How to calculate the integral $\int \frac{a\tan^2{x}+b}{a^2\tan^2{x}+b^2} dx$?

Note that: $$\begin{align} \left(a+b\right)\left(a\tan^2{x}+b\right) & =a^2\tan^2{x}+b^2+ab\left(1+\tan^2{x} \right) \\[5pt] & = \color{blue}{a^2\tan^2{x}+b^2}+ab\sec^2x\end{align}$$ So we then have: $$\begin{align}\int \frac{a\tan^2{x}+b}{\color{blue}{a^2\tan^2{x}+b^2}} \,\mbox{d}x & = \frac{1}{a+b}\int\frac{\color{blue}{a^2\tan^2{x}+b^2}+ab\sec^2x}{\color{blue}{a^2\tan^2{x}+b^2}} \,\mbox{d}x \\[8pt] & = \frac{1}{a+b}\int\left( 1 + \frac{ab\sec^2x}{a^2\tan^2{x}+b^2} \right) \,\mbox{d}x \tag{$\star$}\\[8pt] & = \frac{1}{a+b}\Bigl( x+\arctan\left( \tfrac{a}{b}\tan x\right) \Bigr) +C \end{align}$$

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Where $(\star)$ is a standard integral, or follows after $t=\tfrac{a}{b}y$, here with $y=\tan x$: $$\int\frac{y(x)'}{a^2y(x)^2+b^2}\,\mbox{d}x = \frac{1}{ab}\arctan\left(\tfrac{a}{b}y\right)+C$$

$$I=\int \dfrac{a\tan^2(x)+b}{a^2\tan^2(x)+b^2}\, dx$$ Use $x=\tan ^{-1}(t)$ to make $$I=\int\frac{a t^2+b}{\left(t^2+1\right) \left(a^2 t^2+b^2\right)}\,dt$$ Use partial fraction decomposition $$\frac{a t^2+b}{\left(t^2+1\right) \left(a^2 t^2+b^2\right)}=\frac{a b}{(a+b) \left(a^2 t^2+b^2\right)}+\frac{1}{(a+b)\left(t^2+1\right) }$$ which looks quite pleasant.