Limit to the Oberth effect in a supermassive Black Hole flyby To make this question clear, here are the details of the situation I wish to entertain.

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*A spacecraft does a powered gravitational assist, where it fires engines in a near-approach to the body


*We should all be aware the rotating black holes can directly act on a passerby, and both because of the focus on supermassive, and for mathematical simplicity, I'd like to assume the standard Schwarzschild black hole:


*Best to assume the spacecraft starts with $v_{infinity}=0$, that is, its incoming speed before entering the gravity well is assumed to be minimal.


*The spacecraft comes as close as possible without falling in, or however close gets the maximum Oberth effect.


*I want to convert $\Delta v$ the spacecraft engines generate to the final $V$ after it has departed the gravity well.
To sum up, I want some expression for a relativistic Oberth effect that would apply in the most extreme case.
Preliminary Thinking
Previous question:
What happens to orbits at small radii in general relativity?
I guess a logical approach would be to follow the same approach as the computation of the Oberth effect for a parabolic orbit based on energy balance. But if you go highly relativistic, the gravitation as well as kinetic energy terms can get quite complex, here is gravitational:
$$ V(r)=-{\frac {GMm}{r}}+{\frac {L^{2}}{2\mu r^{2}}}-{\frac {G(M+m)L^{2}}{c^{2}\mu r^{3}}}. $$
I could also guess that the optimal approach is at the IBCO radius of 3/2 times the Schwarzschild radius. But this still leaves quite a few things to plug in, and I'm doubtful about the validity of the approach overall.
Heck, just to put it out there, let's say I use the non-relativistic Oberth equation assuming the IBCO approach distance:
$$ V=\Delta v{\sqrt {1+{\frac {2V_{\text{esc}}}{\Delta v}}}} =\Delta v{\sqrt {1+{ \sqrt{\frac{GM}{3/2 r_s}} \frac {2}{\Delta v}}}} =\Delta v{\sqrt {1+ {  \frac {2 c}{3 \sqrt{3} \Delta v}}}}. $$
This would give a multiplier of something like a factor of 100 for a 10 km/s burn. But this is almost certainly wrong, applied outside its range of applicability.
 A: I'm also interested in the answer to this question, this is how far I got:
The spacecraft follows a geodesic, and if it does an impulsive boost at a point it will now follow a different geodesic from that point but with a different 4-velocity. The incoming trajectory starts with velocity $v_0$ at infinity and the new one ends at velocity $v_1$, so the overall Oberth boost is $|v_1-v_0|$. 
The standard textbook equations for time-like Schwarzschild geodesics are:
$$\frac{dt}{d\tau}=\frac{E}{mc^2}\frac{1}{1-\frac{r_s}{r}}$$
$$\frac{d\theta}{d\tau} = \frac{L}{M}\frac{1}{r^2}$$
$$\left(\frac{dr}{d\tau}\right)^2 = \frac{E^2}{m^2c^2} - \left(1-\frac{r_s}{r}\right)\left(c^2+\frac{L^2}{M^2}\frac{1}{r^2}\right)$$ where $E$ is the energy of the craft, $L$ its angular momentum, $r_s$ the Schwarzschild radius, $M$ the mass of the central body and $\tau$ proper time. The spacecraft mass $m\ll M$.
The effective potential is $$V(r)=-\frac{GMm}{r}+\frac{L^2}{2GMr^2} -\frac{L^2}{c^2r^3}:$$ the particle moves as $$\frac{1}{2}m\left(\frac{dr}{d\tau}\right)^2=\left[\frac{E^2}{2mc^2} - \frac{mc^2}{2}\right] + V(r).$$ It allows different orbit types depending on ($E,L$). The ones we care about are the ones that are unbounded in the past or future. $E$ must be larger than $mc^2$ (otherwise it cannot escape to infinity ). 
So, to do the maneouvre we  drop a craft from infinity towards the hole. It starts with velocity $$v_0=\sqrt{\frac{E^2}{m^2c^2}-c^2}$$ at $r=\infty$. It approaches until the righthand side of the motion equation becomes zero at $r_{turn}(E,L)$. At this point we change velocity to get $E',L'$ and the craft recedes to infinity; we calculate its velocity 
$$v_1 = \sqrt{\frac{E'^2}{m^2c^2}-c^2}$$ and will have our answer $v_1-v_2$.
The part where I get stuck is how to calculate what $E',L'$ different boosts imply. Also, realistic boosts will change $m$ to $m'$ if the ejected mass is significant.
