Integral $\int \tan^{5}(x)\text{ d}x$

Short answer: the answers are equivalent, you have nothing to worry about. :)

Long answer: \begin{align} \frac{\tan^4 x}{4} - \frac{\tan^2 x}{2} &= \frac{\tan^4 x + 2\tan^2 x}{4} - \tan^2 x\\ &= \frac{\sec^4 x- 1}{4} - \tan^2 x\\ &= \frac{\sec^4x}{4} - \tan^2 x + C \end{align}

Your answer is correct. We have $tan^2x=sec^2x-1$ so $tan^4x=sec^4x-2sec^2x+1$ Replacing your $tan^4x$ accordingly gives $0.25sec^4x-0.5sec^2x+0.25$ Replacing that $0.25sec^2x$ with $tan^2x+1$ results into Stewart's solution. The answers do differ by some constant though