Antiderivative of totally real polynomials Let us say that a polynomial with real coefficients is totally real if all its complex roots are real and distinct. Let $P \in \Bbb R [X]$ be totally real. Is it true that
$$Q(X)=\int_0^XP(t)\,dt+aP(X)$$
is also totally real for all real $a$? If not, is this true
if one adds the condition $P(0)=0$? Or some other additional condition?
As an additional question, can one give an upper bound on the roots of
$Q$ in terms of those of $P$?

EDIT: Sorry, this is false in general, it is in fact easy to give counterexamples. My question is then: what reasonable additional condition should one add ?
Sorry if this is an elementary question.
 A: Nope. Lets look at the case a=0.
If you have a positive quadratic polynomial, at its second root the antiderivative has a minimum. It is easy to construct an example where this minimum is positive.
As an example take $P(x)=(x-3)^2-1=x^2-6x+8$.
Its antiderivative is $Q(x)=\frac{1}{3}x^3-3x^2+8x=x(\frac{1}{3}x^2-3x+8)$. It has only one real root at $x=0$, because the discriminant of the quadratic term is negative.
Adding the condition $P(0)=0$ doesn't help. Firstly, in that case the antiderivative has a double root at $x=0$, since $Q(0)=0$ by construction and $Q'(0)=P(0)=0$. So $Q$ isn't totally real.
Secondly you can constuct a $P$, for which the antiderivative has non-real roots: Let $R(x)=x(x-1)(x-2)$. $R$ has real roots at $x=0,1,2$. Its antiderivative has a minimum and double root at $x=2$. Now for some small enough $\epsilon>0$, $P(x)=R(x)+\epsilon x$ still has three distinct real roots, but the minimum of the antiderivative goes above $0$.
These constructions also hold for all $a$, that are small enough that the minimum of the antiderivative stays above 0.

EDIT: I overlooked your EDIT, sorry
A: Not an answer but there are linear operators sending any totally real $p(x)$ of degree $n$ to another one with degree $n+1$ with interlacing roots.  If $a,t$ are real numbers with $t>0$ and $p(x)=\prod_{j=1}^n(x-c_j)$ is totally real, and we let $q(x):=(x+a-t\frac{d}{dx})p(x)$, then $q/p$ has partial fraction
$f(x):=q(x)/p(x)=x+a-\sum_{j=1}^n\frac{t}{x-c_j}$. The graph of $f$ has $n+1$ vertical monotonically increasing branches asymptotic to $y=x$ at $\pm \infty$. For any real $c$, the numerator of  $f(x)-c$  defines a degree $n+1$ totally real polynomial interlacing $p$ (in particular the polynomial $q$ for $c=0$) with roots insides the intervals delimited by $c_j$. If we have real sequence $a_j,t_j>0,j=1,2,3...$, starting with $p(x)$ of any degree, one define in this way infinitely many interlacing sequences
$\prod_{j=1}^m(x+a_j-t_j\frac{d}{dx})p(x),m=1,2,...$ of totally real polynomials. This construction generalizes the classical (probabilist's) Hermite polynomial which is just $H_n(x)=(x-\frac{d}{dx})^n[1]$. In particular, this prove $H_n(x)$ are totally real with interlacing roots.
