Is the total energy of the universe constant?
No. The universe is dominated by dark energy, which is consistent with a cosmological constant $\Lambda$. In other words, as the universe expands, the energy density stays roughly the same. So the (energy density)*volume is growing exponentially at late times.
Although the total energy is not well defined (as the volume of the universe may be infinite), the fractional rate of growth is certainly nonzero.
You might wonder how the total energy can grow without violating energy conservation. The answer is that in general relativity, we just need $\boldsymbol{\nabla} \cdot \boldsymbol{T} = 0$, so a cosmological constant is perfectly consistent as $\boldsymbol{\nabla} \cdot \Lambda \boldsymbol{g} = 0$
For a nice explanation by Sean Carroll, see http://blogs.discovermagazine.com/cosmicvariance/2010/02/22/energy-is-not-conserved/
Energy conservation stems from Noether's theorem applied to time (i.e., time-invariance leads to energy conservation, similarly to how spatial-invariance leads to momentum conservation). Since the universe is expanding (and accelerating at that), the state of the universe today is different than it was yesterday and will be tomorrow, hence energy conservation cannot be established for the whole universe.
Locally, however, the stress-energy tensor, $$T^{\mu\nu}=\left(p+\rho\right)u^\mu u^\nu - pg^{\mu\nu},$$ will satisfy the conservation law (of energy and momentum), $$ T^{\mu\nu}{}_{;\nu}=0 $$ (derived through the Bianchi identity, the $;\nu$ subscript denotes the covariant derivatve).
Wald states (Amazon link, emphasis are his) in Chapter 4
The issue of energy in general relativity is a rather delicate one. In general relativity there is no known meaningful notion of local energy density of the gravitational field. The basic reason for this is closely related to the fact that the spacetime metric, $g_{\mu\nu}$, describes both the background spacetime structure and the dynamical aspects of the gravitational field, but no natural way is known to decompose it into its "background" and "dynamical" parts. Since one would expect to attribute energy to the dynamical aspect of gravity but not to the background spacetime structure, it seems unlikely that a notion of local energy density could be obtained without a corresponding decomposition of the spacetime metric. However, for an isolated system, the total energy can be defined by examining the gravitational field at large distances from the system. In addition, for an isolated system the flux of energy carried away from the system by gravitational radiation also is well defined.
Later, in Chapter 11,
...the most likely candidate for the energy density of the gravitational field in general relativity would be an expression quadratic in the first derivatives of the metric. However, since no tensor other than $g_{\mu\nu}$ itself can be constructed locally from only the coordinate basis components of $g_{\mu\nu}$ and their first derivatives, a meaningful expression quadratic in first derivatives of the metric can be obtained only if one has additional structure on spacetime, such as a preferred coordinate system or a decomposition of the spacetime metric into a "background part" and a "dynamical part" (so that, say one could take derivatives of the "dynamical part" of the metric with respect to the derivative operator associated with the background part). Such additional structure would be completely counter to the spirit of general relativity, which views the spacetime metric as fully describing all aspects of spacetime structure and the gravitational field.
Your question is tagged as general-relativity and cosmology, and as textbooks remark (e.g. Peebles [1]) "there is not a general global energy conservation law in general relativity theory.”
Therefore: ”The conclusion, whether we like it or not, is obvious: energy in the universe is not conserved” [2].
[1] Peebles P. J. E., 1993, Principles of Physical Cosmology (Princeton Univ. Press).
[2] Harrison E., 1981, Cosmology ( Cambridge University Press)