Calculate an indefinite integral Given $\lambda \in [-1,1]$ is it possible to compute explicitly the following integral?
\begin{align*}
 \int \frac{1}{\sqrt{y^2(y^2 - 2) - \lambda^2(\lambda^2 - 2)}}\,dy
\end{align*}
I have tried to rewrite the above primitive as
\begin{align*}
 \int \frac{1}{\sqrt{(y^2 - 1)^2 - (\lambda^2 - 1)^2}}\,dy
\end{align*}
and apply trigonometric substitution. But, unless that I've got some mistake, it didn't work. Can someone help me with this? I thank you in advance.
 A: For $\lambda \neq 0$, this integral can be put into a Legendre normal form,  realizing it in terms of an (incomplete) Elliptic Integral of the First Kind,
$$F(x, k) := \int_0^x \frac{dt}{\sqrt{1 - t^2} \sqrt{1 - k^2 t^2}} .$$ Explicitly (for $|y| < |\lambda|$),
$$\int \frac{dy}{\sqrt{(y^2 - 1)^2 - (\lambda^2 - 1)^2}} = \frac{\sqrt{2 - \lambda^2}}{|\lambda|} F \left(\frac{y}{\sqrt{2 - \lambda^2}}, \frac{\sqrt{2 - \lambda^2}}{|\lambda|}\right) + C$$
For generic values of $\lambda$ this expression cannot be written in closed form in terms of elementary functions. For $\lambda = \pm 1$, however, the integral simplifies to
$$\int \frac{dy}{1 - y^2} = \operatorname{artanh} y + C$$ (for $|y| < 1$).
Finally, for $\lambda = 0$, the integral simplifies to
$$\int\frac{dy}{|y|\sqrt{y^2 - 2}} = -\frac{1}{\sqrt{2}}\operatorname{arccsc} \frac{y}{\sqrt{2}} + C .$$
A: There are two parts where the square root is defined, one being $y\in(-\lambda,\lambda)$.
At least for this interval you can set $y=\lambda\sin(t)$ and arrive to an elliptic integral:
$$\int\frac{dt}{\sqrt{\lambda^2\cos(t)^2+2(1-\lambda^2)}}=\frac 1{\sqrt{2-\lambda^2}}\int\frac{dt}{\sqrt{1-\frac{\lambda^2}{(2-\lambda^2)}\sin(t)^2}}$$
