$$\int_{1}^{x}\frac{dt}{\sqrt{t^3-1}}$$ 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 $$\int\limits_{0}^{\sec^{-1}x^{\frac{3}{2}}}\sec^{\frac{2}{3}}tdt$$.
State without proof:
$$\int_{1}^{x} \frac{dt}{\sqrt{t^{3}-1}}= \frac{1}{\sqrt[4]{3}} \operatorname{cn}^{-1} \left( \frac{\sqrt{3}+1-x}{\sqrt{3}-1+x}, \frac{\sqrt{6}-\sqrt{2}}{4} \right)$$
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.