Integrand size = 93, antiderivative size = 25 \[ \int \frac {e^{-x} \left (3 e^{1+x}+x^{\frac {1}{3} e^{-x} \left (-2 x+2 e^x x\right )} \left (-6 x+e^x (3+6 x)+\left (-8 x+8 e^x x+6 x^2\right ) \log (x)+\left (-2 x+2 e^x x+2 x^2\right ) \log ^2(x)\right )\right )}{3 x} \, dx=\left (e+x^{\frac {1}{3} \left (2 x-2 e^{-x} x\right )}\right ) (3+\log (x)) \] Output:
(exp(1)+exp(ln(x)*(2/3*x-2/3*x/exp(x))))*(3+ln(x))
Time = 5.13 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {e^{-x} \left (3 e^{1+x}+x^{\frac {1}{3} e^{-x} \left (-2 x+2 e^x x\right )} \left (-6 x+e^x (3+6 x)+\left (-8 x+8 e^x x+6 x^2\right ) \log (x)+\left (-2 x+2 e^x x+2 x^2\right ) \log ^2(x)\right )\right )}{3 x} \, dx=\frac {1}{3} \left (3 e \log (x)+x^{-\frac {2}{3} \left (-1+e^{-x}\right ) x} (9+3 \log (x))\right ) \] Input:
Integrate[(3*E^(1 + x) + x^((-2*x + 2*E^x*x)/(3*E^x))*(-6*x + E^x*(3 + 6*x ) + (-8*x + 8*E^x*x + 6*x^2)*Log[x] + (-2*x + 2*E^x*x + 2*x^2)*Log[x]^2))/ (3*E^x*x),x]
Output:
(3*E*Log[x] + (9 + 3*Log[x])/x^((2*(-1 + E^(-x))*x)/3))/3
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.
\(\displaystyle \int \frac {e^{-x} \left (x^{\frac {1}{3} e^{-x} \left (2 e^x x-2 x\right )} \left (\left (2 x^2+2 e^x x-2 x\right ) \log ^2(x)+\left (6 x^2+8 e^x x-8 x\right ) \log (x)-6 x+e^x (6 x+3)\right )+3 e^{x+1}\right )}{3 x} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \int \frac {e^{-x} \left (3 e^{x+1}-x^{-\frac {2}{3} e^{-x} \left (x-e^x x\right )} \left (2 \left (-x^2-e^x x+x\right ) \log ^2(x)+2 \left (-3 x^2-4 e^x x+4 x\right ) \log (x)+6 x-3 e^x (2 x+1)\right )\right )}{x}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{3} \int \left (e^{-x} \left (2 \log ^2(x) x^2+6 \log (x) x^2+6 e^x x+2 e^x \log ^2(x) x-2 \log ^2(x) x+8 e^x \log (x) x-8 \log (x) x-6 x+3 e^x\right ) x^{-\frac {2}{3} \left (-1+e^{-x}\right ) x-1}+\frac {3 e}{x}\right )dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \frac {1}{3} \int \left (e^{-x} \left (2 \log ^2(x) x^2+6 \log (x) x^2+6 e^x x+2 e^x \log ^2(x) x-2 \log ^2(x) x+8 e^x \log (x) x-8 \log (x) x-6 x+3 e^x\right ) x^{-\frac {2}{3} \left (-1+e^{-x}\right ) x-1}+\frac {3 e}{x}\right )dx\) |
Input:
Int[(3*E^(1 + x) + x^((-2*x + 2*E^x*x)/(3*E^x))*(-6*x + E^x*(3 + 6*x) + (- 8*x + 8*E^x*x + 6*x^2)*Log[x] + (-2*x + 2*E^x*x + 2*x^2)*Log[x]^2))/(3*E^x *x),x]
Output:
$Aborted
Time = 12.45 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04
method | result | size |
risch | \({\mathrm e} \ln \left (x \right )+\frac {\left (9+3 \ln \left (x \right )\right ) x^{-\frac {2 \left ({\mathrm e}^{-x}-1\right ) x}{3}}}{3}\) | \(26\) |
parallelrisch | \({\mathrm e} \ln \left (x \right )+\ln \left (x \right ) {\mathrm e}^{\frac {2 x \left ({\mathrm e}^{x}-1\right ) \ln \left (x \right ) {\mathrm e}^{-x}}{3}}+3 \,{\mathrm e}^{\frac {2 x \left ({\mathrm e}^{x}-1\right ) \ln \left (x \right ) {\mathrm e}^{-x}}{3}}\) | \(40\) |
Input:
int(1/3*(((2*exp(x)*x+2*x^2-2*x)*ln(x)^2+(8*exp(x)*x+6*x^2-8*x)*ln(x)+(6*x +3)*exp(x)-6*x)*exp(1/3*(2*exp(x)*x-2*x)*ln(x)/exp(x))+3*exp(1)*exp(x))/ex p(x)/x,x,method=_RETURNVERBOSE)
Output:
exp(1)*ln(x)+1/3*(9+3*ln(x))*x^(-2/3*(exp(-x)-1)*x)
Time = 0.07 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40 \[ \int \frac {e^{-x} \left (3 e^{1+x}+x^{\frac {1}{3} e^{-x} \left (-2 x+2 e^x x\right )} \left (-6 x+e^x (3+6 x)+\left (-8 x+8 e^x x+6 x^2\right ) \log (x)+\left (-2 x+2 e^x x+2 x^2\right ) \log ^2(x)\right )\right )}{3 x} \, dx=e \log \left (x\right ) + \frac {\log \left (x\right ) + 3}{x^{\frac {2}{3} \, {\left (x e - x e^{\left (x + 1\right )}\right )} e^{\left (-x - 1\right )}}} \] Input:
integrate(1/3*(((2*exp(x)*x+2*x^2-2*x)*log(x)^2+(8*exp(x)*x+6*x^2-8*x)*log (x)+(6*x+3)*exp(x)-6*x)*exp(1/3*(2*exp(x)*x-2*x)*log(x)/exp(x))+3*exp(1)*e xp(x))/exp(x)/x,x, algorithm="fricas")
Output:
e*log(x) + (log(x) + 3)/x^(2/3*(x*e - x*e^(x + 1))*e^(-x - 1))
Time = 105.99 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {e^{-x} \left (3 e^{1+x}+x^{\frac {1}{3} e^{-x} \left (-2 x+2 e^x x\right )} \left (-6 x+e^x (3+6 x)+\left (-8 x+8 e^x x+6 x^2\right ) \log (x)+\left (-2 x+2 e^x x+2 x^2\right ) \log ^2(x)\right )\right )}{3 x} \, dx=\left (\log {\left (x \right )} + 3\right ) e^{\left (\frac {2 x e^{x}}{3} - \frac {2 x}{3}\right ) e^{- x} \log {\left (x \right )}} + e \log {\left (x \right )} \] Input:
integrate(1/3*(((2*exp(x)*x+2*x**2-2*x)*ln(x)**2+(8*exp(x)*x+6*x**2-8*x)*l n(x)+(6*x+3)*exp(x)-6*x)*exp(1/3*(2*exp(x)*x-2*x)*ln(x)/exp(x))+3*exp(1)*e xp(x))/exp(x)/x,x)
Output:
(log(x) + 3)*exp((2*x*exp(x)/3 - 2*x/3)*exp(-x)*log(x)) + E*log(x)
Time = 0.12 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {e^{-x} \left (3 e^{1+x}+x^{\frac {1}{3} e^{-x} \left (-2 x+2 e^x x\right )} \left (-6 x+e^x (3+6 x)+\left (-8 x+8 e^x x+6 x^2\right ) \log (x)+\left (-2 x+2 e^x x+2 x^2\right ) \log ^2(x)\right )\right )}{3 x} \, dx={\left (\log \left (x\right ) + 3\right )} e^{\left (-\frac {2}{3} \, x e^{\left (-x\right )} \log \left (x\right ) + \frac {2}{3} \, x \log \left (x\right )\right )} + e \log \left (x\right ) \] Input:
integrate(1/3*(((2*exp(x)*x+2*x^2-2*x)*log(x)^2+(8*exp(x)*x+6*x^2-8*x)*log (x)+(6*x+3)*exp(x)-6*x)*exp(1/3*(2*exp(x)*x-2*x)*log(x)/exp(x))+3*exp(1)*e xp(x))/exp(x)/x,x, algorithm="maxima")
Output:
(log(x) + 3)*e^(-2/3*x*e^(-x)*log(x) + 2/3*x*log(x)) + e*log(x)
\[ \int \frac {e^{-x} \left (3 e^{1+x}+x^{\frac {1}{3} e^{-x} \left (-2 x+2 e^x x\right )} \left (-6 x+e^x (3+6 x)+\left (-8 x+8 e^x x+6 x^2\right ) \log (x)+\left (-2 x+2 e^x x+2 x^2\right ) \log ^2(x)\right )\right )}{3 x} \, dx=\int { \frac {{\left ({\left (2 \, {\left (x^{2} + x e^{x} - x\right )} \log \left (x\right )^{2} + 3 \, {\left (2 \, x + 1\right )} e^{x} + 2 \, {\left (3 \, x^{2} + 4 \, x e^{x} - 4 \, x\right )} \log \left (x\right ) - 6 \, x\right )} x^{\frac {2}{3} \, {\left (x e^{x} - x\right )} e^{\left (-x\right )}} + 3 \, e^{\left (x + 1\right )}\right )} e^{\left (-x\right )}}{3 \, x} \,d x } \] Input:
integrate(1/3*(((2*exp(x)*x+2*x^2-2*x)*log(x)^2+(8*exp(x)*x+6*x^2-8*x)*log (x)+(6*x+3)*exp(x)-6*x)*exp(1/3*(2*exp(x)*x-2*x)*log(x)/exp(x))+3*exp(1)*e xp(x))/exp(x)/x,x, algorithm="giac")
Output:
integrate(1/3*((2*(x^2 + x*e^x - x)*log(x)^2 + 3*(2*x + 1)*e^x + 2*(3*x^2 + 4*x*e^x - 4*x)*log(x) - 6*x)*x^(2/3*(x*e^x - x)*e^(-x)) + 3*e^(x + 1))*e ^(-x)/x, x)
Time = 1.72 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {e^{-x} \left (3 e^{1+x}+x^{\frac {1}{3} e^{-x} \left (-2 x+2 e^x x\right )} \left (-6 x+e^x (3+6 x)+\left (-8 x+8 e^x x+6 x^2\right ) \log (x)+\left (-2 x+2 e^x x+2 x^2\right ) \log ^2(x)\right )\right )}{3 x} \, dx={\mathrm {e}}^{\frac {2\,x\,\ln \left (x\right )}{3}-\frac {2\,x\,{\mathrm {e}}^{-x}\,\ln \left (x\right )}{3}}\,\left (\ln \left (x\right )+3\right )+\mathrm {e}\,\ln \left (x\right ) \] Input:
int((exp(-x)*((exp(-(exp(-x)*log(x)*(2*x - 2*x*exp(x)))/3)*(log(x)*(8*x*ex p(x) - 8*x + 6*x^2) - 6*x + exp(x)*(6*x + 3) + log(x)^2*(2*x*exp(x) - 2*x + 2*x^2)))/3 + exp(1)*exp(x)))/x,x)
Output:
exp((2*x*log(x))/3 - (2*x*exp(-x)*log(x))/3)*(log(x) + 3) + exp(1)*log(x)
Time = 0.21 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.88 \[ \int \frac {e^{-x} \left (3 e^{1+x}+x^{\frac {1}{3} e^{-x} \left (-2 x+2 e^x x\right )} \left (-6 x+e^x (3+6 x)+\left (-8 x+8 e^x x+6 x^2\right ) \log (x)+\left (-2 x+2 e^x x+2 x^2\right ) \log ^2(x)\right )\right )}{3 x} \, dx=\frac {e^{\frac {2 \,\mathrm {log}\left (x \right ) x}{3 e^{x}}} \mathrm {log}\left (x \right ) e +x^{\frac {2 x}{3}} \mathrm {log}\left (x \right )+3 x^{\frac {2 x}{3}}}{e^{\frac {2 \,\mathrm {log}\left (x \right ) x}{3 e^{x}}}} \] Input:
int(1/3*(((2*exp(x)*x+2*x^2-2*x)*log(x)^2+(8*exp(x)*x+6*x^2-8*x)*log(x)+(6 *x+3)*exp(x)-6*x)*exp(1/3*(2*exp(x)*x-2*x)*log(x)/exp(x))+3*exp(1)*exp(x)) /exp(x)/x,x)
Output:
(e**((2*log(x)*x)/(3*e**x))*log(x)*e + x**((2*x)/3)*log(x) + 3*x**((2*x)/3 ))/e**((2*log(x)*x)/(3*e**x))