Looking for a way to perform this integral related to the error function. I am thinking an answer in closed form cannot be done, but hoping I missed something.
$$ \int x\operatorname{erf}^{\,3}(x)\,e^{-x^2}dx $$
Edit:
Yuriy : It comes from calculations related to linear moments.
Ives : Thanks, following integration by parts, we have
$\int u\,dv = uv-\int v\,du$
$u=x\,\operatorname{erf}(x)$
$du=\operatorname{erf}(x)+\frac{2x}{\sqrt{\pi}}e^{-x^2}dx$
$v=\frac{\pi}{6}\operatorname{erf}^3(x)$
$dv=\operatorname{erf}^2(x)\,e^{-x^2}dx$
$\int x\operatorname{erf}^{\,3}(x)\,e^{-x^2}\,dx = \frac{\sqrt{\pi}}{6}\,x\,e^{-x^2}\operatorname{erf}^{\,3}(x)-\frac{\sqrt{\pi}}{6}\,\int \operatorname{erf}^3(x)[\operatorname{erf}(x)+\frac{2}{\sqrt{\pi}}xe^{-x^2}]dx$
$=\frac{\sqrt{\pi}}{4}x\operatorname{erf}^4(x)-\frac{\sqrt{\pi}}{4}\int \operatorname{erf}^{\,4}(x)dx$
Then the challenge becomes how to solve:
$$ \int \operatorname{erf}^{\,4}(x)\,dx $$
Edit 2:
Able to simplify(?) a bit more:
$\int \operatorname{erf}^{\,4}(x)\,dx = x\operatorname{erf}^{\,4}(x)+\frac{4}{\sqrt{\pi}}e^{-x^2}\operatorname{erf}^{\,3}(x)-\frac{24}{\pi}\int \operatorname{erf}^{\,2}(x)e^{-2x^2}dx$
Now the challenge becomes how to solve:
$$ \int \operatorname{erf}^{\,2}(x)e^{-2x^2}dx $$