Closed form solution for this integral? I'm hitting a road block in finding an expression (closed form preferably) for the following integral:
\begin{equation}
\int^{+\infty}_0 x^b \left ( 1-\frac{x}{u} \right )^c \exp(-a x^3) dx
\end{equation}
where $a,b$ are positive constants; $b>1$ is an odd multiple of $0.5$, while $c$ is a positive or negative odd multiple of $0.5$; $u$ is a (positive) parameter.
Things I have considered or tried:


*

*look up in tables (Gradshsteyn and Ryzhik): there are very few explicit results for integrals involving $\exp(-a x^3)$ (or for the other factors after transforming via $y=x^3$). Also, tabulated results involving $\exp(-a x^p)$ for more general $p$ do not include the other factors $x^b (1-x/u)^c$. One exception is (3.478.3): 
\begin{equation}
\int^{u}_0 x^b (u-x)^c \exp(-a x^3) dx, 
\end{equation}
but the limits of integration do not match with my case;

*there is a closed form solution (3.478.1) for the simpler integral
\begin{equation}
\int^{+\infty}_0 x^{d-1} \exp(-a x^3) dx = \frac{a^{-d/3}}{3} \Gamma(d/3).
\end{equation}
(NB: there is also an expression for the indefinite integral.)
A binomial expansion of $[1-(x/u)]^n$ for integer $n$ would produce a solution in series form. However, in my case, the exponents $b$ and $c$ are strictly half-integer. For the same reason, integration by parts does not lead to a simpler integral without the factor $[1-(x/u)]^c$;

*Wolfram Math online did not produce a result;

*the integral is an intermediate step in a longer analysis, so numerical solution (with given values for the parameter) is not practical.
Grateful for any pointers or solution.
 A: Long Comment: 
Since it was explicitly asked for, Mathematica 11.3 when provided with
$$\text{Integrate}\left[e^{-a x^3} x^b \left(1-\frac{x}{u}\right)^c,\{x,0,\infty \},\text{Assumptions}\to b>1\land a>0\land u>0\right]$$
gives up the monster
$$\text{ConditionalExpression}\left[\frac{1}{6} \left(2 a^{\frac{1}{3} (-b-c)} \left(\frac{\left(-\frac{1}{u}\right)^c \Gamma \left(\frac{1}{3} (b+c+1)\right) \, _3F_3\left(\frac{1}{3}-\frac{c}{3},\frac{2}{3}-\frac{c}{3},-\frac{c}{3};\frac{1}{3},\frac{2}{3},-\frac{b}{3}-\frac{c}{3}+\frac{2}{3};-a u^3\right)}{\sqrt[3]{a}}-c u \left(-\frac{1}{u}\right)^c \Gamma \left(\frac{b+c}{3}\right) \, _3F_3\left(\frac{1}{3}-\frac{c}{3},\frac{2}{3}-\frac{c}{3},1-\frac{c}{3};\frac{2}{3},\frac{4}{3},-\frac{b}{3}-\frac{c}{3}+1;-a u^3\right)+\frac{\pi  \sqrt[3]{a} (-1)^c 3^{\frac{3}{2}-c} u^{2-c} \Gamma (c+1) \Gamma \left(\frac{1}{3} (b+c-1)\right) \, _3F_3\left(\frac{2}{3}-\frac{c}{3},1-\frac{c}{3},\frac{4}{3}-\frac{c}{3};\frac{4}{3},\frac{5}{3},-\frac{b}{3}-\frac{c}{3}+\frac{4}{3};-a u^3\right)}{\Gamma \left(\frac{c-1}{3}\right) \Gamma \left(\frac{c}{3}\right) \Gamma \left(\frac{c+1}{3}\right)}\right)+\frac{3^{-b-c-\frac{1}{2}} u^{b+1} \Gamma (c+1) \left(\frac{4 \pi ^2 \Gamma (b+1)}{\Gamma \left(\frac{1}{3} (b+c+2)\right) \Gamma \left(\frac{1}{3} (b+c+3)\right) \Gamma \left(\frac{1}{3} (b+c+4)\right)}+\frac{(-1)^c \Gamma \left(-\frac{b}{3}-\frac{c}{3}\right) \Gamma \left(\frac{1}{3} (-b-c-1)\right) \Gamma \left(\frac{1}{3} (-b-c+1)\right)}{\Gamma (-b)}\right) \, _3F_3\left(\frac{b}{3}+\frac{1}{3},\frac{b}{3}+\frac{2}{3},\frac{b}{3}+1;\frac{b}{3}+\frac{c}{3}+\frac{2}{3},\frac{b}{3}+\frac{c}{3}+1,\frac{b}{3}+\frac{c}{3}+\frac{4}{3};-a u^3\right)}{\pi }\right),\Re(c)>-1\right] $$,
including the condition that the only half integer negative value $c$ can take is $-\frac{1}{2}$ given the above assumptions.
Further Thoughts...
By differentiating $e^{-a x^3} x^b \left(1-\frac{x}{u}\right)^c$ w.r.t $x$ and then integrating between $0$ and $\infty$ we obtain the integral identity
$$\frac{c}{u} \int_0^\infty e^{-a x^3} x^b \left(1-\frac{x}{u}\right)^{c-1} \, dx =b \int_0^\infty e^{-a x^3} x^{b-1} \left(1-\frac{x}{u}\right)^c \, dx-3 a \int_0^\infty e^{-a x^3} x^{b+2} \left(1-\frac{x}{u}\right)^c \, dx$$
since $\left[e^{-a x^3} x^b \left(1-\frac{x}{u}\right)^c \right]_0^\infty=0$
Mathematica again gives the condition that $c>-1$ with the following assumptions
$$\text{Integrate}\left[e^{-a x^3} x^{b} \left(1-\frac{x}{u}\right)^c,\{x,0,\infty \},\text{Assumptions}\to b>0\land a>0\land u>0\right]$$.
which implies $c$ must be positive for your integral to exist. 
A: An integral of a product of two Meijer G-functions of rational powers of the argument gives a G-function:
$$\int_0^\infty x^b \left( 1 - \frac x u \right)^c e^{-a x^3} dx = \\
\int_0^u x^b \left( 1 - \frac x u \right)^c e^{-a x^3} dx +
 (-1)^c \int_u^\infty x^b \left( \frac x u - 1 \right)^c e^{-a x^3} dx = \\
\Gamma(c + 1) \int_0^\infty x^b
 G_{1, 1}^{1, 0} \left( \frac x u \middle| {c + 1 \atop 0} \right)
 G_{0, 1}^{1, 0} \left( a x^3 \middle| {- \atop 0} \right) dx + \\
 (-1)^c \Gamma(c + 1) \int_0^\infty x^b
 G_{1, 1}^{0, 1} \left( \frac x u \middle| {c + 1 \atop 0} \right)
 G_{0, 1}^{1, 0} \left( a x^3 \middle| {- \atop 0} \right) dx = \\
3^{-c - 1} \Gamma(c + 1) u^{b + 1}
 G_{3, 4}^{1, 3} \left(a u^3 \middle|
  { \frac {-b} 3, \frac {-b + 1} 3, \frac {-b + 2} 3 \atop
   0, \frac {-b - c - 1} 3, \frac {-b - c} 3, \frac {-b - c + 1} 3} \right) + \\
3^{-c - 1} (-1)^c \Gamma(c + 1) u^{b + 1}
 G_{3, 4}^{4, 0} \left(a u^3 \middle|
  { \frac {-b} 3, \frac {-b + 1} 3, \frac {-b + 2} 3 \atop
   0, \frac {-b - c - 1} 3, \frac {-b - c} 3, \frac {-b - c + 1} 3} \right).$$
When $b$ and $c$ are half-integers, one of the three numbers $(-b - c - 1)/3, (-b - c)/3, (-b - c + 1)/3$ is an integer. Therefore, the ratio of the gamma functions in the $G_{3, 4}^{4, 0}$ term always has an infinite sequence of double poles inside the integration contour, and the G-function cannot be converted to a sum of hypergeometric functions. The sum will be valid only as a limit, which will give derivatives of ${_pF_q}$ wrt parameters.
