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.
My approach:
Let
$$I(t) = \int_{0}^{\infty} \frac{xt - \sin(xt)}{x^3\left(x^2 + 4\right)} \:dx$$
Where $I = I(1)$
Taking the first derivative:
$$ \frac{dI}{dt} = \int_{0}^{\infty} \frac{x - x\cos(xt)}{x^3\left(x^2 + 4\right)} \:dx = \int_{0}^{\infty} \frac{1 - \cos(xt)}{x^2\left(x^2 + 4\right)} \:dx$$
Taking the second derivative:
$$ \frac{d^2I}{dt^2} = \int_{0}^{\infty} \frac{x\sin(xt)}{x^2\left(x^2 + 4\right)} \:dx = \int_{0}^{\infty} \frac{\sin(xt)}{x\left(x^2 + 4\right)} \:dx$$
Now, take the Laplace Transform w.r.t $t$:
\begin{align}
\mathscr{L}\left[ \frac{d^2I}{dt^2}\right] &= \int_{0}^{\infty} \frac{\mathscr{L}\left[\sin(xt)\right]}{x\left(x^2 + 4\right)} \:dx \\
&= \int_{0}^{\infty} \frac{x}{\left(s^2 + x^2\right)x\left(x^2 + 4\right)}\:dx \\
&= \int_{0}^{\infty} \frac{1}{\left(s^2 + x^2\right)\left(x^2 + 4\right)}\:dx
\end{align}
Applying the Partial Fraction Decomposition we may find the integral
\begin{align}
\mathscr{L}\left[ \frac{d^2I}{dt^2}\right] &= \int_{0}^{\infty} \frac{1}{\left(s^2 + x^2\right)\left(x^2 + 4\right)}\:dx \\
&= \frac{1}{s^2 - 4} \int_{0}^{\infty} \left[\frac{1}{x^2 + 4} - \frac{1}{x^2 + s^2} \right]\:dx \\
&= \frac{1}{s^2 - 4} \left[\frac{1}{2}\arctan\left(\frac{x}{2}\right) - \frac{1}{s}\arctan\left(\frac{x}{s}\right)\right]_{0}^{\infty} \\
&= \frac{1}{s^2 - 4} \left[\frac{1}{2}\frac{\pi}{2} - \frac{1}{s}\frac{\pi}{2} \right] \\
&= \frac{\pi}{4s\left(s + 2\right)}
\end{align}
We now take the inverse Laplace Transform:
$$ \frac{d^2I}{dt^2} = \mathscr{L}^{-1}\left[\frac{\pi}{4s\left(s + 2\right)} \right] = \frac{\pi}{8}\left(1 - e^{-2t} \right) $$
We now integrate with respect to $t$:
$$ \frac{dI}{dt} = \int \frac{\pi}{8}\left(1 - e^{-2t} \right)\:dt = \frac{\pi}{8}\left(t + \frac{e^{-2t}}{2} \right) + C_1$$
Now
$$ \frac{dI}{dt}(0) = \int_{0}^{\infty} \frac{1 - \cos(x\cdot 0)}{x^2\left(x^2 + 4\right)} \:dx = 0 = \frac{\pi}{8}\left(\frac{1}{2} \right) + C_1 \rightarrow C_1 = -\frac{\pi}{16}$$
Thus,
$$ \frac{dI}{dt} = \int \frac{\pi}{8}\left(1 - e^{-2t} \right)\:dt = \frac{\pi}{8}\left(t + \frac{e^{-2t}}{2} \right) - \frac{\pi}{16}$$
We now integrate again w.r.t $t$
$$ I(t) = \int \left[\frac{\pi}{8}\left(t + \frac{e^{-2t}}{2} \right) - \frac{\pi}{16} \right] \:dt = \frac{\pi}{8}\left(\frac{t^2}{2} - \frac{e^{-2t}}{4} \right) - \frac{\pi}{16}t + C_2 $$
Now
$$I(0) = \int_{0}^{\infty} \frac{x\cdot0 - \sin(x\cdot0)}{x^3\left(x^2 + 4\right)} \:dx = 0 = \frac{\pi}{8}\left( -\frac{1}{4} \right) + C_2 \rightarrow C_2 = \frac{\pi}{32}$$
And so we arrive at our expression for $I(t)$
$$I(t)= \frac{\pi}{8}\left(\frac{t^2}{2} - \frac{e^{-2t}}{4} \right) - \frac{\pi}{16}t + \frac{\pi}{32}$$
Thus,
$$I = I(1) = \frac{\pi}{8}\left(\frac{1}{2} - \frac{e^{-2}}{4} \right) - \frac{\pi}{16} + \frac{\pi}{32} = \frac{\pi}{32}\left(1 - e^{-2}\right)$$
Best Answer
Let $\mathcal{I}$ denote the value of the definite integral in question:
$$\mathcal{I}:=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,\frac{\theta+\sin{\left(\theta\right)}\ln{\left(\tan{\left(\frac{\theta}{2}\right)}\right)}}{\left[1+\sin{\left(\theta\right)}\right]\sqrt{\sin{\left(\theta\right)}}}\approx0.141216.$$
$$\begin{align} \mathcal{I} &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\,\frac{\theta+\sin{\left(\theta\right)}\ln{\left(\tan{\left(\frac{\theta}{2}\right)}\right)}}{\left[1+\sin{\left(\theta\right)}\right]\sqrt{\sin{\left(\theta\right)}}}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{2}{1+t^{2}}\cdot\frac{2\arctan{\left(t\right)}+\left(\frac{2t}{1+t^{2}}\right)\ln{\left(t\right)}}{\left(1+\frac{2t}{1+t^{2}}\right)\sqrt{\frac{2t}{1+t^{2}}}};~~~\small{\left[\theta=2\arctan{\left(t\right)}\right]}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{2\left[2\arctan{\left(t\right)}+\left(\frac{2t}{1+t^{2}}\right)\ln{\left(t\right)}\right]}{\left(1+t\right)^{2}\sqrt{\frac{2t}{1+t^{2}}}}\\ &=\int_{0}^{1}\mathrm{d}u\,\frac{2\arctan{\left(\frac{1-u}{1+u}\right)}+\left(\frac{1-u^{2}}{1+u^{2}}\right)\ln{\left(\frac{1-u}{1+u}\right)}}{\sqrt{\frac{1-u^{2}}{1+u^{2}}}};~~~\small{\left[t=\frac{1-u}{1+u}\right]}\\ &=\int_{0}^{1}\mathrm{d}u\,\frac{2\left[\arctan{\left(1\right)}-\arctan{\left(u\right)}\right]-2\left(\frac{1-u^{2}}{1+u^{2}}\right)\operatorname{artanh}{\left(u\right)}}{\sqrt{\frac{1-u^{2}}{1+u^{2}}}}\\ &=2\int_{0}^{1}\mathrm{d}u\,\frac{\arctan{\left(1\right)}}{\sqrt{\frac{1-u^{2}}{1+u^{2}}}}-2\int_{0}^{1}\mathrm{d}u\,\frac{\arctan{\left(u\right)}}{\sqrt{\frac{1-u^{2}}{1+u^{2}}}}-2\int_{0}^{1}\mathrm{d}u\,\sqrt{\frac{1-u^{2}}{1+u^{2}}}\operatorname{artanh}{\left(u\right)}\\ &=\frac{\pi}{2}\int_{0}^{1}\mathrm{d}u\,\frac{1+u^{2}}{\sqrt{1-u^{4}}}-2\int_{0}^{1}\mathrm{d}u\,\sqrt{\frac{1+u^{2}}{1-u^{2}}}\arctan{\left(u\right)}\\ &~~~~~-2\int_{0}^{1}\mathrm{d}u\,\sqrt{\frac{1-u^{2}}{1+u^{2}}}\operatorname{artanh}{\left(u\right)}\\ &=\frac{\pi}{2}\int_{0}^{1}\mathrm{d}x\,\frac{1+\sqrt{x}}{4x^{3/4}\sqrt{1-x}};~~~\small{\left[u=\sqrt[4]{x}\right]}\\ &~~~~~-2\int_{0}^{1}\mathrm{d}u\,\left[\sqrt{\frac{1+u^{2}}{1-u^{2}}}\arctan{\left(u\right)}+\sqrt{\frac{1-u^{2}}{1+u^{2}}}\operatorname{artanh}{\left(u\right)}\right]\\ &=\frac{\pi}{8}\left[\int_{0}^{1}\mathrm{d}x\,\frac{1}{x^{3/4}\sqrt{1-x}}+\int_{0}^{1}\mathrm{d}x\,\frac{1}{\sqrt[4]{x}\sqrt{1-x}}\right]\\ &~~~~~-2\int_{0}^{1}\mathrm{d}u\,\left[\sqrt{\frac{1+u^{2}}{1-u^{2}}}\int_{0}^{1}\mathrm{d}x\,\frac{u}{1+x^{2}u^{2}}+\sqrt{\frac{1-u^{2}}{1+u^{2}}}\int_{0}^{1}\mathrm{d}x\,\frac{u}{1-x^{2}u^{2}}\right]\\ &=\frac{\pi}{8}\left[\int_{0}^{1}\mathrm{d}x\,x^{-3/4}\left(1-x\right)^{-1/2}+\int_{0}^{1}\mathrm{d}x\,x^{-1/4}\left(1-x\right)^{-1/2}\right]\\ &~~~~~-\int_{0}^{1}\mathrm{d}u\int_{0}^{1}\mathrm{d}x\,(2u)\left[\frac{1}{1+x^{2}u^{2}}\sqrt{\frac{1+u^{2}}{1-u^{2}}}+\frac{1}{1-x^{2}u^{2}}\sqrt{\frac{1-u^{2}}{1+u^{2}}}\right]\\ &=\frac{\pi}{8}\left[\operatorname{B}{\left(\frac14,\frac12\right)}+\operatorname{B}{\left(\frac34,\frac12\right)}\right]\\ &~~~~~-\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}u\,(2u)\left[\frac{1}{1+x^{2}u^{2}}\sqrt{\frac{1+u^{2}}{1-u^{2}}}+\frac{1}{1-x^{2}u^{2}}\sqrt{\frac{1-u^{2}}{1+u^{2}}}\right]\\ &=\frac{\pi}{8}\operatorname{B}{\left(\frac14,\frac12\right)}+\frac{\pi}{8}\operatorname{B}{\left(\frac34,\frac12\right)}\\ &~~~~~-\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}v\,\left[\frac{1}{1+x^{2}v}\sqrt{\frac{1+v}{1-v}}+\frac{1}{1-x^{2}v}\sqrt{\frac{1-v}{1+v}}\right];~~~\small{\left[u^{2}=v\right]}\\ &=\frac{\pi}{8}\operatorname{B}{\left(\frac14,\frac12\right)}+\frac{\pi}{8}\operatorname{B}{\left(\frac34,\frac12\right)}-\int_{0}^{1}\mathrm{d}x\int_{1}^{0}\mathrm{d}w\,\frac{(-2)}{\left(1+w\right)^{2}}\\ &~~~~~\times\left[\frac{1}{1+x^{2}\left(\frac{1-w}{1+w}\right)}\sqrt{\frac{1}{w}}+\frac{1}{1-x^{2}\left(\frac{1-w}{1+w}\right)}\sqrt{w}\right];~~~\small{\left[v=\frac{1-w}{1+w}\right]}\\ &=\frac{\pi}{8}\operatorname{B}{\left(\frac14,\frac12\right)}+\frac{\pi}{8}\operatorname{B}{\left(\frac34,\frac12\right)}-\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}w\,\frac{2}{\left(1+w\right)\sqrt{w}}\\ &~~~~~\times\left[\frac{1}{\left(1+w\right)+x^{2}\left(1-w\right)}+\frac{w}{\left(1+w\right)-x^{2}\left(1-w\right)}\right],\\ \end{align}$$
and then,
$$\begin{align} \mathcal{I} &=\frac{\pi}{8}\operatorname{B}{\left(\frac14,\frac12\right)}+\frac{\pi}{8}\operatorname{B}{\left(\frac34,\frac12\right)}-\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}w\,\frac{2}{\left(1+w\right)\sqrt{w}}\\ &~~~~~\times\left[\frac{1}{\left(1+x^{2}\right)+\left(1-x^{2}\right)w}+\frac{w}{\left(1-x^{2}\right)+\left(1+x^{2}\right)w}\right]\\ &=\frac{\pi}{8}\operatorname{B}{\left(\frac14,\frac12\right)}+\frac{\pi}{8}\operatorname{B}{\left(\frac34,\frac12\right)}\\ &~~~~~-\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}w\,\frac{1}{\left(1+x^{2}\right)}\left[\frac{1}{1+\left(\frac{1-x^{2}}{1+x^{2}}\right)w}+\frac{w}{\left(\frac{1-x^{2}}{1+x^{2}}\right)+w}\right]\frac{2}{\left(1+w\right)\sqrt{w}}\\ &=\frac{\pi}{8}\operatorname{B}{\left(\frac14,\frac12\right)}+\frac{\pi}{8}\operatorname{B}{\left(\frac34,\frac12\right)}\\ &~~~~~-\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}w\,\frac{1}{x^{2}}\left[\frac{2}{1+w}-\frac{\left(\frac{1-x^{2}}{1+x^{2}}\right)}{1+\left(\frac{1-x^{2}}{1+x^{2}}\right)w}-\frac{\left(\frac{1-x^{2}}{1+x^{2}}\right)}{\left(\frac{1-x^{2}}{1+x^{2}}\right)+w}\right]\frac{1}{\sqrt{w}}\\ &=\frac{\pi}{8}\operatorname{B}{\left(\frac14,\frac12\right)}+\frac{\pi}{8}\operatorname{B}{\left(\frac34,\frac12\right)}\\ &~~~~~-\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{2}{x^{2}}\left[\frac{2}{1+y^{2}}-\frac{\left(\frac{1-x^{2}}{1+x^{2}}\right)}{1+\left(\frac{1-x^{2}}{1+x^{2}}\right)y^{2}}-\frac{\left(\frac{1-x^{2}}{1+x^{2}}\right)}{\left(\frac{1-x^{2}}{1+x^{2}}\right)+y^{2}}\right];~~~\small{\left[\sqrt{w}=y\right]}\\ &=\frac{\pi}{8}\operatorname{B}{\left(\frac14,\frac12\right)}+\frac{\pi}{8}\operatorname{B}{\left(\frac34,\frac12\right)}\\ &~~~~~-\int_{0}^{1}\mathrm{d}x\,\frac{2}{x^{2}}\left[\int_{0}^{1}\mathrm{d}y\,\frac{2}{1+y^{2}}-\int_{0}^{1}\mathrm{d}y\,\frac{\left(\frac{1-x^{2}}{1+x^{2}}\right)}{1+\left(\frac{1-x^{2}}{1+x^{2}}\right)y^{2}}-\int_{0}^{1}\mathrm{d}y\,\frac{\left(\frac{1-x^{2}}{1+x^{2}}\right)}{\left(\frac{1-x^{2}}{1+x^{2}}\right)+y^{2}}\right]\\ &=\frac{\pi}{8}\operatorname{B}{\left(\frac14,\frac12\right)}+\frac{\pi}{8}\operatorname{B}{\left(\frac34,\frac12\right)}\\ &~~~~~-\int_{0}^{1}\mathrm{d}x\,\frac{2}{x^{2}}\left[\frac{\pi}{2}-\sqrt{\frac{1-x^{2}}{1+x^{2}}}\arctan{\left(\sqrt{\frac{1-x^{2}}{1+x^{2}}}\right)}-\sqrt{\frac{1-x^{2}}{1+x^{2}}}\arctan{\left(\frac{1}{\sqrt{\frac{1-x^{2}}{1+x^{2}}}}\right)}\right]\\ &=\frac{\pi}{8}\operatorname{B}{\left(\frac14,\frac12\right)}+\frac{\pi}{8}\operatorname{B}{\left(\frac34,\frac12\right)}-\int_{0}^{1}\mathrm{d}x\,\frac{2}{x^{2}}\left[\frac{\pi}{2}-\frac{\pi}{2}\sqrt{\frac{1-x^{2}}{1+x^{2}}}\right]\\ &=\frac{\pi}{8}\operatorname{B}{\left(\frac14,\frac12\right)}+\frac{\pi}{8}\operatorname{B}{\left(\frac34,\frac12\right)}-\pi\int_{0}^{1}\mathrm{d}x\,\frac{1}{x^{2}}\left[1-\sqrt{\frac{1-x^{2}}{1+x^{2}}}\right].\\ \end{align}$$
Evaluating the remaining integral by parts, we find
$$\begin{align} \int_{0}^{1}\mathrm{d}x\,\frac{1}{x^{2}}\left[1-\sqrt{\frac{1-x^{2}}{1+x^{2}}}\right] &=\int_{0}^{1}\mathrm{d}x\,\left[1-\sqrt{\frac{1-x^{2}}{1+x^{2}}}\right]\frac{d}{dx}\left[-\frac{1}{x}\right]\\ &=\left[-\frac{1-\sqrt{\frac{1-x^{2}}{1+x^{2}}}}{x}\right]_{x=0}^{x=1}-\int_{0}^{1}\mathrm{d}x\,\left[-\frac{1}{x}\right]\frac{d}{dx}\left[1-\sqrt{\frac{1-x^{2}}{1+x^{2}}}\right];~~~\small{I.B.P.}\\ &=-1+\int_{0}^{1}\mathrm{d}x\,\frac{1}{x}\cdot\frac{2x}{\left(1+x^{2}\right)^{2}\sqrt{\frac{1-x^{2}}{1+x^{2}}}}\\ &=-1+\int_{0}^{1}\mathrm{d}x\,\frac{2}{\left(1+x^{2}\right)\sqrt{1-x^{4}}}\\ &=-1+\int_{0}^{1}\mathrm{d}x\,\frac{1-2x^{2}-x^{4}}{\left(1+x^{2}\right)\sqrt{1-x^{4}}}+\int_{0}^{1}\mathrm{d}x\,\frac{1+x^{2}}{\sqrt{1-x^{4}}}\\ &=-1+\int_{0}^{1}\mathrm{d}x\,\frac{d}{dx}\left[x\sqrt{\frac{1-x^{2}}{1+x^{2}}}\right]+\int_{0}^{1}\mathrm{d}x\,\frac{1+x^{2}}{\sqrt{1-x^{4}}}\\ &=-1+\int_{0}^{1}\mathrm{d}x\,\frac{1+x^{2}}{\sqrt{1-x^{4}}}\\ &=-1+\frac14\left[\operatorname{B}{\left(\frac14,\frac12\right)}+\operatorname{B}{\left(\frac34,\frac12\right)}\right].\\ \end{align}$$
Hence,
$$\mathcal{I}=\pi-\frac{\pi}{8}\left[\operatorname{B}{\left(\frac14,\frac12\right)}+\operatorname{B}{\left(\frac34,\frac12\right)}\right].\blacksquare$$