jmc

Let us first define $y=x^2$. We can then plug this value into the inequality and add 4 to $-4$, $x^4+4x^2$, and 21, yielding $$0<y^2+4y+4<25.$$We can factor $y^2+4y+4$ to obtain $$0<(y+2{)}^2<25.$$Taking the square root gives us $0<|y+2|<5$, which in turn gives us two intervals for solutions of $y$: $-2<y<3$, or $-7<y<-2$.

However, $y$ must be non-negative since $y=x^2$, so we have $0\le y<3$. This means that $-\sqrt{3}<x<\sqrt{3}$ satisfies the original inequality. In interval notation, this is $\boxed{(-\sqrt{3},\sqrt{3})}$.