I am trying to get a analytical solution for a integral,
$$\int{x^a \text{erf}(bx)dx}\text{, when }a<1$$
usually, if $a$ is an integer there are solution which are
$$\int{r\cdot\text{erf}(br) dr}=\frac{1}{4}\Biggl(\frac{2r\cdot\exp(-b^2r^2)}{b\sqrt \pi}+\left(2r^2-\frac{1}{b^2}\right)\text{erf}(br)\biggr)+\text{constant}$$
$$\int{r^2 \text{erf}(br) dr}=\frac{1}{3}\Biggl(\frac{(b^2r^2+1)\exp(-b^2r^2)}{b^3\sqrt \pi}+r^3\text{erf}(br)\biggr)+\text{constant}$$
and so on.
is there a way to solve when $a$ is fraction, for example $a=2/3$?
By integrating by parts,$\newcommand{error}{\operatorname{erf}}$
\begin{align}\int x^a\error(bx)~{\rm d}x&=\frac{x^{a+1}}{a+1}\error(bx)-\frac{2b}{(a+1)\sqrt\pi}\int x^{a+1}e^{(bx)^2}~{\rm d}x\\&=\frac{x^{a+1}}{a+1}\error(bx)+\frac{(-1)^{(a+1)/2}}{(a+1)b^{a+1}\sqrt\pi}\int u^{(a+1)/2}e^{-u}~{\rm d}x\\&=\frac{x^{a+1}}{a+1}\error(bx)+\frac{(-1)^{(a+1)/2}}{(a+1)b^{a+1}\sqrt\pi}\Gamma\left(\frac{a+1}2,-(bx)^2\right)+C\end{align}
where $\Gamma(s,x)$ is the incomplete Gamma function and $u=-(bx)^2$.