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!
 A: 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}
A: *

*Substitute $x = u^2$. You end up with a denominator that contains $\sqrt{u + u^2}$. 

*Complete the square under the radical to $\sqrt{ u^2 + u + \frac{1}{4} - \frac{1}{4}} = \sqrt{ (u+\frac{1}{2})^2 - \frac{1}{4}}$.

*Substitute $z = 2(u+\frac{1}{2})$, so that $u + \frac{1}{2} = \frac{z}{2}$, and simplify. 
Then do a trig substitution on the resulting term under the radical to get rid of the radical, and you should be on your way. 
Post-comment addendum: 
To convert the result (say from @Hrhm's answer) to gamma-function form, use Euler's Reflection Formula, which tells you that $\Gamma(\frac{1}{5}) \Gamma(\frac{4}{5}) = \frac{\pi}{\sin \frac{
\pi}{5}}$.
A: I used the following site: 
http://www.integral-calculator.com/
First substitute $u=x+\sqrt{x}$ 
Note that $\displaystyle \frac{du}{dx}=\frac{1}{2\sqrt{x}}+1$
The integral then becomes $\displaystyle\frac{\pi}{\sqrt{10}}\int_{0}^{5+\sqrt{5}}\frac{du}{\sqrt{u}}$
This is equal to $\displaystyle \frac{2\pi\sqrt{u}}{\sqrt{10}}\Bigg|^{5+\sqrt{5}}_{0}=\frac{2\pi\sqrt{5+\sqrt{5}}}{\sqrt{10}}\approx 5.34$
A: $u=\sqrt{x}$, we have
$$
\int_{0}^{5}  \frac{\pi(1+\frac{1}{2\sqrt{x}} )}{\sqrt{10}\sqrt{\sqrt{x}+x}} 
\,\,\mathrm{d}x=\frac{\pi}{\sqrt{10}}\int_{0}^{\sqrt{5}}  \frac{2u+1} {\sqrt{u+u^2}}du=\frac{2\pi}{\sqrt{10}}\sqrt{u^2+u}\Big{|}_{0}^{\sqrt{5}}
$$
