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

The integral ∫ sec²x/((sec x+tan x)^(9/2)) dx equals (for some arbitrary constant K) $I = \displaystyle\int \frac{\sec^{2}x}{(\sec x + \tan x)^{9/2}}\,dx$ $\dfrac{d}{dx}(\sec x) = \sec x \tan x, \qquad \dfrac{d}{dx}(\tan x) = \sec^{2}x$ $\text{Evaluate } I \text{ and match one of the options (1)–(4)}$ $t = \sec x + \tan x$ $\dfrac{dt}{dx} = \sec x\tan x + \sec^{2}x = \sec x(\sec x + \tan x) = t\sec x$ $dt = t\sec x\,dx \Rightarrow \sec x\,dx = \dfrac{dt}{t}$ $\sec x = \dfrac{1}{2}\left(t + \dfrac{1}{t}\right)$ $\sec^{2}x\,dx = \sec x\,(\sec x\,dx) = \dfrac{1}{2}\left(t+\dfrac{1}{t}\right)\dfrac{dt}{t} = \dfrac{1}{2}\left(1 + \dfrac{1}{t^{2}}\right)dt$ $I = \dfrac{1}{2}\int \left(t^{-9/2} + t^{-13/2}\right)dt$ $I = \dfrac{1}{2}\left[-\dfrac{2}{7}t^{-7/2} - \dfrac{2}{11}t^{-11/2}\right] + K$ $= -\dfrac{1}{7}t^{-7/2} - \dfrac{1}{11}t^{-11/2} + K$ $I = -t^{-11/2}\left(\dfrac{1}{11} + \dfrac{1}{7}t^{2}\right) + K = -\dfrac{1}{t^{11/2}}\left\{\dfrac{1}{11} + \dfrac{1}{7}t^{2}\right\} + K$ $I = -\dfrac{1}{(\sec x+\tan x)^{11/2}}\left\{\dfrac{1}{11} + \dfrac{1}{7}(\sec x+\tan x)^{2}\right\} + K$ $I = -\dfrac{1}{(\sec x+\tan x)^{11/2}}\left\{\dfrac{1}{11} + \dfrac{1}{7}(\sec x+\tan x)^{2}\right\} + K \ \Rightarrow (3)$