Let $X$ be a (smooth, connected) aspherical manifold. Let $LX:=Map(S^1,X)$ be the free loop space of $X$. Pick $x_0\in X$ and let $\Omega_{x_0}(X)$ be the based loop space of $X$ (based at $x_0$).

One has $$LX=\coprod_{\alpha\in A}L_\alpha X,$$ where $A$ is the set of conjugacy classes in $\pi_1(X,x_0)$, and $L_\alpha X$ is the space of loops in the free homotopy class corresponding to $\alpha$. Using the fibration $\Omega_b(X)\hookrightarrow LX\to X$ one can show that, since $X$ is aspherical, $$L_\alpha X\simeq K(C_{[\gamma_\alpha]},1),$$ where $[\gamma_\alpha]$ is any representative of $\alpha$ in $\pi_1(X,x_0)$, and $C_{[\gamma_\alpha]}$ is the centralizer of $[\gamma_\alpha]$.

It follows that for each $\alpha$, $L_\alpha X$ is homotopy equivalent to a covering space of $X$. There are two extremal cases:

 - If $X$ admits a Riemannian metric with negative sectional curvature, then for any non-contractible class $\alpha$, $C_{[\gamma_\alpha]}$ is infinite cyclic, generated by $[\gamma_\alpha]$. So $L_\alpha X\simeq S^1$.
 - If $\pi_1(X)$ is abelian, for example if $X$ is a torus, then for any $\alpha$, $L_\alpha X\simeq X$.

My question is: 

Are there computable examples, other than the two cases above (and products of these two cases), of the homotopy type (or just homology) of $LX$? (I want the topology of $LX$ to be as simple as possible.) Also, I $X$ of even dimensions.

This might be a question in Riemannian geometry, algebraic topology, group cohomology... The homotopy type of $LX\simeq LB\Gamma$ is determined by $\Gamma=\pi_1(X,x_0)$.

Many thanks!