Yes; it involves a couple of integrations by parts and some manipulation with trig identities.
Let $I=\int\sec^3\theta~d\theta$. Let $u=\sec\theta$ and $dv=\sec^2\theta~d\theta$; then $du=\sec\theta\tan\theta~d\theta$ and $v=\tan\theta$, so
$$I=\sec\theta\tan\theta-\int\sec\theta\tan^2\theta~d\theta\;.\tag{1}$$
Now $\tan^2\theta=\sec^2\theta-1$, so
$$\int\sec\theta\tan^2\theta~d\theta=\int\left(\sec^3\theta-\sec\theta\right)d\theta=I-\int\sec\theta~d\theta\;.\tag{2}$$
Combining $(1)$ and $(2)$ yields
$$I=\sec\theta\tan\theta-\left(I-\int\sec\theta~d\theta\right)=\sec\theta\tan\theta-I+\int\sec\theta~d\theta\;,$$ so
$$I=\frac12\left(\sec\theta\tan\theta+\int\sec\theta~d\theta\right)\;.$$
But $$\int\sec\theta~d\theta=\int\frac{\sec\theta(\sec\theta+\tan\theta)}{\sec\theta+\tan\theta}d\theta=\int\frac{d(\sec\theta+\tan\theta)}{\sec\theta+\tan\theta}=\ln(\sec\theta+\tan\theta)+C\;,$$ so finally
$$I=\frac12\sec\theta\tan\theta+\frac12\ln(\sec\theta+\tan\theta)+C\;.$$