The integral $\int \frac{{\sec }^{2}x}{{(\sec x+\tan x)}^{9/2}}dx$ equals (for some arbitrary constant  $K)$

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

Let $\sec x+\tan x=t$

Or $\sec x-\tan x=1/t$

Now,$(\sec x\tan x+{\sec }^{2}x)dx=dt$

Or $\sec x(\sec x+\tan x)dx=dt$

Or $\sec xdx=\frac{dt}{t},\frac{1}{2}(t+\frac{1}{t})=\sec x$

 $\therefore I=\frac{1}{2}\int \frac{(t+\frac{1}{t})}{t^{9/2}}\frac{dt}{t}$

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

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

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

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

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

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

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