Is it more efficient to copy a vector by reserving and copying, or by creating and swapping?

They aren't the same though, are they? One is a copy, the other is a swap. Hence the function names.

My favourite is:

a = b;

Where a and b are vectors.

Your second example does not work if you send the argument by reference. Did you mean

void copyVecFast(vec<int> original) // no reference
{

  vector<int> new_;
  new_.swap(original); 
}

That would work, but an easier way is

vector<int> new_(original);

This is another valid way to make a copy of a vector, just use its constructor:

std::vector<int> newvector(oldvector);

This is even simpler than using std::copy to walk the entire vector from start to finish to std::back_insert them into the new vector.

That being said, your .swap() one is not a copy, instead it swaps the two vectors. You would modify the original to not contain anything anymore! Which is not a copy.