Elliptic Integral $$\int_{1}^{x}\frac{dt}{\sqrt{t^3-1}}$$ does this have a closed form involving jacobi elliptic functions of parameter $k$?
N.B I tried with the change of variables $t=1+k\frac{1-u}{1+u}$. But this leads no where. http://mathworld.wolfram.com/JacobiEllipticFunctions.html
update the above integral is equivalent to $$\int\limits_{0}^{\sec^{-1}x^{\frac{3}{2}}}\sec^{\frac{2}{3}}tdt$$.
 A: State without proof:
$$\int_{1}^{x} \frac{dt}{\sqrt{t^{3}-1}}=
\frac{1}{\sqrt[4]{3}} \operatorname{cn}^{-1}
\left(
  \frac{\sqrt{3}+1-x}{\sqrt{3}-1+x},
  \frac{\sqrt{6}-\sqrt{2}}{4}
\right)$$
A: Here is the proof of Ng Chung Tak's solution.
The integral is of the form of the polynomial under the radical having 1 real and 2 complex roots and 
the lower limit of integration equal to the real root. So we have
\begin{equation}
\int\limits_{a}^{x} \left( (t-a)[(t-r)^{2} + s^{2}] \right)^{-1/2}\, dt
= \frac{1}{\sqrt{m}} \mathrm{cn}^{-1}\left(\frac{m-(x-a)}{m+(x-a)},k \right)
\end{equation}
where
\begin{equation}
m^{2} = (r - a)^{2} + s^{2} \quad \mathrm{and} \quad k^{2} = \frac{1}{2} - \frac{a-r}{2m}
\end{equation}
$\mathrm{cn}^{-1}z$ is the inverse of one of the Jacobi elliptic functions and $k$ is the modulus.
To factor the polynomial inside of the radical, we note that the roots are the cube roots of 1, thus
\begin{align}
t^{3}-1 &= (t-t_{1})(t-t_{2})(t-t_{3}) \\
&= (t-1) \left(\Big[t+\frac{1}{2}\Big]-i\frac{\sqrt{3}}{2} \right) \left(\Big[t+\frac{1}{2}\Big]+i\frac{\sqrt{3}}{2} \right) \\
&= (t-1) \left(\Big[t+\frac{1}{2}\Big]^{2} + \frac{3}{4}\right)
\end{align}
We now have
\begin{equation}
a = 1 \quad r = -\frac{1}{2} \quad s = \frac{\sqrt{3}}{2} \quad m^{2} = 3 \quad k^{2} = \frac{1}{2} - \frac{3}{4\sqrt{3}}
\end{equation}
and thus
\begin{equation}
\int\limits_{1}^{x} \frac{dt}{\sqrt{t^{3}-1}}
= 3^{-1/4} \, \mathrm{cn}^{-1}\left(\frac{\sqrt{3} + 1 - x}{\sqrt{3} \, - 1 + x},\sqrt{\frac{1}{2} - \frac{3}{4\sqrt{3}}} \right)
\end{equation}
Note that the value of $k$ here is equal to that of Ng Chung Tak's solution.
