Would it be correct to find the RMS for the voltage and multiply with
the current because otherwise the power out will turn on and off as
the voltage does?
No that would not be correct. Look at this extract from your waveforms: -
I've focussed on the middle/maximum point of the blue voltage waveform and, I've looked at the phase difference between it (if it were a sinewave) and the current waveform. There is a 30° phase shift. That's not surprising if you have a 3 phase wye connected load; comparing line voltage and phase current for a resistive wye load always produces a 30° phase shift like this.
So, if you calculated \$V_{RMS} \times I_{RMS}\$ you would get an error.
The only way to find the power is by emulating the two-wattmeter method: -
Image from here. It measures true power put into a load on a 3 phase system.
To emulate it you need to calculate power twice (a la 2 wattmeter method). Power can be calculated by taking instantaneous values of the voltage waveform and multiplying each by the instantaneous value of the current waveform. Then average the resulting power waveform to obtain power. You need to respect that fact that the sampling of each waveform needs to be done at a rate that is several times faster than your 3 phase fundamental frequency and, given that the voltage waveform is basically rectangular in nature, it's probably sensible to go for something like 100 samples per cycle if not more).
Repeat for the "other" wattmeter and add the two powers to get true power consumed by the load. Of course, if this is just a simulation then it's a lot easier.
