A conjectural closed form for $\sum\limits_{n=0}^\infty\frac{n!\,(2n)!}{(3n+2)!}$ Let
$$S=\sum\limits_{n=0}^\infty\frac{n!\,(2n)!}{(3n+2)!},\tag1$$
its numeric value is approximately $S \approx 0.517977853388534047...$${}^{[more\ digits]}$
$S$ can be represented in terms of the generalized hypergeometric function:
$$S={_3F_2}\left(\frac12,1,1;\ \frac43,\frac53;\ \frac4{27}\right)\cdot\frac12.\tag2$$

Let $\sigma$ be the closed-form expression constructed from integers and elementary functions as follows:
$$\sigma=3\,\alpha\,\ln(2\,\alpha+1)-\sqrt{\beta\,}\arccos\gamma,\tag3$$
where
$$\alpha=\frac{\sqrt[3]{3\,}}{6\,\sqrt[3]{2\,}}\left(\sqrt[3]{9-\sqrt{69}}+\sqrt[3]{9+\sqrt{69}}\right),\tag4$$
$$\beta=\frac1{4\,\sqrt[3]{2\,}}\left(\sqrt[3]{25+3\,\sqrt{69}}+\sqrt[3]{25-3\,\sqrt{69}}\right)-\frac12,\tag5$$
$$\gamma=\frac1{6\,\sqrt[3]{2\,}}\left(\sqrt[3]{57\,\sqrt{69}-459}-\sqrt[3]{57\,\sqrt{69}+459}\right)+\frac12\tag6$$
are the unique real roots of the following cubic equations:
$$8\,\alpha^3-2\,\alpha-1=0,\tag7$$
$$64\,\beta^3+96\,\beta^2+36\,\beta-23=0,\tag8$$
$$8\,\gamma^3-12\,\gamma^2+16\,\gamma+11=0.\tag9$$
Equivalently,
$$\sigma = \frac{3\,p}{2}\,\ln\big(p+1\big)-\frac{1}{2}\sqrt{\frac{3-p}{p}}\arccos\Big( \frac{p-6}{6p+2}\Big)\tag{10}$$
where $p$ is the plastic constant or the real root of
$$p^3-p-1=0\tag{11}$$

It can be numerically checked that the following inequality holds:
$$\Big|S-\sigma\Big|<10^{-10^5},\tag{12}$$
I conjecture that the actual difference is the exact zero, and thus $S$ has an elementary closed form:
$$\sum\limits_{n=0}^\infty\frac{n!\,(2n)!}{(3n+2)!}\stackrel?=3\,\alpha\,\ln(2\,\alpha+1)-\sqrt{\beta\,}\arccos\gamma,\tag{13}$$
I am asking for you help in proving this conjecture.
 A: Alternatively, one can find a simple and elementary solution in logarithms as,
$$\sum\limits_{n=0}^\infty\frac{n!\,(2n)!}{(3n+2)!}=  -x_1\ln(1-x_1)-x_2\ln(1-x_2)-x_3\ln(1-x_3) = 0.5179778\dots$$
where the $x_n$ are the three roots of $x^3-x+1=0$
and which is the minpoly of the negated plastic constant. See also this post.
A: We show that the sum equals
$$
\int_0^1 \frac{2-3x}{1-x^2+x^3} dx.
$$
This integral is "elementary", but requires expanding the integrand
in partial fractions, which in turn requires all the solutions of
the cubic polynomial in the denominator; so if one insists on
writing everything in radicals then the answer is bound to be complicated.
The "conjecture" is surely correct ($10^5$ digits is more than enough
for moral certainty, especially since $\alpha,\beta,\gamma$ are all
in the field generated by the real root of $1-x^2+x^3$), though
it may be an unpleasant and unrewarding exercise to check that
the partial-fraction integration yields an equivalent answer.
(One also wonders how one could possibly "conjecture" such an answer
without some sense of where to look...)
The key is to write each term $n! (2n)! / (3n+2)!$ in terms of the
beta integral
$a!b!/(a+b+1)! = B(a+1,b+1) = \int_0^1 x^a (1-x)^b dx$.  Here we write
$n! (2n)! / (3n+2)! = B(2n+1, n+2) / (n+1)$, and sum over $n$ to get
$$
\sum_{n=0}^\infty \frac{n! (2n)!} {(3n+2)!}
= \int_0^1 \sum_{n=0}^\infty \frac{(x^2-x^3)^{n+1}}{n+1} \frac{dx}{x^2}
= -\int_0^1 \log(1-x^2+x^3) \frac{dx}{x^2}.
$$
(We easily justify the interchange of infinite sum and definite integral
because all integrands are positive on $0<x<1$.)
We can now integrate by parts to remove the logarithm:
$$
-\int_0^1 \log(1-x^2+x^3) \frac{dx}{x^2}
= \int_0^1 \log(1-x^2+x^3) \phantom.  d\left(\frac{1}{x}\right)
= \int_0^1 \frac1x d(\log(1-x^2+x^3)),
$$
in which the integrand simplifies to $(2-3x)/(1-x^2+x^3)$,
QED.
A: partial answer 
Use the method of LINK
Since
$$
\int_0^1 t^n (1-t)^{2n}\;dt = \frac{n!(2n)!}{(3n+1)!} ,
$$
If we write
$$
S(x) = \sum_{n=0}^\infty \frac{n!(2n)!}{(3n+2)!}\;x^{3n+2}
$$
then
$$
S(x) = \int_0^1 \left(\sum_{n=0}^\infty t^n(1-t)^{2n} \frac{x^{3n+2}}{3n+2}\right)dt
$$
and $S = S(1)$.
But the derivative with respect to $x$ of $\sum_{n=0}^\infty t^n(1-t)^{2n} \frac{x^{3n+2}}{3n+2}$ is a geometric series.  Its sum is a rational function, which may be integrated (with some work or a CAS).  Plug in $x=1$.  Result (if I copied right):
$$
S = \int_0^1\frac{F(t)}{18(1-t)^{4/3}t^{2/3}}\;dt
$$
where
$$
F(t) = \pi \,\sqrt {3}-6\,\sqrt {3}\arctan
 \left( 2/\sqrt {3}\cdot \left( 1-t \right) ^{2/3}{t}^{1/3}+1/
\sqrt {3} \right) \\
-6\,\ln  \left( ({1-t})^{1/3}
{t}^{2/3} -\left( 1-t \right) t \right) +3\,\ln  \left( t \left( 1-t \right) ^{2}+ \left( 1-t \right) ^{4/3}{t}^{2/3}+ \left( 1-t \right) ^{2/3}{t}^{1/3}
 \right) \\
+6\,\ln  \left( ({1-t})^{1/3}{t}^{2/3} \right) -3\,\ln 
 \left(  \left( 1-t \right) ^{2/3}{t}^{1/3} \right) 
$$
