Boyer-Lindquist coordinates to Cartesian coordinates The Kerr metric in Boyer-Lindquist coordinates is given by
\begin{align}
  ds^2 &= -\left[ \frac{r^2 + a^2 \cos^2(\theta) - 2mr}{r^2+ a^2 \cos^2(\theta)} \right] dt^2 -\frac{4mra \sin^2(\theta)}{r^2 + a^2 \cos^2(\theta)} dt d\phi   \\
  &  + \left[ \frac{r^2 + a^2 \cos^2(\theta)}{r^2 - 2mr + a^2} \right] dr^2 + \left(r^2 + a^2 \cos^2(\theta)  \right) d\theta^2   \\
  & + \left[ r^2 + a^2 +\frac{2mra^2 \sin^2(\theta)}{r^2 + a^2 \cos^2(\theta)} \right] \sin^2(\theta) d\phi^2
 \end{align}
I want to find proper transformation to Cartesian
\begin{align}
  ds^2 &= -dt^2 + dx^2 + dy^2 + dz^2    \\
  & + \frac{2mr^3}{r^4+ a^2 z^2} \left[ dt + \frac{r(xdx+ydy)}{a^2+ r^2} + \frac{a(ydx - xdy)}{a^2+ r^2} + \frac{z}{r}dz \right]^2  
\end{align}
First of all I know how to deal with in the case of $m=0$. i.e.,
\begin{align}
  ds^2 &= -dt^2 + \frac{r^2+ a^2 \cos^2(\theta)}{r^2 + a^2} dr^2 + (r^2+ a^2 \cos^2(\theta)) d\theta^2 + (r^2+ a^2) \sin^2(\theta) d\phi^2
 \end{align}
In this case taking spheroidal coordinates $x=\sqrt{r^2+ a^2} \sin(\theta) \cos(\phi), y=\sqrt{r^2+ a^2} \sin(\theta)\sin(\phi), z=r\cos(\theta)$ this reduces to
\begin{align}
ds^2 = - dt^2 + dx^2 + dy^2 + dz^2
\end{align}
What about $m\neq 0$? In this case, after plugging spherodical coordinates into the metric, I have different forms.
How to convert Boyer-Lindquist coordinates to Cartesian coordinates in the case of $m\neq 0$?
I know after some consecutive transformation: Cartesin -> Original Kerr(Edington) -> time shift -> Boyer-Lindquist...
Here the reason I post this question is I want to write transformation rule $\frac{dx'^{\mu}}{dx^{\nu}}$ on these two coordinates.

With the relation between two coordinates is given by
\begin{align}
& d\tilde{t} = dt + \frac{2mr}{r^2+a^2  -2mr} dr  \\
  &dx= \sin(\theta)\left( \cos(\bar{\phi})    - \frac{a   \left( r \sin(\bar{\phi}) + a \cos(\bar{\phi}) \right)}{r^2+a^2 -2mr}
    \right)dr   \\
  &\qquad \qquad
  + \cos(\theta) \left( r \cos(\bar{\phi})
   - a \sin(\bar{\phi}) \right) d\theta - \left( r\sin(\bar{\phi}) + a \cos(\bar{\phi}) \right) \sin(\theta)d \phi     \\
  & dy =\sin(\theta) \left(  \sin(\bar{\phi}) + \frac{a  \left( r \cos(\bar{\phi}) - a \sin(\bar{\phi}) \right)}{r^2+ a^2 -2mr}
    \right)dr   \\
    & \qquad \qquad + \cos(\theta) \left( r \sin(\bar{\phi}) + a \cos(\bar{\phi}) \right) d\theta + \sin(\theta) \left(r\cos(\bar{\phi}) - a \sin(\bar{\phi}) \right) d {\phi}  \\
  & dz = \cos(\theta) dr -r\sin(\theta) d\theta   
\end{align}
where $\bar{\phi} = \phi + \int\frac{a}{r^2+a^2 -2mr} dr$, we have desired results.
