We have

$$ I=\int \frac{\sec ^2 x}{(\sec x+\tan x)^{9 / 2}} d x $$

Let $\sec x+\tan x=t $

$\Rightarrow \sec x-\tan x=1 / t$.

$$ \Rightarrow $$ $\frac{1}{2}\left(t+\frac{1}{t}\right)=\sec x$

Now, $\left(\sec x \tan x+\sec ^2 x\right) d x=d t$

$$ \Rightarrow $$ $\sec x(\sec x+\tan x) d x=d t$

$$ \Rightarrow $$ $\sec x d x=\frac{d t}{t} $

$ I =\frac{1}{2} \int \frac{\left(t+\frac{1}{t}\right)}{t^{9 / 2}} \frac{d t}{t}$

$=\frac{1}{2} \int\left(t^{-9 / 2}+t^{-13 / 2}\right) d t $

$ =\frac{1}{2}\left[\frac{t^{-\frac{9}{2}+1}}{-\frac{9}{2}+1}+\frac{t^{-\frac{13}{2}+1}}{-\frac{13}{2}+1}\right] $

$=\frac{1}{2}\left[\frac{t^{-7 / 2}}{-\frac{7}{2}}+\frac{t^{-11 / 2}}{-\frac{11}{2}}\right]$

$=-\frac{1}{7} t^{-7 / 2}-\frac{1}{11} t^{-11 / 2}$

$=-\frac{1}{7} \frac{1}{t^{7 / 2}}-\frac{1}{11} \frac{1}{t^{11 / 2}}$

$=-\frac{1}{t^{11 / 2}}\left(\frac{1}{11}+\frac{t^2}{7}\right)$

$=-\frac{1}{(\sec x+\tan x)^{11 / 2}}\left\{\frac{1}{11}+\frac{1}{7}(\sec x+\tan x)^2\right\}+K$