While back, I was calculating a physics problem involving an integral of the form
$$\int\frac{dx}{\sqrt{x^2+a}+b}$$
Naturally, one would begin with the trig substitution $x=a\tan\theta$. So, the integral can be written as
\begin{align*} \int\frac{dx}{\sqrt{x^2+a}+b}&=\int\frac{a\sec^2\theta d\theta}{a\sec\theta+b}\\ &=\frac{1}{a}\int\frac{a^2\sec^2\theta-b^2+b^2}{a\sec\theta+b}d\theta\\ &=\int\sec\theta d\theta-\frac{b}{a}\theta+\frac{b^2}{a}\int\frac{d\theta}{a\sec\theta+b}\\ &=\ln|\sec\theta+\tan\theta|-\frac{b}{\sqrt{b^2-a^2}}\ln\frac{\sqrt{b^2-a^2}\tan\left(\frac{\theta}{2}\right)+a+b}{\sqrt{b^2-a^2}\tan\left(\frac{\theta}{2}\right)-(a+b)}\\ &=\ln\left|\frac{\sqrt{x^2+a^2}+x}{a}\right|-\frac{b}{\sqrt{b^2-a^2}}\ln\left[\frac{\frac{\sqrt{b^2-a^2}x}{a+\sqrt{x^2+a^2}}+a+b}{\frac{\sqrt{b^2-a^2}x}{a+\sqrt{x^2+a^2}}-(a+b)}\right] \end{align*}
Here,
\begin{align*} \int\frac{d\theta}{a\sec\theta+b}&=\int\frac{\cos\theta d\theta}{a+b\cos\theta}\\ &=\frac{1}{b}\int d\theta-\frac{a}{b}\int\frac{d\theta}{a+b\cos\theta}\\ &=\frac{1}{b}\theta-\frac{a}{b}\frac{1}{\sqrt{b^2-a^2}}\ln\frac{\sqrt{b^2-a^2}\tan\left(\frac{\theta}{2}\right)+a+b}{\sqrt{b^2-a^2}\tan\left(\frac{\theta}{2}\right)-(a+b)} \end{align*}
For the evaluation of $\int\frac{d\theta}{a+b\cos\theta}$, I used the formula from here.