$\int \frac{{\sec }^{2}x}{{(\sec x+\tan x)}^5}dx=$

$$\begin{gathered} I=\int \frac{{\sec }^{2}x}{{(\sec x+\tan x)}^5}dx \\ put secx+\tan x=t then\left( secx\tan x+se{c}^{2}x\right)dx=dt \\ \Rightarrow secx dx=\frac{dt}{t} \end{gathered}$$$$\begin{gathered} \sin ce secx+\tan x=t \Rightarrow secx-\tan x=\frac{1}{t} \\ add the above equations we get 2secx =t+\frac{1}{t} \\ \Rightarrow secx=\frac{1}{2}\left(t+\frac{1}{t}\right) \end{gathered}$$
$$\begin{gathered} now I=\int \frac{\frac{1}{2}\left(t+\frac{1}{t}\right) {\displaystyle \frac{dt}{t}}}{t^5} \\ =\frac{1}{2}\int \left(\frac{1}{t^5}+\frac{1}{t^7}\right)dt=\frac{1}{2}\left[\frac{-1}{4t^4}-\frac{1}{6t^6}\right]+c \\ =\frac{-1}{4t^4}\left[\frac{1}{2}+\frac{1}{3t^2}\right]+c \end{gathered}$$

$=\frac{-1}{4{(\sec x+\tan x)}^4}\left[\frac{{\displaystyle 1}}{{\displaystyle 2}}+\frac{{\displaystyle 1}}{{\displaystyle 3{(\sec x+\tan x)}^2}}\right]+C$