This question is a follow-up to MSE#3142989.
Two seemingly innocent hypergeometric series ($\phantom{}_3 F_2$) $$ \sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^2\frac{(-1)^n}{2n+1}\qquad \sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^2\frac{1}{2^n(2n+1)}$$ can be reduced to the logarithmic integrals $$ \int_{0}^{1}\frac{\text{arctanh}(\sqrt{x})}{\sqrt{(1-x)(2-x)}}\,dx,\qquad\int_{0}^{1}\frac{\frac{1-x}{1+x}\log(x)}{\sqrt{x(1+x^2)}}\,dx\tag{A,B}$$ by FL-expansions and Fourier series, or directly by semi-integration by parts.
The issue is that neither (A) or (B) seem to be manageable through standard substitutions, the help of computer algebra systems or prayers to special deities, so I hope human beings are able to provide better insights. A promising substitution for (A) is $x=\frac{3-\cosh z}{2}$.
Long Comment: Notes on evaluating $I_A$:
I first found the equivalent integral $$I_A=2 \int_0^1 \frac{\sinh ^{-1}(x)}{x \sqrt{1-x^2}} \, dx\tag{1}$$ Integrating (1) by parts I then found $$I_A=2 \int_0^1 \frac{\log \left(\frac{\sqrt{1-x^2}+1}{x}\right)}{\sqrt{x^2+1}} \, dx\tag{2}$$
Then since $$\frac{1}{\sqrt{x^2+1}}=\sum _{n=0}^{\infty } \frac{(-1)^n \binom{2 n}{n}}{4^n}\, x^{2 n}$$
we obtain using (2) $$I_A=2 \sum _{n=0}^{\infty } \frac{(-1)^n \binom{2 n}{n}}{4^n}\, \int_0^1 x^{2 n}\log \left(\frac{\sqrt{1-x^2}+1}{x}\right) \, dx $$
Then using Mathematica
$$\int_0^1 x^{2 n} \log \left(\frac{\sqrt{1-x^2}+1}{x}\right) \, dx=\frac{\sqrt{\pi }\, \Gamma \left(n+\frac{3}{2}\right)}{(2 n+1)^2\, \Gamma (n+1)}$$
$$I_A=2 \sum _{n=0}^{\infty } \frac{(-1)^n \binom{2 n}{n}}{4^n}\,\frac{\sqrt{\pi }\, \Gamma \left(n+\frac{3}{2}\right)}{(2 n+1)^2\, \Gamma (n+1)}=\pi \, _3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};1,\frac{3}{2};-1\right)$$
UPDATE 23/03/2019
In thinking about Jack D'Aurizio's comment below: I think the relationship between the integrals and Euler sums can perhaps be found by proving and then making use of (3) and (4) $$\int_0^1 x^n \tanh ^{-1}\left(\sqrt{x}\right) \, dx=\frac{1}{n+1}\sum _{k=1}^{n+1} \frac{1}{2 k-1}\tag{3}$$
$$\int_0^1 x^{2n} \tanh ^{-1}\left(\sqrt{x}\right) \, dx=\frac{1}{2n+1}\sum _{k=1}^{2n+1} \frac{1}{2 k-1}\tag{4}$$
for Integrals A and B respectively.
In regard to integral A this looks difficult because it involves simplifying something like:
$$I_A=\frac{1}{\sqrt{2}} \int_0^1 \left(\sum _{n=0}^{\infty } \binom{2 n}{n} \frac{x^n}{4^n}\right) \left(\sum _{n=0}^{\infty } \binom{2 n}{n} \frac{x^n}{8^n} \right) \tanh ^{-1}\left(\sqrt{x}\right) \,dx$$
I get stuck trying to simplify the Cauchy Product.
The proof may be easier in regard to integral B since
$$\frac{1}{\sqrt{1-x^2}}=\sum _{n=0}^{\infty } \binom{2 n}{n} \frac{x^{2 n}} {4^n} $$
giving
$$I_B=\sqrt{2} \int_0^1\frac{ \tanh ^{-1}\left(\sqrt{x}\right)}{\sqrt{(1-x) (x+1)}}\; dx =\sqrt{2}\; \sum _{n=0}^{\infty } \frac{\binom{2 n}{n} \sum _{k=1}^{2 n+1} \frac{1}{2 k-1}}{4^n (2 n+1)}$$