Integrand size = 63, antiderivative size = 23 \[ \int \frac {e^{\frac {-x \log (9)+\left (-5+e^{5 e^4 x} x\right ) \log (x)}{\log (x)}} \left (\log (9)-\log (9) \log (x)+e^{5 e^4 x} \left (1+5 e^4 x\right ) \log ^2(x)\right )}{\log ^2(x)} \, dx=e^{-5+e^{5 e^4 x} x-\frac {x \log (9)}{\log (x)}} \] Output:
exp(x*exp(5*x*exp(4))-5-2*x*ln(3)/ln(x))
Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {e^{\frac {-x \log (9)+\left (-5+e^{5 e^4 x} x\right ) \log (x)}{\log (x)}} \left (\log (9)-\log (9) \log (x)+e^{5 e^4 x} \left (1+5 e^4 x\right ) \log ^2(x)\right )}{\log ^2(x)} \, dx=9^{-\frac {x}{\log (x)}} e^{-5+e^{5 e^4 x} x} \] Input:
Integrate[(E^((-(x*Log[9]) + (-5 + E^(5*E^4*x)*x)*Log[x])/Log[x])*(Log[9] - Log[9]*Log[x] + E^(5*E^4*x)*(1 + 5*E^4*x)*Log[x]^2))/Log[x]^2,x]
Output:
E^(-5 + E^(5*E^4*x)*x)/9^(x/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.
\(\displaystyle \int \frac {\left (e^{5 e^4 x} \left (5 e^4 x+1\right ) \log ^2(x)-\log (9) \log (x)+\log (9)\right ) \exp \left (\frac {\left (e^{5 e^4 x} x-5\right ) \log (x)-x \log (9)}{\log (x)}\right )}{\log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\left (5 e^4 x+1\right ) \exp \left (5 e^4 x+\frac {\left (e^{5 e^4 x} x-5\right ) \log (x)-x \log (9)}{\log (x)}\right )-\frac {\log (9) (\log (x)-1) \exp \left (\frac {\left (e^{5 e^4 x} x-5\right ) \log (x)-x \log (9)}{\log (x)}\right )}{\log ^2(x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \log (9) \int \frac {\exp \left (\frac {\left (e^{5 e^4 x} x-5\right ) \log (x)-x \log (9)}{\log (x)}\right )}{\log ^2(x)}dx+\int \exp \left (\frac {e^{5 e^4 x} \log (x) x+5 e^4 \log (x) x-\log (9) x-5 \log (x)}{\log (x)}\right )dx+5 \int \exp \left (\frac {e^{5 e^4 x} \log (x) x+5 e^4 \log (x) x-\log (9) x-\log (x)}{\log (x)}\right ) xdx-\log (9) \int \frac {\exp \left (\frac {\left (e^{5 e^4 x} x-5\right ) \log (x)-x \log (9)}{\log (x)}\right )}{\log (x)}dx\) |
Input:
Int[(E^((-(x*Log[9]) + (-5 + E^(5*E^4*x)*x)*Log[x])/Log[x])*(Log[9] - Log[ 9]*Log[x] + E^(5*E^4*x)*(1 + 5*E^4*x)*Log[x]^2))/Log[x]^2,x]
Output:
$Aborted
Time = 7.36 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91
method | result | size |
risch | \(\left (\frac {1}{9}\right )^{\frac {x}{\ln \left (x \right )}} {\mathrm e}^{x \,{\mathrm e}^{5 x \,{\mathrm e}^{4}}-5}\) | \(21\) |
parallelrisch | \({\mathrm e}^{\frac {\ln \left (x \right ) {\mathrm e}^{5 x \,{\mathrm e}^{4}} x -2 x \ln \left (3\right )-5 \ln \left (x \right )}{\ln \left (x \right )}}\) | \(27\) |
Input:
int(((5*x*exp(4)+1)*exp(5*x*exp(4))*ln(x)^2-2*ln(3)*ln(x)+2*ln(3))*exp(((x *exp(5*x*exp(4))-5)*ln(x)-2*x*ln(3))/ln(x))/ln(x)^2,x,method=_RETURNVERBOS E)
Output:
(1/9)^(x/ln(x))*exp(x*exp(5*x*exp(4))-5)
Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {e^{\frac {-x \log (9)+\left (-5+e^{5 e^4 x} x\right ) \log (x)}{\log (x)}} \left (\log (9)-\log (9) \log (x)+e^{5 e^4 x} \left (1+5 e^4 x\right ) \log ^2(x)\right )}{\log ^2(x)} \, dx=e^{\left (-\frac {2 \, x \log \left (3\right ) - {\left (x e^{\left (5 \, x e^{4}\right )} - 5\right )} \log \left (x\right )}{\log \left (x\right )}\right )} \] Input:
integrate(((5*x*exp(4)+1)*exp(5*x*exp(4))*log(x)^2-2*log(3)*log(x)+2*log(3 ))*exp(((x*exp(5*x*exp(4))-5)*log(x)-2*x*log(3))/log(x))/log(x)^2,x, algor ithm="fricas")
Output:
e^(-(2*x*log(3) - (x*e^(5*x*e^4) - 5)*log(x))/log(x))
Time = 0.31 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {e^{\frac {-x \log (9)+\left (-5+e^{5 e^4 x} x\right ) \log (x)}{\log (x)}} \left (\log (9)-\log (9) \log (x)+e^{5 e^4 x} \left (1+5 e^4 x\right ) \log ^2(x)\right )}{\log ^2(x)} \, dx=e^{\frac {- 2 x \log {\left (3 \right )} + \left (x e^{5 x e^{4}} - 5\right ) \log {\left (x \right )}}{\log {\left (x \right )}}} \] Input:
integrate(((5*x*exp(4)+1)*exp(5*x*exp(4))*ln(x)**2-2*ln(3)*ln(x)+2*ln(3))* exp(((x*exp(5*x*exp(4))-5)*ln(x)-2*x*ln(3))/ln(x))/ln(x)**2,x)
Output:
exp((-2*x*log(3) + (x*exp(5*x*exp(4)) - 5)*log(x))/log(x))
Time = 0.21 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {e^{\frac {-x \log (9)+\left (-5+e^{5 e^4 x} x\right ) \log (x)}{\log (x)}} \left (\log (9)-\log (9) \log (x)+e^{5 e^4 x} \left (1+5 e^4 x\right ) \log ^2(x)\right )}{\log ^2(x)} \, dx=e^{\left (x e^{\left (5 \, x e^{4}\right )} - \frac {2 \, x \log \left (3\right )}{\log \left (x\right )} - 5\right )} \] Input:
integrate(((5*x*exp(4)+1)*exp(5*x*exp(4))*log(x)^2-2*log(3)*log(x)+2*log(3 ))*exp(((x*exp(5*x*exp(4))-5)*log(x)-2*x*log(3))/log(x))/log(x)^2,x, algor ithm="maxima")
Output:
e^(x*e^(5*x*e^4) - 2*x*log(3)/log(x) - 5)
\[ \int \frac {e^{\frac {-x \log (9)+\left (-5+e^{5 e^4 x} x\right ) \log (x)}{\log (x)}} \left (\log (9)-\log (9) \log (x)+e^{5 e^4 x} \left (1+5 e^4 x\right ) \log ^2(x)\right )}{\log ^2(x)} \, dx=\int { \frac {{\left ({\left (5 \, x e^{4} + 1\right )} e^{\left (5 \, x e^{4}\right )} \log \left (x\right )^{2} - 2 \, \log \left (3\right ) \log \left (x\right ) + 2 \, \log \left (3\right )\right )} e^{\left (-\frac {2 \, x \log \left (3\right ) - {\left (x e^{\left (5 \, x e^{4}\right )} - 5\right )} \log \left (x\right )}{\log \left (x\right )}\right )}}{\log \left (x\right )^{2}} \,d x } \] Input:
integrate(((5*x*exp(4)+1)*exp(5*x*exp(4))*log(x)^2-2*log(3)*log(x)+2*log(3 ))*exp(((x*exp(5*x*exp(4))-5)*log(x)-2*x*log(3))/log(x))/log(x)^2,x, algor ithm="giac")
Output:
integrate(((5*x*e^4 + 1)*e^(5*x*e^4)*log(x)^2 - 2*log(3)*log(x) + 2*log(3) )*e^(-(2*x*log(3) - (x*e^(5*x*e^4) - 5)*log(x))/log(x))/log(x)^2, x)
Time = 3.21 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {-x \log (9)+\left (-5+e^{5 e^4 x} x\right ) \log (x)}{\log (x)}} \left (\log (9)-\log (9) \log (x)+e^{5 e^4 x} \left (1+5 e^4 x\right ) \log ^2(x)\right )}{\log ^2(x)} \, dx=\frac {{\mathrm {e}}^{-5}\,{\mathrm {e}}^{x\,{\mathrm {e}}^{5\,x\,{\mathrm {e}}^4}}}{3^{\frac {2\,x}{\ln \left (x\right )}}} \] Input:
int((exp((log(x)*(x*exp(5*x*exp(4)) - 5) - 2*x*log(3))/log(x))*(2*log(3) - 2*log(3)*log(x) + exp(5*x*exp(4))*log(x)^2*(5*x*exp(4) + 1)))/log(x)^2,x)
Output:
(exp(-5)*exp(x*exp(5*x*exp(4))))/3^((2*x)/log(x))
Time = 0.15 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {e^{\frac {-x \log (9)+\left (-5+e^{5 e^4 x} x\right ) \log (x)}{\log (x)}} \left (\log (9)-\log (9) \log (x)+e^{5 e^4 x} \left (1+5 e^4 x\right ) \log ^2(x)\right )}{\log ^2(x)} \, dx=\frac {e^{e^{5 e^{4} x} x}}{e^{\frac {2 \,\mathrm {log}\left (3\right ) x}{\mathrm {log}\left (x \right )}} e^{5}} \] Input:
int(((5*x*exp(4)+1)*exp(5*x*exp(4))*log(x)^2-2*log(3)*log(x)+2*log(3))*exp (((x*exp(5*x*exp(4))-5)*log(x)-2*x*log(3))/log(x))/log(x)^2,x)
Output:
e**(e**(5*e**4*x)*x)/(e**((2*log(3)*x)/log(x))*e**5)