Here is a more formal complete solution going off the approach suggested from the comments.
$I=\int e^{-\alpha|\mathbf r-\mathbf R_a|^2} {1 \over |\mathbf r-\mathbf R_b|} e^{-\beta|\mathbf r-\mathbf R_b|^2}dV $
First changing the arrangement of the equation we can define $\mathbf \Delta \mathbf R$ and $\mathbf r'$as:
$\mathbf \Delta \mathbf R = \mathbf R_a - \mathbf R_b$
$\mathbf r'=\mathbf r - \mathbf R_b$
Substitution leads to:
$I=\int e^{-\alpha|\mathbf r'-\mathbf \Delta \mathbf R|^2} {1 \over |\mathbf r'|} e^{-\beta|\mathbf r'|^2}dV $
Aligning the $z'$ axis along the direction of $\mathbf \Delta \mathbf R$
$\mathbf r'=r'sin(\phi')cos(\theta')\hat x'+r'sin(\phi')sin(\theta')\hat y'+r'cos(\phi')\hat z'$
$\mathbf \Delta \mathbf R=0\hat x' + 0\hat y' + \Delta R\hat z'$
$|\mathbf r' -\mathbf \Delta \mathbf R|=\sqrt {(r'sin(\phi')cos(\theta')\hat x')^2+(r'sin(\phi')sin(\theta')\hat y')^2+(r'cos(\phi')\hat z'-\Delta R\hat z')^2}$
Returning to the integral in spherical coordinates and plugging in:
$I=\int^\infty_0 r' \int^{\pi}_0 sin(\phi')\int^{2\pi}_0 e^{-\alpha((r'sin(\phi')cos(\theta')\hat x')^2+(r'sin(\phi')sin(\theta')\hat y')^2+(r'cos(\phi')\hat z'-\Delta R\hat z')^2)} e^{-\beta r'^2} d\theta' d\phi' dr' $
This leads to
$I=2\pi\int^\infty_0 r' \int^{\pi}_0 sin(\phi') e^{-\alpha(r'^2-2r'\Delta R cos(\phi')+\Delta R^2)} e^{-\beta r'^2} d\phi' dr' $
$I=2\pi\int^\infty_0 e^{-\alpha(\Delta R+r')^2} {{e^{4\alpha \Delta R r'} -1} \over 2\alpha\Delta R} e^{-\beta r'^2} dr' $'
Finally, if I did everything correctly
$I = {\pi^{3 \over 2} e^{{-\alpha \beta\Delta R^2 \over \alpha+\beta}}Erf({\Delta R \alpha \over \sqrt{\alpha+\beta}}) \over {\Delta R \alpha \sqrt{\alpha+\beta}} }$'