Integrand size = 66, antiderivative size = 22 \[ \int \frac {-1+x^{e^4 x+4 e^{5-x} x} \left (e^4 x+4 e^{5-x} x+\left (e^4 x+4 e^{5-x} \left (x-x^2\right )\right ) \log (x)\right )}{x} \, dx=x^{\left (e^4+4 e^{5-x}\right ) x}-\log (x) \] Output:
exp(ln(x)*x*(exp(2*ln(2)+5-x)+exp(4)))-ln(x)
Time = 0.75 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {-1+x^{e^4 x+4 e^{5-x} x} \left (e^4 x+4 e^{5-x} x+\left (e^4 x+4 e^{5-x} \left (x-x^2\right )\right ) \log (x)\right )}{x} \, dx=x^{e^4 \left (1+4 e^{1-x}\right ) x}-\log (x) \] Input:
Integrate[(-1 + x^(E^4*x + 4*E^(5 - x)*x)*(E^4*x + 4*E^(5 - x)*x + (E^4*x + 4*E^(5 - x)*(x - x^2))*Log[x]))/x,x]
Output:
x^(E^4*(1 + 4*E^(1 - x))*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.
\(\displaystyle \int \frac {x^{4 e^{5-x} x+e^4 x} \left (\left (4 e^{5-x} \left (x-x^2\right )+e^4 x\right ) \log (x)+4 e^{5-x} x+e^4 x\right )-1}{x} \, dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (e^{4-x} x^{e^{4-x} \left (e^x+4 e\right ) x} \left (e^x+e^x \log (x)-4 e x \log (x)+4 e \log (x)+4 e\right )-\frac {1}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle e^4 \int x^{e^{4-x} \left (4 e+e^x\right ) x}dx+4 \int e^{5-x} x^{e^{4-x} \left (4 e+e^x\right ) x}dx-e^4 \int \frac {\int x^{e^{4-x} \left (4 e+e^x\right ) x}dx}{x}dx-4 \int \frac {\int e^{5-x} x^{e^{4-x} \left (4 e+e^x\right ) x}dx}{x}dx+4 \int \frac {\int e^{5-x} x^{4 e^{5-x} x+e^4 x+1}dx}{x}dx+e^4 \log (x) \int x^{e^{4-x} \left (4 e+e^x\right ) x}dx+4 \log (x) \int e^{5-x} x^{e^{4-x} \left (4 e+e^x\right ) x}dx-4 \log (x) \int e^{5-x} x^{e^{4-x} \left (4 e+e^x\right ) x+1}dx-\log (x)\) |
Input:
Int[(-1 + x^(E^4*x + 4*E^(5 - x)*x)*(E^4*x + 4*E^(5 - x)*x + (E^4*x + 4*E^ (5 - x)*(x - x^2))*Log[x]))/x,x]
Output:
$Aborted
Time = 3.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95
method | result | size |
risch | \(x^{\left ({\mathrm e}^{4}+4 \,{\mathrm e}^{5-x}\right ) x}-\ln \left (x \right )\) | \(21\) |
parallelrisch | \({\mathrm e}^{\ln \left (x \right ) x \left ({\mathrm e}^{2 \ln \left (2\right )+5-x}+{\mathrm e}^{4}\right )}-\ln \left (x \right )\) | \(24\) |
default | \({\mathrm e}^{\left (x \,{\mathrm e}^{2 \ln \left (2\right )+5-x}+x \,{\mathrm e}^{4}\right ) \ln \left (x \right )}-\ln \left (x \right )\) | \(27\) |
parts | \({\mathrm e}^{\left (x \,{\mathrm e}^{2 \ln \left (2\right )+5-x}+x \,{\mathrm e}^{4}\right ) \ln \left (x \right )}-\ln \left (x \right )\) | \(27\) |
Input:
int(((((-x^2+x)*exp(2*ln(2)+5-x)+x*exp(4))*ln(x)+x*exp(2*ln(2)+5-x)+x*exp( 4))*exp((x*exp(2*ln(2)+5-x)+x*exp(4))*ln(x))-1)/x,x,method=_RETURNVERBOSE)
Output:
x^((exp(4)+4*exp(5-x))*x)-ln(x)
Time = 0.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {-1+x^{e^4 x+4 e^{5-x} x} \left (e^4 x+4 e^{5-x} x+\left (e^4 x+4 e^{5-x} \left (x-x^2\right )\right ) \log (x)\right )}{x} \, dx=x^{x e^{4} + x e^{\left (-x + 2 \, \log \left (2\right ) + 5\right )}} - \log \left (x\right ) \] Input:
integrate(((((-x^2+x)*exp(2*log(2)+5-x)+x*exp(4))*log(x)+x*exp(2*log(2)+5- x)+x*exp(4))*exp((x*exp(2*log(2)+5-x)+x*exp(4))*log(x))-1)/x,x, algorithm= "fricas")
Output:
x^(x*e^4 + x*e^(-x + 2*log(2) + 5)) - log(x)
Time = 0.23 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {-1+x^{e^4 x+4 e^{5-x} x} \left (e^4 x+4 e^{5-x} x+\left (e^4 x+4 e^{5-x} \left (x-x^2\right )\right ) \log (x)\right )}{x} \, dx=e^{\left (4 x e^{5 - x} + x e^{4}\right ) \log {\left (x \right )}} - \log {\left (x \right )} \] Input:
integrate(((((-x**2+x)*exp(2*ln(2)+5-x)+x*exp(4))*ln(x)+x*exp(2*ln(2)+5-x) +x*exp(4))*exp((x*exp(2*ln(2)+5-x)+x*exp(4))*ln(x))-1)/x,x)
Output:
exp((4*x*exp(5 - x) + x*exp(4))*log(x)) - log(x)
Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {-1+x^{e^4 x+4 e^{5-x} x} \left (e^4 x+4 e^{5-x} x+\left (e^4 x+4 e^{5-x} \left (x-x^2\right )\right ) \log (x)\right )}{x} \, dx=e^{\left (x e^{4} \log \left (x\right ) + 4 \, x e^{\left (-x + 5\right )} \log \left (x\right )\right )} - \log \left (x\right ) \] Input:
integrate(((((-x^2+x)*exp(2*log(2)+5-x)+x*exp(4))*log(x)+x*exp(2*log(2)+5- x)+x*exp(4))*exp((x*exp(2*log(2)+5-x)+x*exp(4))*log(x))-1)/x,x, algorithm= "maxima")
Output:
e^(x*e^4*log(x) + 4*x*e^(-x + 5)*log(x)) - log(x)
\[ \int \frac {-1+x^{e^4 x+4 e^{5-x} x} \left (e^4 x+4 e^{5-x} x+\left (e^4 x+4 e^{5-x} \left (x-x^2\right )\right ) \log (x)\right )}{x} \, dx=\int { \frac {{\left (x e^{4} + x e^{\left (-x + 2 \, \log \left (2\right ) + 5\right )} + {\left (x e^{4} - {\left (x^{2} - x\right )} e^{\left (-x + 2 \, \log \left (2\right ) + 5\right )}\right )} \log \left (x\right )\right )} x^{x e^{4} + x e^{\left (-x + 2 \, \log \left (2\right ) + 5\right )}} - 1}{x} \,d x } \] Input:
integrate(((((-x^2+x)*exp(2*log(2)+5-x)+x*exp(4))*log(x)+x*exp(2*log(2)+5- x)+x*exp(4))*exp((x*exp(2*log(2)+5-x)+x*exp(4))*log(x))-1)/x,x, algorithm= "giac")
Output:
integrate(((x*e^4 + x*e^(-x + 2*log(2) + 5) + (x*e^4 - (x^2 - x)*e^(-x + 2 *log(2) + 5))*log(x))*x^(x*e^4 + x*e^(-x + 2*log(2) + 5)) - 1)/x, x)
Time = 2.90 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {-1+x^{e^4 x+4 e^{5-x} x} \left (e^4 x+4 e^{5-x} x+\left (e^4 x+4 e^{5-x} \left (x-x^2\right )\right ) \log (x)\right )}{x} \, dx=x^{x\,{\mathrm {e}}^4}\,x^{4\,x\,{\mathrm {e}}^{5-x}}-\ln \left (x\right ) \] Input:
int((exp(log(x)*(x*exp(4) + x*exp(2*log(2) - x + 5)))*(x*exp(4) + x*exp(2* log(2) - x + 5) + log(x)*(x*exp(4) + exp(2*log(2) - x + 5)*(x - x^2))) - 1 )/x,x)
Output:
x^(x*exp(4))*x^(4*x*exp(5 - x)) - log(x)
Time = 0.21 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {-1+x^{e^4 x+4 e^{5-x} x} \left (e^4 x+4 e^{5-x} x+\left (e^4 x+4 e^{5-x} \left (x-x^2\right )\right ) \log (x)\right )}{x} \, dx=x^{e^{4} x} e^{\frac {4 \,\mathrm {log}\left (x \right ) e^{5} x}{e^{x}}}-\mathrm {log}\left (x \right ) \] Input:
int(((((-x^2+x)*exp(2*log(2)+5-x)+x*exp(4))*log(x)+x*exp(2*log(2)+5-x)+x*e xp(4))*exp((x*exp(2*log(2)+5-x)+x*exp(4))*log(x))-1)/x,x)
Output:
x**(e**4*x)*e**((4*log(x)*e**5*x)/e**x) - log(x)