Integrand size = 100, antiderivative size = 27 \[ \int \frac {e^{\frac {-5-14 x-5 x^2-x^3-x^4+\left (-2 x^2-x^3\right ) \log (3)}{1+2 x+x^2}} \left (-4+4 x-3 x^2-5 x^3-2 x^4+\left (-4 x-3 x^2-x^3\right ) \log (3)\right )}{1+3 x+3 x^2+x^3} \, dx=e^{-5+\frac {x (2+x) (-2+x (1-x-\log (3)))}{(1+x)^2}} \] Output:
exp((2+x)/(1+x)^2*(x*(1-x-ln(3))-2)*x-5)
Time = 4.51 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.52 \[ \int \frac {e^{\frac {-5-14 x-5 x^2-x^3-x^4+\left (-2 x^2-x^3\right ) \log (3)}{1+2 x+x^2}} \left (-4+4 x-3 x^2-5 x^3-2 x^4+\left (-4 x-3 x^2-x^3\right ) \log (3)\right )}{1+3 x+3 x^2+x^3} \, dx=3^{-\frac {x^2 (2+x)}{(1+x)^2}} e^{-\frac {5+14 x+5 x^2+x^3+x^4}{(1+x)^2}} \] Input:
Integrate[(E^((-5 - 14*x - 5*x^2 - x^3 - x^4 + (-2*x^2 - x^3)*Log[3])/(1 + 2*x + x^2))*(-4 + 4*x - 3*x^2 - 5*x^3 - 2*x^4 + (-4*x - 3*x^2 - x^3)*Log[ 3]))/(1 + 3*x + 3*x^2 + x^3),x]
Output:
1/(3^((x^2*(2 + x))/(1 + x)^2)*E^((5 + 14*x + 5*x^2 + x^3 + x^4)/(1 + x)^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.
\(\displaystyle \int \frac {\left (-2 x^4-5 x^3-3 x^2+\left (-x^3-3 x^2-4 x\right ) \log (3)+4 x-4\right ) \exp \left (\frac {-x^4-x^3-5 x^2+\left (-x^3-2 x^2\right ) \log (3)-14 x-5}{x^2+2 x+1}\right )}{x^3+3 x^2+3 x+1} \, dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {\left (-2 x^4-5 x^3-3 x^2+\left (-x^3-3 x^2-4 x\right ) \log (3)+4 x-4\right ) \exp \left (\frac {-x^4-x^3-5 x^2+\left (-x^3-2 x^2\right ) \log (3)-14 x-5}{x^2+2 x+1}\right )}{(x+1)^3}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\left (-2 x^4-x^3 (5+\log (3))-3 x^2 (1+\log (3))+4 x (1-\log (3))-4\right ) \exp \left (\frac {-x^4-x^3 (1+\log (3))-x^2 (5+\log (9))-14 x-5}{x^2+2 x+1}\right )}{(x+1)^3}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-2 x \exp \left (\frac {-x^4-x^3 (1+\log (3))-x^2 (5+\log (9))-14 x-5}{x^2+2 x+1}\right )+\frac {(\log (9)-8) \exp \left (\frac {-x^4-x^3 (1+\log (3))-x^2 (5+\log (9))-14 x-5}{x^2+2 x+1}\right )}{(x+1)^3}+\frac {(3-\log (3)) \exp \left (\frac {-x^4-x^3 (1+\log (3))-x^2 (5+\log (9))-14 x-5}{x^2+2 x+1}\right )}{(x+1)^2}+(1-\log (3)) \exp \left (\frac {-x^4-x^3 (1+\log (3))-x^2 (5+\log (9))-14 x-5}{x^2+2 x+1}\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle (1-\log (3)) \int \exp \left (\frac {-x^4-(1+\log (3)) x^3-(5+\log (9)) x^2-14 x-5}{x^2+2 x+1}\right )dx-2 \int \exp \left (\frac {-x^4-(1+\log (3)) x^3-(5+\log (9)) x^2-14 x-5}{x^2+2 x+1}\right ) xdx-(8-\log (9)) \int \frac {\exp \left (\frac {-x^4-(1+\log (3)) x^3-(5+\log (9)) x^2-14 x-5}{x^2+2 x+1}\right )}{(x+1)^3}dx+(3-\log (3)) \int \frac {\exp \left (\frac {-x^4-(1+\log (3)) x^3-(5+\log (9)) x^2-14 x-5}{x^2+2 x+1}\right )}{(x+1)^2}dx\) |
Input:
Int[(E^((-5 - 14*x - 5*x^2 - x^3 - x^4 + (-2*x^2 - x^3)*Log[3])/(1 + 2*x + x^2))*(-4 + 4*x - 3*x^2 - 5*x^3 - 2*x^4 + (-4*x - 3*x^2 - x^3)*Log[3]))/( 1 + 3*x + 3*x^2 + x^3),x]
Output:
$Aborted
Time = 0.32 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41
method | result | size |
risch | \({\mathrm e}^{-\frac {x^{3} \ln \left (3\right )+x^{4}+2 x^{2} \ln \left (3\right )+x^{3}+5 x^{2}+14 x +5}{\left (1+x \right )^{2}}}\) | \(38\) |
gosper | \({\mathrm e}^{-\frac {x^{3} \ln \left (3\right )+x^{4}+2 x^{2} \ln \left (3\right )+x^{3}+5 x^{2}+14 x +5}{x^{2}+2 x +1}}\) | \(43\) |
parallelrisch | \({\mathrm e}^{\frac {\left (-x^{3}-2 x^{2}\right ) \ln \left (3\right )-x^{4}-x^{3}-5 x^{2}-14 x -5}{x^{2}+2 x +1}}\) | \(47\) |
norman | \(\frac {x^{2} {\mathrm e}^{\frac {\left (-x^{3}-2 x^{2}\right ) \ln \left (3\right )-x^{4}-x^{3}-5 x^{2}-14 x -5}{x^{2}+2 x +1}}+2 x \,{\mathrm e}^{\frac {\left (-x^{3}-2 x^{2}\right ) \ln \left (3\right )-x^{4}-x^{3}-5 x^{2}-14 x -5}{x^{2}+2 x +1}}+{\mathrm e}^{\frac {\left (-x^{3}-2 x^{2}\right ) \ln \left (3\right )-x^{4}-x^{3}-5 x^{2}-14 x -5}{x^{2}+2 x +1}}}{\left (1+x \right )^{2}}\) | \(153\) |
Input:
int(((-x^3-3*x^2-4*x)*ln(3)-2*x^4-5*x^3-3*x^2+4*x-4)*exp(((-x^3-2*x^2)*ln( 3)-x^4-x^3-5*x^2-14*x-5)/(x^2+2*x+1))/(x^3+3*x^2+3*x+1),x,method=_RETURNVE RBOSE)
Output:
exp(-(x^3*ln(3)+x^4+2*x^2*ln(3)+x^3+5*x^2+14*x+5)/(1+x)^2)
Time = 0.08 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.52 \[ \int \frac {e^{\frac {-5-14 x-5 x^2-x^3-x^4+\left (-2 x^2-x^3\right ) \log (3)}{1+2 x+x^2}} \left (-4+4 x-3 x^2-5 x^3-2 x^4+\left (-4 x-3 x^2-x^3\right ) \log (3)\right )}{1+3 x+3 x^2+x^3} \, dx=e^{\left (-\frac {x^{4} + x^{3} + 5 \, x^{2} + {\left (x^{3} + 2 \, x^{2}\right )} \log \left (3\right ) + 14 \, x + 5}{x^{2} + 2 \, x + 1}\right )} \] Input:
integrate(((-x^3-3*x^2-4*x)*log(3)-2*x^4-5*x^3-3*x^2+4*x-4)*exp(((-x^3-2*x ^2)*log(3)-x^4-x^3-5*x^2-14*x-5)/(x^2+2*x+1))/(x^3+3*x^2+3*x+1),x, algorit hm="fricas")
Output:
e^(-(x^4 + x^3 + 5*x^2 + (x^3 + 2*x^2)*log(3) + 14*x + 5)/(x^2 + 2*x + 1))
Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44 \[ \int \frac {e^{\frac {-5-14 x-5 x^2-x^3-x^4+\left (-2 x^2-x^3\right ) \log (3)}{1+2 x+x^2}} \left (-4+4 x-3 x^2-5 x^3-2 x^4+\left (-4 x-3 x^2-x^3\right ) \log (3)\right )}{1+3 x+3 x^2+x^3} \, dx=e^{\frac {- x^{4} - x^{3} - 5 x^{2} - 14 x + \left (- x^{3} - 2 x^{2}\right ) \log {\left (3 \right )} - 5}{x^{2} + 2 x + 1}} \] Input:
integrate(((-x**3-3*x**2-4*x)*ln(3)-2*x**4-5*x**3-3*x**2+4*x-4)*exp(((-x** 3-2*x**2)*ln(3)-x**4-x**3-5*x**2-14*x-5)/(x**2+2*x+1))/(x**3+3*x**2+3*x+1) ,x)
Output:
exp((-x**4 - x**3 - 5*x**2 - 14*x + (-x**3 - 2*x**2)*log(3) - 5)/(x**2 + 2 *x + 1))
Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (23) = 46\).
Time = 0.32 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.04 \[ \int \frac {e^{\frac {-5-14 x-5 x^2-x^3-x^4+\left (-2 x^2-x^3\right ) \log (3)}{1+2 x+x^2}} \left (-4+4 x-3 x^2-5 x^3-2 x^4+\left (-4 x-3 x^2-x^3\right ) \log (3)\right )}{1+3 x+3 x^2+x^3} \, dx=e^{\left (-x^{2} - x \log \left (3\right ) + x - \frac {\log \left (3\right )}{x^{2} + 2 \, x + 1} + \frac {\log \left (3\right )}{x + 1} + \frac {4}{x^{2} + 2 \, x + 1} - \frac {3}{x + 1} - 6\right )} \] Input:
integrate(((-x^3-3*x^2-4*x)*log(3)-2*x^4-5*x^3-3*x^2+4*x-4)*exp(((-x^3-2*x ^2)*log(3)-x^4-x^3-5*x^2-14*x-5)/(x^2+2*x+1))/(x^3+3*x^2+3*x+1),x, algorit hm="maxima")
Output:
e^(-x^2 - x*log(3) + x - log(3)/(x^2 + 2*x + 1) + log(3)/(x + 1) + 4/(x^2 + 2*x + 1) - 3/(x + 1) - 6)
Leaf count of result is larger than twice the leaf count of optimal. 106 vs. \(2 (23) = 46\).
Time = 0.13 (sec) , antiderivative size = 106, normalized size of antiderivative = 3.93 \[ \int \frac {e^{\frac {-5-14 x-5 x^2-x^3-x^4+\left (-2 x^2-x^3\right ) \log (3)}{1+2 x+x^2}} \left (-4+4 x-3 x^2-5 x^3-2 x^4+\left (-4 x-3 x^2-x^3\right ) \log (3)\right )}{1+3 x+3 x^2+x^3} \, dx=e^{\left (-\frac {x^{4}}{x^{2} + 2 \, x + 1} - \frac {x^{3} \log \left (3\right )}{x^{2} + 2 \, x + 1} - \frac {x^{3}}{x^{2} + 2 \, x + 1} - \frac {2 \, x^{2} \log \left (3\right )}{x^{2} + 2 \, x + 1} - \frac {5 \, x^{2}}{x^{2} + 2 \, x + 1} - \frac {14 \, x}{x^{2} + 2 \, x + 1} - \frac {5}{x^{2} + 2 \, x + 1}\right )} \] Input:
integrate(((-x^3-3*x^2-4*x)*log(3)-2*x^4-5*x^3-3*x^2+4*x-4)*exp(((-x^3-2*x ^2)*log(3)-x^4-x^3-5*x^2-14*x-5)/(x^2+2*x+1))/(x^3+3*x^2+3*x+1),x, algorit hm="giac")
Output:
e^(-x^4/(x^2 + 2*x + 1) - x^3*log(3)/(x^2 + 2*x + 1) - x^3/(x^2 + 2*x + 1) - 2*x^2*log(3)/(x^2 + 2*x + 1) - 5*x^2/(x^2 + 2*x + 1) - 14*x/(x^2 + 2*x + 1) - 5/(x^2 + 2*x + 1))
Time = 3.93 (sec) , antiderivative size = 98, normalized size of antiderivative = 3.63 \[ \int \frac {e^{\frac {-5-14 x-5 x^2-x^3-x^4+\left (-2 x^2-x^3\right ) \log (3)}{1+2 x+x^2}} \left (-4+4 x-3 x^2-5 x^3-2 x^4+\left (-4 x-3 x^2-x^3\right ) \log (3)\right )}{1+3 x+3 x^2+x^3} \, dx={\left (\frac {1}{3}\right )}^{\frac {x^3+2\,x^2}{x^2+2\,x+1}}\,{\mathrm {e}}^{-\frac {x^3}{x^2+2\,x+1}}\,{\mathrm {e}}^{-\frac {x^4}{x^2+2\,x+1}}\,{\mathrm {e}}^{-\frac {5\,x^2}{x^2+2\,x+1}}\,{\mathrm {e}}^{-\frac {5}{x^2+2\,x+1}}\,{\mathrm {e}}^{-\frac {14\,x}{x^2+2\,x+1}} \] Input:
int(-(exp(-(14*x + log(3)*(2*x^2 + x^3) + 5*x^2 + x^3 + x^4 + 5)/(2*x + x^ 2 + 1))*(log(3)*(4*x + 3*x^2 + x^3) - 4*x + 3*x^2 + 5*x^3 + 2*x^4 + 4))/(3 *x + 3*x^2 + x^3 + 1),x)
Output:
(1/3)^((2*x^2 + x^3)/(2*x + x^2 + 1))*exp(-x^3/(2*x + x^2 + 1))*exp(-x^4/( 2*x + x^2 + 1))*exp(-(5*x^2)/(2*x + x^2 + 1))*exp(-5/(2*x + x^2 + 1))*exp( -(14*x)/(2*x + x^2 + 1))
Time = 0.50 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.48 \[ \int \frac {e^{\frac {-5-14 x-5 x^2-x^3-x^4+\left (-2 x^2-x^3\right ) \log (3)}{1+2 x+x^2}} \left (-4+4 x-3 x^2-5 x^3-2 x^4+\left (-4 x-3 x^2-x^3\right ) \log (3)\right )}{1+3 x+3 x^2+x^3} \, dx=\frac {e^{\frac {\mathrm {log}\left (3\right ) x +x^{3}+2 x^{2}+x +1}{x^{2}+2 x +1}}}{e^{\frac {x^{4}+2 x^{3}+x^{2}+3 x}{x^{2}+2 x +1}} 3^{x} e^{6}} \] Input:
int(((-x^3-3*x^2-4*x)*log(3)-2*x^4-5*x^3-3*x^2+4*x-4)*exp(((-x^3-2*x^2)*lo g(3)-x^4-x^3-5*x^2-14*x-5)/(x^2+2*x+1))/(x^3+3*x^2+3*x+1),x)
Output:
e**((log(3)*x + x**3 + 2*x**2 + x + 1)/(x**2 + 2*x + 1))/(e**((x**4 + 2*x* *3 + x**2 + 3*x)/(x**2 + 2*x + 1))*3**x*e**6)