\(\int \frac {e^{-x+\frac {e^{-x} x}{15}} (15 e^x+e^{3+3 x} (-x-45 e^x x+x^2)+(x-x^2) \log (x))}{15 x \log (2)} \, dx\) [1579]

Optimal result

Integrand size = 64, antiderivative size = 31 \[ \int \frac {e^{-x+\frac {e^{-x} x}{15}} \left (15 e^x+e^{3+3 x} \left (-x-45 e^x x+x^2\right )+\left (x-x^2\right ) \log (x)\right )}{15 x \log (2)} \, dx=5+\frac {e^{\frac {e^{-x} x}{15}} \left (-e^{3+3 x}+\log (x)\right )}{\log (2)} \] Output:

(ln(x)-exp(3*x+3))*exp(1/15*x/exp(x))/ln(2)+5
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.29 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {e^{-x+\frac {e^{-x} x}{15}} \left (15 e^x+e^{3+3 x} \left (-x-45 e^x x+x^2\right )+\left (x-x^2\right ) \log (x)\right )}{15 x \log (2)} \, dx=\frac {e^{\frac {e^{-x} x}{15}} \left (-15 e^{3+3 x}+15 \log (x)\right )}{15 \log (2)} \] Input:

Integrate[(E^(-x + x/(15*E^x))*(15*E^x + E^(3 + 3*x)*(-x - 45*E^x*x + x^2) 
 + (x - x^2)*Log[x]))/(15*x*Log[2]),x]
 

Output:

(E^(x/(15*E^x))*(-15*E^(3 + 3*x) + 15*Log[x]))/(15*Log[2])
 

Rubi [F]

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.

\(\displaystyle \int \frac {e^{\frac {e^{-x} x}{15}-x} \left (e^{3 x+3} \left (x^2-45 e^x x-x\right )+\left (x-x^2\right ) \log (x)+15 e^x\right )}{15 x \log (2)} \, dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {e^{\frac {e^{-x} x}{15}-x} \left (-e^{3 x+3} \left (-x^2+45 e^x x+x\right )+15 e^x+\left (x-x^2\right ) \log (x)\right )}{x}dx}{15 \log (2)}\)

\(\Big \downarrow \) 7293

\(\displaystyle \frac {\int \left (e^{\frac {e^{-x} x}{15}+2 x+3} (x-1)-e^{\frac {e^{-x} x}{15}-x} \log (x) (x-1)-45 e^{\frac {e^{-x} x}{15}+3 x+3}+\frac {15 e^{\frac {e^{-x} x}{15}}}{x}\right )dx}{15 \log (2)}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\int e^{\frac {e^{-x} x}{15}+2 x+3}dx-45 \int e^{\frac {e^{-x} x}{15}+3 x+3}dx+15 \int \frac {e^{\frac {e^{-x} x}{15}}}{x}dx+\int e^{\frac {e^{-x} x}{15}+2 x+3} xdx-\int \frac {\int e^{\frac {1}{15} \left (-15+e^{-x}\right ) x}dx}{x}dx+\int \frac {\int e^{\frac {1}{15} \left (-15+e^{-x}\right ) x} xdx}{x}dx+\log (x) \int e^{\frac {e^{-x} x}{15}-x}dx-\log (x) \int e^{\frac {e^{-x} x}{15}-x} xdx}{15 \log (2)}\)

Input:

Int[(E^(-x + x/(15*E^x))*(15*E^x + E^(3 + 3*x)*(-x - 45*E^x*x + x^2) + (x 
- x^2)*Log[x]))/(15*x*Log[2]),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90

method	result	size
risch	\(\frac {\left (-15 \,{\mathrm e}^{3 x +3}+15 \ln \left (x \right )\right ) {\mathrm e}^{\frac {{\mathrm e}^{-x} x}{15}}}{15 \ln \left (2\right )}\)	\(28\)
parallelrisch	\(\frac {15 \,{\mathrm e}^{\frac {{\mathrm e}^{-x} x}{15}} \ln \left (x \right )-15 \,{\mathrm e}^{\frac {{\mathrm e}^{-x} x}{15}} {\mathrm e}^{3 x +3}}{15 \ln \left (2\right )}\)	\(36\)

Input:

int(1/15*((-x^2+x)*ln(x)+(-45*exp(x)*x+x^2-x)*exp(3*x+3)+15*exp(x))*exp(1/ 
15*x/exp(x))/x/ln(2)/exp(x),x,method=_RETURNVERBOSE)
 

Output:

1/15/ln(2)*(-15*exp(3*x+3)+15*ln(x))*exp(1/15*exp(-x)*x)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.13 \[ \int \frac {e^{-x+\frac {e^{-x} x}{15}} \left (15 e^x+e^{3+3 x} \left (-x-45 e^x x+x^2\right )+\left (x-x^2\right ) \log (x)\right )}{15 x \log (2)} \, dx=\frac {{\left (e^{x} \log \left (x\right ) - e^{\left (4 \, x + 3\right )}\right )} e^{\left (-\frac {1}{15} \, {\left (15 \, x e^{x} - x\right )} e^{\left (-x\right )}\right )}}{\log \left (2\right )} \] Input:

integrate(1/15*((-x^2+x)*log(x)+(-45*exp(x)*x+x^2-x)*exp(3*x+3)+15*exp(x)) 
*exp(1/15*x/exp(x))/x/log(2)/exp(x),x, algorithm="fricas")
 

Output:

(e^x*log(x) - e^(4*x + 3))*e^(-1/15*(15*x*e^x - x)*e^(-x))/log(2)
 

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {e^{-x+\frac {e^{-x} x}{15}} \left (15 e^x+e^{3+3 x} \left (-x-45 e^x x+x^2\right )+\left (x-x^2\right ) \log (x)\right )}{15 x \log (2)} \, dx=\frac {\left (- e^{3} + e^{- 3 x} \log {\left (x \right )}\right ) e^{3 x} e^{\frac {x e^{- x}}{15}}}{\log {\left (2 \right )}} \] Input:

integrate(1/15*((-x**2+x)*ln(x)+(-45*exp(x)*x+x**2-x)*exp(3*x+3)+15*exp(x) 
)*exp(1/15*x/exp(x))/x/ln(2)/exp(x),x)
 

Output:

(-exp(3) + exp(-3*x)*log(x))*exp(3*x)*exp(x*exp(-x)/15)/log(2)
 

Maxima [F]

\[ \int \frac {e^{-x+\frac {e^{-x} x}{15}} \left (15 e^x+e^{3+3 x} \left (-x-45 e^x x+x^2\right )+\left (x-x^2\right ) \log (x)\right )}{15 x \log (2)} \, dx=\int { \frac {{\left ({\left (x^{2} - 45 \, x e^{x} - x\right )} e^{\left (3 \, x + 3\right )} - {\left (x^{2} - x\right )} \log \left (x\right ) + 15 \, e^{x}\right )} e^{\left (\frac {1}{15} \, x e^{\left (-x\right )} - x\right )}}{15 \, x \log \left (2\right )} \,d x } \] Input:

integrate(1/15*((-x^2+x)*log(x)+(-45*exp(x)*x+x^2-x)*exp(3*x+3)+15*exp(x)) 
*exp(1/15*x/exp(x))/x/log(2)/exp(x),x, algorithm="maxima")
 

Output:

-1/15*(15*e^(1/15*x*e^(-x) + 3*x + 3) - integrate(-((x^2 - x)*log(x) - 15* 
e^x)*e^(1/15*x*e^(-x) - x)/x, x))/log(2)
 

Giac [F]

\[ \int \frac {e^{-x+\frac {e^{-x} x}{15}} \left (15 e^x+e^{3+3 x} \left (-x-45 e^x x+x^2\right )+\left (x-x^2\right ) \log (x)\right )}{15 x \log (2)} \, dx=\int { \frac {{\left ({\left (x^{2} - 45 \, x e^{x} - x\right )} e^{\left (3 \, x + 3\right )} - {\left (x^{2} - x\right )} \log \left (x\right ) + 15 \, e^{x}\right )} e^{\left (\frac {1}{15} \, x e^{\left (-x\right )} - x\right )}}{15 \, x \log \left (2\right )} \,d x } \] Input:

integrate(1/15*((-x^2+x)*log(x)+(-45*exp(x)*x+x^2-x)*exp(3*x+3)+15*exp(x)) 
*exp(1/15*x/exp(x))/x/log(2)/exp(x),x, algorithm="giac")
 

Output:

integrate(1/15*((x^2 - 45*x*e^x - x)*e^(3*x + 3) - (x^2 - x)*log(x) + 15*e 
^x)*e^(1/15*x*e^(-x) - x)/(x*log(2)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{-x+\frac {e^{-x} x}{15}} \left (15 e^x+e^{3+3 x} \left (-x-45 e^x x+x^2\right )+\left (x-x^2\right ) \log (x)\right )}{15 x \log (2)} \, dx=\int \frac {{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{-x}}{15}-x}\,\left (15\,{\mathrm {e}}^x+\ln \left (x\right )\,\left (x-x^2\right )-{\mathrm {e}}^{3\,x+3}\,\left (x+45\,x\,{\mathrm {e}}^x-x^2\right )\right )}{15\,x\,\ln \left (2\right )} \,d x \] Input:

int((exp(-x)*exp((x*exp(-x))/15)*(15*exp(x) + log(x)*(x - x^2) - exp(3*x + 
 3)*(x + 45*x*exp(x) - x^2)))/(15*x*log(2)),x)
 

Output:

int((exp((x*exp(-x))/15 - x)*(15*exp(x) + log(x)*(x - x^2) - exp(3*x + 3)* 
(x + 45*x*exp(x) - x^2)))/(15*x*log(2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-x+\frac {e^{-x} x}{15}} \left (15 e^x+e^{3+3 x} \left (-x-45 e^x x+x^2\right )+\left (x-x^2\right ) \log (x)\right )}{15 x \log (2)} \, dx=\frac {e^{\frac {x}{15 e^{x}}} \left (-e^{3 x} e^{3}+\mathrm {log}\left (x \right )\right )}{\mathrm {log}\left (2\right )} \] Input:

int(1/15*((-x^2+x)*log(x)+(-45*exp(x)*x+x^2-x)*exp(3*x+3)+15*exp(x))*exp(1 
/15*x/exp(x))/x/log(2)/exp(x),x)
 

Output:

(e**(x/(15*e**x))*( - e**(3*x)*e**3 + log(x)))/log(2)