The AdS-Schwarzschild black hole metric which is in the $(t,z,x)$ coordinates is given by,
$$ds^2 = \frac{1}{z^2} \left( -f(z) dt^2 + \frac{dz^2}{f(z)} + dx^2 \right)$$
where $t$ is time, $z$ is the radial direction, $x$ is a transverse spatial direction, $f(z) = 1-m z^4$, and $m$ is a constant mass. This metric is clearly time-independent, i.e., independent of $t$.
Another black hole metric which is time-dependent is given by the AdS-Vaidya black hole metric,
$$ds^2 = \frac{1}{z^2} \left( -f(u,z) du^2 - 2 du dz + dx^2 \right)$$
Here, $u$ acts like the new time coordinate. The transformation that allows this is given by,
$$t = u + \int \frac{dz}{f(u,z)}$$
where $f(u,z) = 1 - m(u) z^4$ and $m(u) = 1000 e^{-3 u}$. The ranges are $t \in [\epsilon, 1]$, $u \in [\epsilon, 1]$, $z \in [\epsilon, 1]$, $x \in [\epsilon, 1]$, and $\epsilon = 10^{-2}$. Take note that the integral is a function of two variables but the integration is only for the z variable.
I want to plot the plane $u=x$, however, I want to see this in the $(t,z,x)$ coordinates instead of the $(u,z,x)$. In this case, we first need to evaluate the integral of the transformation, using some software like Maple, Mathematica, etc. we get,
$$t = u + \frac{2 \tan^{-1}\left(m(u)^{1/4}z\right)+\ln\left(\frac{1 + m(u)^{1/4}z}{1 - m(u)^{1/4}z}\right)}{4 m(u)^{1/4}}$$
Looking at the plot, it seems kind of weird, it may have hit a branch cut. The upper left region is kind of torn out. Also, evaluating $\ln\left(\frac{1 + m(u)^{1/4}z}{1 - m(u)^{1/4}z}\right)$, say for $u=0.5$ and $z=0.5$ gives an imaginary value which kind of hints to an issue.
Plotting $f(u,z)$ shows where the integrand blows up,
Any help/guide on how to properly do the integration in this case?