Crazy $\int_0^\infty{_3F_2}\left(\begin{array}c\tfrac58,\tfrac58,\tfrac98\\\tfrac12,\tfrac{13}8\end{array}\middle|\ {-x}\right)^2\frac{dx}{\sqrt x}$

$$I=\frac{50\,\pi^{3/2}}{3\,\Gamma^2\left(\frac14\right)}\Big(\ln\left(3+\sqrt8\right)-\sqrt2\Big)$$

Looks like we have a more general result: $$I(a)=\int_0^\infty{_3F_2}\left(\begin{array}ca,a,a+\tfrac12\\\tfrac12,a+1\end{array}\middle|\ {-x}\right)^2\frac{dx}{\sqrt x}\\=\frac{2^{4a-2}a^2}{\Gamma(2a)^2}G^{3~3}_{4~4}\left(1\middle|\begin{array}c\tfrac12,1,1;2a+\tfrac12\\2a-\tfrac12,2a-\tfrac12,2a;0\end{array}\right)\\ =\frac{2^{4a-2}a^2}{\Gamma(2a)^2}G^{4~4}_{6~6}\left(1\middle|\begin{array}c\tfrac12,\tfrac12,1,1;2a,2a+\tfrac12\\2a-\tfrac12,2a-\tfrac12,2a,2a;0,\tfrac12\end{array}\right)\\=\frac{4\pi a^2}{\Gamma(2a)^2}G^{2~2}_{3~3}\left(1\middle|\begin{array}c1,1;4a\\4a-1,4a-1;0\end{array}\right).$$

To prove Cleo's form is correct, we need to prove that $$G^{2~2}_{3~3}\left(1\middle|\begin{array}c1,1;\tfrac52\\\tfrac32,\tfrac32;0\end{array}\right)\stackrel?=\frac{2\sqrt\pi}{3}(2\log(1+\sqrt2)-\sqrt2).$$

Here Mathematica gives $$G^{2~2}_{3~3}\left(z\middle|\begin{array}c1,1;\tfrac52\\\tfrac32,\tfrac32;0\end{array}\right)=\frac{2\sqrt\pi}{3}\left(\log(\sqrt z+\sqrt{z+1})+z^{3/2}\log(1+\sqrt{z+1})-z^{3/2}\log(\sqrt z)-\sqrt{z}\sqrt{z+1})\right).$$