Integral $\int \frac{1}{\sqrt{1 - ix^3}}$ dx using complex calculus.

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I have recently embarked on learning some complex calculus. However, I am getting a bit stumped with some contour integrals involving branch cuts. I am trying to recover some of the results of Carl Bender in his complex classical mechanics papers and in particular, the time period of complex orbits. I would like to solve the integral $\int_{x^{-}}^{x^{+}} \cfrac{dx}{\sqrt{1 - ix^3}}$, whereby we are interested in the paths connecting the symmetric turning points, i.e. those encompassing $x^{+} = \exp(- i \pi/6)$ and $x^{-} = \exp( - 5 i \pi/6)$. We also have a turning point at $i$ and so a branch is needed to span up the positive imaginary axis. I have done this in the case of $x^2$ being in the denominator by using a dogbone contour and recovered the correct answer. However, my method is failing here. Any guidance/ advice is much appreciated to get me started.

UPDATE: I think I have done this by using the analytic continuation of the Beta function and the Pochhammer contour.

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Note that $1/\sqrt{1-iz^2}$ has branch points only at $z^2=-i$, as the singularity at infinity is instead a first-order pole. By contrast, $1/\sqrt{1-iz^3}$ has branch points at infinity along with $z^3=-i$, so a dogbone that goes around the finite singularity points may still hit the portion of the cut that has to go to infinity. You may have to see which way the cut goes to infinity and "counter-drive" it.