I assume the waveform is the voltage from the bottom IGBT collector to GND? What is the gate signal under the same conditions? It appears that there are two fairly solid turn-ons, but one is about 20 uSec while the other is about 30 uSec. And these have 5 volts drop, so if you are running 50 amps that's 250 watts! Then there are the second turn-ons 6-12 uSec later, but there is a much higher voltage drop of 20-15 volts, which is an even higher amount of power.
You may be exceeding the safe operating area (SOA) of the IGBT, which can happen when there is a very high power dissipation for even a short time. It is generally destructive, but maybe you've been lucky so far.
Apparently older devices were more susceptible to second breakdown, so if you have IGBTs from old drives that may be a problem
It would be very helpful to show your complete schematic, especially the base drive components, FWDs, and any snubbers or capacitors or inductors you have in the circuit.
If you have a very high snubber capacitance across the IGBT, the turn-off may cause a transient due to the motor inductance which could result in a breakdown of the IGBT, especially if the gate drive is too "soft". Thus the IGBT will act as part of the snubber and it will dissipate the inductive energy rather than the FWD. And the FWD also might be too slow, or defective.
So if you can show exactly what you have, I might be able to help.
Oh, and another clue is that the saturation voltage seems to be decreasing with time, which also hints at slow gate drive. If the IGBT is turned on full, the voltage should start very low and then rise as the inductance saturates. And it should be no more than 1 or 2 volts. If this is to be used for a motor that will draw 500 amps, you want to limit the peak dissipation to maybe 500W. If the IGBT has a 5V drop on a little starter motor, it will surely blow up on an automotive traction motor.