From Kerr metric, we do know that there exist a function with the form of:
$$\Delta = r^2 - 2 M r + a^2 \tag{1}.$$
Following $[1]$, I did understand the coordinate transformation from Boyer-Lindquist (BL) to Kerr-Schild (KS) coordinates, (BL$\to$ KS), for the Kerr spacetime:
$$ dt' = dt + \frac{2 M r}{r^2 - 2 M r + a^2} dr = dt + \frac{2 M r}{\Delta} dr , d\phi'=d\phi + \frac{a}{r^2 - 2 M r + a^2} dr = d\phi + \frac{a}{\Delta} dr. \tag{2}. $$
Now, suposse we have a general version of $\Delta$ function, such as:
$$\Delta' = Kr^2 - 2 M r + a^2 + f(r) \tag{3}.$$
Where, $K$ is a constant. My question is:
The tranformation (BL$\to$KS) would have its form precisely as: $$ dt' = dt + \frac{2 M r}{\Delta'} dr , d\phi'= d\phi + \frac{a}{\Delta'} dr \tag{4}? $$
$[1]$ Coordinate transform of Kerr metric to Kerr-Schild coordinates
