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.

Generally, consider the real function
$$f(x)=\frac{1}{a+b\cos x+c\sin x}$$
With $a^2>b^2+c^2$ so that $f$ is defined on $\mathbb{R}$.
It is not hard to check that the derivative of 
$$F(x)=\frac{x}{d}+\frac{2}{d}\arctan\left(\frac{c\cos x-b\sin x}{d+a+b\cos x+c\sin x}\right)$$
with $d=\sqrt{a^2-b^2-c^2}$, is $f(x)$. So $\int f(x)dx=F(x)+k$. The advantage
of this expression of $F$ is that it is also defined on $\mathbb{R}$. In particular,
$$\int \frac{1}{7+5\cos x+\sin x}=\frac{x}{\sqrt{23}}+\frac{2}{\sqrt{23}}\arctan\left(\frac{\cos x-5\sin x}{\sqrt{23}+7+5\cos x+\sin x}\right)$$
