Is there an alternate form for \begin{equation} \int e^{c q(x)+p(x)} \,\mathrm{d}x, \end{equation} where the constant $c$ is not within an integral?
I have tried rewriting the exponential as follows, and using integration by parts:
\begin{align} I&=\int e^{c q(x)} e^{p(x)} \,\mathrm{d}x \\ I&=e^{c q(x)} \int e^{p(x)} \,\mathrm{d}x - c \int \left( q'(x) e^{c q(x)} \int e^{p(x)} \,\mathrm{d}x \right) \,\mathrm{d}x \end{align}
However, I can't quite manage to find a form where there constant $c$ is not stuck inside the integral. (I am happy for everything else to be inside integrals.)
EDIT:
For those wondering why I want to do this, here is some background:
I am trying to solve the following PDE using the method of characteristics:
\begin{equation} \left[ 2Dig c_{3} j(t) + c_{5} \right] \frac{\partial u}{\partial x} + c_{3} \frac{\partial u}{\partial t} = u \left[ -ig c_{5} j(t) + c_{6} + Dg^2 c_{3} j(t)^2 - ig c_{3} j'(t) x \right] + c_{7} e^{-D g^2 \frac{t^3}{3}-igxt} \end{equation}
PDEs of the form $P(x,t,u) \frac{\partial u}{\partial x} + Q(x,t,u) \frac{\partial u}{\partial t} = R(x,t,u)$ can be solved using the method of characteristics by equating and solving any two of the following ODEs:
\begin{equation} \frac{\mathrm{d}x}{P(x,t,u)} = \frac{\mathrm{d}t}{Q(x,t,u)} = \frac{\mathrm{d}u}{R(x,t,u)} \end{equation}
If the constants introduced are given the labels $\alpha_{1}$ and $\alpha_{2}$, then the method of characteristics requires that $\alpha_{1}$ be written as an arbitrary function of $\alpha_{2}$. e.g. $\alpha_{1} = F(\alpha_{2})$.
Finally, the form of the arbitrary function $F$ may be determined by differentiating the solution, and substituting back into the original equation.
In my case, I have taken \begin{align} P(x,t,u) &= 2Dig c_{3} j(t) + c_{5} \\ Q(x,t,u) &= c_{3} \\ R(x,t,u) &= u \left[ -ig c_{5} j(t) + c_{6} + Dg^2 c_{3} j(t)^2 - ig c_{3} j'(t) x \right] + c_{7} e^{-D g^2 \frac{t^3}{3}-igxt} \end{align}
First taking $\frac{\mathrm{d}x}{P(x,t,u)} = \frac{\mathrm{d}t}{Q(x,t,u)}$, I determined that \begin{equation} x = \left[ \alpha_{1} + c_{5}t + 2Digc_{3} \int j(t) \, \mathrm{d} t \right] \Big / c_{3} \end{equation} or, alternatively, \begin{equation} \alpha_{1} = c_{3}x - c_{5} t - 2Digc_{3} \int j(t) \,\mathrm{d}t \end{equation}
Second I took $\frac{\mathrm{d}t}{Q(x,t,u)} = \frac{\mathrm{d}u}{R(x,t,u)}$ (Actually, in this case $\frac{\mathrm{d}u}{\mathrm{d}t} = \frac{R(x,t,u)}{Q(x,t,u)}$). Upon substituting $x = \left[ \alpha_{1} + c_{5}t + 2Digc_{3} \int j(t) \, \mathrm{d} t \right] \Big / c_{3}$ , this becomes a first order linear ODE: \begin{align} c_3 \frac{\mathrm{d}u}{\mathrm{d}t} - u \left[ -ig c_{5} j(t) + c_{6} + Dg^2 c_{3} j(t)^2 - ig c_{3} j'(t) x \right] &= c_{7} e^{-D g^2 \frac{t^3}{3}-igxt} \\ c_3 \frac{\mathrm{d}u}{\mathrm{d}t} - u \left[ -ig c_{5} j(t) + c_{6} + Dg^2 c_{3} j(t)^2 - ig c_{3} j'(t) \left( \alpha_{1} + c_{5}t + 2Digc_{3} \int j(t) \, \mathrm{d} t \right) \Big / c_{3} \right] &= c_{7} e^{-D g^2 \frac{t^3}{3}-igt\left( \alpha_{1} + c_{5}t + 2Digc_{3} \int j(t) \, \mathrm{d} t \right) \big / c_{3}} \end{align}
Now, when we solve this, using the integrating factor method for first order linear ODEs, which I will not do here, we gain a second constant $\alpha_{2}$. Once all the integration is done, we can substitute $c_{3}x - c_{5} t - 2Digc_{3} \int j(t) \,\mathrm{d}t$ back in for $\alpha_{1}$. Note that there is no point performing this substitution before the integration is done, because there are unhelpful $x$ terms, and the integrations are all in terms of $t$.
In my case, this step is a little complicated, due to the arbitrary function $j(t)$. Due to this arbitrary function, the solution tends to remain in terms of integrals. However, if $\alpha_{1}$ is contained within any of the solution integrals, we can't finish the method of characteristics, because we can't substitute back for $\alpha_{1}$ until after the integration step. Hence my desire to help a constant escape from an integral.
I have managed to sneak the constant $\alpha_{1}$ outside of integrals at every point, except for one. To avoid clutter, the solution is essentially of the form \begin{equation} u = e^{f(t)} e^{\alpha_{1} g(t)} \int e^{\alpha_{1} q(t)} e^{p(t)} \,\mathrm{d}x \end{equation} where $f(t), g(t), q(t), p(t)$ are messy functions of $t$.
If I can help $\alpha_{1}$ to escape the integral, then I can find a solution in terms of the arbitrary function $j(t)$. If I can't, then I will need to individually find solutions for particular forms of $j(t)$.