The case of $n=2$ possesses an elementary closed form, which Mathematica 9 failed to find. This gives me an inkling of hope for the specific cases of the general form, as I have determined the existence of a recurrence relation between three consecutive integers.
$(n+1)I_{n+1} + I_n + nI_{n-1} = e^{\tan^{-1}{x}} x^n \sqrt{x^2+1} $
So I pose the following question: Does a closed form exist for $n \in \mathbb{Z}$ ?
If the answer is no, then a follow up is: Does a closed form exist for $n \in \mathbb{N}$ ?
And finally, if the answers to both of those are no, then does a closed form exist for either from $-1$ to $1$, or from $0$ to $1$?
Those borders are simple values, and if either exist, I can constuct a recurrence relation for the general definite integral.
For the curious reader, the integral for $n=2$ is $\frac{1}{2}e^{\tan ^{-1}(x)}(x-1) \sqrt{x^2+1} $
This is not an answer but it is too long for a comment.
Using a CAS, the result is $$I_n=\left(\frac{1}{2}-\frac{i}{2}\right) \sqrt{x^2+1} x^n e^{\tan ^{-1}(x)} \left(1-e^{2 i \tan ^{-1}(x)}\right)^{-n} \left(1+e^{2 i \tan ^{-1}(x)}\right)^{n+1}$$ $$ F_1\left(\frac{1}{2}-\frac{i}{2};n+1,-n;\frac{3}{2}-\frac{i}{2};-e^{2 i \tan ^{-1}(x)},e^{2 i \tan ^{-1}(x)}\right)$$ where appears the Appell hypergeometric function of two variables (as already mentioned by mickep.
A further simplication could be done taking into account $$\frac{1+e^{2 i \tan ^{-1}(x)}}{1-e^{2 i \tan ^{-1}(x)}}=\frac ix$$ This goes in the same direction as mickep commented about a transform of the integrand.