There is not. In fact, there isn't even an inefficient way!
That is, the problem of determining whether two given CFGs represent the same language is undecidable. In fact, an even stronger statement is true: the problem of determining whether a given CFG accepts all strings on its alphabet is undecidable.
The proof of this can be found in chapter 5 of Sipser's Introduction to the Theory of Computation. The basic idea is that, for any Turing machine $M$, we can obtain a context-free grammar which accepts all strings that do not encode a proof that $M$ halts (under some specific encoding that's more complicated than I really want to get into here). So determining whether this grammar accepts all strings is equivalent to solving the halting problem for $M$.