Integrand size = 53, antiderivative size = 20 \[ \int \frac {1}{3} e^{\frac {1}{16} \left (-9-24 e^5-16 e^{10}\right )} x^{\frac {2}{3} e^{\frac {1}{16} \left (-9-24 e^5-16 e^{10}\right )} x} (2+2 \log (x)) \, dx=x^{\frac {2}{3} e^{-\left (\frac {3}{4}+e^5\right )^2} x} \] Output:
exp(1/3*ln(x)*x/exp((exp(5)+3/4)^2))^2
Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {1}{3} e^{\frac {1}{16} \left (-9-24 e^5-16 e^{10}\right )} x^{\frac {2}{3} e^{\frac {1}{16} \left (-9-24 e^5-16 e^{10}\right )} x} (2+2 \log (x)) \, dx=x^{\frac {2}{3} e^{-\frac {1}{16} \left (3+4 e^5\right )^2} x} \] Input:
Integrate[(E^((-9 - 24*E^5 - 16*E^10)/16)*x^((2*E^((-9 - 24*E^5 - 16*E^10) /16)*x)/3)*(2 + 2*Log[x]))/3,x]
Output:
x^((2*x)/(3*E^((3 + 4*E^5)^2/16)))
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.
\(\displaystyle \int \frac {1}{3} e^{\frac {1}{16} \left (-9-24 e^5-16 e^{10}\right )} x^{\frac {2}{3} e^{\frac {1}{16} \left (-9-24 e^5-16 e^{10}\right )} x} (2 \log (x)+2) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} e^{-\frac {1}{16} \left (3+4 e^5\right )^2} \int 2 x^{\frac {2}{3} e^{-\frac {1}{16} \left (3+4 e^5\right )^2} x} (\log (x)+1)dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2}{3} e^{-\frac {1}{16} \left (3+4 e^5\right )^2} \int x^{\frac {2}{3} e^{-\frac {1}{16} \left (3+4 e^5\right )^2} x} (\log (x)+1)dx\) |
\(\Big \downarrow \) 7281 |
\(\displaystyle 2 e^{-\frac {1}{16} \left (3+4 e^5\right )^2} \int x^{\frac {2}{3} e^{-\frac {1}{16} \left (3+4 e^5\right )^2} x} (\log (x)+1)d\frac {x}{3}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 2 e^{-\frac {1}{16} \left (3+4 e^5\right )^2} \int \left (\log (x) x^{\frac {2}{3} e^{-\frac {1}{16} \left (3+4 e^5\right )^2} x}+x^{\frac {2}{3} e^{-\frac {1}{16} \left (3+4 e^5\right )^2} x}\right )d\frac {x}{3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 e^{-\frac {1}{16} \left (3+4 e^5\right )^2} \left (\int x^{\frac {2}{3} e^{-\frac {1}{16} \left (3+4 e^5\right )^2} x}d\frac {x}{3}-\int \frac {3 \int x^{\frac {2}{3} e^{-\frac {1}{16} \left (3+4 e^5\right )^2} x}d\frac {x}{3}}{x}d\frac {x}{3}+\log (x) \int x^{\frac {2}{3} e^{-\frac {1}{16} \left (3+4 e^5\right )^2} x}d\frac {x}{3}\right )\) |
Input:
Int[(E^((-9 - 24*E^5 - 16*E^10)/16)*x^((2*E^((-9 - 24*E^5 - 16*E^10)/16)*x )/3)*(2 + 2*Log[x]))/3,x]
Output:
$Aborted
Time = 0.40 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95
method | result | size |
risch | \(x^{\frac {2 x \,{\mathrm e}^{-{\mathrm e}^{10}-\frac {3 \,{\mathrm e}^{5}}{2}-\frac {9}{16}}}{3}}\) | \(19\) |
default | \({\mathrm e}^{x \,{\mathrm e}^{-{\mathrm e}^{10}-\frac {3 \,{\mathrm e}^{5}}{2}-\frac {9}{16}} \ln \left (x^{\frac {2}{3}}\right )}\) | \(22\) |
norman | \({\mathrm e}^{x \,{\mathrm e}^{-{\mathrm e}^{10}-\frac {3 \,{\mathrm e}^{5}}{2}-\frac {9}{16}} \ln \left (x^{\frac {2}{3}}\right )}\) | \(22\) |
parallelrisch | \({\mathrm e}^{x \,{\mathrm e}^{-{\mathrm e}^{10}-\frac {3 \,{\mathrm e}^{5}}{2}-\frac {9}{16}} \ln \left (x^{\frac {2}{3}}\right )}\) | \(22\) |
Input:
int(1/3*(2*ln(x)+2)*exp(1/3*x*ln(x)/exp(exp(5)^2+3/2*exp(5)+9/16))^2/exp(e xp(5)^2+3/2*exp(5)+9/16),x,method=_RETURNVERBOSE)
Output:
(x^(1/3*x*exp(-exp(10)-3/2*exp(5)-9/16)))^2
Time = 0.09 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {1}{3} e^{\frac {1}{16} \left (-9-24 e^5-16 e^{10}\right )} x^{\frac {2}{3} e^{\frac {1}{16} \left (-9-24 e^5-16 e^{10}\right )} x} (2+2 \log (x)) \, dx=x^{\frac {2}{3} \, x e^{\left (-e^{10} - \frac {3}{2} \, e^{5} - \frac {9}{16}\right )}} \] Input:
integrate(1/3*(2*log(x)+2)*exp(1/3*x*log(x)/exp(exp(5)^2+3/2*exp(5)+9/16)) ^2/exp(exp(5)^2+3/2*exp(5)+9/16),x, algorithm="fricas")
Output:
x^(2/3*x*e^(-e^10 - 3/2*e^5 - 9/16))
Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {1}{3} e^{\frac {1}{16} \left (-9-24 e^5-16 e^{10}\right )} x^{\frac {2}{3} e^{\frac {1}{16} \left (-9-24 e^5-16 e^{10}\right )} x} (2+2 \log (x)) \, dx=e^{\frac {2 x \log {\left (x \right )}}{3 e^{\frac {9}{16} + \frac {3 e^{5}}{2} + e^{10}}}} \] Input:
integrate(1/3*(2*ln(x)+2)*exp(1/3*x*ln(x)/exp(exp(5)**2+3/2*exp(5)+9/16))* *2/exp(exp(5)**2+3/2*exp(5)+9/16),x)
Output:
exp(2*x*exp(-exp(10) - 3*exp(5)/2 - 9/16)*log(x)/3)
Time = 0.08 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {1}{3} e^{\frac {1}{16} \left (-9-24 e^5-16 e^{10}\right )} x^{\frac {2}{3} e^{\frac {1}{16} \left (-9-24 e^5-16 e^{10}\right )} x} (2+2 \log (x)) \, dx=x^{\frac {2}{3} \, x e^{\left (-e^{10} - \frac {3}{2} \, e^{5} - \frac {9}{16}\right )}} \] Input:
integrate(1/3*(2*log(x)+2)*exp(1/3*x*log(x)/exp(exp(5)^2+3/2*exp(5)+9/16)) ^2/exp(exp(5)^2+3/2*exp(5)+9/16),x, algorithm="maxima")
Output:
x^(2/3*x*e^(-e^10 - 3/2*e^5 - 9/16))
Time = 0.47 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {1}{3} e^{\frac {1}{16} \left (-9-24 e^5-16 e^{10}\right )} x^{\frac {2}{3} e^{\frac {1}{16} \left (-9-24 e^5-16 e^{10}\right )} x} (2+2 \log (x)) \, dx=x^{\frac {2}{3} \, x e^{\left (-e^{10} - \frac {3}{2} \, e^{5} - \frac {9}{16}\right )}} \] Input:
integrate(1/3*(2*log(x)+2)*exp(1/3*x*log(x)/exp(exp(5)^2+3/2*exp(5)+9/16)) ^2/exp(exp(5)^2+3/2*exp(5)+9/16),x, algorithm="giac")
Output:
x^(2/3*x*e^(-e^10 - 3/2*e^5 - 9/16))
Time = 4.33 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {1}{3} e^{\frac {1}{16} \left (-9-24 e^5-16 e^{10}\right )} x^{\frac {2}{3} e^{\frac {1}{16} \left (-9-24 e^5-16 e^{10}\right )} x} (2+2 \log (x)) \, dx=x^{\frac {2\,x\,{\mathrm {e}}^{-\frac {3\,{\mathrm {e}}^5}{2}-{\mathrm {e}}^{10}-\frac {9}{16}}}{3}} \] Input:
int((exp(- (3*exp(5))/2 - exp(10) - 9/16)*exp((2*x*exp(- (3*exp(5))/2 - ex p(10) - 9/16)*log(x))/3)*(2*log(x) + 2))/3,x)
Output:
x^((2*x*exp(- (3*exp(5))/2 - exp(10) - 9/16))/3)
Time = 0.20 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {1}{3} e^{\frac {1}{16} \left (-9-24 e^5-16 e^{10}\right )} x^{\frac {2}{3} e^{\frac {1}{16} \left (-9-24 e^5-16 e^{10}\right )} x} (2+2 \log (x)) \, dx=e^{\frac {2 \,\mathrm {log}\left (x \right ) x}{3 e^{e^{10}+\frac {3}{2} e^{5}+\frac {9}{16}}}} \] Input:
int(1/3*(2*log(x)+2)*exp(1/3*x*log(x)/exp(exp(5)^2+3/2*exp(5)+9/16))^2/exp (exp(5)^2+3/2*exp(5)+9/16),x)
Output:
e**((2*log(x)*x)/(3*e**((16*e**10 + 24*e**5 + 9)/16)))