If only $\delta$ depends on the scale factor, then
\[\rho_{dm}=\rho_{dm0}a^{-3}e^{\int\delta(a)d\ln a}.\]
One more time use the assumption
\[r\equiv\frac{\rho_{dm}}{\rho_{de}}=\frac{\rho_{dm0}}{\rho_{de0}}a^{-\xi}=A^{-1}a^{-\xi},\quad A\equiv\frac{\rho_{dm0}}{\rho_{de0}}=\frac{\Omega_{dm0}}{\Omega_{de0}}.\]
Consequently,
\[\rho_{de}=\frac{Aa^\xi}{1+Aa^\xi}\rho_{tot},\quad \rho_{dm}=\frac{1}{1+Aa^\xi}\rho_{tot}.\]
Then the total energy density satisfies the equation
\[\frac{d\rho_{tot}}{da}+\frac3a\frac{1+(1+w_{de})Aa^\xi}{1+Aa^\xi}\rho_{tot}=0.\]
Integrate this expression to obtain
\[\rho_{tot}=\rho_{tot0}a^{-3}\left[1-\Omega_{de0}(1-a^\xi)\right]^{-3w_{de}/\xi},\quad \rho_{tot0}=\rho_{de0}+\rho_{dm0}.\]
Consequently the first Friedmann equation can be written as
\[H^2=H_0^2 a^{-3}\left[1-\Omega_{de0}(1-a^\xi)\right]^{-3w_{de}/\xi}.\]
Using the conservation equation for DM one can get the coupling function
\[\delta=3+\frac1H\frac{\dot\rho_{dm}}{\rho_{dm}}=-\frac{(3+w_{de})Aa^\xi}{1+Aa^\xi}= -(3+w_{de}) \frac{\rho_{de}}{\rho_{tot}}.\]
This relation can be expressed as
\[\delta=\frac{\delta_0}{\Omega_{de0}+(1-\Omega_{de0})a^{-\xi}}, \quad \delta_0\equiv-\Omega_{de0}(\xi+3w_{de}).\]
Let us analyze the obtained expression. Asymptote of the interaction $\delta$ represents a constant, $\delta(a\to\infty)=\delta_0/\Omega_{de0}$. Therefore, if the expansion dynamics is such that $\xi>-3w_{de}$, then $\delta<0$, which implies that the energy flow is from the dark matter to dark energy. On the contrary, when $0<\xi<-3w_{de}$, the energy flow is from the phantom dark energy to dark matter. Further, we can see that there is no coupling between the dark energy and dark matter at $\xi=-3w_{de}$. Specifically, there is no coupling in SCM, for which $\xi=3$, $w_{de}=-1$.