Finding $\int\frac{x^2-1}{\sqrt{x^4+x^2+1}}$ Finding
$$\int\frac{x^2-1}{\sqrt{x^4+x^2+1}}$$
Try: I have tried it to convert it into $\displaystyle \left(x+\frac{1}{x}\right)=t$ and $\displaystyle \left(1-\frac{1}{x^2}\right)$  but nothing happen.
I felt that it must be in Elliptical Integral of first kind, second kind or in third  kind.
Can someone explain me how to write in elliptical form

The other answers to this question use complex numbers. Here is a "real-only" answer.
Let us first work out
$$\int_y^\infty\frac{x^2-1}{\sqrt{x^4+x^2+1}}\,dx\qquad y\ge0$$
"But wait a minute, isn't this always infinite?" Yes, but the integral from $0$ to $y$ (the generally implied bounds of the indefinite integral) can be obtained by a simple subtraction once we work out the antiderivative, so there's no problem.
The denominator quartic can be written as $(x^2+z^2)(x^2+\overline z^2)$ where $z=e^{i\pi/6}$. Thus Byrd and Friedman 225.09 applies:
$$\int_y^\infty\frac{x^2-1}{\sqrt{x^4+x^2+1}}\,dx=\frac12\int_0^{u_1}\left(\frac{1+\operatorname{cn}u}{1-\operatorname{cn}u}-1\right)\,du=\int_0^{u_1}\frac{\operatorname{cn}u}{1-\operatorname{cn}u}\,du$$
where $u_1=F(\varphi,m)$, $m=\frac14$ is the parameter and $\operatorname{cn}u_1=\cos\varphi=\frac{y^2-1}{y^2+1}$. B&F 361.61 provides the solution to this second integral:
$$\int_0^{u_1}\frac{\operatorname{cn}u}{1-\operatorname{cn}u}\,du=-E(\varphi,m)-\frac{\operatorname{sn}u_1\operatorname{dn}u_1}{1-\operatorname{cn}u_1}$$
We can work out $\operatorname{sn}u_1$ and $\operatorname{dn}u_1$ using the known value of $\operatorname{cn}u_1$ and the fundamental relations between Jacobian elliptic functions, yielding
$$\int_y^\infty\frac{x^2-1}{\sqrt{x^4+x^2+1}}\,dx=-E\left(\varphi,\frac14\right)-\frac{y\sqrt{y^4+y^2+1}}{y^2+1}=J(y)$$
The integral from $0$ to $y$ is then $J(0)-J(y)$ and $J(0)=-2E(1/4)$ (a complete integral), so
$$\int_0^y\frac{x^2-1}{\sqrt{x^4+x^2+1}}\,dx=-2E\left(\frac14\right)+E\left(\cos^{-1}\frac{y^2-1}{y^2+1},\frac14\right)+\frac{y\sqrt{y^4+y^2+1}}{y^2+1}$$
The quasiperiodicity of $E$ allows us to absorb $2E(1/4)$. Applying a few trigonometric identities then allows us to simplify the amplitude, in the process extending the domain of validity to all real numbers:
$$\int_0^y\frac{x^2-1}{\sqrt{x^4+x^2+1}}\,dx=E\left(\cos^{-1}\frac{y^2-1}{y^2+1}-\pi,\frac14\right)+\frac{y\sqrt{y^4+y^2+1}}{y^2+1}$$
$$\color{red}{\int_0^y\frac{x^2-1}{\sqrt{x^4+x^2+1}}\,dx=\frac{y\sqrt{y^4+y^2+1}}{y^2+1}-E\left(2\tan^{-1}y,\frac14\right)}$$
