Integrand size = 55, antiderivative size = 24 \[ \int \frac {e^{\frac {1}{e^2}} (50-10 x)+e^x (10-2 x)+\left (5 e^{\frac {1}{e^2}} x+e^x \left (-4 x+x^2\right )\right ) \log (x)}{5 x \log ^3(x)} \, dx=-\frac {\left (e^{\frac {1}{e^2}}+\frac {e^x}{5}\right ) (5-x)}{\log ^2(x)} \] Output:
-(5-x)/ln(x)^2*(1/5*exp(x)+exp(exp(-2)))
Time = 0.14 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\frac {1}{e^2}} (50-10 x)+e^x (10-2 x)+\left (5 e^{\frac {1}{e^2}} x+e^x \left (-4 x+x^2\right )\right ) \log (x)}{5 x \log ^3(x)} \, dx=\frac {\left (5 e^{\frac {1}{e^2}}+e^x\right ) (-5+x)}{5 \log ^2(x)} \] Input:
Integrate[(E^E^(-2)*(50 - 10*x) + E^x*(10 - 2*x) + (5*E^E^(-2)*x + E^x*(-4 *x + x^2))*Log[x])/(5*x*Log[x]^3),x]
Output:
((5*E^E^(-2) + E^x)*(-5 + x))/(5*Log[x]^2)
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.
\(\displaystyle \int \frac {\left (e^x \left (x^2-4 x\right )+5 e^{\frac {1}{e^2}} x\right ) \log (x)+e^{\frac {1}{e^2}} (50-10 x)+e^x (10-2 x)}{5 x \log ^3(x)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \frac {2 e^x (5-x)+10 e^{\frac {1}{e^2}} (5-x)+\left (5 e^{\frac {1}{e^2}} x-e^x \left (4 x-x^2\right )\right ) \log (x)}{x \log ^3(x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{5} \int \left (\frac {5 e^{\frac {1}{e^2}} (\log (x) x-2 x+10)}{x \log ^3(x)}+\frac {e^x \left (\log (x) x^2-4 \log (x) x-2 x+10\right )}{x \log ^3(x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (-2 \int \frac {e^x}{\log ^3(x)}dx+10 \int \frac {e^x}{x \log ^3(x)}dx-4 \int \frac {e^x}{\log ^2(x)}dx+\int \frac {e^x x}{\log ^2(x)}dx+\frac {5 e^{\frac {1}{e^2}} x}{\log ^2(x)}-\frac {25 e^{\frac {1}{e^2}}}{\log ^2(x)}\right )\) |
Input:
Int[(E^E^(-2)*(50 - 10*x) + E^x*(10 - 2*x) + (5*E^E^(-2)*x + E^x*(-4*x + x ^2))*Log[x])/(5*x*Log[x]^3),x]
Output:
$Aborted
Time = 0.51 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12
method | result | size |
risch | \(\frac {5 x \,{\mathrm e}^{{\mathrm e}^{-2}}+{\mathrm e}^{x} x -25 \,{\mathrm e}^{{\mathrm e}^{-2}}-5 \,{\mathrm e}^{x}}{5 \ln \left (x \right )^{2}}\) | \(27\) |
parallelrisch | \(\frac {5 x \,{\mathrm e}^{{\mathrm e}^{-2}}+{\mathrm e}^{x} x -25 \,{\mathrm e}^{{\mathrm e}^{-2}}-5 \,{\mathrm e}^{x}}{5 \ln \left (x \right )^{2}}\) | \(31\) |
Input:
int(1/5*(((x^2-4*x)*exp(x)+5*x*exp(1/exp(2)))*ln(x)+(-2*x+10)*exp(x)+(-10* x+50)*exp(1/exp(2)))/x/ln(x)^3,x,method=_RETURNVERBOSE)
Output:
1/5*(5*x*exp(exp(-2))+exp(x)*x-25*exp(exp(-2))-5*exp(x))/ln(x)^2
Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {1}{e^2}} (50-10 x)+e^x (10-2 x)+\left (5 e^{\frac {1}{e^2}} x+e^x \left (-4 x+x^2\right )\right ) \log (x)}{5 x \log ^3(x)} \, dx=\frac {{\left (x - 5\right )} e^{x} + 5 \, {\left (x - 5\right )} e^{\left (e^{\left (-2\right )}\right )}}{5 \, \log \left (x\right )^{2}} \] Input:
integrate(1/5*(((x^2-4*x)*exp(x)+5*x*exp(1/exp(2)))*log(x)+(-2*x+10)*exp(x )+(-10*x+50)*exp(1/exp(2)))/x/log(x)^3,x, algorithm="fricas")
Output:
1/5*((x - 5)*e^x + 5*(x - 5)*e^(e^(-2)))/log(x)^2
Time = 0.09 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {e^{\frac {1}{e^2}} (50-10 x)+e^x (10-2 x)+\left (5 e^{\frac {1}{e^2}} x+e^x \left (-4 x+x^2\right )\right ) \log (x)}{5 x \log ^3(x)} \, dx=\frac {\left (x - 5\right ) e^{x}}{5 \log {\left (x \right )}^{2}} + \frac {x e^{e^{-2}} - 5 e^{e^{-2}}}{\log {\left (x \right )}^{2}} \] Input:
integrate(1/5*(((x**2-4*x)*exp(x)+5*x*exp(1/exp(2)))*ln(x)+(-2*x+10)*exp(x )+(-10*x+50)*exp(1/exp(2)))/x/ln(x)**3,x)
Output:
(x - 5)*exp(x)/(5*log(x)**2) + (x*exp(exp(-2)) - 5*exp(exp(-2)))/log(x)**2
\[ \int \frac {e^{\frac {1}{e^2}} (50-10 x)+e^x (10-2 x)+\left (5 e^{\frac {1}{e^2}} x+e^x \left (-4 x+x^2\right )\right ) \log (x)}{5 x \log ^3(x)} \, dx=\int { -\frac {2 \, {\left (x - 5\right )} e^{x} + 10 \, {\left (x - 5\right )} e^{\left (e^{\left (-2\right )}\right )} - {\left ({\left (x^{2} - 4 \, x\right )} e^{x} + 5 \, x e^{\left (e^{\left (-2\right )}\right )}\right )} \log \left (x\right )}{5 \, x \log \left (x\right )^{3}} \,d x } \] Input:
integrate(1/5*(((x^2-4*x)*exp(x)+5*x*exp(1/exp(2)))*log(x)+(-2*x+10)*exp(x )+(-10*x+50)*exp(1/exp(2)))/x/log(x)^3,x, algorithm="maxima")
Output:
2*e^(e^(-2))*gamma(-2, -log(x)) + e^(e^(-2))*integrate(1/log(x), x) - 1/5* (5*x*e^(e^(-2))*log(x) - (x - 5)*e^x)/log(x)^2 - 5*e^(e^(-2))/log(x)^2
Time = 0.12 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{\frac {1}{e^2}} (50-10 x)+e^x (10-2 x)+\left (5 e^{\frac {1}{e^2}} x+e^x \left (-4 x+x^2\right )\right ) \log (x)}{5 x \log ^3(x)} \, dx=\frac {x e^{x} + 5 \, x e^{\left (e^{\left (-2\right )}\right )} - 5 \, e^{x} - 25 \, e^{\left (e^{\left (-2\right )}\right )}}{5 \, \log \left (x\right )^{2}} \] Input:
integrate(1/5*(((x^2-4*x)*exp(x)+5*x*exp(1/exp(2)))*log(x)+(-2*x+10)*exp(x )+(-10*x+50)*exp(1/exp(2)))/x/log(x)^3,x, algorithm="giac")
Output:
1/5*(x*e^x + 5*x*e^(e^(-2)) - 5*e^x - 25*e^(e^(-2)))/log(x)^2
Time = 3.10 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {e^{\frac {1}{e^2}} (50-10 x)+e^x (10-2 x)+\left (5 e^{\frac {1}{e^2}} x+e^x \left (-4 x+x^2\right )\right ) \log (x)}{5 x \log ^3(x)} \, dx=\frac {\left (5\,{\mathrm {e}}^{{\mathrm {e}}^{-2}}+{\mathrm {e}}^x\right )\,\left (x-5\right )}{5\,{\ln \left (x\right )}^2} \] Input:
int(-((exp(exp(-2))*(10*x - 50))/5 + (exp(x)*(2*x - 10))/5 - (log(x)*(5*x* exp(exp(-2)) - exp(x)*(4*x - x^2)))/5)/(x*log(x)^3),x)
Output:
((5*exp(exp(-2)) + exp(x))*(x - 5))/(5*log(x)^2)
Time = 0.16 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {e^{\frac {1}{e^2}} (50-10 x)+e^x (10-2 x)+\left (5 e^{\frac {1}{e^2}} x+e^x \left (-4 x+x^2\right )\right ) \log (x)}{5 x \log ^3(x)} \, dx=\frac {5 e^{\frac {1}{e^{2}}} x -25 e^{\frac {1}{e^{2}}}+e^{x} x -5 e^{x}}{5 \mathrm {log}\left (x \right )^{2}} \] Input:
int(1/5*(((x^2-4*x)*exp(x)+5*x*exp(1/exp(2)))*log(x)+(-2*x+10)*exp(x)+(-10 *x+50)*exp(1/exp(2)))/x/log(x)^3,x)
Output:
(5*e**(1/e**2)*x - 25*e**(1/e**2) + e**x*x - 5*e**x)/(5*log(x)**2)