<s>[INST] Evaluating $\int \frac{\sec^2 x}{(\sec x + \tan x )^{{9}/{2}}}\,\mathrm dx$ How do I evaluate
$$\int \frac{\sec^2 x}{(\sec x + \tan x )^{{9}/{2}}}\,\mathrm dx$$
I've started doing this problem by taking $u=\tan x$.
 [/INST] Given $$\displaystyle \int\frac{\sec^2 x}{(\sec x+\tan x)^{\frac{9}{2}}}dx = \frac{1}{2}\int\frac{\sec x(\sec x+\tan x)+\sec x(\sec x-\tan x)}{(\sec x+\tan x)^{\frac{9}{2}}}dx$$
$$\displaystyle = \frac{1}{2}\int\frac{\sec x\cdot (\sec x+\tan x)}{(\sec x+\tan x)^{\frac{9}{2}}}dx+\frac{1}{2}\int\frac{\sec x\cdot (\sec x-\tan x)}{(\sec x+\tan x)^{\frac{9}{2}}}dx$$
Now Use $\sec^2x-\tan^2 x= 1$ For Second Integral, We Get
$$\displaystyle = \frac{1}{2}\int\frac{\sec x\cdot (\sec x+\tan x)}{(\sec x+\tan x)^{\frac{9}{2}}}dx+\frac{1}{2}\int\frac{\sec x\cdot (\sec x+\tan x)}{(\sec x+\tan x)^{\frac{13}{2}}}dx$$
Now For both  Integral , Let $\sec x+\tan x=t\;,$ Then $\sec x(\sec x+\tan x)dx = dt$
We Get $$\displaystyle = \frac{1}{2}\int\frac{1}{t^{\frac{9}{2}}}dt+\frac{1}{2}\int\frac{1}{t^{\frac{13}{2}}}dt = -\frac{1}{2}\cdot \frac{2}{7}t^{-\frac{7}{2}}-\frac{1}{2}\cdot \frac{2}{11}t^{-\frac{11}{2}}+\mathcal{C}$$
So We Get $$\displaystyle \int\frac{\sec^2 x}{(\sec x+\tan x)^{\frac{9}{2}}}dx = -\frac{1}{7}(\sec x+\tan x)^{-\frac{9}{2}}-\frac{1}{11}(\sec x+\tan x)^{-\frac{11}{2}}+\mathcal{C}$$
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