How to calculate ILT of an expression with terms $e^{\sqrt{s}}$ by residue theorem?

Question

In order to realize the inverse Laplace transform of $$H(s)=\frac{1}{\sqrt{s}(-m+b\ s)}\frac{ \exp{[(r+r_0)\sqrt{s}/\sqrt{D}]} + \exp{[(-r+r_0+2 r_x)\sqrt{s}/\sqrt{D}}]}{\exp{(2r_0\sqrt{s}/\sqrt{D})}-\exp{(2r_x\sqrt{s}/\sqrt{D})}}$$, the contour integral and the residue theorem are applied, $$\frac{1}{2\pi i}\oint_{\gamma_1+\gamma_2+\gamma_3+\gamma_4+\gamma_5+\gamma_6}H(s)e^{st} ds=Res[H(s)e^{st}, m/b]$$. $$Res[H(s)e^{st}, m/b]=\frac{1}{\sqrt{mb}} \frac{\exp((-r+r_0)\sqrt{m/b}/\sqrt{D}+mt/b)(\exp(2r\sqrt{m/b}/\sqrt{D})+\exp(2r_x\sqrt{m/b}/\sqrt{D})}{\exp(2r_0\sqrt{m/b}/\sqrt{D})-\exp(2r_x\sqrt{m/b}/\sqrt{D})}$$. $$\gamma_3: s=s_1\ e^{i\pi}, s_1:R\to 0, \frac{1}{2\pi i}\int_{\gamma_3}H(s)e^{st}ds=\frac{1}{2\pi i}\int_R^0 \frac{1}{(-m-b s_1)\sqrt{s_1}e^{i\pi/2}}\frac{ \exp{[(r+r_0)\sqrt{s_1}e^{i\pi/2}/\sqrt{D}]} + \exp{[(-r+r_0+2 r_x)\sqrt{s_1}e^{i\pi/2}/\sqrt{D}}]}{\exp{(2r_0\sqrt{s_1}e^{i\pi/2}/\sqrt{D})}-\exp{(2r_x\sqrt{s_1}e^{i\pi/2}/\sqrt{D})}}e^{-s_{1} t} e^{i\pi}ds_1$$. $$\gamma_5: s=s_1\ e^{-i\pi}, s_1:0\to R, \frac{1}{2\pi i}\int_{\gamma_5}H(s)e^{st}ds=\frac{1}{2\pi i}\int^R_0 \frac{1}{(-m-b s_1)\sqrt{s_1}e^{-i\pi/2}}\frac{ \exp{[(r+r_0)\sqrt{s_1}e^{-i\pi/2}/\sqrt{D}]} + \exp{[(-r+r_0+2 r_x)\sqrt{s_1}e^{-i\pi/2}/\sqrt{D}}]}{\exp{(2r_0\sqrt{s_1}e^{-i\pi/2}/\sqrt{D})}-\exp{(2r_x\sqrt{s_1}e^{-i\pi/2}/\sqrt{D})}}e^{-s_1 t} e^{-i\pi}ds_1$$. $$\gamma_4: s=\epsilon\ e^{i\theta}, \theta:\pi \to -\pi, \frac{1}{2\pi i}\int_{\gamma_4}H(s)e^{st}ds =\frac{1}{2\pi i}\int_{\pi}^{-\pi}\frac{1}{\sqrt{\epsilon} e^{i\theta/2}(-m+b \epsilon e^{i\theta})} \frac{ \exp{[(r+r_0)\sqrt{\epsilon} e^{i\theta/2}\sqrt{D}]} + \exp{[(-r+r_0+2 r_x)\sqrt{\epsilon}e^{i\theta/2}\sqrt{D}]}}{\exp{(2 r_0\sqrt{\epsilon}e^{i\theta/2}/\sqrt{D})}-\exp{(2 r_x\sqrt{\epsilon}e^{i\theta/2}/\sqrt{D})}}e^{\epsilon e^{i\theta}t}\epsilon e^{i\theta}id\theta = -\frac{1}{2\pi m}\int_{-\pi}^{\pi}\frac{\sqrt{D}}{2(r_0-r_x)}[2+2(\frac{r_0+r_x}{\sqrt{D}})\sqrt{\epsilon}e^{i\theta/2}+o(\sqrt{\epsilon})]d\theta=\frac{\sqrt{D}}{\pi m (r_x-r_0)}$$. The sum of $\int_{\gamma_3}$ and $\int_{\gamma_5}$ is $0$ in my calculation. This calculation is wrong, but I couldn't find the error.