$\int_{0}^{\frac{\pi}{2}}\cos{x}\sqrt{1+\tan{x}}dx$ Find this integral
$$I=\int_{0}^{\frac{\pi}{2}}\cos{x}\sqrt{1+\tan{x}}dx$$
My try:I find this wolf can't find it. **
then I try: let 
$$\sqrt{1+\tan{x}}=t\Longrightarrow x=\arctan{(t^2-1)}$$
so
$$I=\int_{1}^{+\infty}\cos{\left(\arctan{(t^2-1)}\right)}\dfrac{2t^2}{(t^2-1)^2}dt$$
Then I can't
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$\ds{\large%
I \equiv \int_{0}^{\pi/2}\cos\pars{x}\root{1 + \tan\pars{x}}\,\dd x:\ {\large ?}}$

\begin{align}
I&=\int_{0}^{\pi/2}\root{\cos^{2}\pars{x} + \sin\pars{x}\cos\pars{x}}\,\dd x
=
\int_{0}^{\pi/2}\root{{1 + \cos\pars{2x} \over 2} + {\sin\pars{2x} \over 2}}\,\dd x
\\[3mm]&=
{1 \over 4}\,\root{2}\int_{0}^{\pi}\root{1 + \cos\pars{x} + \sin\pars{x}}\,\dd x
=
{1 \over 4}\,\root{2}
\int_{0}^{\pi}\root{1 + \root{2}\sin\pars{x + {\pi \over 4}}}\,\dd x
\\[3mm]&=
{1 \over 4}\,\root{2}
\int_{\pi/4}^{5\pi/4}\root{1 + \root{2}\sin\pars{x}}\,\dd x
\end{align}

The $\it\underline{last\ integral}$ is evaluated here in terms of a Second Kind Elliptic function.
A: This is an awful integral ! The formula for the antiderivative write in several pages. The numerical value is : 1.3571445175439115954095406.
A: As I wrote earlier, its value is non-elementary, and expressible in terms of the Meijer G function:

The expression of its anti-derivative can be found here. It involves incomplete elliptic integrals of the first and second kind.
