where a constant external force is included
In the vertical take off case, the external force is not constant, as the gravitational force decreases somewhat with altitude. But as you only require it to be valid near the surface, I am going to assume you actually want the force to be constant1.
We have to sources of change in velocity here.
- The rocket engine, given by the Tsiolkovsky equation and
- The acceleration due to the external acceleration, that is simply acceleration times duration, in this case $g \cdot t_{burn}$
So, if the direction of the external force is exactly opposite of the direction we thrust, like gravity, it is not more complicated than subtracting:
$$\Delta v=\left|ln\left(\frac{m_{initial}}{m_{final}}\right) \cdot v_{exhaust} -g \cdot t_{burn}\right|$$
We need to take the absolute value, as the external force can be stronger than our own rocket's acceleration, but $\Delta v$ is a scalar.
In general, if you have an angle $u$ between the two forces, you can express it as
$$\Delta v=\sqrt{\left(ln\left(\frac{m_{initial}}{m_{final}}\right) \cdot v_{exhaust}+cos(u) \cdot g \cdot t_{burn}\right)^2+(sin(u)\cdot g \cdot t_{burn})^2}$$
The derivation should be pretty self explanatory, as this is only adding two vectors by decomposing the $g \cdot t_{brun}$ component, and get the length of the resulting vector.
(I am still representing the acceleration of the outside force as $g$)
I am not sure how this is going to be useful though, as launch analysis usually requires numerical simulation anyway.
1Err, acceleration really, as the system has a changing mass.