calculus - Integrating $sec^5x$ using integration by parts

I'd like to know what I did wrong in my solution using this particular $u$ and $v$

$\int\sec^5x dx$

$u=\sec x$

$du=\sec x\tan x$

$v=\int \sec^4x dx = (\tan^3x)/3+\tan x$

After completing the integral, I ended up with $\int\sec^5x dx= \frac{\tan^3x\sec x}4+\frac{3\tan x\sec x}8+\frac{3\ln(\tan x+\sec x)}8+C$

Which is very close to the correct answer but not quite. I can't figure out where I went wrong at all.