~r_P ^{d-1}}{4 \pi G_{N}} \log \frac{\mathfrak{a}^2}{|f(r_P)|}.$$ The contribution of the null-joint term at point $M$ is given by . Using it is not hard to find that for the nonsymmetric case we have $$\frac{d r_M}{d \mu^+} = -f(r_M).$$ Hence we find $$\begin{aligned}
\frac{d I_{J}^\text{null}(M)}{d \mu^+} = \frac{(d-2) ~\omega^{d-2} ~\Omega_{d-1}}{8 \pi G_{N}}-\frac{r_M^{d-2}~\Omega_{d-1}}{8 \pi G_{N}} ~\left((d-1)f(r_M) \log \frac{-\mathfrak{a}^2}{f(r_M)}\right).\end{aligned}$$ Finally using we find the rate of complexity growth as $$\begin{aligned}
\nonumber\dot{\mathcal{C}} =  \frac{1}{\pi} ~\Big[ \frac{4 M}{1+\gamma}  &- \frac{r_M^{d-2}~ \Omega_{d-1}}{8 \pi G_{N}} \frac{\gamma}{1+\gamma}~(d-1)f(r_M) \log \frac{-\mathfrak{a}^2}{f(r_M)} 
\\
&+\frac{1-\gamma}{1+\gamma} \frac{\Omega_{d-1} ~(d-1)~r_P^{d-2}}{8 \pi G_{N}} f(r_P) \log \frac{\mathfrak{a}    ^2}{f(r_P)} \Big].\end{aligned}$$ For $\gamma=1$ the result is exactly the same as the symmetric case. Moreover, in the nonsymmetric case, the complexity growth depends not only on $r_M$ but also on $r_P$.

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