Integrand size = 85, antiderivative size = 32 \[ \int \frac {e^{\frac {8-e^{-1-x+x^2}+4 x}{80+80 x^2+20 x^4}} \left (8-32 x-12 x^2+e^{-1-x+x^2} \left (2+x^2-2 x^3\right )\right )}{160+240 x^2+120 x^4+20 x^6} \, dx=-10+e^{\frac {2-\frac {1}{4} e^{-1-x+x^2}+x}{5 \left (2+x^2\right )^2}} \] Output:
exp(1/5*(2-1/4*exp(x^2-x-1)+x)/(x^2+2)^2)-10
Time = 5.06 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \[ \int \frac {e^{\frac {8-e^{-1-x+x^2}+4 x}{80+80 x^2+20 x^4}} \left (8-32 x-12 x^2+e^{-1-x+x^2} \left (2+x^2-2 x^3\right )\right )}{160+240 x^2+120 x^4+20 x^6} \, dx=e^{\frac {8-e^{-1-x+x^2}+4 x}{20 \left (2+x^2\right )^2}} \] Input:
Integrate[(E^((8 - E^(-1 - x + x^2) + 4*x)/(80 + 80*x^2 + 20*x^4))*(8 - 32 *x - 12*x^2 + E^(-1 - x + x^2)*(2 + x^2 - 2*x^3)))/(160 + 240*x^2 + 120*x^ 4 + 20*x^6),x]
Output:
E^((8 - E^(-1 - x + x^2) + 4*x)/(20*(2 + x^2)^2))
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.
\(\displaystyle \int \frac {e^{\frac {-e^{x^2-x-1}+4 x+8}{20 x^4+80 x^2+80}} \left (-12 x^2+e^{x^2-x-1} \left (-2 x^3+x^2+2\right )-32 x+8\right )}{20 x^6+120 x^4+240 x^2+160} \, dx\) |
\(\Big \downarrow \) 2070 |
\(\displaystyle \int \frac {e^{\frac {-e^{x^2-x-1}+4 x+8}{20 x^4+80 x^2+80}} \left (-12 x^2+e^{x^2-x-1} \left (-2 x^3+x^2+2\right )-32 x+8\right )}{\left (2^{2/3} \sqrt [3]{5} x^2+2\ 2^{2/3} \sqrt [3]{5}\right )^3}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {\left (2 x^3-x^2-2\right ) \exp \left (x^2+\frac {-e^{x^2-x-1}+4 x+8}{20 x^4+80 x^2+80}-x-1\right )}{20 \left (x^2+2\right )^3}-\frac {3 e^{\frac {-e^{x^2-x-1}+4 x+8}{20 x^4+80 x^2+80}} x^2}{5 \left (x^2+2\right )^3}-\frac {8 e^{\frac {-e^{x^2-x-1}+4 x+8}{20 x^4+80 x^2+80}} x}{5 \left (x^2+2\right )^3}+\frac {2 e^{\frac {-e^{x^2-x-1}+4 x+8}{20 x^4+80 x^2+80}}}{5 \left (x^2+2\right )^3}\right )dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {\left (e^{x^2} \left (-2 x^3+x^2+2\right )-4 e^{x+1} \left (3 x^2+8 x-2\right )\right ) \exp \left (-\frac {20 x^5+20 x^4+80 x^3+80 x^2+e^{x^2-x-1}+76 x+72}{20 \left (x^2+2\right )^2}\right )}{20 \left (x^2+2\right )^3}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{20} \int \frac {\exp \left (-\frac {20 x^5+20 x^4+80 x^3+80 x^2+76 x+e^{x^2-x-1}+72}{20 \left (x^2+2\right )^2}\right ) \left (4 e^{x+1} \left (-3 x^2-8 x+2\right )+e^{x^2} \left (-2 x^3+x^2+2\right )\right )}{\left (x^2+2\right )^3}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{20} \int \left (-\frac {4 \exp \left (x-\frac {20 x^5+20 x^4+80 x^3+80 x^2+76 x+e^{x^2-x-1}+72}{20 \left (x^2+2\right )^2}+1\right ) \left (3 x^2+8 x-2\right )}{\left (x^2+2\right )^3}-\frac {\exp \left (x^2-\frac {20 x^5+20 x^4+80 x^3+80 x^2+76 x+e^{x^2-x-1}+72}{20 \left (x^2+2\right )^2}\right ) \left (2 x^3-x^2-2\right )}{\left (x^2+2\right )^3}\right )dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \frac {1}{20} \int \left (-\frac {4 \exp \left (\frac {e^{-x-1} \left (4 e^{x+1} x-e^{x^2}+8 e^{x+1}\right )}{20 \left (x^2+2\right )^2}\right ) \left (3 x^2+8 x-2\right )}{\left (x^2+2\right )^3}-\frac {\exp \left (x^2-\frac {20 x^5+20 x^4+80 x^3+80 x^2+76 x+e^{x^2-x-1}+72}{20 \left (x^2+2\right )^2}\right ) \left (2 x^3-x^2-2\right )}{\left (x^2+2\right )^3}\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{20} \int \left (\frac {4 \exp \left (\frac {e^{-x-1} \left (4 e^{x+1} x-e^{x^2}+8 e^{x+1}\right )}{20 \left (x^2+2\right )^2}\right ) \left (-3 x^2-8 x+2\right )}{\left (x^2+2\right )^3}+\frac {\exp \left (\frac {e^{-x-1} \left (20 e^{x+1} x^6-20 e^{x+1} x^5+60 e^{x+1} x^4-80 e^{x+1} x^3-76 e^{x+1} x-e^{x^2}-72 e^{x+1}\right )}{20 \left (x^2+2\right )^2}\right ) \left (-2 x^3+x^2+2\right )}{\left (x^2+2\right )^3}\right )dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \frac {1}{20} \int \left (\frac {4 \exp \left (\frac {e^{-x-1} \left (4 e^{x+1} x-e^{x^2}+8 e^{x+1}\right )}{20 \left (x^2+2\right )^2}\right ) \left (-3 x^2-8 x+2\right )}{\left (x^2+2\right )^3}+\frac {\exp \left (\frac {e^{-x-1} \left (20 e^{x+1} x^6-20 e^{x+1} x^5+60 e^{x+1} x^4-80 e^{x+1} x^3-76 e^{x+1} x-e^{x^2}-72 e^{x+1}\right )}{20 \left (x^2+2\right )^2}\right ) \left (-2 x^3+x^2+2\right )}{\left (x^2+2\right )^3}\right )dx\) |
Input:
Int[(E^((8 - E^(-1 - x + x^2) + 4*x)/(80 + 80*x^2 + 20*x^4))*(8 - 32*x - 1 2*x^2 + E^(-1 - x + x^2)*(2 + x^2 - 2*x^3)))/(160 + 240*x^2 + 120*x^4 + 20 *x^6),x]
Output:
$Aborted
Time = 1.64 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84
method | result | size |
risch | \({\mathrm e}^{\frac {-{\mathrm e}^{x^{2}-x -1}+4 x +8}{20 \left (x^{2}+2\right )^{2}}}\) | \(27\) |
parallelrisch | \({\mathrm e}^{-\frac {{\mathrm e}^{x^{2}-x -1}-4 x -8}{20 \left (x^{4}+4 x^{2}+4\right )}}\) | \(30\) |
norman | \(\frac {x^{4} {\mathrm e}^{\frac {-{\mathrm e}^{x^{2}-x -1}+4 x +8}{20 x^{4}+80 x^{2}+80}}+4 x^{2} {\mathrm e}^{\frac {-{\mathrm e}^{x^{2}-x -1}+4 x +8}{20 x^{4}+80 x^{2}+80}}+4 \,{\mathrm e}^{\frac {-{\mathrm e}^{x^{2}-x -1}+4 x +8}{20 x^{4}+80 x^{2}+80}}}{\left (x^{2}+2\right )^{2}}\) | \(117\) |
Input:
int(((-2*x^3+x^2+2)*exp(x^2-x-1)-12*x^2-32*x+8)*exp((-exp(x^2-x-1)+4*x+8)/ (20*x^4+80*x^2+80))/(20*x^6+120*x^4+240*x^2+160),x,method=_RETURNVERBOSE)
Output:
exp(1/20*(-exp(x^2-x-1)+4*x+8)/(x^2+2)^2)
Time = 0.10 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {e^{\frac {8-e^{-1-x+x^2}+4 x}{80+80 x^2+20 x^4}} \left (8-32 x-12 x^2+e^{-1-x+x^2} \left (2+x^2-2 x^3\right )\right )}{160+240 x^2+120 x^4+20 x^6} \, dx=e^{\left (\frac {4 \, x - e^{\left (x^{2} - x - 1\right )} + 8}{20 \, {\left (x^{4} + 4 \, x^{2} + 4\right )}}\right )} \] Input:
integrate(((-2*x^3+x^2+2)*exp(x^2-x-1)-12*x^2-32*x+8)*exp((-exp(x^2-x-1)+4 *x+8)/(20*x^4+80*x^2+80))/(20*x^6+120*x^4+240*x^2+160),x, algorithm="frica s")
Output:
e^(1/20*(4*x - e^(x^2 - x - 1) + 8)/(x^4 + 4*x^2 + 4))
Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {e^{\frac {8-e^{-1-x+x^2}+4 x}{80+80 x^2+20 x^4}} \left (8-32 x-12 x^2+e^{-1-x+x^2} \left (2+x^2-2 x^3\right )\right )}{160+240 x^2+120 x^4+20 x^6} \, dx=e^{\frac {4 x - e^{x^{2} - x - 1} + 8}{20 x^{4} + 80 x^{2} + 80}} \] Input:
integrate(((-2*x**3+x**2+2)*exp(x**2-x-1)-12*x**2-32*x+8)*exp((-exp(x**2-x -1)+4*x+8)/(20*x**4+80*x**2+80))/(20*x**6+120*x**4+240*x**2+160),x)
Output:
exp((4*x - exp(x**2 - x - 1) + 8)/(20*x**4 + 80*x**2 + 80))
Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (28) = 56\).
Time = 0.27 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.91 \[ \int \frac {e^{\frac {8-e^{-1-x+x^2}+4 x}{80+80 x^2+20 x^4}} \left (8-32 x-12 x^2+e^{-1-x+x^2} \left (2+x^2-2 x^3\right )\right )}{160+240 x^2+120 x^4+20 x^6} \, dx=e^{\left (\frac {x}{5 \, {\left (x^{4} + 4 \, x^{2} + 4\right )}} - \frac {e^{\left (x^{2} - x\right )}}{20 \, {\left (x^{4} e + 4 \, x^{2} e + 4 \, e\right )}} + \frac {2}{5 \, {\left (x^{4} + 4 \, x^{2} + 4\right )}}\right )} \] Input:
integrate(((-2*x^3+x^2+2)*exp(x^2-x-1)-12*x^2-32*x+8)*exp((-exp(x^2-x-1)+4 *x+8)/(20*x^4+80*x^2+80))/(20*x^6+120*x^4+240*x^2+160),x, algorithm="maxim a")
Output:
e^(1/5*x/(x^4 + 4*x^2 + 4) - 1/20*e^(x^2 - x)/(x^4*e + 4*x^2*e + 4*e) + 2/ 5/(x^4 + 4*x^2 + 4))
Time = 0.14 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.69 \[ \int \frac {e^{\frac {8-e^{-1-x+x^2}+4 x}{80+80 x^2+20 x^4}} \left (8-32 x-12 x^2+e^{-1-x+x^2} \left (2+x^2-2 x^3\right )\right )}{160+240 x^2+120 x^4+20 x^6} \, dx=e^{\left (\frac {x}{5 \, {\left (x^{4} + 4 \, x^{2} + 4\right )}} - \frac {e^{\left (x^{2} - x - 1\right )}}{20 \, {\left (x^{4} + 4 \, x^{2} + 4\right )}} + \frac {2}{5 \, {\left (x^{4} + 4 \, x^{2} + 4\right )}}\right )} \] Input:
integrate(((-2*x^3+x^2+2)*exp(x^2-x-1)-12*x^2-32*x+8)*exp((-exp(x^2-x-1)+4 *x+8)/(20*x^4+80*x^2+80))/(20*x^6+120*x^4+240*x^2+160),x, algorithm="giac" )
Output:
e^(1/5*x/(x^4 + 4*x^2 + 4) - 1/20*e^(x^2 - x - 1)/(x^4 + 4*x^2 + 4) + 2/5/ (x^4 + 4*x^2 + 4))
Time = 8.89 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.97 \[ \int \frac {e^{\frac {8-e^{-1-x+x^2}+4 x}{80+80 x^2+20 x^4}} \left (8-32 x-12 x^2+e^{-1-x+x^2} \left (2+x^2-2 x^3\right )\right )}{160+240 x^2+120 x^4+20 x^6} \, dx={\mathrm {e}}^{\frac {4\,x}{20\,x^4+80\,x^2+80}}\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^{-x}\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{-1}}{20\,x^4+80\,x^2+80}}\,{\mathrm {e}}^{\frac {8}{20\,x^4+80\,x^2+80}} \] Input:
int(-(exp((4*x - exp(x^2 - x - 1) + 8)/(80*x^2 + 20*x^4 + 80))*(32*x - exp (x^2 - x - 1)*(x^2 - 2*x^3 + 2) + 12*x^2 - 8))/(240*x^2 + 120*x^4 + 20*x^6 + 160),x)
Output:
exp((4*x)/(80*x^2 + 20*x^4 + 80))*exp(-(exp(-x)*exp(x^2)*exp(-1))/(80*x^2 + 20*x^4 + 80))*exp(8/(80*x^2 + 20*x^4 + 80))
Time = 0.25 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.81 \[ \int \frac {e^{\frac {8-e^{-1-x+x^2}+4 x}{80+80 x^2+20 x^4}} \left (8-32 x-12 x^2+e^{-1-x+x^2} \left (2+x^2-2 x^3\right )\right )}{160+240 x^2+120 x^4+20 x^6} \, dx=\frac {e^{\frac {x +2}{5 x^{4}+20 x^{2}+20}}}{e^{\frac {e^{x^{2}}}{20 e^{x} e \,x^{4}+80 e^{x} e \,x^{2}+80 e^{x} e}}} \] Input:
int(((-2*x^3+x^2+2)*exp(x^2-x-1)-12*x^2-32*x+8)*exp((-exp(x^2-x-1)+4*x+8)/ (20*x^4+80*x^2+80))/(20*x^6+120*x^4+240*x^2+160),x)
Output:
e**((x + 2)/(5*x**4 + 20*x**2 + 20))/e**(e**(x**2)/(20*e**x*e*x**4 + 80*e* *x*e*x**2 + 80*e**x*e))