v_h}{16\sqrt{3}\alpha k^2} \frac{[2\sqrt{9 - 40\kappa^2\alpha\Lambda} + 3]}
 {[\sqrt{9 - 40\kappa^2\alpha\Lambda} - 2]^{\frac{2}{3}}}*\nonumber\\
 &\bigg[&\frac{\exp{\big[2(2+\nu)ky-4\nu kr_c\pi\big]}}{m_{\Phi}^2r_c^2 + 8kr_c(2+\nu)-4(2+\nu)^2}\nonumber\\ 
 &+&\frac{2v_v}{v_h}\frac{\exp{\big[4ky-(2+\nu)kr_c\pi\big]}}{m_{\Phi}^2r_c^2 + 16kr_c- 16}\bigg]
 \label{sol_scalar_final}\end{aligned}$$ and

$$\begin{aligned}
 A(y) = ky + \frac{\kappa^2v_h^2}{12}\bigg[e^{-4\nu kr_c\pi} e^{-2(2+\nu)ky} + e^{2(2-\nu)ky} \big(1+\frac{v_v}{v_h}e^{-(2+\nu)kr_c\pi}\big)\bigg]
 \label{sol_warp_final}\end{aligned}$$

Thus due to the inclusion of the bulk scalar field backreaction, the warp factor gets modified and the correction term is proportional to $\kappa^2v_h^2$ which is indeed small for $\kappa v_h< 1$.\
However from the solutions of scalar-tensor theory, one can extract the solution in F(R) model by inverse conformal transformation as indicated earlier (see eqn.(\[conformal\])) : $$\begin{aligned}
 ds^2&=&\exp{\big[-\frac{\kappa}{\sqrt{3}}(<\Phi>+\xi(y))\big]} \bigg[e^{- 2 A(y)} \eta_{\mu\nu} dx^{\mu} dx^{\nu} - dy^2\bigg]
 \label{grav.sol1.F(R)_final}\end{aligned}$$ where $ds^2$ is the line element in $F(R)$ model and $\xi(y)$, $A(y)$ are given in eqn.(\[sol\_scalar\_final\]) and eqn.(\[sol\_warp\_final\]) respectively.\
At this point, we need to verify whether the above solution of $G_{MN}$ (in eqn.(\[grav.sol1.F(R)\_final\])) satisfies the field equations of the original $F(R)$ theory. The five dimensional gravitational field equation for $F(R)$ theory is given by, $$\begin{aligned}
 \frac{1}{2}G_{MN}F(R) - R_{MN}F'(R) - G_{MN}\Box F'(R) + \nabla_{M}\nabla_{N}F'(R) = -\frac{1}{2}\Lambda G_{MN}
 \label{field1}\end{aligned}$$

In the present context, we take the form of $F(R)$ as $F(R) = R + \alpha R^2$ and thus the above field equation is simplified to the form : $$\begin{aligned}
 