Integrand size = 100, antiderivative size = 27 \[ \int \frac {e^{\frac {-5-14 x-5 x^2-x^3-x^4+\left (-2 x^2-x^3\right ) \log (3)}{1+2 x+x^2}} \left (-4+4 x-3 x^2-5 x^3-2 x^4+\left (-4 x-3 x^2-x^3\right ) \log (3)\right )}{1+3 x+3 x^2+x^3} \, dx=e^{-5+\frac {x (2+x) (-2+x (1-x-\log (3)))}{(1+x)^2}} \]
Time = 7.97 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.52 \[ \int \frac {e^{\frac {-5-14 x-5 x^2-x^3-x^4+\left (-2 x^2-x^3\right ) \log (3)}{1+2 x+x^2}} \left (-4+4 x-3 x^2-5 x^3-2 x^4+\left (-4 x-3 x^2-x^3\right ) \log (3)\right )}{1+3 x+3 x^2+x^3} \, dx=3^{-\frac {x^2 (2+x)}{(1+x)^2}} e^{-\frac {5+14 x+5 x^2+x^3+x^4}{(1+x)^2}} \]
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]
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 |
\(\displaystyle (1-\log (3)) \int \exp \left (\frac {-x^4-(1+\log (3)) x^3-(5+\log (9)) x^2-14 x-5}{x^2+2 x+1}\right )dx-2 \int \exp \left (\frac {-x^4-(1+\log (3)) x^3-(5+\log (9)) x^2-14 x-5}{x^2+2 x+1}\right ) xdx-(8-\log (9)) \int \frac {\exp \left (\frac {-x^4-(1+\log (3)) x^3-(5+\log (9)) x^2-14 x-5}{x^2+2 x+1}\right )}{(x+1)^3}dx+(3-\log (3)) \int \frac {\exp \left (\frac {-x^4-(1+\log (3)) x^3-(5+\log (9)) x^2-14 x-5}{x^2+2 x+1}\right )}{(x+1)^2}dx\) |
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]
3.17.28.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
Time = 0.25 (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 | \(\frac {x^{2} {\mathrm e}^{\frac {\left (-x^{3}-2 x^{2}\right ) \ln \left (3\right )-x^{4}-x^{3}-5 x^{2}-14 x -5}{x^{2}+2 x +1}}+2 x \,{\mathrm e}^{\frac {\left (-x^{3}-2 x^{2}\right ) \ln \left (3\right )-x^{4}-x^{3}-5 x^{2}-14 x -5}{x^{2}+2 x +1}}+{\mathrm e}^{\frac {\left (-x^{3}-2 x^{2}\right ) \ln \left (3\right )-x^{4}-x^{3}-5 x^{2}-14 x -5}{x^{2}+2 x +1}}}{\left (1+x \right )^{2}}\) | \(153\) |
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)
Time = 0.25 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.52 \[ \int \frac {e^{\frac {-5-14 x-5 x^2-x^3-x^4+\left (-2 x^2-x^3\right ) \log (3)}{1+2 x+x^2}} \left (-4+4 x-3 x^2-5 x^3-2 x^4+\left (-4 x-3 x^2-x^3\right ) \log (3)\right )}{1+3 x+3 x^2+x^3} \, dx=e^{\left (-\frac {x^{4} + x^{3} + 5 \, x^{2} + {\left (x^{3} + 2 \, x^{2}\right )} \log \left (3\right ) + 14 \, x + 5}{x^{2} + 2 \, x + 1}\right )} \]
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=\
Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44 \[ \int \frac {e^{\frac {-5-14 x-5 x^2-x^3-x^4+\left (-2 x^2-x^3\right ) \log (3)}{1+2 x+x^2}} \left (-4+4 x-3 x^2-5 x^3-2 x^4+\left (-4 x-3 x^2-x^3\right ) \log (3)\right )}{1+3 x+3 x^2+x^3} \, dx=e^{\frac {- x^{4} - x^{3} - 5 x^{2} - 14 x + \left (- x^{3} - 2 x^{2}\right ) \log {\left (3 \right )} - 5}{x^{2} + 2 x + 1}} \]
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)
Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (23) = 46\).
Time = 0.50 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.04 \[ \int \frac {e^{\frac {-5-14 x-5 x^2-x^3-x^4+\left (-2 x^2-x^3\right ) \log (3)}{1+2 x+x^2}} \left (-4+4 x-3 x^2-5 x^3-2 x^4+\left (-4 x-3 x^2-x^3\right ) \log (3)\right )}{1+3 x+3 x^2+x^3} \, dx=e^{\left (-x^{2} - x \log \left (3\right ) + x - \frac {\log \left (3\right )}{x^{2} + 2 \, x + 1} + \frac {\log \left (3\right )}{x + 1} + \frac {4}{x^{2} + 2 \, x + 1} - \frac {3}{x + 1} - 6\right )} \]
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=\
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 = 0.27 (sec) , antiderivative size = 106, normalized size of antiderivative = 3.93 \[ \int \frac {e^{\frac {-5-14 x-5 x^2-x^3-x^4+\left (-2 x^2-x^3\right ) \log (3)}{1+2 x+x^2}} \left (-4+4 x-3 x^2-5 x^3-2 x^4+\left (-4 x-3 x^2-x^3\right ) \log (3)\right )}{1+3 x+3 x^2+x^3} \, dx=e^{\left (-\frac {x^{4}}{x^{2} + 2 \, x + 1} - \frac {x^{3} \log \left (3\right )}{x^{2} + 2 \, x + 1} - \frac {x^{3}}{x^{2} + 2 \, x + 1} - \frac {2 \, x^{2} \log \left (3\right )}{x^{2} + 2 \, x + 1} - \frac {5 \, x^{2}}{x^{2} + 2 \, x + 1} - \frac {14 \, x}{x^{2} + 2 \, x + 1} - \frac {5}{x^{2} + 2 \, x + 1}\right )} \]
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=\
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 = 0.46 (sec) , antiderivative size = 98, normalized size of antiderivative = 3.63 \[ \int \frac {e^{\frac {-5-14 x-5 x^2-x^3-x^4+\left (-2 x^2-x^3\right ) \log (3)}{1+2 x+x^2}} \left (-4+4 x-3 x^2-5 x^3-2 x^4+\left (-4 x-3 x^2-x^3\right ) \log (3)\right )}{1+3 x+3 x^2+x^3} \, dx={\left (\frac {1}{3}\right )}^{\frac {x^3+2\,x^2}{x^2+2\,x+1}}\,{\mathrm {e}}^{-\frac {x^3}{x^2+2\,x+1}}\,{\mathrm {e}}^{-\frac {x^4}{x^2+2\,x+1}}\,{\mathrm {e}}^{-\frac {5\,x^2}{x^2+2\,x+1}}\,{\mathrm {e}}^{-\frac {5}{x^2+2\,x+1}}\,{\mathrm {e}}^{-\frac {14\,x}{x^2+2\,x+1}} \]
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)