How is Wolfram Alpha and the reduction formula arriving at a different result for the integral of $\int \sec^4 x\,dx$ than naive $u$-substitution?

This is because you made a slight error in $$\frac13\tan^3x+\tan x+C=\frac13\tan x(\tan^2x+3)+C\ne\frac13\tan x(\tan^2x+1)+C.$$ Then you get \begin{align}\frac13\tan x(\tan^2x+3)+C&=\frac13\tan x(\tan^2 x+1)+\frac23\tan x+C\\&=\frac13\tan x\sec^2x+\frac23\tan x\cos^2x\sec^2x+C\\&=\frac13\tan x\sec^2x+\frac23\tan x\cdot\frac{1+\cos2x}2\sec^2x+C\\&=\frac13\tan x\sec^2x(2+\cos2x)\end{align} as given in W|A.

You made an arithmetic error. It should be $$ \frac13\tan^3 x+\tan x=\frac13\tan(x)(\tan^2x+3) $$ not $\frac13\tan(x)(\tan^2x+1)$.