Symbolic computation with partial derivatives and a double integral

[fluidistic]

I have an expression of a vector field $\vec J = J_r \hat r + J_\theta \hat \theta$ in polar coordinates, in a region that's a quarter of an annulus, where theta ranges between 0 and pi/2 and r goes from ri to ro.
I need to compute $-\int _\Omega T(S_{xx}-S_{yy}) \frac{\partial J_x}{\partial x}d\Omega =-(S_{xx}-S_{yy})\int_0^{\pi/2} \int_{ri}^{ro} [T_0 + \Delta T \ln(r/ro) / \ln(ro/ri)] \frac{\partial J_x}{\partial x} rdrd\theta$ where all the parameters in the last expressions are known constants.
The thing is that I do not have the $\hat x$ component of $\vec J$, I have the $\hat r$ and $\hat \theta$ components only.
Therefore I need to obtain $J_x$ first. I get that $J_x = J_r \cos(\theta)-J_\theta \sin(\theta)$.
Using the chain rule I get that $\frac{\partial }{\partial x}=\cos(\theta)\frac{\partial}{\partial r}-\frac{\sin(\theta)}{r}\frac{\partial}{\partial \theta}$.
Having derived this, I can now write the Simpy code that should give me the sought integral.

import simpy as sp

# Define the variables
r, theta, H0, ri, ro, rho, T_0, Delta_T, S_xx, S_yy = sp.symbols('r theta H0 ri ro rho T_0 Delta_T S_xx S_yy')

# Define H0
H0_expr = -(S_xx - S_yy) / (4 * rho) * Delta_T / sp.log(ro/ri)

# Define the components of the vector field J_r and J_theta
J_r = 2 * H0 * (1/r - 1/(ri**2 + ro**2)*(r + ri**2*ro**2/r**3)) * sp.cos(2*theta)
J_theta = H0 * (1/(ri**2 + ro**2)*(2*r - 2*ri**2*ro**2/r**3)) * sp.sin(2*theta)

J_x = J_r * sp.cos(theta) - J_theta * sp.sin(theta)
dJ_x_dx = sp.diff(J_x, r) * sp.cos(theta) - 1/r * sp.sin(theta) * sp.diff(J_x, theta) * sp.sin(theta)

T = T_0 + Delta_T * sp.log(r/ro) / sp.log(ro/ri)

# Set up the integral for Q
Q = sp.integrate(r * T * (S_xx - S_yy) * dJ_x_dx, (theta, 0, sp.pi/2), (r, ri, ro))

# Substitute H0 in Q
Q_sub = Q.subs(H0, H0_expr)
Q_sub = sp.simplify(Q_sub)
Q_sub = sp.collect(Q_sub, [ri, ro, sp.log(ro/ri)])
Q_sub = sp.factor(Q_sub)
print(f"This is Q: {Q}")

The code yields
This is Q: -Delta_T*(S_xx - S_yy)**2*(-16*Delta_T*ri**2*log(ri/ro)**2 + 15*pi*Delta_T*ri**2*log(ri/ro)**2 + 80*Delta_T*ri**2*log(ri/ro) + 30*pi*Delta_T*ri**2*log(ri/ro) - 64*Delta_T*ri**2 - 15*pi*Delta_T*ri**2 - 16*Delta_T*ro**2*log(ri/ro)**2 + 15*pi*Delta_T*ro**2*log(ri/ro)**2 + 48*Delta_T*ro**2*log(ri/ro) + 15*pi*Delta_T*ro**2 + 64*Delta_T*ro**2 - 32*T_0*ri**2*log(ri)*log(ro/ri) + 30*pi*T_0*ri**2*log(ri)*log(ro/ri) - 30*pi*T_0*ri**2*log(ro)*log(ro/ri) + 32*T_0*ri**2*log(ro)*log(ro/ri) + 32*T_0*ri**2*log(ro/ri) + 30*pi*T_0*ri**2*log(ro/ri) - 32*T_0*ro**2*log(ri)*log(ro/ri) + 30*pi*T_0*ro**2*log(ri)*log(ro/ri) - 30*pi*T_0*ro**2*log(ro)*log(ro/ri) + 32*T_0*ro**2*log(ro)*log(ro/ri) - 30*pi*T_0*ro**2*log(ro/ri) - 32*T_0*ro**2*log(ro/ri))/(480*rho*(ri**2 + ro**2)*log(ro/ri)**2).

I wonder what Mathematica (or a better Simpy code) can yield, I strongly suspect the expression can be further simplified. Can someone confirm my result or show me a shoter one, which I could then compare?