Evaluating $ \int_{-\infty}^{\infty}x\exp\left(-b^{2}\left(x-c\right)^{2}\right)\mathrm{erf}^{2}\left(a\left(x-d\right)\right)\,\mathrm{d}x $

A suggestion instead of a complete answer: taking $c=0$ for ease of typing and integrating by parts $$\begin{align*} \int_{-\infty}^\infty xe^{-b^2x^2}\text{erf}^2(a(x-d))\ \mathrm dx &= -\text{erf}^2(a(x-d))\left.\frac{e^{-b^2x^2}}{2b^2}\right|_{-\infty}^\infty\\ &\qquad\qquad +\int_{-\infty}^\infty2\text{erf}(a(x-d))\frac{2}{\sqrt{\pi}} e^{-a^2(x-d)^2}\frac{e^{-b^2x^2}}{2b^2}\ \mathrm dx\\ &=\frac{2}{b^2\sqrt{\pi}}\int_{-\infty}^\infty\text{erf}(a(x-d)) e^{-a^2(x-d)^2-b^2x^2}\ \mathrm dx\\ &=\frac{2}{b^2\sqrt{\pi}}\int_{-\infty}^\infty\text{erf}(a(x-d)) e^{-(a^2+b^2)x^2 + 2a^2dx-a^2x^2}\ \mathrm dx \end{align*}$$ to which, after completing the square in the exponent, we can apply the OP's given integral formula $$\int_{-\infty}^{\infty}\exp\left(-b^{2}\left(x-c\right)^{2}\right)\mathrm{erf}\left(a\left(x-d\right)\right)\,\mathrm{d}x= {\frac{\sqrt\pi}{b}}\mathrm{erf}\left(\frac{ab\left(c-d\right)}{\sqrt{a^{2}+b^{2}}}\right)\,.$$