$\int_{-\infty}^{\infty} |\psi(x)|^2 ~ dx = 1$ when $\psi(x) = C\exp\left(\frac{x^2}{2a^2} + \frac{ix^3}{3a^3}\right)$ The information given is:
Consider a state $|\psi\rangle $ describing a quantum particle on a line, whose position representation $\langle x|\psi\rangle = \psi(x)$ is given by:
\begin{gather*}
    \psi(x) = C\exp\left(-\frac{x^2}{2a^2} + \frac{ix^3}{3a^3}\right),
\end{gather*}
where $a$ and $C$ are positive constants.
Find the constant $C$ by requiring that this state be normalized.
My progress so far:
For the state to be normalized it must satisfy $\int_{-\infty}^{\infty} |\psi(x)|^2 \text{d} x = 1$.
Calculate the integral $\int_{-\infty}^{\infty} |\psi(x)|^2 \text{d}x$.
\begin{align*}
    \int_{-\infty}^{\infty} |\psi(x)|^2 \text{d}x &= 1\\
    \int_{-\infty}^{\infty} \left| C\exp(-\frac{x^2}{2a^2} + \frac{ix^3}{3a^3})\right|^2 \text{d}x &= 1\\
    C^2 \int_{-\infty}^{\infty} \left|\exp(-\frac{x^2}{2a^2} + \frac{ix^3}{3a^3})\right|^2 \text{d}x &= 1\\
\end{align*}
I know that the result should be:
\begin{align*}
    C^2 \int_{-\infty}^\infty e^{-x^2/a^2} \text{d}x &= 1\\
    C^2 = \frac{1}{\sqrt{\pi a}}
\end{align*}
But I fail to see how. Help is greatly appreciated, thanks.

You can write
$$
\exp \left(-\frac{x^2}{2a^2} + \frac{ix^3}{3a^3}\right) = \exp \left(-\frac{x^2}{2a^2}\right)\exp\left(\frac{ix^3}{3a^3}\right).
$$
When you take the magnitude of this complex number, you drop the term with the $i$ as $|\exp(i\times \text{anything})|=1$. Therefore,
$$
|C|^2 \int_{-\infty}^\infty \exp\left(-\frac{x^2}{a^2}\right)= 1.
$$
