Evaluate $\sum _{n=0}^{\infty } \frac{16^n}{(2n+1)^3 \binom{2 n}{n}^2}$ In this  post give the result of $$\sum _{n=1}^{\infty } \frac{16^n}{n^3 \binom{2 n}{n}^2}=8\pi\text{G}-14 \zeta (3)$$
Then i wonder if i change $n^3$ to $(2n+1)^3$ so what is the result. I check with CAS give the result: $$\sum _{n=0}^{\infty } \frac{16^n}{(2n+1)^3 \binom{2 n}{n}^2}=-\pi\text{G}+\frac{7}{2} \zeta (3)\tag1$$.
My attemp is: $\frac{2}{(2n+1)^3}=\int_{0}^{1}x^{2n}\log^{2}{x}dx$, then (1) become: $\sum _{n=0}^{\infty } \frac{16^n}{(2n+1)^3 \binom{2 n}{n}^2}=\frac{1}{2}\sum _{n=0}^{\infty } \frac{16^n}{\binom{2 n}{n}^2}\int_{0}^{1}x^{2n}\log^{2}{x}dx=\int_{0}^{1}\log^{2}{x}dx\sum _{n=0}^{\infty } x^{2n}\frac{16^n}{\binom{2 n}{n}^2}$. But the sum is seem harder. I don't have much experience with central binomial coefficient. Can you give some hints? Thank you.

Other than @Frank W's elegant solution, here is a different approach.
It is well-known that:
$\displaystyle \tag*{}\frac {\arcsin x}{\sqrt{1-x^2}}=\sum\limits_{n\ge 0}\frac {4^nx^{2n+1}}{(2n+1)\binom {2n}n}$
Dividing by $x$ and doing $\int_0^x \,dx$ on both sides gives:
$\displaystyle \tag{1}\int_0^x\frac {\arcsin x}{x\sqrt{1-x^2}}dx=\sum\limits_{n\ge 0}\frac {4^nx^{2n+1}}{(2n+1)^2\binom {2n}n}$
Now, we use the fact that
$$  \int_0^{\pi/2} \sin^{2n+1}(\theta) \, d\theta = \frac {4^n}{(2n+1)\binom {2n}n} $$
This tells us that if we let $x = \sin (\theta)$ and taking $\int_0^{\pi /2} \, d\theta$ both sides of $(1)$ gives our sum. Thus,
$$ \sum\limits_{n\ge 0}\frac {16^n}{(2n+1)^3\binom {2n}n} = \underbrace{\int_0^{\pi /2} \int_0^\theta \frac{x}{\sin x} \, dx \, d\theta}_{I}$$
Integrating by parts with $u=x$ and $v' = \log(\tan\frac x2)$, we get
$$I = \int_0^{\pi /2} \theta \log \tan \frac \theta 2 \, d\theta -\int_0^{\pi/2}\int_0^\theta \log \tan \frac x2 \, dx \, d\theta $$
Now, with the help of Fourier series, we have $\forall \ x \in (0, \pi)$
$$\begin{align} \tag{2} \log \sin \frac x2 &= -\log 2 - \sum_{n \ge 1} \frac{\cos nx}{n} \\ \tag{3}\log \cos \frac x2 &= -\log 2 + \sum_{n \ge 1} \frac{(-1)^{n-1}\cos nx}{n}\end{align}$$
Doing $(2)-(3)$ gives the series for $\log \tan \frac x2$. Now,plugging in the value, we get
$$ \begin{align}  I &= \frac \pi 2\sum_{n\ge 1} \frac{(-1)^n \sin (n \pi/2)}{n^2} - \underbrace{\sum_{n\ge 1} \frac{(-1)^n}{n^3}}_{\dfrac{-3\zeta(3)}{4}} - \frac \pi 2 \sum_{n \ge 1} \frac{\sin (n \pi /2)}{n^2} + \underbrace{\sum_{n \ge 1} \frac{1}{n^3}}_{\displaystyle \zeta(3)} + \underbrace{2\sum_{n \ge 1} \frac{1}{(2n+1)^3} + 2}_{\dfrac{7 \zeta(3)}{4}} \\\\ &= \frac{7 \zeta(3)}{2} - \frac \pi 2 \sum_{n \ge 1} \frac{\sin(n\pi /2)((-1)^n-1)}{n^2} \\ &=  \frac{7 \zeta(3)}{2} -  \pi \sum_{n \ge 1} \frac{(-1)^n}{(2n+1)^2} \\ &= {\color{Red}{\frac{7 \zeta(3)}{2} - \pi G}}\end{align}$$
