Good upper-bound for $\mathbb E[|X-np|^r]$ where $X \sim \text{Binomial}(n,p)$ and $r \ge 1$ Disclaimer. Question moved from SE.
Setup
Let $X \sim \text{Binomial}(p, n)$, and $r \ge 1$.
Question
What is a good upper-bound for $\mathbb E[|X-np|^r]$ ?
Solution for small $r$


*

*If $r=2$, then $\mathbb E[|X-np|^2] = np(1-p)$.

*If $1 \le r < 2$, then Jensen's inequality gives
$$
\mathbb E[|X-np|^r] =(\mathbb E[|X-np|^r])^{2/r})^{(r/2)} \le (\mathbb E[|X-np|^2])^{r/2} = (np(1-p))^{(r/2)}.
$$
Notes
I'm ultimately interested in bounding (via McDiamid's inequality) the $\ell_r$-distance between a distribution $P$ and it's empirical version $\hat{P}_n$.
 A: By the main result of the paper Exact Rosenthal-type bounds, we have 
$$E|X-np|^r\le c^r E|\Pi_\lambda-\lambda|^r
$$
for real $r\in(2,\infty)\setminus(3,4)\setminus(4,5)$, where real $c>0$ and $\lambda>0$ are defined by the conditions 
$$c^r\lambda=n(q^rp+p^rq)\quad \text{and}\quad c^2\lambda=npq;
$$
$q:=1-p$; and $\Pi_\lambda$ is a Poisson random variable with parameter $\lambda$. 
Other results on Rosenthal-type bounds can be found e.g. in this paper or its  arXiv version, and in references therein. 
Added: In a comment, the OP stated that it may be assumed that $1\le r\le 2$. This simplifies the matter a great deal. Indeed, in this case we have 
\begin{equation}
 E|X-np|^r\le\min((npq)^{r/2},2npq(q^{r-1}+p^{r-1}))\ll s^r\wedge s^2,\tag{1}
\end{equation}
where $(npq)^{r/2}$ is the bound the OP obtained by using Jensen's inequality; and $2npq(q^{r-1}+p^{r-1})$ is a bound immediately obtained by using the von Bahr--Esseen inequality -- see e.g. this paper or its arXiv version; $s:=\sqrt{npq}$. For positive expressions $e_1$ and $e_2$, we write $e_1\ll e_2$ or, equivalently, $e_2\gg e_1$ if $e_1\le C e_2$ for some universal positive real constant $C$. 
The upper bound on $E|X-np|^r$ in (1) is optimal up to a universal constant factor: for $r\in[1,2)$
\begin{equation}
 E|X-np|^r\gg s^r\wedge s^2.\tag{2}
\end{equation}
This lower bound on $E|X-np|^r$ is obtained by using the log-convexity of $m_t:=E|X-np|^t$ in $t>0$ and an upper Rosenthal-type bound -- as follows: By (say) Theorem 1.5 in the already cited paper Exact Rosenthal-type bounds, we have $m_3\ll s^3\wedge s^2$. Now using the log-convexity of $m_t:=E|X-np|^t$ in $t>0$ or, equivalently, the Hölder inequality, we have 
\begin{equation*}
 m_2\le m_r^{1/a}m_3^{1-1/a},
\end{equation*}
where $a:=3-r$. Hence,
\begin{equation*}
 E|X-np|^r=m_r\ge m_2^{3-r}m_3^{r-2}\gg s^{6-2r}(s^{3r-6}\wedge s^{2r-4})
=s^r\wedge s^2. 
\end{equation*}
Thus, (2) indeed holds. 
