Using

$$f(r_{ij})= \frac{ar_{ij}}{1+\beta r_{ij}},$$

and $g'(r_{kj})=dg(r_{kj})/dr_{kj}$ and $g''(r_{kj})=d^2g(r_{kj})/dr_{kj}^2$ we find that for particle $k$ we have

$$\frac{\nabla^2_k \Psi_C}{\Psi_C }= \sum_{ij\ne k}\frac{(\mathbf{r}_k-\mathbf{r}_i)(\mathbf{r}_k-\mathbf{r}_j)}{r_{ki}r_{kj}}\frac{a}{(1+\beta r_{ki})^2} \frac{a}{(1+\beta r_{kj})^2}+ \sum_{j\ne k}\left(\frac{2a}{r_{kj}(1+\beta r_{kj})^2}-\frac{2a\beta}{(1+\beta r_{kj})^3}\right)$$