Integral $\int \frac{dx}{\tan x + \cot x + \csc x + \sec x}$ $$\int \frac{dx}{\tan x + \cot x + \csc x + \sec x}$$
$$\tan x + \cot x + \csc x + \sec x=\frac{\sin x + 1}{\cos x} +\frac{\cos x + 1}{\sin x} $$
$$= \frac{\sin x +\cos x +1}{\sin x \cos x}$$
$$t= \tan {\frac{x}{2}}$$
On solving ,
$$\frac{1}{\tan x + \cot x + \csc x + \sec x}=\frac{t(1- t)}{1+ t^2}$$
$$\implies \int \frac{\tan {\frac{x}{2}}(1-\tan {\frac{x}{2}})}{1+\tan^2 {\frac{x}{2}}}{dx}$$
I think, I have made the things more difficult. How can I proceed further? Is there any better substitution for it?

A trigonometric formula can be used：
\begin{align}
&(\sin x+\cos x+1)(\sin x+\cos x-1)=\sin 2x\\
I&=\int\frac{\sin x\cos x}{\sin x+\cos x+1}dx=\int\frac{1}{2}(\sin x+\cos x-1)dx=\frac{1}{2}(-\cos x+\sin x-x)+C\\
\end{align}
