Need some help expressing this as a series, if possible

This is just for your curiosity

$U(x)$ is already expressed as a linear combination of two Taylor series (for which you give only the first terms but in which the patterns of the coefficients is quite clear). I suppose that this is the power series solution of a second oreder differential equation which could be intresting to know.

So, what I think is that the problem would better be to find the analytic functions to which correspond these Taylor series.

The first one is "amazing" since it seems that we have $$f(x)=\sum_{n=0}^\infty (-1)^n \frac {c_n}{(2n)!} x^{2n}\quad \text{with} \quad c_n=\prod_{i=0}^{2n-1} (a+i)=a \,(a+1)_{2 n-1}$$ where appear Pochhammer symbols. This makes $$f(x)=\, _2F_1\left(\frac{a}{2},\frac{a+1}{2};\frac{1}{2};-x^2\right)$$ where appears the gaussian hypergeometric function. But, guess what ? $$\color{blue}{f(x)=\left(1+x^2\right)^{-\frac a2} \cos \left(a \tan ^{-1}(x)\right)}$$

For the second series, it seems to be much more complex since we have $$g(x)=\sum_{n=0}^\infty (-1)^n \frac {d_n}{(2n+1)!} x^{2n+1}$$ with $$d_n=\Bigg[\prod_{i=0}^{n} (a+2i)\Bigg]\,\, \Bigg[\prod_{i=0}^{n-1} (a+2i-1)\Bigg]=(-1)^{n+1}(a-1) 2^{2 n-1} \left(\frac{3-a}{2}\right)_{n-1} \left(\frac{a+2}{2}\right)_n$$ $$\color{blue}{g(x)=x \, _2F_1\left(\frac{1-a}{2},\frac{a+2}{2};\frac{3}{2};x^2\right)}$$ which, unfortunately, does not simplify but explains why was required a power series solution.

For integer values of $a$, we have quite nice and simple expressions $$\left( \begin{array}{cc} a & \left(x^2+1\right)^a\,f(x) \\ 1 & 1 \\ 2 & 1-x^2 \\ 3 & 1-3 x^2 \\ 4 & x^4-6 x^2+1 \\ 5 & 5 x^4-10 x^2+1 \\ 6 & -x^6+15 x^4-15 x^2+1 \end{array} \right)$$

For $g(x)$, it is a bit more complex but it write $$g(x)= x Q(x)+ R(x) \tanh^{-1}(x)$$ $$\left( \begin{array}{ccc} a & Q(x) & R(x) \\ 1 & 1 & 0 \\ 2 & \frac{3}{4} & -\frac{3 x^2}{4}+\frac{1}{4} \\ 3 & -\frac{5 x^2}{3}+1 & 0 \\ 4 & -\frac{105 x^2}{64}+\frac{55}{64} & \frac{105 x^4}{64}-\frac{45 x^2}{32}+\frac{9}{64} \\ 5 & \frac{21 x^4}{5}-\frac{14 x^2}{3}+1 & 0 \\ 6 & \frac{1155 x^4}{256}-\frac{595 x^2}{128}+\frac{231}{256} & -\frac{1155 x^6}{256}+\frac{1575 x^4}{256}-\frac{525 x^2}{256}+\frac{25}{256} \end{array} \right)$$