Is It Improper Integral?

Relevant wiki: Differentiation Under the Integral Sign

Let $F(a)=\int_0^{\frac\pi 2}\frac{\arctan(a\tan x)}{\tan x}\ dx,a\ge 0$ we have

$\begin{aligned} F'(a)&=\int_0^{\frac\pi 2}\frac\partial{\partial a}\frac{\arctan(a\tan x)}{\tan x}\ dx&\small\color{#3D99F6}\text{Using differentiation under the integral sign} \\&=\int_0^{\frac\pi 2}\frac{dx}{1+a^2\tan^2x}&\small\color{#3D99F6}\text{Let }t=\tan x\Rightarrow dx=\frac {dt}{1+t^2} \\&=\int_0^\infty\frac{dt}{(1+a^2t^2)(1+t^2)} \\&=\frac 1{a^2-1}\int_0^\infty\left(\frac 1{t^2+\frac 1{a^2}}-\frac 1{t^2+1}\right)\ dt \\&=\frac 1{a^2-1}(a\arctan(ax)-\arctan x)\Big|_0^\infty \\&=\frac\pi{2(a+1)}. \end{aligned}$

Hence $F(a)=F(0)+\int_0^a\frac\pi{2(1+x)}\ dx=\frac{\pi\ln(1+a)}2.$ Set $a=2$ , we get $F(2)=\int_0^{\frac\pi 2}\frac{\arctan(2\tan x)}{\tan x}\ dx=\frac{\pi\ln 3}2,$ making the answer $3+2=\boxed 5$ .