A horrid-looking integral $\int_{0}^{5} \frac{\pi(1+\frac{1}{2+\sqrt{x}} )}{\sqrt{10}\sqrt{\sqrt{x}+x}} $ 
$$
\mathbf{\mbox{Evaluate:}}\qquad
\int_{0}^{5}  \frac{\pi(1+\frac{1}{2\sqrt{x}} )}{\sqrt{10}\sqrt{\sqrt{x}+x}} 
\,\,\mathrm{d}x
$$

This is a very ugly integral, but appears to have a very simple closed form of: $$\Gamma(\frac15)\Gamma(\frac45)$$ Mathematica can evaluate this integral, but WolframAlpha doesn't even give a correct numerical answer. I have tried many techniques on this integral but have not been able to crack it at all.
Any help on this integral would be greatly appreciated. Thank you!

Let $u = \sqrt{x} + x$.  Then we have
$$ \frac\pi{\sqrt{10}} \int_0^5 \frac{1 + \frac{1}{2\sqrt x}}{\sqrt{\sqrt x + x}} \, dx = \frac\pi{\sqrt{10}} \int_0^{\sqrt{5}+5} \frac{1}{\sqrt{u}} \, du$$
If you want to be pedantic (and who doesn't?!) then we need to note that this is an improper integral because the integrand is not defined at the lower limit of integration.  Therefore:
\begin{align}
  \frac\pi{\sqrt{10}} \int_0^{\sqrt{5}+5} \frac{1}{\sqrt{u}} \, du
    &= \lim_{B\to0^+}\frac\pi{\sqrt{10}} \int_B^{\sqrt{5}+5} \frac{1}{\sqrt{u}}\, du \\[0.3cm]
    &= \frac{2\pi}{\sqrt{10}} \lim_{B \to 0^+} \sqrt{u}\bigg|_B^{\sqrt5 + 5} \\[0.3cm]
    &= \frac{2\pi}{\sqrt{10}} \lim_{B\to0^+} \left(\sqrt{\sqrt5 + 5} - \sqrt B\right)\\[0.3cm]
    &= \frac{2\pi}{\sqrt{10}} \left(\sqrt{\sqrt5 + 5} - 0\right)\\[0.3cm]
    &= \frac{2\pi\sqrt{\sqrt5 + 5}}{\sqrt{10}}
\end{align}
