The following result is claimed to be easy in the paper by Foxby, Bounded complexes of flat modules, 2.1:
Let $R$ be a commutative ring and $X,Y$ complexes in $D(R)$ which are cohomologically bounded above (I would assume them moreover to be perfect, i.e. bounded complexes of finitely generated free modules, up to quasi-isomorphism). If we denote by $s(X)$ and $s(Y)$ the maximal indexes $k$ such that respectively $H^k(X)\neq 0$ and $H^k(Y)\neq 0$ and if $H^{s(X)}(X)\otimes H^{s(Y)}(Y)\neq 0$, then it holds $s(X\otimes^L Y)=s(X)+s(Y)$.
Since $X$ and $Y$ are perfect, I think that we can compute the derived tensor product $X\otimes^L Y$ just as the total complex $\mathrm{Tot}(X\otimes Y)$, i.e. the complex having in degree $n$ the direct sum $\underset{i+j=n}\bigoplus X^i\otimes Y^j$, and differentials $1\otimes d+ (-1)^i d\otimes 1$. This should be since it's not really needed to take a flat quasi-isomorphic complex if we are working yet with free (and hence projective) modules (please, correct me if I'm wrong!).
Then I sould prove that when $k>s(X)+s(Y)$ one has $H^k(X\otimes^L Y)=0$, and when $k=s(X)+s(Y)$ one has $H^k(X\otimes^L Y)\neq 0$. But how this cohomology (the $k$-th cohomology of $X\otimes^LY$) is related to $H^{s(X)}(X)\otimes H^{s(Y)}(Y)$, the module that I have supposed to be non-zero? For a principal ideal domain I remember an explicit relation given by Kunneth Theorem, but I guess it should be something easier and more general...
Any help is appreciated!