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,
$$\mathcal{I}=\pi-\frac{\pi}{8}\left[\operatorname{B}{\left(\frac14,\frac12\right)}+\operatorname{B}{\left(\frac34,\frac12\right)}\right].\blacksquare$$
Best Answer
Finally I was able to compute $(2)$ via Beta function. $$I=\sqrt{3}\int_{0}^{\pi/2}\frac{\sqrt{3+\sqrt{\cos^2x+8}}}{\sqrt{8+\cos^2x}}dx \overset{x=\arccos{y}} = \\ \sqrt{3}\int_{0}^{1}\frac{1}{\sqrt{1-y^2}}\sqrt{\frac{1}{\sqrt{y^2+8}}+\frac{3}{y^2+8}}dy \overset{y=\sqrt{1-x^2}}= \\ \sqrt{3}\int_{0}^{1}\frac{\sqrt{3\sqrt{9-x^2}-x^2+9}}{\sqrt{1-x^2}(9-x^2)^{3/4}}dx \overset{x=3y}= \\ 3\int_{0}^{1/3}\frac{\sqrt{\sqrt{1-y^2}-y^2+1}}{\sqrt{1-9y^2}(1-y^2)^{3/4}}dy \overset{y=\sqrt{x}}= \\ \frac{3}{2}\int_{0}^{1/9}\frac{\sqrt{\sqrt{1-x}-x+1}}{\sqrt{x(1-9x)}(1-x)^{3/4}}dx \overset{x=1-y}= \\ \frac{3}{2}\int_{8/9}^{1}\frac{\sqrt{\sqrt{y}+1}}{\sqrt{y(1-y)(9y-8)}}dy \overset{y=x^2}=\\ 3\int_{\frac{2\sqrt{2}}{3}}^{1}\frac{dx}{\sqrt{1-x}\sqrt{9x^2-8}} \overset{x=\frac{y}{3}}= \\ \sqrt{3}\int_{2\sqrt{2}}^{3}\frac{dy}{\sqrt{3-y}\sqrt{y^2-8}} \overset{y=3-x}= \\ \sqrt{3}\int_{0}^{3-2\sqrt{2}}\frac{dx}{\sqrt{x}\sqrt{x^2-6x+1}} \overset{x=\frac{y-1}{y+1}}= \\ \sqrt{3}\int_{1}^{\sqrt{2}}\frac{dy}{\sqrt{2-y^2}\sqrt{y^2-1}} \overset{y=\sqrt{2}x}= \\ \sqrt{3}\int_{\frac{1}{\sqrt{2}}}^{1}\frac{dx}{\sqrt{1-x^2}\sqrt{2x^2-1}} \overset{x=\sqrt{y}}= \\ \frac{\sqrt{3}}{2}\int_{1/2}^{1}\frac{dy}{\sqrt{y}\sqrt{1-y}\sqrt{2y-1}} \overset{y=x+1/2}= \\ \frac{\sqrt{6}}{2}\int_{0}^{1/2}\frac{dx}{\sqrt{x}\sqrt{1-4x^2}} \overset{x=y/2}= \\ \frac{\sqrt{3}}{2}\int_{0}^{1}\frac{dy}{\sqrt{y}\sqrt{1-y^2}} \overset{y=x^2}= \\ \sqrt{3}\int_{0}^{1}\frac{dx}{\sqrt{1-x^4}}. $$ And it is well known that this last integral is evaluated using Beta function.