"Efficiency" is perhaps not quite the word you are looking for. Proton-proton fusion actually releases more energy per mass than fusion of heavier nuclei because there is no binding energy in the source nuclei. However, an individual proton, which has only a fraction of the strong-force attraction at intermediate distances than a larger nucleus would have, requires greater pressures and temperatures to impart the energies necessary to get the particles close enough to actually react. Thus P-P fusion involves a complicated pathway (the P-P chain) in which two hydrogens interact to form a "deuteron" (a proton-neutron combination). In the core of a lightweight star that generates most of its energy through the P-P method, this is the rate-limiting step. The deuterium can then (very quickly -- in the core of a star, typically one second) add additional hydrogens to form a Helium 3 nucleus. Several reactions are possible here -- a typical one would be the addition of two He-3 nuclei to product helium-4 plus two hydrogen atoms.
You see a diagram and learn more about this at the Contemporary Physics Education Project.
The other possible reactions you mentioned proceed directly with a substantially smaller energy barrier, and are thus much, much faster at a temperature in which the typical energy of particles are below this regime (even in a star, it's the extraordinarily rare collision which actually leads to a result, lest the entire star consume its fuel in a few minutes).
So the critical factor, particularly for ease of construction of a practical reactor, is the energy necessary to get the nuclei close enough to fuse in the rate-limiting step -- this determines the temperatures and pressures which the fusion device must achieve in its interval of operation (inertial confinement fusion must obviously reach higher temperatures and pressures because the confinement is so much shorter).
Other factors to consider are -- are the end products charged or neutral? (i.e., are neutrons present?) If so, these will easily react with the wall of the reaction vessel, which can produce long-lived radioactive waste in the case of steel, or can be used to breed more fuel in the case of lithium (--> tritium). If the end products are charged, the energy in their motion (but only their motion) can be tapped at relatively high efficiency through magnetic fields (magnetohydrodynamic conversion) -- if neutral, then thermal cycles with their associated inefficiency must be used.
You can do the neutrino calculation yourself. Just look at each particle on the source side of the equation and determine whether it has an odd or even number of nucleons, and therefore is of odd or even spin. Look at the results and make the same observations. Electrons and positrons count as odd spin. See if the mixup of odd and even match up.
For example: H + H --> D + positron + ??
Notice that we have in terms of spin, odd + odd (equals even) --> even + odd (equals odd). So we've left something out -- that's the neutrino, to balance things. The reaction, already balanced for the number of heavy nucleons and for charge, still needs balancing for odd or even spin, thus you throw in a neutrino as an extra product.
Try this with the other reactions -- you'll be able to figure it out yourself.
making a small donation to science.ca.