Integrand size = 97, antiderivative size = 30 \[ \int \frac {e^{-x+\frac {e^{-x} \left (-3 x^3+3 e^x x^4+e^x \left (x-x^2\right ) \log (x)\right )}{\log (x)}} \left (3 x^2-3 e^x x^3+\left (-9 x^2+3 x^3+12 e^x x^3\right ) \log (x)+e^x (1-2 x) \log ^2(x)\right )}{\log ^2(x)} \, dx=e^{x-x^2+\frac {3 x^2 \left (-e^{-x} x+x^2\right )}{\log (x)}} \] Output:
exp(3*x^2*(x^2-x/exp(x))/ln(x)+x-x^2)
Time = 5.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-x+\frac {e^{-x} \left (-3 x^3+3 e^x x^4+e^x \left (x-x^2\right ) \log (x)\right )}{\log (x)}} \left (3 x^2-3 e^x x^3+\left (-9 x^2+3 x^3+12 e^x x^3\right ) \log (x)+e^x (1-2 x) \log ^2(x)\right )}{\log ^2(x)} \, dx=e^{x-x^2-\frac {3 \left (e^{-x}-x\right ) x^3}{\log (x)}} \] Input:
Integrate[(E^(-x + (-3*x^3 + 3*E^x*x^4 + E^x*(x - x^2)*Log[x])/(E^x*Log[x] ))*(3*x^2 - 3*E^x*x^3 + (-9*x^2 + 3*x^3 + 12*E^x*x^3)*Log[x] + E^x*(1 - 2* x)*Log[x]^2))/Log[x]^2,x]
Output:
E^(x - x^2 - (3*(E^(-x) - x)*x^3)/Log[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.
\(\displaystyle \int \frac {\left (-3 e^x x^3+3 x^2+\left (12 e^x x^3+3 x^3-9 x^2\right ) \log (x)+e^x (1-2 x) \log ^2(x)\right ) \exp \left (\frac {e^{-x} \left (3 e^x x^4-3 x^3+e^x \left (x-x^2\right ) \log (x)\right )}{\log (x)}-x\right )}{\log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\left (-3 e^x x^3+3 x^2+\left (12 e^x x^3+3 x^3-9 x^2\right ) \log (x)+e^x (1-2 x) \log ^2(x)\right ) \exp \left (\frac {e^{-x} x^2 \left (3 e^x x^2-3 x-e^x \log (x)\right )}{\log (x)}\right )}{\log ^2(x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {3 x^2 (x \log (x)-3 \log (x)+1) \exp \left (\frac {e^{-x} x^2 \left (3 e^x x^2-3 x-e^x \log (x)\right )}{\log (x)}\right )}{\log ^2(x)}-\frac {\left (3 x^3-12 x^3 \log (x)+2 x \log ^2(x)-\log ^2(x)\right ) \exp \left (\frac {e^{-x} x^2 \left (3 e^x x^2-3 x-e^x \log (x)\right )}{\log (x)}+x\right )}{\log ^2(x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 3 \int \frac {\exp \left (\frac {e^{-x} x^2 \left (3 e^x x^2-3 x-e^x \log (x)\right )}{\log (x)}\right ) x^2}{\log ^2(x)}dx+\int \exp \left (\frac {e^{-x} \left (3 e^x x^2-3 x-e^x \log (x)\right ) x^2}{\log (x)}+x\right )dx-2 \int \exp \left (\frac {e^{-x} \left (3 e^x x^2-3 x-e^x \log (x)\right ) x^2}{\log (x)}+x\right ) xdx-9 \int \frac {\exp \left (\frac {e^{-x} x^2 \left (3 e^x x^2-3 x-e^x \log (x)\right )}{\log (x)}\right ) x^2}{\log (x)}dx-3 \int \frac {\exp \left (\frac {e^{-x} \left (3 e^x x^2-3 x-e^x \log (x)\right ) x^2}{\log (x)}+x\right ) x^3}{\log ^2(x)}dx+3 \int \frac {\exp \left (\frac {e^{-x} x^2 \left (3 e^x x^2-3 x-e^x \log (x)\right )}{\log (x)}\right ) x^3}{\log (x)}dx+12 \int \frac {\exp \left (\frac {e^{-x} \left (3 e^x x^2-3 x-e^x \log (x)\right ) x^2}{\log (x)}+x\right ) x^3}{\log (x)}dx\) |
Input:
Int[(E^(-x + (-3*x^3 + 3*E^x*x^4 + E^x*(x - x^2)*Log[x])/(E^x*Log[x]))*(3* x^2 - 3*E^x*x^3 + (-9*x^2 + 3*x^3 + 12*E^x*x^3)*Log[x] + E^x*(1 - 2*x)*Log [x]^2))/Log[x]^2,x]
Output:
$Aborted
Time = 1.85 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20
method | result | size |
parallelrisch | \({\mathrm e}^{\frac {\left (\left (-x^{2}+x \right ) {\mathrm e}^{x} \ln \left (x \right )+3 \,{\mathrm e}^{x} x^{4}-3 x^{3}\right ) {\mathrm e}^{-x}}{\ln \left (x \right )}}\) | \(36\) |
risch | \({\mathrm e}^{-\frac {x \left (-3 \,{\mathrm e}^{x} x^{3}+x \,{\mathrm e}^{x} \ln \left (x \right )-{\mathrm e}^{x} \ln \left (x \right )+3 x^{2}\right ) {\mathrm e}^{-x}}{\ln \left (x \right )}}\) | \(38\) |
Input:
int(((1-2*x)*exp(x)*ln(x)^2+(12*exp(x)*x^3+3*x^3-9*x^2)*ln(x)-3*exp(x)*x^3 +3*x^2)*exp(((-x^2+x)*exp(x)*ln(x)+3*exp(x)*x^4-3*x^3)/exp(x)/ln(x))/exp(x )/ln(x)^2,x,method=_RETURNVERBOSE)
Output:
exp(((-x^2+x)*exp(x)*ln(x)+3*exp(x)*x^4-3*x^3)/exp(x)/ln(x))
Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {e^{-x+\frac {e^{-x} \left (-3 x^3+3 e^x x^4+e^x \left (x-x^2\right ) \log (x)\right )}{\log (x)}} \left (3 x^2-3 e^x x^3+\left (-9 x^2+3 x^3+12 e^x x^3\right ) \log (x)+e^x (1-2 x) \log ^2(x)\right )}{\log ^2(x)} \, dx=e^{\left (x + \frac {{\left (3 \, x^{4} e^{x} - x^{2} e^{x} \log \left (x\right ) - 3 \, x^{3}\right )} e^{\left (-x\right )}}{\log \left (x\right )}\right )} \] Input:
integrate(((1-2*x)*exp(x)*log(x)^2+(12*exp(x)*x^3+3*x^3-9*x^2)*log(x)-3*ex p(x)*x^3+3*x^2)*exp(((-x^2+x)*exp(x)*log(x)+3*exp(x)*x^4-3*x^3)/exp(x)/log (x))/exp(x)/log(x)^2,x, algorithm="fricas")
Output:
e^(x + (3*x^4*e^x - x^2*e^x*log(x) - 3*x^3)*e^(-x)/log(x))
Time = 0.36 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-x+\frac {e^{-x} \left (-3 x^3+3 e^x x^4+e^x \left (x-x^2\right ) \log (x)\right )}{\log (x)}} \left (3 x^2-3 e^x x^3+\left (-9 x^2+3 x^3+12 e^x x^3\right ) \log (x)+e^x (1-2 x) \log ^2(x)\right )}{\log ^2(x)} \, dx=e^{\frac {\left (3 x^{4} e^{x} - 3 x^{3} + \left (- x^{2} + x\right ) e^{x} \log {\left (x \right )}\right ) e^{- x}}{\log {\left (x \right )}}} \] Input:
integrate(((1-2*x)*exp(x)*ln(x)**2+(12*exp(x)*x**3+3*x**3-9*x**2)*ln(x)-3* exp(x)*x**3+3*x**2)*exp(((-x**2+x)*exp(x)*ln(x)+3*exp(x)*x**4-3*x**3)/exp( x)/ln(x))/exp(x)/ln(x)**2,x)
Output:
exp((3*x**4*exp(x) - 3*x**3 + (-x**2 + x)*exp(x)*log(x))*exp(-x)/log(x))
Time = 0.21 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x+\frac {e^{-x} \left (-3 x^3+3 e^x x^4+e^x \left (x-x^2\right ) \log (x)\right )}{\log (x)}} \left (3 x^2-3 e^x x^3+\left (-9 x^2+3 x^3+12 e^x x^3\right ) \log (x)+e^x (1-2 x) \log ^2(x)\right )}{\log ^2(x)} \, dx=e^{\left (\frac {3 \, x^{4}}{\log \left (x\right )} - \frac {3 \, x^{3} e^{\left (-x\right )}}{\log \left (x\right )} - x^{2} + x\right )} \] Input:
integrate(((1-2*x)*exp(x)*log(x)^2+(12*exp(x)*x^3+3*x^3-9*x^2)*log(x)-3*ex p(x)*x^3+3*x^2)*exp(((-x^2+x)*exp(x)*log(x)+3*exp(x)*x^4-3*x^3)/exp(x)/log (x))/exp(x)/log(x)^2,x, algorithm="maxima")
Output:
e^(3*x^4/log(x) - 3*x^3*e^(-x)/log(x) - x^2 + x)
\[ \int \frac {e^{-x+\frac {e^{-x} \left (-3 x^3+3 e^x x^4+e^x \left (x-x^2\right ) \log (x)\right )}{\log (x)}} \left (3 x^2-3 e^x x^3+\left (-9 x^2+3 x^3+12 e^x x^3\right ) \log (x)+e^x (1-2 x) \log ^2(x)\right )}{\log ^2(x)} \, dx=\int { -\frac {{\left (3 \, x^{3} e^{x} + {\left (2 \, x - 1\right )} e^{x} \log \left (x\right )^{2} - 3 \, x^{2} - 3 \, {\left (4 \, x^{3} e^{x} + x^{3} - 3 \, x^{2}\right )} \log \left (x\right )\right )} e^{\left (-x + \frac {{\left (3 \, x^{4} e^{x} - 3 \, x^{3} - {\left (x^{2} - x\right )} e^{x} \log \left (x\right )\right )} e^{\left (-x\right )}}{\log \left (x\right )}\right )}}{\log \left (x\right )^{2}} \,d x } \] Input:
integrate(((1-2*x)*exp(x)*log(x)^2+(12*exp(x)*x^3+3*x^3-9*x^2)*log(x)-3*ex p(x)*x^3+3*x^2)*exp(((-x^2+x)*exp(x)*log(x)+3*exp(x)*x^4-3*x^3)/exp(x)/log (x))/exp(x)/log(x)^2,x, algorithm="giac")
Output:
integrate(-(3*x^3*e^x + (2*x - 1)*e^x*log(x)^2 - 3*x^2 - 3*(4*x^3*e^x + x^ 3 - 3*x^2)*log(x))*e^(-x + (3*x^4*e^x - 3*x^3 - (x^2 - x)*e^x*log(x))*e^(- x)/log(x))/log(x)^2, x)
Time = 0.55 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {e^{-x+\frac {e^{-x} \left (-3 x^3+3 e^x x^4+e^x \left (x-x^2\right ) \log (x)\right )}{\log (x)}} \left (3 x^2-3 e^x x^3+\left (-9 x^2+3 x^3+12 e^x x^3\right ) \log (x)+e^x (1-2 x) \log ^2(x)\right )}{\log ^2(x)} \, dx={\mathrm {e}}^{-\frac {3\,x^3\,{\mathrm {e}}^{-x}}{\ln \left (x\right )}}\,{\mathrm {e}}^{-x^2}\,{\mathrm {e}}^x\,{\mathrm {e}}^{\frac {3\,x^4}{\ln \left (x\right )}} \] Input:
int(-(exp((exp(-x)*(3*x^4*exp(x) - 3*x^3 + exp(x)*log(x)*(x - x^2)))/log(x ))*exp(-x)*(3*x^3*exp(x) - log(x)*(12*x^3*exp(x) - 9*x^2 + 3*x^3) - 3*x^2 + exp(x)*log(x)^2*(2*x - 1)))/log(x)^2,x)
Output:
exp(-(3*x^3*exp(-x))/log(x))*exp(-x^2)*exp(x)*exp((3*x^4)/log(x))
Time = 0.29 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.57 \[ \int \frac {e^{-x+\frac {e^{-x} \left (-3 x^3+3 e^x x^4+e^x \left (x-x^2\right ) \log (x)\right )}{\log (x)}} \left (3 x^2-3 e^x x^3+\left (-9 x^2+3 x^3+12 e^x x^3\right ) \log (x)+e^x (1-2 x) \log ^2(x)\right )}{\log ^2(x)} \, dx=\frac {e^{\frac {\mathrm {log}\left (x \right ) x +3 x^{4}}{\mathrm {log}\left (x \right )}}}{e^{\frac {e^{x} \mathrm {log}\left (x \right ) x^{2}+3 x^{3}}{e^{x} \mathrm {log}\left (x \right )}}} \] Input:
int(((1-2*x)*exp(x)*log(x)^2+(12*exp(x)*x^3+3*x^3-9*x^2)*log(x)-3*exp(x)*x ^3+3*x^2)*exp(((-x^2+x)*exp(x)*log(x)+3*exp(x)*x^4-3*x^3)/exp(x)/log(x))/e xp(x)/log(x)^2,x)
Output:
e**((log(x)*x + 3*x**4)/log(x))/e**((e**x*log(x)*x**2 + 3*x**3)/(e**x*log( x)))