<s>[INST] Ordinary generating function for $\binom{3n}{n}$ The ordinary generating function for the central binomial coefficients, that is, $$\displaystyle \sum_{n=0}^{\infty} \binom{2n}{n} x^{n} = \frac{1}{\sqrt{1-4x}} \, , \quad |x| < \frac{1}{4},$$
can be derived by using the duplication formula for the gamma function and the generalized binomial theorem.
But what about the ordinary generating function for $ \displaystyle  \binom{3n}{n}$?
According to Wolfram Alpha,   $$ \sum_{n=0}^{\infty} \binom{3n}{n} x^{n} = \frac{2\cos \left(\frac{1}{3} \arcsin \left(\frac{3 \sqrt{3x}}{2} \right)\right)}{\sqrt{4-27x}} \, , \quad |x| < \frac{4}{27}. $$
Any suggestions on how to prove this?
EDIT:
Approaching this problem using the fact that $$ \text{Res} \Big[ \frac{(1+z)^{3n}}{z^{n+1}},0 \Big] = \binom{3n}{n},$$
I get $$ \sum_{n=0}^{\infty} \binom{3n}{n} x^{n} = -\frac{1}{2 \pi i x} \int_{C} \frac{dz}{z^{3}+3z^{2}+3z - \frac{z}{x}+1},$$
where $C$ is a circle centered at $z=0$ such that every point on the circle satisfies $ \displaystyle\Big|\frac{x(1+z)^{3}}{z} \Big| < 1$.
Evaluating that contour integral would appear to be quite difficult.
 [/INST] 
Proposition : $$ f(z,a) = \int_{0}^{\infty} \dfrac{x^z}{x^2 +2ax +1} \mathrm{d}x = \dfrac{\pi}{\sin \pi z} \dfrac{\sin((1-z) \cos^{-1}(a))}{\sin(\cos^{-1}(a))}$$

Proof : Note that,
$$ \sum_{n=0}^{\infty}{{U}_{n}(a) {(-x)}^{n}} = \dfrac{1}{x^2 +2ax+1} $$
where $ U_{n} (x) $ is the Chebyshev Polynomial of the second kind.
$$ \implies \sum_{n=0}^{\infty}{{U}_{n}(a) \Gamma(n+1) \frac{{(-x)}^{n}}{n!}} = \frac{1}{x^2 +2ax +1} $$
Using Ramanujan Master Theorem, we have,
$$ f(z,a) = \dfrac{\pi}{\sin \pi z} U_{-s} (a) $$
$$ = \dfrac{\pi}{\sin \pi z} \dfrac{\sin((1-z) \cos^{-1}(a))}{\sin(\cos^{-1}(a))} \quad \square $$
Now, using Gamma Triplication Formula,
$$ (3n)! \; = \; \Gamma(3n+1) \; = \; \dfrac{1}{2\pi} 3^{2n+\frac12} \Gamma \left(n+\dfrac13\right)\Gamma \left(n+\dfrac23\right)\Gamma(n+1) $$
and hence
$$ {3n \choose n} \; = \; \frac{3^{3n+\frac12}}{2\pi} \times \frac{\Gamma \left(n+\dfrac13\right)\Gamma \left(n+\dfrac23 \right)}{\Gamma(2n+1)} \; = \;\frac{3^{2n+\frac12}}{2\pi} \operatorname{B} \left(n+\dfrac23,n+\dfrac13\right) $$
Thus,
$$ \begin{array}{rcl}\displaystyle \text{S} \; =\; \sum_{n=0}^\infty {3n \choose n}x^n & = & \displaystyle\frac{\sqrt{3}}{2\pi}\sum_{n=0}^\infty 3^{2n} x^n \int_0^1 u^{n-\frac13}(1-u)^{n-\frac23}\,du \\ & = & \displaystyle \frac{\sqrt{3}}{2\pi} \int_0^1 \frac{u^{-\frac13}(1-u)^{-\frac23}}{1 - 27xu(1-u)}\,du \; =\; \frac{\sqrt{3}}{2\pi}\int_0^1 \frac{u^{-\frac13}(1-u)^{-\frac23}}{1 - 4a^2\, u(1-u)}\,du \end{array} $$
where $a = \dfrac{3}{2} \sqrt{3x} $. Substituting $u = \sin^2\theta$ and then $\tan\theta = t$, we have,
$$\text{S} = \dfrac{\sqrt{3}}{2 \pi} \int_0^\infty \frac{t^{\frac{1}{3}}(1 + t^2)}{(t^2 + 2at + 1)(t^2 - 2at + 1)} \mathrm{d}t  $$
Using Partial Fraction and the Proposition, we have,
$$\text{S} = \dfrac{\sqrt{3}}{2\pi} \left[ \dfrac{1}{2} f \left( \dfrac{1}{3} , -a \right) + \dfrac{1}{2} f\left( \dfrac{7}{3} , -a \right) - \dfrac{1}{4a} f\left( \dfrac{4}{3} , -a \right) - \dfrac{1}{4a} f\left( \dfrac{10}{3} , -a \right) + \dfrac{1}{2} f \left( \dfrac{1}{3} , a \right) + \dfrac{1}{2} f\left( \dfrac{7}{3} , a \right) + \dfrac{1}{4a} f\left( \dfrac{4}{3} , a \right) + \dfrac{1}{4a} f\left( \dfrac{10}{3} , a \right)\right]  $$
After simplification, we have,
$$ \text{S} = \dfrac{2\cos \left(\frac{1}{3} \sin^{-1} \left(\dfrac{3\sqrt{3x}}{2} \right)\right)}{\sqrt{4-27x}} \quad \square $$
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