Integrand size = 62, antiderivative size = 29 \[ \int \frac {e^{\frac {1}{3} (-4-15 x)} \left (-75 \log ^2(x)+e^{\frac {2}{3} \left (-1+e^{\frac {x}{\log (x)}}\right )} \left (-75 \log ^2(x)+e^{\frac {x}{\log (x)}} (-10+10 \log (x))\right )\right )}{\log ^2(x)} \, dx=15 e^{-\frac {4}{3}-5 x} \left (1+e^{\frac {2}{3} \left (-1+e^{\frac {x}{\log (x)}}\right )}\right ) \] Output:
15*(exp(1/3*exp(x/ln(x))-1/3)^2+1)/exp(5*x+4/3)
Time = 0.96 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {e^{\frac {1}{3} (-4-15 x)} \left (-75 \log ^2(x)+e^{\frac {2}{3} \left (-1+e^{\frac {x}{\log (x)}}\right )} \left (-75 \log ^2(x)+e^{\frac {x}{\log (x)}} (-10+10 \log (x))\right )\right )}{\log ^2(x)} \, dx=15 e^{-\frac {4}{3}-5 x}+15 e^{-2+\frac {2}{3} e^{\frac {x}{\log (x)}}-5 x} \] Input:
Integrate[(E^((-4 - 15*x)/3)*(-75*Log[x]^2 + E^((2*(-1 + E^(x/Log[x])))/3) *(-75*Log[x]^2 + E^(x/Log[x])*(-10 + 10*Log[x]))))/Log[x]^2,x]
Output:
15*E^(-4/3 - 5*x) + 15*E^(-2 + (2*E^(x/Log[x]))/3 - 5*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.
\(\displaystyle \int \frac {e^{\frac {1}{3} (-15 x-4)} \left (e^{\frac {2}{3} \left (e^{\frac {x}{\log (x)}}-1\right )} \left (e^{\frac {x}{\log (x)}} (10 \log (x)-10)-75 \log ^2(x)\right )-75 \log ^2(x)\right )}{\log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{-5 x-\frac {4}{3}} \left (e^{\frac {2}{3} \left (e^{\frac {x}{\log (x)}}-1\right )} \left (e^{\frac {x}{\log (x)}} (10 \log (x)-10)-75 \log ^2(x)\right )-75 \log ^2(x)\right )}{\log ^2(x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {10 e^{-5 x+\frac {x}{\log (x)}+\frac {2}{3} e^{\frac {x}{\log (x)}}-2} (\log (x)-1)}{\log ^2(x)}-75 e^{-5 x-2} \left (e^{\frac {2}{3} e^{\frac {x}{\log (x)}}}+e^{2/3}\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -10 \int \frac {e^{\frac {x}{\log (x)}-5 x+\frac {2}{3} e^{\frac {x}{\log (x)}}-2}}{\log ^2(x)}dx-75 \int e^{\frac {1}{3} \left (-15 x+2 e^{\frac {x}{\log (x)}}-6\right )}dx+10 \int \frac {e^{\frac {x}{\log (x)}-5 x+\frac {2}{3} e^{\frac {x}{\log (x)}}-2}}{\log (x)}dx+15 e^{-5 x-\frac {4}{3}}\) |
Input:
Int[(E^((-4 - 15*x)/3)*(-75*Log[x]^2 + E^((2*(-1 + E^(x/Log[x])))/3)*(-75* Log[x]^2 + E^(x/Log[x])*(-10 + 10*Log[x]))))/Log[x]^2,x]
Output:
$Aborted
Time = 0.91 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93
method | result | size |
risch | \(15 \,{\mathrm e}^{-5 x -\frac {4}{3}}+15 \,{\mathrm e}^{\frac {2 \,{\mathrm e}^{\frac {x}{\ln \left (x \right )}}}{3}-2-5 x}\) | \(27\) |
parallelrisch | \(\left (15+15 \,{\mathrm e}^{\frac {2 \,{\mathrm e}^{\frac {x}{\ln \left (x \right )}}}{3}-\frac {2}{3}}\right ) {\mathrm e}^{-5 x -\frac {4}{3}}\) | \(28\) |
Input:
int((((10*ln(x)-10)*exp(x/ln(x))-75*ln(x)^2)*exp(1/3*exp(x/ln(x))-1/3)^2-7 5*ln(x)^2)/exp(5*x+4/3)/ln(x)^2,x,method=_RETURNVERBOSE)
Output:
15*exp(-5*x-4/3)+15*exp(2/3*exp(x/ln(x))-2-5*x)
Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\frac {1}{3} (-4-15 x)} \left (-75 \log ^2(x)+e^{\frac {2}{3} \left (-1+e^{\frac {x}{\log (x)}}\right )} \left (-75 \log ^2(x)+e^{\frac {x}{\log (x)}} (-10+10 \log (x))\right )\right )}{\log ^2(x)} \, dx=15 \, e^{\left (-5 \, x + \frac {2}{3} \, e^{\frac {x}{\log \left (x\right )}} - 2\right )} + 15 \, e^{\left (-5 \, x - \frac {4}{3}\right )} \] Input:
integrate((((10*log(x)-10)*exp(x/log(x))-75*log(x)^2)*exp(1/3*exp(x/log(x) )-1/3)^2-75*log(x)^2)/exp(5*x+4/3)/log(x)^2,x, algorithm="fricas")
Output:
15*e^(-5*x + 2/3*e^(x/log(x)) - 2) + 15*e^(-5*x - 4/3)
Exception generated. \[ \int \frac {e^{\frac {1}{3} (-4-15 x)} \left (-75 \log ^2(x)+e^{\frac {2}{3} \left (-1+e^{\frac {x}{\log (x)}}\right )} \left (-75 \log ^2(x)+e^{\frac {x}{\log (x)}} (-10+10 \log (x))\right )\right )}{\log ^2(x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((((10*ln(x)-10)*exp(x/ln(x))-75*ln(x)**2)*exp(1/3*exp(x/ln(x))-1 /3)**2-75*ln(x)**2)/exp(5*x+4/3)/ln(x)**2,x)
Output:
Exception raised: TypeError >> '>' not supported between instances of 'Pol y' and 'int'
Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\frac {1}{3} (-4-15 x)} \left (-75 \log ^2(x)+e^{\frac {2}{3} \left (-1+e^{\frac {x}{\log (x)}}\right )} \left (-75 \log ^2(x)+e^{\frac {x}{\log (x)}} (-10+10 \log (x))\right )\right )}{\log ^2(x)} \, dx=15 \, e^{\left (-5 \, x + \frac {2}{3} \, e^{\frac {x}{\log \left (x\right )}} - 2\right )} + 15 \, e^{\left (-5 \, x - \frac {4}{3}\right )} \] Input:
integrate((((10*log(x)-10)*exp(x/log(x))-75*log(x)^2)*exp(1/3*exp(x/log(x) )-1/3)^2-75*log(x)^2)/exp(5*x+4/3)/log(x)^2,x, algorithm="maxima")
Output:
15*e^(-5*x + 2/3*e^(x/log(x)) - 2) + 15*e^(-5*x - 4/3)
\[ \int \frac {e^{\frac {1}{3} (-4-15 x)} \left (-75 \log ^2(x)+e^{\frac {2}{3} \left (-1+e^{\frac {x}{\log (x)}}\right )} \left (-75 \log ^2(x)+e^{\frac {x}{\log (x)}} (-10+10 \log (x))\right )\right )}{\log ^2(x)} \, dx=\int { \frac {5 \, {\left ({\left (2 \, {\left (\log \left (x\right ) - 1\right )} e^{\frac {x}{\log \left (x\right )}} - 15 \, \log \left (x\right )^{2}\right )} e^{\left (\frac {2}{3} \, e^{\frac {x}{\log \left (x\right )}} - \frac {2}{3}\right )} - 15 \, \log \left (x\right )^{2}\right )} e^{\left (-5 \, x - \frac {4}{3}\right )}}{\log \left (x\right )^{2}} \,d x } \] Input:
integrate((((10*log(x)-10)*exp(x/log(x))-75*log(x)^2)*exp(1/3*exp(x/log(x) )-1/3)^2-75*log(x)^2)/exp(5*x+4/3)/log(x)^2,x, algorithm="giac")
Output:
undef
Time = 2.50 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72 \[ \int \frac {e^{\frac {1}{3} (-4-15 x)} \left (-75 \log ^2(x)+e^{\frac {2}{3} \left (-1+e^{\frac {x}{\log (x)}}\right )} \left (-75 \log ^2(x)+e^{\frac {x}{\log (x)}} (-10+10 \log (x))\right )\right )}{\log ^2(x)} \, dx=15\,{\mathrm {e}}^{-5\,x}\,{\mathrm {e}}^{-2}\,\left ({\mathrm {e}}^{\frac {2\,{\mathrm {e}}^{\frac {x}{\ln \left (x\right )}}}{3}}+{\mathrm {e}}^{2/3}\right ) \] Input:
int(-(exp(- 5*x - 4/3)*(75*log(x)^2 + exp((2*exp(x/log(x)))/3 - 2/3)*(75*l og(x)^2 - exp(x/log(x))*(10*log(x) - 10))))/log(x)^2,x)
Output:
15*exp(-5*x)*exp(-2)*(exp((2*exp(x/log(x)))/3) + exp(2/3))
Time = 0.19 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {e^{\frac {1}{3} (-4-15 x)} \left (-75 \log ^2(x)+e^{\frac {2}{3} \left (-1+e^{\frac {x}{\log (x)}}\right )} \left (-75 \log ^2(x)+e^{\frac {x}{\log (x)}} (-10+10 \log (x))\right )\right )}{\log ^2(x)} \, dx=\frac {15 e^{\frac {2 e^{\frac {x}{\mathrm {log}\left (x \right )}}}{3}+\frac {1}{3}}+15 e}{e^{5 x +\frac {1}{3}} e^{2}} \] Input:
int((((10*log(x)-10)*exp(x/log(x))-75*log(x)^2)*exp(1/3*exp(x/log(x))-1/3) ^2-75*log(x)^2)/exp(5*x+4/3)/log(x)^2,x)
Output:
(15*(e**((2*e**(x/log(x)) + 1)/3) + e))/(e**((15*x + 1)/3)*e**2)