I've heard from Zhi-Wei Sun that he recently considered this question. In a post a few days ago to OEIS 	 [Least integer b>2n+1 such that the numbers written as [1,3,...,2n-1,2n+1] and [2n+1,2n-1,...,3,1] in base b are both prime.][1] He gives the first of what he conjectures are infinitely many bases (for each fixed $n$) with the named property. Other fairly specific conjectures concerning Galois groups, reducibility over $\mathbb{Z}_p$ and the like can be found on that page. Any one of the conjectures would imply that $1+3x+5x^2+\cdots+(2n+1)x^{n}$ is always irreducible over the integers. A [similar post][2] a few days earlier than that concerns $1+2x+\cdots+(n+1)x^{n}$ which he would also conjecture is always irreducible over the integers.

Of course there are integer arithmetic progressions such that $f(a,b,n)=\sum_0^n(a+bk)x^k$ does factor (with $a \ne 0$ of course). At least there is $1+x+x^2+\cdots+x^n$  which is irreducible when and only when $n+1$ is prime. That is not the only case but other examples are no so easy to come up with. A fairly simple minded search over small parameters turns up 

 - $-n+(2-n)x+(4-n)x^2+\cdots+nx^n$ which has $(x-1)$ as a factor (and $(x+1)$ for even $n$) but no other factors up to $n=42$
 - After some manipulation, the integer quadratic examples can written $s(t-2s)+(t^2-s^2)x+t(2t-s)x^2$ with linear factor $(s+tx)$
 - One can first work over $\mathbb{Q}$ , stipulate a factor $x-r=x-t/s$ , set $c_{n-1}=1$ and then solve for $c_0,\cdots,c_{n-2}$ such that  $(x-t/s)(c_0+c_1x+\cdots+c_{n-1}x^{n-1})=f(a,b,n)$ for $a=c_0-rc_1$ and $b=r+1-c_{n-2}$. The solutions will have $c_i$ rational functions in $r$ with denominator $n+(n-1)r+\cdots+r^{n-1}$. Then one can scale to integer examples. However it is not immediate how to parameterize for pleasant integer solutions.
 - Perhaps there are nice solutions which are reducible but without a linear factor.

  [1]: http://oeis.org/A218465
  [2]: http://oeis.org/A217785