Integrand size = 98, antiderivative size = 27 \[ \int \frac {e^x (-16-16 x)+e^{x+\frac {2}{3} \left (2 x+2 x^2\right )} (-1-x)+e^{x+\frac {1}{3} \left (2 x+2 x^2\right )} \left (10+6 x-8 x^2\right )}{4-4 e^{\frac {1}{3} \left (2 x+2 x^2\right )}+e^{\frac {2}{3} \left (2 x+2 x^2\right )}} \, dx=e^x \left (-x+\frac {6 x}{-2+e^{\frac {1}{3} x (2+2 x)}}\right ) \]
Time = 3.82 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {e^x (-16-16 x)+e^{x+\frac {2}{3} \left (2 x+2 x^2\right )} (-1-x)+e^{x+\frac {1}{3} \left (2 x+2 x^2\right )} \left (10+6 x-8 x^2\right )}{4-4 e^{\frac {1}{3} \left (2 x+2 x^2\right )}+e^{\frac {2}{3} \left (2 x+2 x^2\right )}} \, dx=-e^x \left (1-\frac {6}{-2+e^{\frac {2}{3} x (1+x)}}\right ) x \]
Integrate[(E^x*(-16 - 16*x) + E^(x + (2*(2*x + 2*x^2))/3)*(-1 - x) + E^(x + (2*x + 2*x^2)/3)*(10 + 6*x - 8*x^2))/(4 - 4*E^((2*x + 2*x^2)/3) + E^((2* (2*x + 2*x^2))/3)),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.
\(\displaystyle \int \frac {e^{\frac {2}{3} \left (2 x^2+2 x\right )+x} (-x-1)+e^{\frac {1}{3} \left (2 x^2+2 x\right )+x} \left (-8 x^2+6 x+10\right )+e^x (-16 x-16)}{-4 e^{\frac {1}{3} \left (2 x^2+2 x\right )}+e^{\frac {2}{3} \left (2 x^2+2 x\right )}+4} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{\frac {2}{3} \left (2 x^2+2 x\right )+x} (-x-1)+e^{\frac {1}{3} \left (2 x^2+2 x\right )+x} \left (-8 x^2+6 x+10\right )+e^x (-16 x-16)}{\left (2-e^{\frac {2}{3} x (x+1)}\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {8 e^{\frac {1}{3} x (2 x+5)} x^2}{\left (e^{\frac {2}{3} x (x+1)}-2\right )^2}-\frac {16 e^x x}{\left (e^{\frac {2}{3} x (x+1)}-2\right )^2}+\frac {6 e^{\frac {1}{3} x (2 x+5)} x}{\left (e^{\frac {2}{3} x (x+1)}-2\right )^2}-\frac {e^{\frac {1}{3} x (4 x+7)} x}{\left (e^{\frac {2}{3} x (x+1)}-2\right )^2}-\frac {16 e^x}{\left (e^{\frac {2}{3} x (x+1)}-2\right )^2}+\frac {10 e^{\frac {1}{3} x (2 x+5)}}{\left (e^{\frac {2}{3} x (x+1)}-2\right )^2}-\frac {e^{\frac {1}{3} x (4 x+7)}}{\left (e^{\frac {2}{3} x (x+1)}-2\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -8 \int \frac {e^{\frac {1}{3} x (2 x+5)} x^2}{\left (-2+e^{\frac {2}{3} x (x+1)}\right )^2}dx-16 \int \frac {e^x}{\left (-2+e^{\frac {2}{3} x (x+1)}\right )^2}dx+10 \int \frac {e^{\frac {1}{3} x (2 x+5)}}{\left (-2+e^{\frac {2}{3} x (x+1)}\right )^2}dx-\int \frac {e^{\frac {1}{3} x (4 x+7)}}{\left (-2+e^{\frac {2}{3} x (x+1)}\right )^2}dx-16 \int \frac {e^x x}{\left (-2+e^{\frac {2}{3} x (x+1)}\right )^2}dx+6 \int \frac {e^{\frac {1}{3} x (2 x+5)} x}{\left (-2+e^{\frac {2}{3} x (x+1)}\right )^2}dx-\int \frac {e^{\frac {1}{3} x (4 x+7)} x}{\left (-2+e^{\frac {2}{3} x (x+1)}\right )^2}dx\) |
Int[(E^x*(-16 - 16*x) + E^(x + (2*(2*x + 2*x^2))/3)*(-1 - x) + E^(x + (2*x + 2*x^2)/3)*(10 + 6*x - 8*x^2))/(4 - 4*E^((2*x + 2*x^2)/3) + E^((2*(2*x + 2*x^2))/3)),x]
3.21.84.3.1 Defintions of rubi rules used
Time = 0.11 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85
method | result | size |
risch | \(-{\mathrm e}^{x} x +\frac {6 x \,{\mathrm e}^{x}}{{\mathrm e}^{\frac {2 \left (1+x \right ) x}{3}}-2}\) | \(23\) |
norman | \(\frac {8 \,{\mathrm e}^{x} x -{\mathrm e}^{x} x \,{\mathrm e}^{\frac {2}{3} x^{2}+\frac {2}{3} x}}{{\mathrm e}^{\frac {2}{3} x^{2}+\frac {2}{3} x}-2}\) | \(37\) |
parallelrisch | \(-\frac {{\mathrm e}^{x} x \,{\mathrm e}^{\frac {2}{3} x^{2}+\frac {2}{3} x}-8 \,{\mathrm e}^{x} x}{{\mathrm e}^{\frac {2}{3} x^{2}+\frac {2}{3} x}-2}\) | \(37\) |
int(((-1-x)*exp(x)*exp(2/3*x^2+2/3*x)^2+(-8*x^2+6*x+10)*exp(x)*exp(2/3*x^2 +2/3*x)+(-16*x-16)*exp(x))/(exp(2/3*x^2+2/3*x)^2-4*exp(2/3*x^2+2/3*x)+4),x ,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (20) = 40\).
Time = 0.25 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.96 \[ \int \frac {e^x (-16-16 x)+e^{x+\frac {2}{3} \left (2 x+2 x^2\right )} (-1-x)+e^{x+\frac {1}{3} \left (2 x+2 x^2\right )} \left (10+6 x-8 x^2\right )}{4-4 e^{\frac {1}{3} \left (2 x+2 x^2\right )}+e^{\frac {2}{3} \left (2 x+2 x^2\right )}} \, dx=-\frac {x e^{\left (\frac {8}{3} \, x^{2} + \frac {17}{3} \, x\right )} - 8 \, x e^{\left (2 \, x^{2} + 5 \, x\right )}}{e^{\left (\frac {8}{3} \, x^{2} + \frac {14}{3} \, x\right )} - 2 \, e^{\left (2 \, x^{2} + 4 \, x\right )}} \]
integrate(((-1-x)*exp(x)*exp(2/3*x^2+2/3*x)^2+(-8*x^2+6*x+10)*exp(x)*exp(2 /3*x^2+2/3*x)+(-16*x-16)*exp(x))/(exp(2/3*x^2+2/3*x)^2-4*exp(2/3*x^2+2/3*x )+4),x, algorithm=\
Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {e^x (-16-16 x)+e^{x+\frac {2}{3} \left (2 x+2 x^2\right )} (-1-x)+e^{x+\frac {1}{3} \left (2 x+2 x^2\right )} \left (10+6 x-8 x^2\right )}{4-4 e^{\frac {1}{3} \left (2 x+2 x^2\right )}+e^{\frac {2}{3} \left (2 x+2 x^2\right )}} \, dx=- x e^{x} + \frac {6 x e^{x}}{e^{\frac {2 x^{2}}{3} + \frac {2 x}{3}} - 2} \]
integrate(((-1-x)*exp(x)*exp(2/3*x**2+2/3*x)**2+(-8*x**2+6*x+10)*exp(x)*ex p(2/3*x**2+2/3*x)+(-16*x-16)*exp(x))/(exp(2/3*x**2+2/3*x)**2-4*exp(2/3*x** 2+2/3*x)+4),x)
Time = 0.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {e^x (-16-16 x)+e^{x+\frac {2}{3} \left (2 x+2 x^2\right )} (-1-x)+e^{x+\frac {1}{3} \left (2 x+2 x^2\right )} \left (10+6 x-8 x^2\right )}{4-4 e^{\frac {1}{3} \left (2 x+2 x^2\right )}+e^{\frac {2}{3} \left (2 x+2 x^2\right )}} \, dx=-\frac {x e^{\left (\frac {2}{3} \, x^{2} + \frac {5}{3} \, x\right )} - 8 \, x e^{x}}{e^{\left (\frac {2}{3} \, x^{2} + \frac {2}{3} \, x\right )} - 2} \]
integrate(((-1-x)*exp(x)*exp(2/3*x^2+2/3*x)^2+(-8*x^2+6*x+10)*exp(x)*exp(2 /3*x^2+2/3*x)+(-16*x-16)*exp(x))/(exp(2/3*x^2+2/3*x)^2-4*exp(2/3*x^2+2/3*x )+4),x, algorithm=\
Timed out. \[ \int \frac {e^x (-16-16 x)+e^{x+\frac {2}{3} \left (2 x+2 x^2\right )} (-1-x)+e^{x+\frac {1}{3} \left (2 x+2 x^2\right )} \left (10+6 x-8 x^2\right )}{4-4 e^{\frac {1}{3} \left (2 x+2 x^2\right )}+e^{\frac {2}{3} \left (2 x+2 x^2\right )}} \, dx=\text {Timed out} \]
integrate(((-1-x)*exp(x)*exp(2/3*x^2+2/3*x)^2+(-8*x^2+6*x+10)*exp(x)*exp(2 /3*x^2+2/3*x)+(-16*x-16)*exp(x))/(exp(2/3*x^2+2/3*x)^2-4*exp(2/3*x^2+2/3*x )+4),x, algorithm=\
Time = 0.44 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^x (-16-16 x)+e^{x+\frac {2}{3} \left (2 x+2 x^2\right )} (-1-x)+e^{x+\frac {1}{3} \left (2 x+2 x^2\right )} \left (10+6 x-8 x^2\right )}{4-4 e^{\frac {1}{3} \left (2 x+2 x^2\right )}+e^{\frac {2}{3} \left (2 x+2 x^2\right )}} \, dx=\frac {6\,x\,{\mathrm {e}}^x}{{\mathrm {e}}^{\frac {2\,x^2}{3}+\frac {2\,x}{3}}-2}-x\,{\mathrm {e}}^x \]