José is correct in his comment. Just to elaborate: in the linear case, one can easily study the equation using Fourier methods. Let $\tilde{u}$ denote the space-time Fourier transform and $\hat{u}$ denote the spatial Fourier transform, the equation with $p=1$ can be written as 
$$ (\tau - \xi^3 + \xi)\tilde{u} = 0$$
or
$$ \partial_t\hat{u} = i(\xi^3 - \xi)\hat{u} $$

The first formulation tells you that the space-time Fourier transform of a solution is a measure supported on the curve $\tau = \xi^3 - \xi$ in frequency space. That the frequency support has curvature implies that the solution should decay in time in physical space, justifying José's comment. (Look up Fourier restriction theorems or Strichartz estimates in the literature for more details.)

We can also see the temporal decay directly from the second formulation using oscillatory integral techniques. The solution can be written as
$$ \hat{u}(t,\xi) = e^{i(\xi^3 - \xi)t}\hat{u}_0(\xi) $$
It is then a standard exercise to show that, given initial data of sufficient decay in frequency space (say, the frequency has compact support), taking the inverse Fourier transform of the above solution gives you something with decay in $L^\infty$. 

If a solution decays in $L^\infty$, it cannot be a soliton. 

We can also see the lack of solitons by posing the traveling wave ansatz $u(t,x) = f(\omega t + x)$. A simple computation shows that the function $f$ must solve
$$ -(\omega + 1)f' = f''' \implies f = \exp [ i \sqrt{\omega + 1}(\omega t + x) ] $$
so that traveling waves cannot be spatially localized. 

In short, I think the reason why this equation is not heavily studied in the literature is because that as a linear PDE in (1+1) dimensions, it is not really all that interesting to look at.