With some experiments, I discover an identity, which eventually evaluates to
$$
\,_3F_2\left ( \frac12,\frac12,1;\frac34,\frac34;-\frac18 \right )
=\frac23+\frac{\Gamma\left ( \frac34 \right )^2}{3\sqrt{\pi} }.
$$
This is special for the factor $-1/8$ and its simple result. I hope that there exist some ways we could attack it, and thank for your creative efforts.
Hypergeometric Function – Proving 3F2(1/2,1/2,1;3/4,3/4;-1/8)
closed-formhypergeometric functionsequences-and-series
Related Solutions
Thank you for posting this question, I enjoyed trying to answer it.
Start with the expression that Mathematica gave you and replace each argument $\frac14$ of a hypergeometric function with $\frac z4$, because we will be taking limits. I will call the two hypergeometric functions $Q_1(z)$ and $Q_2(z)$. Each term can be brought to a closed form by using identity 16.6.2 from the DLMF.
Setting $a=\frac12$, $b=1-\frac\nu2$, we get $$ Q_1(z) = (1-z)^{-\frac12} F\left(\frac16, \frac36,\frac 56; 1-\frac\nu2, 1+\frac\nu2; \frac{-27 z}{4(1-z)^3} \right), $$ and setting $a=\frac{1+3\nu}{2}$, $b=1+\frac\nu2$, we get $$ Q_2(z) = (1-z)^{-\frac{1+3\nu}{2}} F\left(\frac{1+3\nu}6, \frac{3+3\nu}{6}, \frac{5+3\nu}{6}; 1+\frac\nu2, 1+\nu; \frac{-27z}{4(1-z)^3} \right). $$ (Note that there are 6 possible identities to try per function, one for each possible choice of $a$ and $b$ from the parameters, so it helps to do this on a computer.)
The reason this works is that now the point $z=1$ is a singular point of the hypergeometric functions on the right hand side, and Mathematica will succeed in finding the limits as $z\to1$. The expression for the whole integral that you have is $$ Q = \frac{2^{\frac43}\pi^{\frac12}}{\Gamma(-\frac16)\Gamma(\frac56+\frac\nu2)\Gamma(\frac56-\frac\nu2)\sin\frac{\nu\pi}2} \left( -1 + 3^{-\frac{3\nu}2}\cos\left(\frac{\nu\pi}{2}\right) \frac{\Gamma(\frac{1+3\nu}{2}) \Gamma(\frac56-\frac\nu2)}{\Gamma(\frac{1+\nu}2)\Gamma(\frac56+\frac\nu2)} \right). $$ Call the large expression in brackets $A$, and then write $$ A = -1 + B 3^{-\frac{3\nu}{2}}\cos\frac{\pi\nu}{2}, \qquad B = \frac{\Gamma(\frac{1+3\nu}{2}) \Gamma(\frac56-\frac\nu2)}{\Gamma(\frac{1+\nu}2)\Gamma(\frac56+\frac\nu2)}. $$
Now, Mathematica will not simplify $A$ or $B$ on its own, so it needs help. Set $x=\frac16+\frac\nu2$, and use the multiplication formula to get $$ \frac{\Gamma(\frac{1+3\nu}2)}{\Gamma(\frac{1+\nu}{2})\Gamma(\frac56+\frac\nu2)} = \frac{\Gamma(3x)}{\Gamma(x+\frac13)\Gamma(x+\frac23)} = \frac{\Gamma(x)}{2\pi} 3^{3x-1/2}. $$ After this, $A$ simplifies to $$ A = -1 + \frac{\Gamma(\frac16+\frac\nu2)\Gamma(\frac56-\frac\nu2)}{2\pi}\cos\frac{\pi\nu}{2} = -1 + \frac{\cos\frac{\pi\nu}{2}}{2\sin(\frac\pi6+\frac{\pi\nu}{2})}, $$ where I've also used the reflection formula for $\Gamma(z)\Gamma(1-z)$ to get rid of the gamma functions. Some further amount of manual trigonometry yields $$ A = -\frac{\sqrt{3}}{2}\frac{\sin\frac{\pi\nu}{2}}{\sin(\frac\pi6 + \frac{\nu\pi}{2})}. $$
Finally, write $$ \frac{1}{\sin(\frac\pi6+\frac{\pi\nu}{2})} = \frac{\Gamma(\frac16+\frac\nu2)\Gamma(\frac56-\frac\nu2)}{\pi}, $$ and substitute back. Lots of things cancel, and the answer is $$ Q = -\frac{3^{1/2}2^{1/3}}{\pi^{1/2}} \frac{\Gamma(\frac16+\frac\nu2)}{\Gamma(-\frac16)\Gamma(\frac56+\frac\nu2)}. $$
This closed form is equivalent to the one you gave through the use of $\Gamma(\frac16)\Gamma(-\frac16)=-12\pi$.
P.S. I would also like to note that the integral $$ I(\nu,c) = \int_0^\infty J_\nu(x)^2 J_\nu(c x)\,dx $$ and its general form $$ \int_0^\infty x^{\rho-1}J_\nu(a x) J_\mu(b x) J_\lambda(c x)\,dx $$ appear in Gradshteyn and Ryzhik, and you can find a paper "Some infinite integrals involving bessel functions, I and II" by W. N. Bailey, which evaluates this integral in terms of Appell functions, but only in the case $c>2$ ($|c|>|a|+|b|$), which is where the $F_4$ Appell function converges. DLMF 16.16.6 actually gives a way to write this integral as $$ \frac{\Gamma(\frac{1+3\nu}{2})c^{-1-2\nu}}{\Gamma(1+\nu)^2\Gamma(\frac{1-\nu}{2})} \,\,\,{}_2F_1\left( \frac{1+\nu}{2}, \frac{1+3\nu}{2}; 1+\nu; x \right)^2, \qquad x = \frac{1-\sqrt{1-4/c^2}}{2}, $$ but the issue is that this is only correct for $c>2$, and the rhs is complex for $c<2$. Appell function would only be defined by analytic continuation in this case anyway, and I didn't find anything useful about non-principal branches of Appell or hypergeometric functions.
For $c>2$, Mathematica also gives the following: $$ I(\nu,c) = \frac{\Gamma(\frac{1+3\nu}{2})c^{-1-2\nu}}{\Gamma(\frac{1-\nu}{2})\Gamma(1+\nu)^2} \,\,\,{}_3F_2\left( \frac{1+\nu}{2}, \frac{1}{2}+\nu, \frac{1+3\nu}{2}; 1+\nu, 1+2\nu; \frac{4}{c^2} \right), $$ but this is incorrect when $c<2$.
$$I=\int_0^\infty \operatorname{actanh}\left(\frac{2x}{(1+x)^2}\right)\frac{dx}{x}=\frac12 \int_0^\infty \frac{\ln(1+4x+x^2)-\ln(1+x^2)}{x}dx$$ Now we will consider the following integral: $$I(a)=\frac12 \int_0^\infty \frac{\ln(1+ax+x^2)-\ln(1+x^2)}{x}dx\Rightarrow I'(a)=\frac12 \int_0^\infty \frac{dx}{1+ax+x^2}$$ $$=\frac{1}{\sqrt{4-a^2}}\arctan\left(\frac{a+2x}{\sqrt{4-a^2}}\right)\bigg|_0^\infty =\frac{1}{\sqrt{4-a^2}}\arctan\left(\frac{\sqrt{4-a^2}}{a}\right)$$ We have $I(0)=0$ and we are looking to find $I(4)$, then: $$I=\int_0^4 \frac{1}{\sqrt{4-a^2}}\arctan\left(\frac{\sqrt{4-a^2}}{a}\right)da=-\frac12 \arctan^2\left(\frac{\sqrt{4-a^2}}{a}\right)\bigg|_0^4$$ $$=\lim_{a\to 4}\frac12 \operatorname{arctanh}^2\left(\frac{\sqrt{a^2-4}}{a}\right)+\lim_{a\to 0}\frac12 \arctan^2\left(\frac{\sqrt{4-a^2}}{a}\right)$$ $$\Rightarrow \boxed{\int_0^\infty \operatorname{actanh}\left(\frac{2x}{(1+x)^2}\right)\frac{dx}{x}=\frac12 \ln^2(2+\sqrt 3)+\frac{\pi^2}{8}}$$
Another way to differentiate under the integral sign: $$I= \int_0^\infty \operatorname{arctanh} \left(\frac{2t}{(1+t)^2} \right)\frac{dt}{t}\overset{t=\tan \frac{x}{2}}=\int_0^\pi \operatorname{arctanh} \left(\frac{\sin x}{\sin x+1}\right)\frac{dx}{\sin x}$$ $$\operatorname{arctanh} x=\frac12 \ln\left(\frac{1+x}{1-x}\right)\Rightarrow I=\frac12 \int_0^\pi \frac{\ln(1+2\sin x)}{\sin x}dx$$ $$I(a)=\frac12 \int_{0}^\pi \frac{\ln(1+\sin a\sin x)}{\sin x}dx\Rightarrow I'(a)=\frac12 \int_0^{\pi}\frac{\cos a}{1+\sin a\sin x}dx$$ $$\overset{\tan \frac{x}{2}=t}=\int_0^\infty \frac{\cos a}{(t+\sin a)^2+\cos^2 a}dt=\arctan\left(\frac{t+\sin a}{\cos a}\right)\bigg|_0^\infty =\frac{\pi}{2}-a$$ $$I(0)=0\Rightarrow I(a)=\int_0^a \left(\frac{\pi}{2}-x\right)dx=\frac{a}{2}(\pi-a)$$ $$\Rightarrow \boxed{I=\frac{\arcsin 2}{2}(\pi-\arcsin 2)=\frac{\pi^2}{8}+\frac12 \ln^2(2+\sqrt 3)}$$
Best Answer
One can give a very short and elementary proof using WZ-pairs.
We have the following analytic fact: if $F,G$ satisfy $$\tag{WZ}F(n+1,k) - F(n,k) = G(n,k+1)-G(n,k)$$ and moreover
then $\lim_{n\to\infty} \sum_{k\geq 0} F(n,k)$ exists and is finite, also $$\tag{1}\sum_{k\geq 0} F(0,k) = \sum_{n\geq 0} G(n,0) + \lim_{n\to\infty} \sum_{k\geq 0} F(n,k)$$
Proof sketch: We apply $\sum_{n= 0}^{N-1} \sum_{k=0}^{K-1}$ on both sides of $(\text{WZ})$, giving $$\sum_{k=0}^{K-1} (F(N,k)-F(0,k)) = \sum_{n= 0}^{N-1} (G(n,K) - G(n,0))$$ by our assumption, we can let $K\to \infty$ while fixing $N$, to obtain $$\sum_{k\geq 0} F(N,k) - \sum_{k\geq 0} F(0,k) = -\sum_{n= 0}^{N-1} G(n,0)$$ then letting $N\to\infty$ proves the claim.
Now let $$F(n,k) = \frac{2^{-2 k-3 n-2} \Gamma (n+1) \Gamma \left(2 k+n-\frac{1}{2}\right)}{\Gamma \left(-n-\frac{1}{2}\right) \Gamma \left(n+\frac{3}{4}\right) \Gamma \left(k+n+\frac{3}{4}\right) \Gamma (k+n+1)}\quad G(n,k) = \frac{2(2 k+3 n+1)}{4 n+3}F(n,k)$$ one checks $(\text{WZ})$ is satisfied. All three bullets are also met, and $\lim_{n\to\infty} \sum_{k\geq 0} F(n,k) = 0$. Therefore we have $$\sum_{k\geq 0} F(0,k) = \sum_{n\geq 0} G(n,0)$$ explicitly: $$\sum_{k\geq 0} \frac{2^{-2 k-1} \left(\frac{1}{2}\right)_{2 k}}{(4 k-1) \left(\frac{3}{4}\right)_k (1)_k} = \sum_{n\geq 0} \frac{(-1/8)^n (2 n+1) (3 n+1) (1/2)_n^2}{(2 n-1) (4 n+3) (3/4)_n^2}$$
LHS can be summed using Gauss's $_2F_1$ theorem, giving $-\frac{\Gamma(3/4)^2}{2 \sqrt{\pi }}$. The summand of RHS is
$$-\Delta_n\left(\frac{(-1/8)^n(4 n-1) (1/2)_n^2}{(2 n-1) (3/4)_n^2}\right)-\frac{3}{2}\frac{(-1/8)^n (1/2)_n^2}{(3/4)_n^2}$$
here $\Delta_n f(n) = f(n+1)-f(n)$. From this we can get the value of $\sum_{n\geq 0} \frac{(-1/8)^n (1/2)_n^2}{(3/4)_n^2}$, which is OP's $_3F_2$ series.