-1}\right]\\ &=\binom{2n}{n}\int_{0}^{1}\mathrm{d}z\int_{0}^{1}\mathrm{d}t\,n\left(1-t\right)^{n}\left[\left(1+zt\right)^{n-1}+\left(1-zt\right)^{n-1}\right]\left[-\ln{\left(z\right)}\right]\\ &=\binom{2n}{n}\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}z\,n\left(1-t\right)^{n}\left[\left(1+zt\right)^{n-1}+\left(1-zt\right)^{n-1}\right]\left[-\ln{\left(z\right)}\right]\\ &=\binom{2n}{n}\int_{0}^{1}\mathrm{d}t\,\left(1-t\right)^{n}\left[-n\int_{0}^{1}\mathrm{d}z\,\left(1+zt\right)^{n-1}\ln{\left(z\right)}-n\int_{0}^{1}\mathrm{d}z\,\left(1-zt\right)^{n-1}\ln{\left(z\right)}\right]\\ &=\binom{2n}{n}\int_{0}^{1}\mathrm{d}t\,\left(1-t\right)^{n}\left[\int_{0}^{1}\mathrm{d}z\,\frac{\left(1+zt\right)^{n}-1}{zt}-\int_{0}^{1}\mathrm{d}z\,\frac{\left(1-zt\right)^{n}-1}{zt}\right];~~~\small{I.B.P.}\\ &=\binom{2n}{n}\int_{0}^{1}\mathrm{d}t\,\left(1-t\right)^{n}\int_{0}^{1}\mathrm{d}z\,\frac{\left(1+zt\right)^{n}-\left(1-zt\right)^{n}}{zt}.\\ \end{align}$$
Define the auxiliary function $f{(x)}$ by the series
$$f{\left(x\right)}:=\sum_{n=1}^{\infty}\frac{\binom{2n}{n}\,x^{n}}{\left(n+1\right)4^{n}};~~~\small{|x|\le1}.$$
For $|x|\le1$, we find
$$\begin{align} x\left[1+f{\left(x\right)}\right] &=x\left[1+\sum_{n=1}^{\infty}\frac{\binom{2n}{n}\,x^{n}}{\left(n+1\right)4^{n}}\right]\\ &=x+x\sum_{n=1}^{\infty}\frac{\binom{2n}{n}\,x^{n}}{\left(n+1\right)4^{n}}\\ &=x\sum_{n=0}^{\infty}\frac{\binom{2n}{n}\,x^{n}}{\left(n+1\right)4^{n}}\\ &=\sum_{n=0}^{\infty}\frac{\binom{2n}{n}\,x^{n+1}}{\left(n+1\right)4^{n}}\\ &=\sum_{n=0}^{\infty}\frac{\binom{2n}{n}}{4^{n}}\int_{0}^{x}\mathrm{d}y\,y^{n}\\ &=\sum_{n=0}^{\infty}\frac{1}{\pi}\operatorname{B}{\left(n+\frac12,\frac12\right)}\int_{0}^{x}\mathrm{d}y\,y^{n}\\ &=\frac{1}{\pi}\int_{0}^{x}\mathrm{d}y\,\sum_{n=0}^{\infty}y^{n}\operatorname{B}{\left(n+\frac12,\frac12\right)}\\ &=\frac{1}{\pi}\int_{0}^{x}\mathrm{d}y\,\sum_{n=0}^{\infty}y^{n}\int_{0}^{1}\mathrm{d}u\,u^{n-1/2}\left(1-u\right)^{-1/2}\\ &=\frac{1}{\pi}\int_{0}^{x}\mathrm{d}y\,\sum_{n=0}^{\infty}y^{n}\int_{0}^{1}\mathrm{d}u\,\frac{u^{n}}{\sqrt{u\left(1-u\right)}}\\ &=\frac{1}{\pi}\int_{0}^{x}\mathrm{d}y\int_{0}^{1}\mathrm{d}u\,\sum_{n=0}^{\infty}\frac{y^{n}u^{n}}{\sqrt{u\left(1-u\right)}}\\ &=\frac{1}{\pi}\int_{0}^{x}\mathrm{d}y\int_{0}^{1}\mathrm{d}u\,\frac{1}{\left(1-yu\right)\sqrt{u\left(1-u\right)}}\\ &=\int_{0}^{x}\mathrm{d}y\,\frac{1}{\sqrt{1-y}}\\ &=2-2\sqrt{1-x},\\ \end{align}$$
$$\implies f{\left(x\right)}=\frac{\left(1-\sqrt{1-x}\right)^{2}}{x}=\frac{1-\sqrt{1-x}}{1+\sqrt{1-x}}.$$
We can then write $\mathcal{S}$ as a double integral in the following way:
$$\begin{align} \mathcal{S} &=\sum_{n=1}^{\infty}\frac{a_{n}}{\left(n+1\right)4^{n}}\\ &=\sum_{n=1}^{\infty}\frac{\binom{2n}{n}}{\left(n+1\right)4^{n}}\int_{0}^{1}\mathrm{d}t\,\left(1-t\right)^{n}\int_{0}^{1}\mathrm{d}z\,\frac{\left(1+zt\right)^{n}-\left(1-zt\right)^{n}}{zt}\\ &=\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}z\,\sum_{n=1}^{\infty}\frac{\binom{2n}{n}}{\left(n+1\right)4^{n}}\cdot\frac{\left(1-t\right)^{n}\left(1+zt\right)^{n}-\left(1-t\right)^{n}\left(1-zt\right)^{n}}{zt}\\ &=\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}z\,\frac{f{\left(\left(1-t\right)\left(1+zt\right)\right)}-f{\left(\left(1-t\right)\left(1-zt\right)\right)}}{zt}\\ &=\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}z\,\frac{1}{zt}\left[\frac{1-\sqrt{1-\left(1-t\right)\left(1+zt\right)}}{1+\sqrt{1-\left(1-t\right)\left(1+zt\right)}}-\frac{1-\sqrt{1-\left(1-t\right)\left(1-zt\right)}}{1+\sqrt{1-\left(1-t\right)\left(1-zt\right)}}\right].\\ \end{align}$$
We can further reduce $\mathcal{S}$ to a single-variable integral as follows:
$$\begin{align} \mathcal{S} &=\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}z\,\frac{1}{zt}\left[\frac{1-\sqrt{1-\left(1-t\right)\left(1+zt\right)}}{1+\sqrt{1-\left(1-t\right)\left(1+zt\right)}}-\frac{1-\sqrt{1-\left(1-t\right)\left(1-zt\right)}}{1+\sqrt{1-\left(1-t\right)\left(1-zt\right)}}\right]\\ &=\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}z\,\frac{1}{tz}\left[\frac{1-\sqrt{t\left(1-\left(1-t\right)z\right)}}{1+\sqrt{t\left(1-\left(1-t\right)z\right)}}-\frac{1-\sqrt{t\left(1+\left(1-t\right)z\right)}}{1+\sqrt{t\left(1+\left(1-t\right)z\right)}}\right]\\ &=\int_{0}^{1}\mathrm{d}t\int_{0}^{1-t}\mathrm{d}y\,\frac{1}{ty}\left[\frac{1-\sqrt{t\left(1-y\right)}}{1+\sqrt{t\left(1-y\right)}}-\frac{1-\sqrt{t\left(1+y\right)}}{1+\sqrt{t\left(1+y\right)}}\right];~~~\small{\left[\left(1-t\right)z=y\right]}\\ &=\int_{0}^{1}\mathrm{d}y\int_{0}^{1-y}\mathrm{d}t\,\frac{1}{yt}\left[\frac{1-\sqrt{t\left(1-y\right)}}{1+\sqrt{t\left(1-y\right)}}-\frac{1-\sqrt{t\left(1+y\right)}}{1+\sqrt{t\left(1+y\right)}}\right]\\ &=\int_{0}^{1}\mathrm{d}y\int_{0}^{\sqrt{1-y}}\mathrm{d}u\,\frac{2}{yu}\left[\frac{1-u\sqrt{1-y}}{1+u\sqrt{1-y}}-\frac{1-u\sqrt{1+y}}{1+u\sqrt{1+y}}\right];~~~\small{\left[\sqrt{t}=u\right]}\\ &=\int_{0}^{1}\mathrm{d}y\,\frac{4}{y}\int_{0}^{\sqrt{1-y}}\mathrm{d}u\,\frac{1}{2u}\left[\frac{1-u\sqrt{1-y}}{1+u\sqrt{1-y}}-\frac{1-u\sqrt{1+y}}{1+u\sqrt{1+y}}\right]\\ &=\int_{0}^{1}\mathrm{d}y\,\frac{4}{y}\int_{0}^{\sqrt{1-y}}\mathrm{d}u\,\left[\frac{\sqrt{1+y}}{1+u\sqrt{1+y}}-\frac{\sqrt{1-y}}{1+u\sqrt{1-y}}\right]\\ &=\int_{0}^{1}\mathrm{d}y\,\frac{4}{y}\int_{0}^{\sqrt{1-y}}\mathrm{d}u\,\frac{d}{du}\ln{\left(\frac{1+u\sqrt{1+y}}{1+u\sqrt{1-y}}\right)}\\ &=\int_{0}^{1}\mathrm{d}y\,\frac{4}{y}\ln{\left(\frac{1+\sqrt{1-y^{2}}}{2-y}\right)}.\\ \end{align}$$
Then,
$$\begin{align} \mathcal{S} &=\int_{0}^{1}\mathrm{d}y\,\frac{4}{y}\ln{\left(\frac{1+\sqrt{1-y^{2}}}{2-y}\right)}\\ &=\int_{0}^{1}\mathrm{d}y\,\frac{4}{y}\ln{\left(\frac{1+\left(1+y\right)\sqrt{\frac{1-y}{1+y}}}{2-y}\right)}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{2}{\left(1+t\right)^{2}}\cdot\frac{4}{\left(\frac{1-t}{1+t}\right)}\ln{\left(\frac{1+\left(\frac{2}{1+t}\right)\sqrt{t}}{2-\left(\frac{1-t}{1+t}\right)}\right)};~~~\small{\left[y=\frac{1-t}{1+t}\right]}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{8}{1-t^{2}}\ln{\left(\frac{(1+t)+2\sqrt{t}}{2(1+t)-\left(1-t\right)}\right)}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{8}{1-t^{2}}\ln{\left(\frac{1+t+2\sqrt{t}}{1+3t}\right)}\\ &=\int_{0}^{1}\mathrm{d}u\,\frac{16u}{1-u^{4}}\ln{\left(\frac{1+2u+u^{2}}{1+3u^{2}}\right)};~~~\small{\left[\sqrt{t}=u\right]}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{2}{\left(1+x\right)^{2}}\cdot\frac{2\left(\frac{1-x}{1+x}\right)\left(1+x\right)^{4}}{x\left(1+x^{2}\right)}\ln{\left(\frac{1}{1-x+x^{2}}\right)};~~~\small{\left[u=\frac{1-x}{1+x}\right]}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{4\left(1-x^{2}\right)}{x\left(1+x^{2}\right)}\ln{\left(\frac{1}{1-x+x^{2}}\right)}\\ &=4\int_{0}^{1}\mathrm{d}x\,\left[\frac{1}{x}-\frac{2x}{1+x^{2}}\right]\ln{\left(\frac{1}{1-x+x^{2}}\right)}\\ &=4\int_{0}^{1}\mathrm{d}x\,\frac{1}{x}\ln{\left(\frac{1}{1-x+x^{2}}\right)}-4\int_{0}^{1}\mathrm{d}x\,\frac{2x}{1+x^{2}}\ln{\left(\frac{1}{1-x+x^{2}}\right)}\\ &=4\int_{0}^{1}\mathrm{d}x\,\frac{1}{x}\ln{\left(\frac{1+x}{1+x^{3}}\right)}+4\int_{0}^{1}\mathrm{d}x\,\frac{2x\ln{\left(1-x+x^{2}\right)}}{1+x^{2}}\\ &=4\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+x\right)}}{x}-4\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+x^{3}\right)}}{x}+4\int_{0}^{1}\mathrm{d}x\,\frac{2x\ln{\left(1-x+x^{2}\right)}}{1+x^{2}}\\ &=4\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+x\right)}}{x}-4\int_{0}^{1}\mathrm{d}u\,\frac{\ln{\left(1+u\right)}}{3u};~~~\small{\left[x^{3}=u\right]}\\ &~~~~~+4\int_{0}^{1}\mathrm{d}x\,\frac{2x\ln{\left(1-x+x^{2}\right)}}{1+x^{2}}\\ &=\frac83\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+x\right)}}{x}+4\int_{0}^{1}\mathrm{d}x\,\frac{2x\ln{\left(1-x+x^{2}\right)}}{1+x^{2}}\\ &=\frac43\zeta{\left(2\right)}+4\int_{0}^{1}\mathrm{d}x\,\frac{2x\ln{\left(1-x+x^{2}\right)}}{1+x^{2}}.\\ \end{align}$$
The integral we need can be evaluated using Feynman's method of differentiating under the integral sign.
Define the function $\mathcal{I}:\mathbb{R}\rightarrow\mathbb{R}$ via the definite integral
$$\mathcal{I}{\left(\alpha\right)}:=\int_{0}^{1}\mathrm{d}x\,\frac{2x\ln{\left(x^{2}-2x\cos{\left(\alpha\right)}+1\right)}}{x^{2}+1}.$$
By the fundamental theorem of calculus, we have
$$\mathcal{I}{\left(\alpha\right)}-\mathcal{I}{\left(\frac{\pi}{2}\right)}=\int_{\frac{\pi}{2}}^{\alpha}\mathrm{d}\varphi\,\mathcal{I}^{\prime}{\left(\varphi\right)};~~~\small{0<\alpha<\pi}.$$
Note that
$$\mathcal{I}{\left(\frac{\pi}{2}\right)}=\frac12\ln^{2}{\left(2\right)}.$$
For $\alpha\in(0,\pi)\land\alpha\neq\frac{\pi}{2}$, we find
$$\begin{align} \mathcal{I}{\left(\alpha\right)} &=\mathcal{I}{\left(\frac{\pi}{2}\right)}+\int_{\frac{\pi}{2}}^{\alpha}\mathrm{d}\varphi\,\mathcal{I}^{\prime}{\left(\varphi\right)}\\ &=\mathcal{I}{\left(\frac{\pi}{2}\right)}+\int_{\frac{\pi}{2}}^{\alpha}\mathrm{d}\varphi\,\frac{d}{d\varphi}\int_{0}^{1}\mathrm{d}x\,\frac{2x\ln{\left(x^{2}-2x\cos{\left(\varphi\right)}+1\right)}}{x^{2}+1}\\ &=\mathcal{I}{\left(\frac{\pi}{2}\right)}+\int_{\frac{\pi}{2}}^{\alpha}\mathrm{d}\varphi\int_{0}^{1}\mathrm{d}x\,\frac{\partial}{\partial\varphi}\frac{2x\ln{\left(x^{2}-2x\cos{\left(\varphi\right)}+1\right)}}{x^{2}+1}\\ &=\mathcal{I}{\left(\frac{\pi}{2}\right)}+\int_{\frac{\pi}{2}}^{\alpha}\mathrm{d}\varphi\int_{0}^{1}\mathrm{d}x\,\frac{4x^{2}\sin{\left(\varphi\right)}}{\left(x^{2}+1\right)\left[x^{2}-2x\cos{\left(\varphi\right)}+1\right]}\\ &=\mathcal{I}{\left(\frac{\pi}{2}\right)}+\int_{\frac{\pi}{2}}^{\alpha}\mathrm{d}\varphi\int_{0}^{1}\mathrm{d}x\,\tan{\left(\varphi\right)}\left[\frac{2x}{x^{2}-2x\cos{\left(\varphi\right)}+1}-\frac{2x}{x^{2}+1}\right]\\ &=\mathcal{I}{\left(\frac{\pi}{2}\right)}+\int_{\frac{\pi}{2}}^{\alpha}\mathrm{d}\varphi\,\tan{\left(\varphi\right)}\int_{0}^{1}\mathrm{d}x\,\left[\frac{2x-2\cos{\left(\varphi\right)}}{x^{2}-2x\cos{\left(\varphi\right)}+1}-\frac{2x}{x^{2}+1}+\frac{2\cos{\left(\varphi\right)}}{x^{2}-2x\cos{\left(\varphi\right)}+1}\right]\\ &=\mathcal{I}{\left(\frac{\pi}{2}\right)}+\int_{\frac{\pi}{2}}^{\alpha}\mathrm{d}\varphi\,\tan{\left(\varphi\right)}\left[\ln{\left(1-\cos{\left(\alpha\right)}\right)}+\int_{0}^{1}\mathrm{d}x\,\frac{2\cos{\left(\varphi\right)}}{x^{2}-2x\cos{\left(\varphi\right)}+1}\right]\\ &=\mathcal{I}{\left(\frac{\pi}{2}\right)}+\int_{\frac{\pi}{2}}^{\alpha}\mathrm{d}\varphi\,\left[\tan{\left(\varphi\right)}\ln{\left(1-\cos{\left(\alpha\right)}\right)}+\int_{0}^{1}\mathrm{d}x\,\frac{2\sin{\left(\varphi\right)}}{x^{2}-2x\cos{\left(\varphi\right)}+1}\right]\\ &=\mathcal{I}{\left(\frac{\pi}{2}\right)}+\int_{\frac{\pi}{2}}^{\alpha}\mathrm{d}\varphi\,\left[\tan{\left(\varphi\right)}\ln{\left(1-\cos{\left(\alpha\right)}\right)}+\left(\pi-\varphi\right)\right]\\ &=\mathcal{I}{\left(\frac{\pi}{2}\right)}+\left[\int_{\frac{\pi}{2}}^{\alpha}\mathrm{d}\varphi\,\left(\pi-\varphi\right)+\int_{\frac{\pi}{2}}^{\alpha}\mathrm{d}\varphi\,\tan{\left(\varphi\right)}\ln{\left(1-\cos{\left(\alpha\right)}\right)}\right]\\ &=\mathcal{I}{\left(\frac{\pi}{2}\right)}+\left[\frac{\pi^{2}}{8}-\frac12\left(\pi-\alpha\right)^{2}+\operatorname{Li}_{2}{\left(\cos{\left(\alpha\right)}\right)}\right]\\ &=\frac12\ln^{2}{\left(2\right)}+\frac{\pi^{2}}{8}-\frac12\left(\pi-\alpha\right)^{2}+\operatorname{Li}_{2}{\left(\cos{\left(\alpha\right)}\right)}.\\ \end{align}$$
In particular, we can use the special value of the dilogarithm at $\frac12$ to obtain the value of $\mathcal{I}$ at $\frac{\pi}{3}$:
$$\mathcal{I}{\left(\frac{\pi}{3}\right)}=\int_{0}^{1}\mathrm{d}x\,\frac{2x\ln{\left(x^{2}-x+1\right)}}{x^{2}+1}=-\frac{\pi^{2}}{72}=-\frac{1}{12}\zeta{\left(2\right)}.$$
Hence,
$$\mathcal{S}=\frac43\zeta{\left(2\right)}+4\int_{0}^{1}\mathrm{d}x\,\frac{2x\ln{\left(1-x+x^{2}\right)}}{1+x^{2}}=\zeta{\left(2\right)}.\blacksquare$$