The part of your question about the non-visible light gives quite surprising (to me at least) results! :  

Compare the power from the sun with the cosmic microwave background ("CMB"), taking both as blackbody radiators:  
The temperatures are 2.725 and 5778 K respectively (wikipedia :))  
The solid angle of the CMB is obviously $4\pi$. The diameter of the sun as seen from the earth is 32' (= 32/60 degrees) according to wikipedia, which gives a solid angle of $6.8\cdot 10^{-5}$. This would give a ratio in the angles of $5.4\cdot 10^{-6}$.  
(To be precise, we need a factor of 1/2 more here, since the sunlight is absorbed only by half of the earth, whereas the CMB is always "shining". So I will use a ratio of $2.7\cdot 10^{-6}$. Ignore the factor 1/2 if this reasoning seems too complicated)

So we get a ratio of the radiation power of $$\frac {P_\mathrm{sun}}{P_\mathrm{CMB}} = \frac {T_\mathrm{sun}^4}{T_\mathrm{CMB}^4} \cdot \frac {\Omega_\mathrm{sun}}{\Omega_\mathrm{CMB}} = 2\cdot 10^{13} * 2.7\cdot 10^{-6} = 54\,\mathrm{million}$$

Comparing with 125 million from SpiderPig (which is definitely the more precise answer), I get the astonishing result, that the CMB "shines" more than twice as powerful than all the other stars combined!