Integrand size = 71, antiderivative size = 25 \[ \int \left (5+e^{e (-35-5 x)+e^x (-35-5 x)+7 x+x^2} \left (14+6 x+7 x^2+2 x^3+e \left (-10-5 x^2\right )+e^x \left (-80-10 x-40 x^2-5 x^3\right )\right )\right ) \, dx=5 x+e^{(7+x) \left (-5 \left (e+e^x\right )+x\right )} \left (2+x^2\right ) \] Output:
5*x+exp((x-5*exp(1)-5*exp(x))*(7+x))*(x^2+2)
Time = 5.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \left (5+e^{e (-35-5 x)+e^x (-35-5 x)+7 x+x^2} \left (14+6 x+7 x^2+2 x^3+e \left (-10-5 x^2\right )+e^x \left (-80-10 x-40 x^2-5 x^3\right )\right )\right ) \, dx=5 x+e^{-\left (\left (5 e+5 e^x-x\right ) (7+x)\right )} \left (2+x^2\right ) \] Input:
Integrate[5 + E^(E*(-35 - 5*x) + E^x*(-35 - 5*x) + 7*x + x^2)*(14 + 6*x + 7*x^2 + 2*x^3 + E*(-10 - 5*x^2) + E^x*(-80 - 10*x - 40*x^2 - 5*x^3)),x]
Output:
5*x + (2 + x^2)/E^((5*E + 5*E^x - x)*(7 + x))
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \left (\left (2 x^3+7 x^2+e \left (-5 x^2-10\right )+e^x \left (-5 x^3-40 x^2-10 x-80\right )+6 x+14\right ) \exp \left (x^2+7 x+e^x (-5 x-35)+e (-5 x-35)\right )+5\right ) \, dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \int e^{-\left (\left (-x+5 e^x+5 e\right ) (x+7)\right )} x^3dx-5 \int e^{x-\left (-x+5 e^x+5 e\right ) (x+7)} x^3dx+7 \int e^{-\left (\left (-x+5 e^x+5 e\right ) (x+7)\right )} x^2dx-5 \int e^{1-\left (-x+5 e^x+5 e\right ) (x+7)} x^2dx-40 \int e^{x-\left (-x+5 e^x+5 e\right ) (x+7)} x^2dx+14 \int e^{-\left (\left (-x+5 e^x+5 e\right ) (x+7)\right )}dx-10 \int e^{1-\left (-x+5 e^x+5 e\right ) (x+7)}dx-80 \int e^{x-\left (-x+5 e^x+5 e\right ) (x+7)}dx+6 \int e^{-\left (\left (-x+5 e^x+5 e\right ) (x+7)\right )} xdx-10 \int e^{x-\left (-x+5 e^x+5 e\right ) (x+7)} xdx+5 x\) |
Input:
Int[5 + E^(E*(-35 - 5*x) + E^x*(-35 - 5*x) + 7*x + x^2)*(14 + 6*x + 7*x^2 + 2*x^3 + E*(-10 - 5*x^2) + E^x*(-80 - 10*x - 40*x^2 - 5*x^3)),x]
Output:
$Aborted
Time = 0.39 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16
method | result | size |
risch | \(\left (x^{2}+2\right ) {\mathrm e}^{-\left (x +7\right ) \left (-x +5 \,{\mathrm e}^{x}+5 \,{\mathrm e}\right )}+5 x\) | \(29\) |
default | \(5 x +x^{2} {\mathrm e}^{\left (-5 x -35\right ) {\mathrm e}^{x}+\left (-5 x -35\right ) {\mathrm e}+x^{2}+7 x}+2 \,{\mathrm e}^{\left (-5 x -35\right ) {\mathrm e}^{x}+\left (-5 x -35\right ) {\mathrm e}+x^{2}+7 x}\) | \(59\) |
norman | \(5 x +x^{2} {\mathrm e}^{\left (-5 x -35\right ) {\mathrm e}^{x}+\left (-5 x -35\right ) {\mathrm e}+x^{2}+7 x}+2 \,{\mathrm e}^{\left (-5 x -35\right ) {\mathrm e}^{x}+\left (-5 x -35\right ) {\mathrm e}+x^{2}+7 x}\) | \(59\) |
parallelrisch | \(5 x +x^{2} {\mathrm e}^{\left (-5 x -35\right ) {\mathrm e}^{x}+\left (-5 x -35\right ) {\mathrm e}+x^{2}+7 x}+2 \,{\mathrm e}^{\left (-5 x -35\right ) {\mathrm e}^{x}+\left (-5 x -35\right ) {\mathrm e}+x^{2}+7 x}\) | \(59\) |
Input:
int(((-5*x^3-40*x^2-10*x-80)*exp(x)+(-5*x^2-10)*exp(1)+2*x^3+7*x^2+6*x+14) *exp((-5*x-35)*exp(x)+(-5*x-35)*exp(1)+x^2+7*x)+5,x,method=_RETURNVERBOSE)
Output:
(x^2+2)*exp(-(x+7)*(-x+5*exp(x)+5*exp(1)))+5*x
Time = 0.10 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \left (5+e^{e (-35-5 x)+e^x (-35-5 x)+7 x+x^2} \left (14+6 x+7 x^2+2 x^3+e \left (-10-5 x^2\right )+e^x \left (-80-10 x-40 x^2-5 x^3\right )\right )\right ) \, dx={\left (x^{2} + 2\right )} e^{\left (x^{2} - 5 \, {\left (x + 7\right )} e - 5 \, {\left (x + 7\right )} e^{x} + 7 \, x\right )} + 5 \, x \] Input:
integrate(((-5*x^3-40*x^2-10*x-80)*exp(x)+(-5*x^2-10)*exp(1)+2*x^3+7*x^2+6 *x+14)*exp((-5*x-35)*exp(x)+(-5*x-35)*exp(1)+x^2+7*x)+5,x, algorithm="fric as")
Output:
(x^2 + 2)*e^(x^2 - 5*(x + 7)*e - 5*(x + 7)*e^x + 7*x) + 5*x
Time = 0.13 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44 \[ \int \left (5+e^{e (-35-5 x)+e^x (-35-5 x)+7 x+x^2} \left (14+6 x+7 x^2+2 x^3+e \left (-10-5 x^2\right )+e^x \left (-80-10 x-40 x^2-5 x^3\right )\right )\right ) \, dx=5 x + \left (x^{2} + 2\right ) e^{x^{2} + 7 x + \left (- 5 x - 35\right ) e^{x} + e \left (- 5 x - 35\right )} \] Input:
integrate(((-5*x**3-40*x**2-10*x-80)*exp(x)+(-5*x**2-10)*exp(1)+2*x**3+7*x **2+6*x+14)*exp((-5*x-35)*exp(x)+(-5*x-35)*exp(1)+x**2+7*x)+5,x)
Output:
5*x + (x**2 + 2)*exp(x**2 + 7*x + (-5*x - 35)*exp(x) + E*(-5*x - 35))
Time = 0.16 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44 \[ \int \left (5+e^{e (-35-5 x)+e^x (-35-5 x)+7 x+x^2} \left (14+6 x+7 x^2+2 x^3+e \left (-10-5 x^2\right )+e^x \left (-80-10 x-40 x^2-5 x^3\right )\right )\right ) \, dx={\left (x^{2} + 2\right )} e^{\left (x^{2} - 5 \, x e - 5 \, x e^{x} + 7 \, x - 35 \, e - 35 \, e^{x}\right )} + 5 \, x \] Input:
integrate(((-5*x^3-40*x^2-10*x-80)*exp(x)+(-5*x^2-10)*exp(1)+2*x^3+7*x^2+6 *x+14)*exp((-5*x-35)*exp(x)+(-5*x-35)*exp(1)+x^2+7*x)+5,x, algorithm="maxi ma")
Output:
(x^2 + 2)*e^(x^2 - 5*x*e - 5*x*e^x + 7*x - 35*e - 35*e^x) + 5*x
Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (25) = 50\).
Time = 0.15 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.72 \[ \int \left (5+e^{e (-35-5 x)+e^x (-35-5 x)+7 x+x^2} \left (14+6 x+7 x^2+2 x^3+e \left (-10-5 x^2\right )+e^x \left (-80-10 x-40 x^2-5 x^3\right )\right )\right ) \, dx={\left (x^{2} e^{\left (x^{2} - 5 \, x e - 5 \, x e^{x} + 7 \, x - 35 \, e - 35 \, e^{x} + 1\right )} + 2 \, e^{\left (x^{2} - 5 \, x e - 5 \, x e^{x} + 7 \, x - 35 \, e - 35 \, e^{x} + 1\right )}\right )} e^{\left (-1\right )} + 5 \, x \] Input:
integrate(((-5*x^3-40*x^2-10*x-80)*exp(x)+(-5*x^2-10)*exp(1)+2*x^3+7*x^2+6 *x+14)*exp((-5*x-35)*exp(x)+(-5*x-35)*exp(1)+x^2+7*x)+5,x, algorithm="giac ")
Output:
(x^2*e^(x^2 - 5*x*e - 5*x*e^x + 7*x - 35*e - 35*e^x + 1) + 2*e^(x^2 - 5*x* e - 5*x*e^x + 7*x - 35*e - 35*e^x + 1))*e^(-1) + 5*x
Time = 4.07 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.80 \[ \int \left (5+e^{e (-35-5 x)+e^x (-35-5 x)+7 x+x^2} \left (14+6 x+7 x^2+2 x^3+e \left (-10-5 x^2\right )+e^x \left (-80-10 x-40 x^2-5 x^3\right )\right )\right ) \, dx=5\,x+2\,{\mathrm {e}}^{-5\,x\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-35\,\mathrm {e}}\,{\mathrm {e}}^{7\,x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{-5\,x\,\mathrm {e}}\,{\mathrm {e}}^{-35\,{\mathrm {e}}^x}+x^2\,{\mathrm {e}}^{-5\,x\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-35\,\mathrm {e}}\,{\mathrm {e}}^{7\,x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{-5\,x\,\mathrm {e}}\,{\mathrm {e}}^{-35\,{\mathrm {e}}^x} \] Input:
int(exp(7*x - exp(x)*(5*x + 35) + x^2 - exp(1)*(5*x + 35))*(6*x - exp(1)*( 5*x^2 + 10) + 7*x^2 + 2*x^3 - exp(x)*(10*x + 40*x^2 + 5*x^3 + 80) + 14) + 5,x)
Output:
5*x + 2*exp(-5*x*exp(x))*exp(-35*exp(1))*exp(7*x)*exp(x^2)*exp(-5*x*exp(1) )*exp(-35*exp(x)) + x^2*exp(-5*x*exp(x))*exp(-35*exp(1))*exp(7*x)*exp(x^2) *exp(-5*x*exp(1))*exp(-35*exp(x))
Time = 0.20 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.92 \[ \int \left (5+e^{e (-35-5 x)+e^x (-35-5 x)+7 x+x^2} \left (14+6 x+7 x^2+2 x^3+e \left (-10-5 x^2\right )+e^x \left (-80-10 x-40 x^2-5 x^3\right )\right )\right ) \, dx=\frac {5 e^{5 e^{x} x +35 e^{x}+5 e x +35 e} x +e^{x^{2}+7 x} x^{2}+2 e^{x^{2}+7 x}}{e^{5 e^{x} x +35 e^{x}+5 e x +35 e}} \] Input:
int(((-5*x^3-40*x^2-10*x-80)*exp(x)+(-5*x^2-10)*exp(1)+2*x^3+7*x^2+6*x+14) *exp((-5*x-35)*exp(x)+(-5*x-35)*exp(1)+x^2+7*x)+5,x)
Output:
(5*e**(5*e**x*x + 35*e**x + 5*e*x + 35*e)*x + e**(x**2 + 7*x)*x**2 + 2*e** (x**2 + 7*x))/e**(5*e**x*x + 35*e**x + 5*e*x + 35*e)