Let $I$ denote the integral and consider the substitution $u= \frac{\pi }{2} - x.$ Then $I = \displaystyle\int_0^{\frac{\pi }{2}} \frac{\sqrt{\cos u}}{\sqrt{\cos u } + \sqrt{\sin u }} du$ and $2I = \displaystyle\int_0^{\frac{\pi }{2}} \frac{\sqrt{\cos u} + \sqrt{\sin u }}{\sqrt{\cos u } + \sqrt{\sin u }} du = \frac{\pi }{2}.$ Hence $I = \frac{\pi }{4}.$
In general, $ \displaystyle\int_0^a f(x) dx = \displaystyle\int_0^a f(a-x) $ $dx$ whenever $f$ is integrable, and $\displaystyle\int_0^{\frac{\pi }{2}} \frac{\cos^a x}{\cos^a x + \sin^a x } dx = \displaystyle\int_0^{\frac{\pi }{2}} \frac{\sin^a x}{\cos^a x + \sin^a x } dx = \frac{\pi }{4}$ for $a>0$ (same trick.)
The general method of attack on an integral like this is to recognize that the $x^2$ could be ignored for the time being upon expressing the integral in terms of a suitable parameter which may then be twice differentiated. To wit, after a little manipulatin, I find:
$$\displaystyle \int_{0}^{\frac{\pi}{2}} dx \: {x}^2 \sqrt{\tan x} \sin \left( 2x \right) = 2 \Im{ \int_{0}^{\frac{\pi}{2}} dx \: x^2 e^{i x} \sqrt{\sin{x} \cos{x}}} $$
So consider the following integral:
$$J(\alpha) = \int_{0}^{\frac{\pi}{2}} dx \: e^{i \alpha x} \sqrt{\sin{x} \cos{x}} $$
Note that the integral we seek is
$$\displaystyle \int_{0}^{\frac{\pi}{2}} dx \: {x}^2 \sqrt{\tan x} \sin \left( 2x \right) = -2 \Im{\left [ \frac{\partial^2}{\partial \alpha^2} J(\alpha) \right ]_{\alpha = 1}} $$
It turns out that there is, in fact, a closed for for $J(\alpha)$:
$$J(\alpha) = -\frac{\left(\frac{1}{16}+\frac{i}{16}\right) \sqrt{\frac{\pi }{2}} \left ( e^{\frac{i \pi a}{2}}-i\right) \Gamma \left(\frac{a-1}{4}\right)}{\Gamma \left ( \frac{a+5}{4} \right )} $$
Plugging this into the above expression, you will find terms including polylogs and harmonic numbers, which I will spare you unless explicitly asked for. But this is how you would get a closed-form expression for your integral.
EDIT
The integral is even nicer when we consider
$$J(\alpha) = \int_{0}^{\frac{\pi}{2}} dx \: \sin{\alpha x} \sqrt{\sin{x} \cos{x}} $$
Then
$$J(\alpha) = \frac{\pi ^{3/2} \sin \left(\frac{\pi a}{4}\right)}{8 \Gamma
\left(\frac{5-a}{4}\right) \Gamma \left(\frac{5+a}{4}\right)}$$
and
$$\left [ \frac{\partial^2}{\partial \alpha^2} J(\alpha) \right ]_{\alpha = 1} = \frac{\pi \left(-5 \pi ^2+6 \pi (\log (4)-2)+3 (8+(\log (4)-4) \log
(4))\right)}{192 \sqrt{2}}$$
The integral we seek is then
$$\int_0^{\frac{\pi}{2}} dx \: x^2 \sqrt{\tan{x}} \sin{(2 x)} = -2 \left [ \frac{\partial^2}{\partial \alpha^2} J(\alpha) \right ]_{\alpha = 1} = \frac{\pi \left(-24+12 \pi +5 \pi ^2-3 \log ^2(4)+12 \log (4)-6 \pi \log
(4)\right)}{96 \sqrt{2}}$$
It turns out that the numerical value of the latter value is about $1.10577$, which agrees with the numerical approximation of the integral mentioned by @mrf and verified in Mathematica.
Best Answer
Some algebra to arrive at a manageable integral
Notice that the $\arccos $ argument is the "$+$" solution of a quadratic equation:
$$\cos x\left(2\sin^2x\pm\sqrt{1+4\sin^4x}\right)=\frac{2\sin x \sin(2x)\pm\sqrt{4\sin^2 x\sin^2(2x)+4\cos^2 x}}{2}$$
More exactly, with $b=-2\sin x\sin(2x)$ and $c=-\cos^2x$ in $y_{\pm}=\frac{-b\pm\sqrt{b^2-4c}}{2}$.
We also have $\arccos y_-\pm\arccos y_+=\arccos\left(y_-y_+\mp\sqrt{1-(y_-+y_+)^2 +2y_-y_+ +y_-^2y_+^2}\right)$, and since Vieta's relations allows us to utilize $y_-+y_+=2\sin x\sin(2x)$ and $\ y_-y_+=-\cos^2 x$, it makes sense to consider the following integrals:
$$I_{\pm}=\int_{\arccos(1/4)}^{\pi/2}\arccos\left(\cos x\left(2\sin^2x\pm\sqrt{1+4\sin^4x}\right)\right) dx$$
The original integral can then be rewritten as $I_+=\frac12\left((I_- + I_+)-(I_- - I_+)\right)$, where:
$$I_- \pm I_+=\int_{\arccos(1/4)}^{\pi/2}\arccos\left(-\cos^2 x\mp\sin^2 x\sqrt{1-16\cos^2 x}\right)dx$$
$$\overset{\sqrt{1-16\cos^2 x}\to x}=\int_0^1 \frac{x\arccos \left(-\frac{1-x^2}{16}\mp x\frac{15+x^2}{16}\right)}{\sqrt{1-x^2}{\sqrt{15+x^2}}}dx$$
Moreover, substituting $x\to -x$ in $I_- - I_+$, gives us:
$$I_+=\frac12\int_{-1}^1 \frac{x\arccos \left(-\frac{1-x^2}{16}- x\frac{15+x^2}{16}\right)}{\sqrt{1-x^2}{\sqrt{15+x^2}}}dx=\int_{-1}^1 \frac{x\operatorname{arctan}\sqrt{\frac{1-x^2}{15+x^2} \left(1+\frac{16}{(1-x)^2}\right)}}{\sqrt{1-x^2}{\sqrt{15+x^2}}}dx$$
In the last step it was utilized that $\arccos x =2\arctan \sqrt{\frac{1-x}{1+x}}$.
Evaluation of the integral
$$\int_{-1}^1 \frac{x\operatorname{arctan}\sqrt{\frac{1-x^2}{15+x^2} \left(1+\frac{16}{(1-x)^2}\right)}}{\sqrt{1-x^2}{\sqrt{15+x^2}}}dx=\int_{-1}^1 \int_0^{\sqrt{1+\frac{16}{(1-x)^2}}} \frac{x}{(15+x^2)+(1-x^2) y^2}dy dx$$
$$\overset{(*)}=16\int_{-1}^1 \int_0^{x} \frac{x}{(15+x^2)+(1-x^2) \left(1+\frac{16}{(1-y)^2}\right)}\frac{1}{(1-y)^2\sqrt{16+(1-y)^2}} dy dx$$
$$=16\int_{-1}^1 \frac{1}{(1-y)^2\sqrt{16+(1-y)^2}}\int_y^1 \frac{x}{(15+x^2)+\left(1+\frac{16}{(1-y)^2}\right)(1-x^2)} dx dy$$
$$=-\frac12\int_{-1}^1 \frac{\ln\left(\frac{1-y}{2}\right)}{\sqrt{16+(1-y)^2}}dy\overset{\frac{1-y}{2}\to y}=-\frac12 \int_0^1 \frac{\ln y}{\sqrt{4+y^2}}dy$$
$$\overset{y \to \frac{1-y^2}{y}}=\frac12 \int_{\large \frac{1}{\phi}}^1 \frac{\ln\left(\frac{y}{1-y^2}\right)}{y}dy =\frac14\operatorname{Li}_2(1) - \frac14\operatorname{Li}_2\left(\frac{1}{\phi^2}\right) - \frac{1}{4}\ln^2\phi \overset{(**)}=\, \boxed{\frac{\pi^2}{40}}$$
In $(**)$, the following dilogarithm values were employed:
$$\operatorname{Li}_2(1)=\frac{\pi^2}{6},\quad \operatorname{Li}_2\left(\frac{1}{\phi^2}\right) = \frac{\pi^2}{15}-\ln^2\phi,\quad \phi=\frac{1+\sqrt 5}{2}$$
Also, the $(*)$ step is not necessary, however I didn't know how to change the order of integration directly, so I had to make the substitution $y^2\to 1+\frac{16}{(1-y)^2}$ after noticing that the $x$-integral is odd which allows to write $\int_{-1}^1 \int_0^{f(x)}dydx = -\int_{-1}^1 \int_{f(x)}^\infty dydx$. Finally, $-\int_{-1}^1 \int_x^1 dydx$ was also rewritten as $\int_{-1}^1 \int_0^x dydx$ in order to change the order of integration easier.