Action and Action integral: Different kinds of variational principles What are the difference between: 


*

*the action $\int_{t_{1}}^{t_{2}}(L+H) dt$ that we use in the principle of least action, and 

*the action integral $\int_{t_{1}}^{t_{2}}L dt$ that we use in Hamilton's variational principle?
References:


*

*H. Goldstein, Classical Mechanics, Section 8.6.

*H. Goldstein, Classical Mechanics, Chapter 2.
 A: As user ACuriousMind correctly writes: 


*

*What Goldstein calls the principle of least action $\int p~\mathrm{d}q$ is usually called Maupertuis' principle or the principle of abbreviated action. 

*What Goldstein calls the Hamilton's variational principle is often also called the the principle of least/extremal/stationary action $\int L~\mathrm{d}t$.
This is also explained in a footnote in Goldstein, Section 8.6. At the physical level, besides the different appearances, the important thing is to realize that different quantities are kept fixed in the two variational principles 1 and 2.
Also note that confusingly the Hamilton's principle is a variational principle in the Lagrangian formulation (as opposed to the Hamiltonian formulation). 
Finally, let us remark that the word action has several different meanings in physics and mathematics.
A: The more common names for what you are talking about are the abbreviated action
$$S_0[q] := \int p \mathrm{d}q$$
versus the action
$$ S[q] := \int_{t_1}^{t_2}L(q,\dot q,t)\mathrm{d}t$$
Both are used in different formulations of classical mechanics, and deliver a different "flavor" of solutions. On both one can do variations calculus and obtains the classical trajectory $q_{cl}$ as the extremum of the action.
The abbreviated action obeys Maupertuis' principle and gives you the classical path provided you know the conserved energy along the trajectory and the start- and endpoints.
The action's extrema are found by the usual Euler-Lagrange equations and gives you the classical path provided you know the start- and endtimes as well as points.
Now, what has the abbreviated action to do with $L + H$? Observe that 
$$\int p \mathrm{d}q =\int p \dot q \mathrm{d}t = \int (L + H) \mathrm{d}t$$
since $L$ and $H$ are Legrendre transforms w.r.t. $\dot q$ resp. $p$ of each other.
