Evaluating $\int\frac{1}{5\cos x+\sin x+7}~dx$ Evaluating  $$\int\frac{1}{5\cos x+\sin x+7}~dx.$$
This can be done by substituting $$\sin x = \frac{2t}{1 + t^2}$$ and $$\cos x = \frac{1 - t^2}{1 + t^2}.$$
However after I substitute it I cannot simplify it to get anything easier to integrate.
After substituting I got: integral $$\frac{1 + t^2}{2(t + 2)(t + 3)}$$ or $$\frac{1 + t^2}{12 + 2t^2 + 10t}.$$
Could someone give me a hint?
Many thanks.

Let $t = x - \tan^{-1}\frac15$ to integrate
\begin{align}
& \int\frac{1}{5\cos x+\sin x+7}~dx 
= \int \frac1{\sqrt{26}\cos t +7}dt\\
=& \int \frac{2d(\tan\frac t2)}{(7+\sqrt{26})+(7-\sqrt{26})\tan^2\frac t2}\\
=& \frac2{\sqrt{23}}\tan^{-1} \left( \sqrt{\frac{ 7-\sqrt{26}}{7+\sqrt{26}}}\tan\frac t2 \right)+C
\end{align}
