Evaluate $int_0^{pi/2} frac{arctan{left(frac{2sin{x}}{2cos{x}-1}right)}sin{left(frac{x}{2}right)}}{sqrt{cos{x}}} , mathrm{d}x$

$$boxed{I=int_0^frac{pi}{2}arctanleft(frac{2sin x}{2cos x -1}right)frac{sinleft(frac{x}{2}right)}{sqrt{cos x}}dx=sqrt 2 pi lnvarphi-frac{pi}{sqrt 2}ln(2+sqrt 3), varphi =frac{1+sqrt 5}{2}}$$ To show this result, we'll do some substitutions until it's clear how to simplify that $arctan $ term. $$I ,overset{cos x=t}=frac1{sqrt 2}int_0^1 arctan left(frac{2sqrt{1-t^2}}{2t-1}right)frac{dt}{sqrt tsqrt{1+t}}overset{large t=frac{1-x}{1+x}}=int_0^1 frac{arctanleft(frac{4sqrt x}{1-3x}right)}{sqrt{1-x}(1+x)}dx$$ $$=int_0^1frac{arctan (sqrt x)+arctan(3sqrt x)}{sqrt{1-x}(1+x)}dx-int_frac13^1frac{pi}{sqrt{1-x}(1+x)}dx=mathcal J-frac{pi}{sqrt2}ln(2+sqrt 3)$$ The second integral appears since for $x>frac13$ we have $sqrt x cdot 3sqrt x>1$.
Now in order to show $mathcal J=sqrt 2piln varphi$ we can differentiate under the integral sign, considering: $$mathcal J(a)=int_0^1frac{arctan(asqrt x)+arctan(3sqrt x)}{sqrt{1-x}(1+x)}dx$$ $$Rightarrow mathcal J'(a)=int_0^1 frac{sqrt x}{sqrt{1-x}(1+x)(1+a^2 x)}dxoverset{frac{1-x}{x}=t}=int_0^infty frac{dt}{sqrt t(2+t)(1+a^2+t)}$$ $$overset{t=x^2}=frac{2}{1-a^2}int_0^infty left(frac{1}{1+a^2+x^2}-frac{1}{2+x^2}right)dx=frac{pi}{1-a^2}left(frac{1}{sqrt{1+a^2}}-frac{1}{sqrt 2}right)$$ We are looking to find $mathcal J(1)=mathcal J$, but $mathcal J(-3)=0$ therefore our integral is: $$mathcal J=int_{-3}^1 frac{pi}{1-a^2}left(frac{1}{sqrt{1+a^2}}-frac{1}{sqrt 2}right)daoverset{-a=frac{1-x}{1+x}}=frac{pi}{2sqrt 2}int_0^2frac{1}{x}left(1-frac{1-x}{sqrt{1+x^2}}right)dx$$ $$=frac{pi}{2sqrt 2}left(lnleft(x+sqrt{1+x^2}right)+lnleft(1+sqrt{1+x^2}right)right)bigg|_0^2=sqrt 2 pi ln varphi$$