What is $\int_{0}^{z} e^{-a^{2} x^{2}} {\rm erf}(bx)\, dx$?

This indefinite integral is a special function, called Owen's T:

$$\int_0^z e^{-a^2 x^2}{\rm erf}\,(bx)\,dx=\frac{\arctan(b/a)}{a\sqrt\pi}-\frac{2\sqrt\pi}{a} T\left(\sqrt{2} az,b/a\right)$$

---

Here is the requested derivation:

$$\int_0^z e^{-a^2x^2}{\rm erf}\,(bx)\,dx=\frac{2}{\sqrt\pi}\int_0^z e^{-a^2x^2}\left(\int_0^{bx}e^{-t^2}dt\right)dx=$$ $$\frac{2}{\sqrt\pi}\int_0^z e^{-a^2x^2}\left(\int_0^{b}e^{-(yx)^2}x\,dy\right)dx=$$ $$\frac{2}{\sqrt\pi}\int_0^b\left(\int_0^z e^{-(a^2+y^2)x^2}xdx\right)dy=\\ \frac{1}{\sqrt\pi}\int_0^b\frac{e^{-(a^2+y^2)z^2}-1}{a^2+y^2}\,dy=$$ $$\frac{1}{a\sqrt\pi}\left[ \arctan(b/a)-2\pi T(\sqrt{2}az,b/a)\right]$$ with the definition $T(z,b)=\frac{1}{2\pi}\int_0^b(1+y^2)^{-1}\exp[-\tfrac{1}{2}z^2(1+y^2)]\,dy$ of Owen's T-function.

Maple does not evaluate this integral. So I suspect there is no known evaluation.

I did find that integration by parts can switch the $a$ and $b$:

$$ \int \!{{\rm e}^{-{a}^{2}{x}^{2}}}{\rm erf} \left(bx\right)\,{\rm d}x= {\frac {{\rm erf} \left(bx\right)\sqrt {\pi }\;{\rm erf} \left(ax \right)}{2 a}}-\frac{b}{a}\int \!{ {{{\rm e}^{-{b}^{2}{x}^{2}}}{\rm erf} \left(ax\right)}}\,{\rm d}x $$