Function equal to infinite series $\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n+3)(2n+1)!}$ I'd like to know if there is a simple function equivalent of $$\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n+3)(2n+1)!}$$
I recognize that it looks similar to $\frac{\sin{x}}{x}$, but with an extra $(2n+3)$ factor in the denominator that I don't know how to account for. I've graphed the sum from $n=0$ to $100$ on an online graphing calculator and tried to find a function that matches it, but I've had no luck. I noticed that $\frac{\sin{x}}{3x}$ looks similar, but it is a very naive guess and I don't have a better one. It may even be the case that such a function does not exist.

$$f(x)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n+3)(2n+1)!}$$
$$x^3f(x)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+3}}{(2n+3)(2n+1)!}$$
$$\big[x^3f(x)\big]'=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+2}}{(2n+1)!}=x\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}=x \sin(x)$$
$$x^3f(x)=\int x\sin(x)\,dx=\sin (x)-x \cos (x)$$
$$f(x)=\frac{\sin (x)-x \cos (x) }{x^3}$$
