Finding the general Taylor Series of a function

Oh yes, substituting $t=-6x^2+x^4$ into this and "simplifying" looks like a lot of fun, indeed. =D Just kidding. Let's remember the generating function of Legendre polynomials: $$\frac1{\sqrt{1-2xt+t^2}}=\sum^\infty_{n=0}P_n(x)\,t^n.$$ Multiplying with $t-x$ gives $$\frac{t-x}{\sqrt{1-2xt+t^2}}=-x+\sum^\infty_{n=1}[P_{n-1}(x)-x\,P_n(x)]\,t^n, \tag{1}$$ and the recurrence relation (http://mathworld.wolfram.com/LegendrePolynomial.html, (43)) gives $$P_{n-1}(x)-x\,P_n(x)=(n+1)\frac{x\,P_n(x)-P_{n-1}(x)}n,$$ so integrating this from $0$ to $t$, we obtain $$\sqrt{1-2xt+t^2}=1-xt+\sum^\infty_{n=2}\frac{x\,P_{n-1}(x)-P_n(x)}{n-1}t^n \tag{2}.$$ Substituting $x=3$ in (2), we arrive at $$\sqrt{1-6t+t^2}=1-3t+\sum^\infty_{n=2}\frac{3\,P_{n-1}(3)-P_n(3)}{n-1}t^n \tag{3}.$$ For the final result, we substitute $t=z^2$ to obtain $$\sqrt{1-6z^2+z^4}=1-3z^2+\sum^\infty_{n=2}\frac{3\,P_{n-1}(3)-P_n(3)}{n-1}z^{2n}.$$ BTW, $P_n(3)$ is always an integer, because (http://mathworld.wolfram.com/LegendrePolynomial.html, (33)) $$P_n(3)=\sum^n_{k=0}\binom{n}{k}^2 2^k.$$ The coefficients in (3) seem to be integers, too, at least that's what numerical results suggest (http://swift.sandbox.bluemix.net/#/repl/597d0432c3917c7502ddfa97).

hint

Put $t=x^4-6x^2$.

we check that when $x\to 0, \;\; t $ goes also to zero.

so we can use the expansion of $$\sqrt {1+t}=1+t/2+.... $$ where we replace $t $ by $x^4-6x^2$.

and use $$t^n=x^{2n}\sum_{k=0}^n\binom {n}{k}x^{2k}(-6)^{n-k} $$

Substitute in the sums and you get

$$\sqrt{x^4-6x^2+1}=\sum_{n=0}^\infty a_{2n}x^{2n}$$

where

$$a_{2n}=\sum_{k=0}^n\binom{1/2}{n-k}\binom{n-k}k(-6)^{n-k}$$