Actually, most of the power will be due to how the ON resistance of the FET effects the driven internal bipolar transistor at any given frequency and power. People seem to think of IGBT's as a simple device, when they are in fact compound devices consistance of the driving FET internal biasing and the BJT itself. The formula to calculate IGBT power losses is probably isanely complex, and would require detailed knowledge of both the materials and structure of the IC itself, IGBT's are IC's not "trasistors" in the classic sense. Much the same as a Darlington pair, only the initial input stage is a FET instead.