<s>[INST] 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
 [/INST] 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:
$$\int_0^y\frac{x^2-1}{\sqrt{x^4+x^2+1}}\,dx=E\left(\cos^{-1}\frac{y^2-1}{y^2+1}-\pi,\frac14\right)+\frac{y\sqrt{y^4+y^2+1}}{y^2+1}$$
$$\color{red}{\int_0^y\frac{x^2-1}{\sqrt{x^4+x^2+1}}\,dx=\frac{y\sqrt{y^4+y^2+1}}{y^2+1}-E\left(2\tan^{-1}y,\frac14\right)}$$
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