Closed form of this type $\sum_{j=0}^{\infty}\frac{2^jj^n}{(2j+1)(2j+3){2j \choose j}}$ Given that, $$\sum_{j=0}^{\infty}\frac{2^j\left(j-\frac{1}{3}\right)^3\left(j^2+j-1\right)}{(2j+1)(2j+3){2j \choose j}}=A\tag1$$
We have $A=2\pi+12+\frac{1}{3}?$
We can generalize the above $(1)$:
$$\sum_{j=0}^{\infty}\frac{2^jj^n}{(2j+1)(2j+3){2j \choose j}}=F(n)\tag2$$ 
I believe the closed for $(2)$ is 
$$F(n)=(-1)^{n}(2n+1)-(-1)^n (2n-1)\cdot \frac{\pi}{2}$$
How can we show that the proposed $(2)$ is correct? 

From the expansion of $\arcsin^2 t$
\begin{equation}
 \arcsin^2 t=\sum_{p=0}^\infty \frac{2^{2p}t^{2p+2}}{(2p+1)(p+1)\binom{2p}{p}}
\end{equation} 
we can obtain by differentiation 
\begin{equation}
 2\frac{\arcsin t}{\sqrt{1-t^2}}=\sum_{p=0}^\infty \frac{2^{2p+1}t^{2p+1}}{(2p+1)\binom{2p}{p}}
\end{equation} 
Multiplying the above identity by $t$ and integrating, one obtains
\begin{equation}
 2\int_0^x\frac{t\arcsin t}{\sqrt{1-t^2}}\,dt=\sum_{p=0}^\infty \frac{2^{2p+1}x^{2p+3}}{(2p+1)(2p+3)\binom{2p}{p}}
\end{equation} 
We choose $x=\sqrt{y}/2$ and divide the identity by $y^{3/2}$ to write
\begin{equation}
 \frac{8}{y^{3/2}}\int_0^{\sqrt{y}/2}\frac{t\arcsin t}{\sqrt{1-t^2}}\,dt=\sum_{p=0}^\infty \frac{y^{p}}{(2p+1)(2p+3)\binom{2p}{p}}
\end{equation} 
Now, applying the operator $y\frac{d}{dy}$ $n$ times and taking the result at $y=2$ gives
\begin{equation}
 \left.\left[ y\frac{d}{dy}\right]^n\frac{8}{y^{3/2}}\int_0^{\sqrt{y}/2}\frac{t\arcsin t}{\sqrt{1-t^2}}\,dt\right|_{y=2}=\sum_{p=0}^\infty \frac{p^{n}2^p}{(2p+1)(2p+3)\binom{2p}{p}}
\end{equation} 
The function can be evaluated as
\begin{equation}
 \frac{8}{y^{3/2}}\int_0^{\sqrt{y}/2}\frac{t\arcsin t}{\sqrt{1-t^2}}\,dt=\frac{4}{y}-8\frac{\sqrt{1-\frac{y}{4}}}{y^{3/2}}\arcsin\left( \frac{\sqrt{y}}{2} \right)
\end{equation} 
The above result can be simplified by taking $z=\sqrt{y}/2$:
 \begin{equation}
 \left.\left[\frac{1}{2} z\frac{d}{dz}\right]^n
 \left[ \frac{1}{z^2}-\frac{\sqrt{1-z^2}}{z^3}\arcsin z\right]
 \right|_{z=\sqrt{2}/2}=\sum_{p=0}^\infty \frac{p^{n}2^p}{(2p+1)(2p+3)\binom{2p}{p}}
\end{equation} 
As $\arcsin \left(\sqrt{2}/2\right)=\pi/4$ and since the successive applications of the operator on $z^{-2}$ and $z^{-3}\left( 1-z^2 \right)^k$ ($k$ is an integer) at $z=1/\sqrt{2}$ give rational results, we expect
\begin{equation}
 F(n)=a_n+b_n\pi
\end{equation} 
where $a_n$ and $b_n$ are rational. I could not find any simple expression for these coefficients  however. First few values are $F(n)=2-\pi/2,-3+\pi,5-3\pi/2,-7+5\pi/2,13-3\pi,-7+17\pi/2, 93+27\pi/2\cdots$ for $n=0,1,2,3,4,5,6\cdots$. They do not correspond to the propose formula as remarked by @Mariusz Iwaniuk in a comment.
With $z=\exp(t/2)$, we have $1/2zd/dz=d/dt$, and we can build the generating function for the $F(n)$:
\begin{equation}
 \sum_{n=0}^\infty F(n)\frac{t^n}{n!}=\phi\left( t-\ln2 \right)
\end{equation} 
where
\begin{equation}
 \phi\left( t-\ln 2 \right)=2e^{-3t/2}\left[ e^{t/2}-\sqrt{2-e^{t}}\arcsin\left(\frac{ e^{t/2}}{\sqrt{2}} \right)\right]
\end{equation} 
