Trigonometric substitution for this tricky integral

Question

Whats the $$\int \sqrt{\frac{x^2+1}{x^2-1}}dx$$ after substituting $x=\sec(t)$ i get $$\int \sqrt{\frac{1}{(-(\cos^2(t)+\csc^2(t))}}.\sec(t)\tan(t)dt$$ i dont know how to proceed from here.

Answer 1

Let's first consider the case: $$|x|<1$$

Then the integral will be:

$$\int \sqrt{\frac{x^2+1}{x^2-1}}dx=-i \int \sqrt{\frac{1+x^2}{1-x^2}}dx$$

$i=\sqrt{-1}$

Incomplete elliptic integral of the second kind is:

$$E(x,k)=\int_0^x \frac{\sqrt{1-k^2 t^2}}{\sqrt{1-t^2}}dt$$

Thus, if we take $k=i$:

$$E(\phi,i)=\int_0^{\sin \phi} \frac{\sqrt{1+ t^2}}{\sqrt{1-t^2}}dt$$

Another notation uses the parameter $m=k^2$, so:

$$E(\phi~ |-1)=\int_0^{\sin \phi} \frac{\sqrt{1+ t^2}}{\sqrt{1-t^2}}dt$$

Thus:

$$\int \sqrt{\frac{x^2+1}{x^2-1}}dx=-i E(\arcsin(x)~ |-1)+C,~~~~~|x|<1$$

$C$ is an arbitrary constant.

As for the second case:

$$|x|>1$$

I'm not sure how to write the closed form, but we can probably use the following definition of $\arcsin(z)$ for all complex $z$:

$$\arcsin(z)=-i \log(i z +\sqrt{1-z^2})$$

WolframAlpha still gives the answer in terms of $E$ though.