Any hint for evaluating $\int_0^\theta \sec^2(\phi)\cdot\sqrt{\sec2\phi}\,d\phi\;$?

Question

How can I evaluate this integral? $$\int_0^\theta \sec^2(\phi)\cdot\sqrt{\sec2\phi}\,d\phi$$

I tried integral by parts using $u = \sqrt{\sec2\phi}$ and $dv =\sec^2(\phi) d\phi$ but ended up falling into a worse integral to solve

If anyone is curious about the origin of the problem, this integral appears when I tried to calculate the length of a hyperbola projected onto a sphere via stereographic projection.

In Wolfram alpha I can get the result but I'm curious how to do it, anyway below is the wolfram result, where $E(\phi,2)$ is the Elliptic Integral of the Second Kind with $k=2$ and $F(\phi,2)$ Elliptic Integral of the First Kind with $k=2$

$$\sqrt{\sec(2\phi)} \left(\sqrt{\cos(2\phi)} E(\phi,2) + \sqrt{\cos(\phi)} F(\phi,2) - \sin(2\phi) + \tan(\phi) \right)$$

Answer 1

The substitution $$\tan \phi = t, \qquad \sec^2 \!\phi \,d\phi = dt$$ transforms the definite integral to $$\int_0^{\tan \theta} \sqrt{\frac{1 + t^2}{1 - t^2}} \,dt = E(t \mid -1)\big\vert_0^{\tan \theta} = E(\tan \theta \mid -1) ,$$ where $E$ is the incomplete elliptic integral of the second kind (N.b. notation conventions for elliptic integrals vary.) The first equality follows from recognizing the integral as the Legendre normal form of $E$ for $k^2 = -1$.

Notice that for $\theta = \frac{\pi}{4}$ the (then improper) integral is equal to $E(i) = \sqrt{2} E\left(\frac{1}{\sqrt{2}}\right)$, where here $E(\,\cdot\,)$ denotes the complete elliptical integral of the second kind.

Answer 2

Using the same steps as @Travis Willse (but using Mathematica notations) $$f(\theta)= \int_0^\theta \sec^2(\phi)\,\sqrt{\sec2\phi}\,d\phi=E\left(\left.\sin ^{-1}(\tan (\theta ))\right|-1\right)$$

If you need it for calculations, you can have quite accurate results using its $[2n+1,2n]$ Padé approximant $P_n$ built around $\theta=0$.

Using for example

$$P_3=\theta\,\, \frac{1-\frac{63456821089 }{29809409916}\theta ^2+\frac{84897035101 }{74523524790}\theta^4-\frac{3354760809383 }{37559856494160}\theta ^6 } {1-\frac{27776587011 }{9936469972}\theta ^2+\frac{36309994555 }{14904704958}\theta ^4-\frac{3572751912491 }{5365693784880} \theta ^6}+O\left(\theta ^{15}\right)$$

$$\Phi=\int_0^{\frac \pi 4} \Big[E\left(\left.\sin ^{-1}(\tan (\theta ))\right|-1\right)-P_3 \Big]^2\,d\theta=1.129\times 10^{-4}$$ Using $P_4$ insteads leads to $\Phi= 2.353\times 10^{-5}$