The integral ∫sec²x(secx+tanx)⁹/²dx equals (for some arbitrary constant k)

I=∫sec²x(secx+tanx)⁹/²dx
Let secx+tanx=t ⇒secx−tanx=1/t
Now (secxtanx+sec²x)dx=dt
secx(secx+tanx)dx=dt
secxdx=dt/t, ½(t+1/t)=secx
I=½∫(t+1/t)t⁹/²dt
=½∫(t¹³/²+t⁷/²)dt
=½[t¹³/²/(13/2+1)+t⁷/²/(7/2+1)]
=⅓[t¹¹/²(11/7+t²/7)]
=⅓[t¹¹/²(111+7t²)]
=⅓(secx+tanx)¹¹/²(111+17(secx+tanx)²)+k