Trying to solve $\int$erfc$(a+bz) dz$, but my answer doesn't seem to match what is given here: https://functions.wolfram.com/GammaBetaErf/Erfc/21/ShowAll.html:
What I've got so far:
$\int\text{erfc}(a+bz) dz$
Let $u=a+bz$, $du=b dz$
$$\frac{1}{b} \int\text{erfc}(u) du = \frac{1}{b}(u\text{erfc}(u)- \frac{e^{u^2}}{\sqrt{\pi}})$$
Substituting back in
$$\frac{1}{b}((a+bz)\text{erfc}(a+bz)- \frac{e^{(a+bz)^2}}{\sqrt{\pi}}) = \frac{a}{b}\text{erfc}(a+bz) + z \text{erfc}(a+bz) - \frac{e^{(a+bz)^2}}{b\sqrt{\pi}}$$
How does the $-\frac{a}{b}\text{erf}(a+bz)$ come in? From what I understand, $-\text{erf}(z) = \text{erfc}(z) - 1$, which I don't have.