To find the arclength of the spiral above from $0$ to $2\pi$, we use
$$s=\int_{0}^{2\pi}\sqrt{\theta^2+1}d\theta$$
After doing and initial trig substitution of $\theta=\tan{x}$ and the I.B.P., we get that
$$2\int_C \sec^3{x}dx=(\sec x \tan x+x)|_{\theta=0}^{\theta=2\pi}$$
Back substituting and dividing by 2 gives
$$\int_C \sec^3{x}dx=\frac{1}{2}\left[\theta\sqrt{\theta^2+1}+\arctan{\theta}\right]_0^{2\pi}=\pi\sqrt{4\pi^2+1}+\frac{1}{2}\arctan{2\pi}\approx 20.69412...$$
However, if I let Wolfram Alpha solve this it gives
$$s=\pi\sqrt{4\pi^2+1}+\frac{1}{2}\sinh^{-1}{2\pi}\approx 21.2563...$$
What gives??
$\displaystyle \int \sec^3 x ~ \mathrm dx \ne \sec x \tan x + x$