Even a low-order symplectic integrator will conserve the phase-space volume of a problem, and as such, never introduce or remove energy systematically. Below is a demo of how a second order symplectic integrator can outperform a fourth order non-symplectic integrator. Beware, of course, that if your Hamiltonian is not formulated to conserve a quantity, a symplectic integrator will not guarantee that quantity is conserved (in this case, the phase angle of our orbit).