Seemingly unsolvable integral for length of parametric curve I have come across a question for the length of the loop of a given parametric curve defined by the equations
$x(t)= 4t-t^3$ ;
$y(t)=-2+2t^2$
The loop is defined about the interval from t=-2 to t=2 or notice the symmetry by taking twice the length from 0 to 2. The problem lies in the calculation of the antiderivative with the traditional formula of
$\int_{-2}^2 \sqrt{(dx/dt)^2+(dy/dt)^2} dt$
This simplifies out to;
$\int_{-2}^2 \sqrt{9t^4-8t^2+16} dt$
As far as I know and have tried, this integral has no solvable real antiderivative as the root is complex. My best guess would be either to eliminate the parameter of the curve and potentially solve in Cartesian or rather find a polar interpretation and solve in polar (both of which I have tried but failed). If someone can help point me in the right direction it would be greatly appreciated!
 A: This is just an exercise with almost no interest.
As I wrote in comments, if the problem was
$$\int_{0}^{\frac 23} \sqrt{9t^4-8t^2+16}\, dt$$ it could be possible to obtain almost the exact solution writing
$$ \sqrt{9t^4-8t^2+16}=\sum_{n=0}^\infty a_n\,t^{2n}$$ where the coefficients, given by
$$a_n=\frac{4(2n-3)\,a_{n-1}-9(n-3)\,a_{n-2}}{16\,n}\quad \quad\text{with} \quad a_0=4 \quad\text{and} \quad a_1=-1$$ decrease very fast to $0$.
Comuting the partial sums
$$S_k=\sum_{n=0}^k \frac {a_n}{2n+1}\left(\frac 23\right)^{2n+1}$$
$$\left(
\begin{array}{cc}
 k  & S_k \\ 
 0  & 2.666666667 \\
 5  & 2.596066206 \\
 10 & 2.596065324 \\
 15 & 2.596065321 \\
\end{array}
\right)$$
Just for your curiosity
$$I=\int_0^x \sqrt{9t^4-8t^2+16}\, dt$$  is
$$I=\frac{\sqrt{9 x^4-8 x^2+16}}{27}\,\,\, \Bigg[\cdots\Bigg]$$
$$\Bigg[\cdots\Bigg]=9 x+32 \left(\sqrt{2}-2 i\right) F\left(\sin ^{-1}\left(\frac{x}{2} \sqrt{1+2 i \sqrt{2}} \right)|-\frac{7+4 i \sqrt{2}}{9}\right)+16
   \left(\sqrt{2}+i\right) E\left(\sin ^{-1}\left(\frac{x}{2} \sqrt{1+2 i \sqrt{2}} \right)|-\frac{7+4 i \sqrt{2}}{9}\right)$$
Make $x=\frac 23$ or whatever and ... enjoy !
