\[\Large\displaystyle \int_0^\infty{_3F_2}\left(\begin{array}c\dfrac58,\dfrac58,\dfrac98\\\dfrac12,\dfrac{{13}}8\end{array}\middle|\ {-x}\right)^2\frac{\mathrm{d}x}{\sqrt x}
\]
\(\Large\mathbf{Solution:}\)
Looks like we have a more general result:
\[\begin{align*}
I(a)&=\int_0^\infty{_3F_2}\left(\begin{array}ca,a,a+\dfrac12\\\dfrac12,a+1\end{array}\middle|\ {-x}\right)^2\frac{\mathrm{d}x}{\sqrt x}\\
&=\frac{2^{4a-2}a^2}{\Gamma(2a)^2}G^{3~3}_{4~4}\left(1\middle|\begin{array}c\dfrac12,1,1;2a+\dfrac12\\2a-\dfrac12,2a-\dfrac12,2a;0\end{array}\right)\\
&=\frac{2^{4a-2}a^2}{\Gamma(2a)^2}G^{4~4}_{6~6}\left(1\middle|\begin{array}c\dfrac12,\dfrac12,1,1;2a,2a+\dfrac12\\2a-\dfrac12,2a-\dfrac12,2a,2a;0,\dfrac12\end{array}\right)\\
&=\frac{4\pi a^2}{\Gamma(2a)^2}G^{2~2}_{3~3}\left(1\middle|\begin{array}c1,1;4a\\4a-1,4a-1;0\end{array}\right)
\end{align*}\]
Here Mathematica gives when \(a=\dfrac58\),
\[\begin{align*}
&G^{2~2}_{3~3}\left(z\middle|\begin{array}c1,1;\dfrac52\\\dfrac32,\dfrac32;0\end{array}\right)\\
&=\frac{2\sqrt\pi}{3}\left(\ln(\sqrt z+\sqrt{z+1})+z^{3/2}\ln(1+\sqrt{z+1})-z^{3/2}\ln(\sqrt z)-\sqrt{z}\sqrt{z+1})\right)
\end{align*}\]
So
\[G^{2~2}_{3~3}\left(1\middle|\begin{array}c1,1;\dfrac52\\\dfrac32,\dfrac32;0\end{array}\right)=\frac{2\sqrt\pi}{3}(2\ln(1+\sqrt2)-\sqrt2)
\]
Now we have:
\[\large\boxed{\displaystyle \int_0^\infty{_3F_2}\left(\begin{array}c\dfrac58,\dfrac58,\dfrac98\\\dfrac12,\dfrac{{13}}8\end{array}\middle|\ {-x}\right)^2\frac{\mathrm{d}x}{\sqrt x}=\color{Blue} {\frac{50\,\pi^{3/2}}{3\,\Gamma^2\left(\dfrac14\right)}\Big(\ln\left(3+\sqrt8\right)-\sqrt2\Big)}}
\]