Smooth four-manifolds with contractible universal cover

Let $X$ be a smooth compact four-manifold with definite non-trivial intersection form. Can the universal cover of $X$ be contractible?

It semms to me that the answer is negative when $X$ is simply connected using results of Freedman and Donaldson. Is anything known when $X$ is not simply connected? Donaldson proved that also in this case the intersection form is diagonalizable over $\mathbb Z$.

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I don't understand your second paragraph: if $X$ is simply connected, and has non-trivial intersection form, then $X$ is its own universal cover, so it is not contractible. There is no need for appealing to Freedman or Donaldson. –  Ian Agol Dec 27 '11 at 3:38
You're right: this second paragraph makes no sense, as it stands. In fact I was thinking of the following question: Let $X$ be a smooth compact four-manifold with definite non-trivial intersection form. Then the intersection form is diagonalizable over $\mathbb Z$. Does it follow that $X$ is homeomorphic to a connected sum of a manifold $Y$ with trivial intersection form and a finite number of $\mathbb C \mathbb P^2$-s (with direct or reversed orientation)? If this was the case, one could relate the universal cover of $X$ to that of $Y$ and try to show that the first is not contractible. –  Matei Jan 5 '12 at 13:28

Thank you very much for your answer! The situation is more complex than I was imagining. I would be interested indeed to know whether also examples with $n$ bigger than $1$ exist. –  Matei Dec 15 '11 at 21:02