\(\int \frac {e^{\frac {-15+3 x}{x+\log (x)}} (15 x^3+12 x^4+3 x^5+e^x (-15-12 x-x^3)+(-15 x^2-12 x^3-2 x^4) \log (15)+(9 x^4+e^x (-3 x-2 x^2)-7 x^3 \log (15)) \log (x)+(-e^x x+3 x^3-2 x^2 \log (15)) \log ^2(x))}{2 x^3+4 x^2 \log (x)+2 x \log ^2(x)} \, dx\) [1134]

Optimal result

Integrand size = 139, antiderivative size = 33 \[ \int \frac {e^{\frac {-15+3 x}{x+\log (x)}} \left (15 x^3+12 x^4+3 x^5+e^x \left (-15-12 x-x^3\right )+\left (-15 x^2-12 x^3-2 x^4\right ) \log (15)+\left (9 x^4+e^x \left (-3 x-2 x^2\right )-7 x^3 \log (15)\right ) \log (x)+\left (-e^x x+3 x^3-2 x^2 \log (15)\right ) \log ^2(x)\right )}{2 x^3+4 x^2 \log (x)+2 x \log ^2(x)} \, dx=\frac {1}{2} e^{\frac {3 (-5+x)}{x+\log (x)}} \left (-e^x+x^2 (x-\log (15))\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int \frac {e^{\frac {-15+3 x}{x+\log (x)}} \left (15 x^3+12 x^4+3 x^5+e^x \left (-15-12 x-x^3\right )+\left (-15 x^2-12 x^3-2 x^4\right ) \log (15)+\left (9 x^4+e^x \left (-3 x-2 x^2\right )-7 x^3 \log (15)\right ) \log (x)+\left (-e^x x+3 x^3-2 x^2 \log (15)\right ) \log ^2(x)\right )}{2 x^3+4 x^2 \log (x)+2 x \log ^2(x)} \, dx=\frac {1}{4} e^{\frac {3 (-5+x)}{x+\log (x)}} \left (-2 e^x+2 x^3-x^2 \log (225)\right ) \] Input:

Integrate[(E^((-15 + 3*x)/(x + Log[x]))*(15*x^3 + 12*x^4 + 3*x^5 + E^x*(-1 
5 - 12*x - x^3) + (-15*x^2 - 12*x^3 - 2*x^4)*Log[15] + (9*x^4 + E^x*(-3*x 
- 2*x^2) - 7*x^3*Log[15])*Log[x] + (-(E^x*x) + 3*x^3 - 2*x^2*Log[15])*Log[ 
x]^2))/(2*x^3 + 4*x^2*Log[x] + 2*x*Log[x]^2),x]
 

Output:

(E^((3*(-5 + x))/(x + Log[x]))*(-2*E^x + 2*x^3 - x^2*Log[225]))/4
 

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.

\(\displaystyle \int \frac {e^{\frac {3 x-15}{x+\log (x)}} \left (3 x^5+12 x^4+15 x^3+e^x \left (-x^3-12 x-15\right )+\left (3 x^3-2 x^2 \log (15)-e^x x\right ) \log ^2(x)+\left (9 x^4-7 x^3 \log (15)+e^x \left (-2 x^2-3 x\right )\right ) \log (x)+\left (-2 x^4-12 x^3-15 x^2\right ) \log (15)\right )}{2 x^3+4 x^2 \log (x)+2 x \log ^2(x)} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {e^{\frac {3 x-15}{x+\log (x)}} \left (3 x^5+12 x^4+15 x^3+e^x \left (-x^3-12 x-15\right )+\left (3 x^3-2 x^2 \log (15)-e^x x\right ) \log ^2(x)+\left (9 x^4-7 x^3 \log (15)+e^x \left (-2 x^2-3 x\right )\right ) \log (x)+\left (-2 x^4-12 x^3-15 x^2\right ) \log (15)\right )}{2 x (x+\log (x))^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \int \frac {e^{-\frac {3 (5-x)}{x+\log (x)}} \left (3 x^5+12 x^4+15 x^3-\left (-3 x^3+2 \log (15) x^2+e^x x\right ) \log ^2(x)-e^x \left (x^3+12 x+15\right )+\left (9 x^4-7 \log (15) x^3-e^x \left (2 x^2+3 x\right )\right ) \log (x)-\left (2 x^4+12 x^3+15 x^2\right ) \log (15)\right )}{x (x+\log (x))^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \frac {1}{2} \int \left (\frac {3 e^{-\frac {3 (5-x)}{x+\log (x)}} x^4}{(x+\log (x))^2}+\frac {12 e^{-\frac {3 (5-x)}{x+\log (x)}} x^3}{(x+\log (x))^2}+\frac {9 e^{-\frac {3 (5-x)}{x+\log (x)}} \log (x) x^3}{(x+\log (x))^2}+\frac {15 e^{-\frac {3 (5-x)}{x+\log (x)}} x^2}{(x+\log (x))^2}+\frac {3 e^{-\frac {3 (5-x)}{x+\log (x)}} \log ^2(x) x^2}{(x+\log (x))^2}-\frac {7 e^{-\frac {3 (5-x)}{x+\log (x)}} \log (15) \log (x) x^2}{(x+\log (x))^2}-\frac {2 e^{-\frac {3 (5-x)}{x+\log (x)}} \log (15) \log ^2(x) x}{(x+\log (x))^2}-\frac {e^{-\frac {3 (5-x)}{x+\log (x)}} \left (2 x^2+12 x+15\right ) \log (15) x}{(x+\log (x))^2}-\frac {e^{x-\frac {3 (5-x)}{x+\log (x)}} \left (x^3+2 \log (x) x^2+\log ^2(x) x+3 \log (x) x+12 x+15\right )}{(x+\log (x))^2 x}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (-3 \int \frac {e^{-\frac {3 (5-x)}{x+\log (x)}} x^4}{(x+\log (x))^2}dx+3 \log (15) \int \frac {e^{-\frac {3 (5-x)}{x+\log (x)}} x^3}{(x+\log (x))^2}dx+12 \int \frac {e^{-\frac {3 (5-x)}{x+\log (x)}} x^3}{(x+\log (x))^2}dx+3 \int \frac {e^{-\frac {3 (5-x)}{x+\log (x)}} x^3}{x+\log (x)}dx+3 \int e^{-\frac {3 (5-x)}{x+\log (x)}} x^2dx-12 \log (15) \int \frac {e^{-\frac {3 (5-x)}{x+\log (x)}} x^2}{(x+\log (x))^2}dx+15 \int \frac {e^{-\frac {3 (5-x)}{x+\log (x)}} x^2}{(x+\log (x))^2}dx-3 \log (15) \int \frac {e^{-\frac {3 (5-x)}{x+\log (x)}} x^2}{x+\log (x)}dx-\int e^{x-\frac {3 (5-x)}{x+\log (x)}}dx-2 \log (15) \int e^{-\frac {3 (5-x)}{x+\log (x)}} xdx-12 \int \frac {e^{x-\frac {3 (5-x)}{x+\log (x)}}}{(x+\log (x))^2}dx-15 \int \frac {e^{x-\frac {3 (5-x)}{x+\log (x)}}}{x (x+\log (x))^2}dx-15 \log (15) \int \frac {e^{-\frac {3 (5-x)}{x+\log (x)}} x}{(x+\log (x))^2}dx+3 \int \frac {e^{x-\frac {3 (5-x)}{x+\log (x)}} x}{(x+\log (x))^2}dx-3 \int \frac {e^{x-\frac {3 (5-x)}{x+\log (x)}}}{x+\log (x)}dx\right )\)

Input:

Int[(E^((-15 + 3*x)/(x + Log[x]))*(15*x^3 + 12*x^4 + 3*x^5 + E^x*(-15 - 12 
*x - x^3) + (-15*x^2 - 12*x^3 - 2*x^4)*Log[15] + (9*x^4 + E^x*(-3*x - 2*x^ 
2) - 7*x^3*Log[15])*Log[x] + (-(E^x*x) + 3*x^3 - 2*x^2*Log[15])*Log[x]^2)) 
/(2*x^3 + 4*x^2*Log[x] + 2*x*Log[x]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.15

\[\left (\frac {x^{3}}{2}-\frac {x^{2} \ln \left (3\right )}{2}-\frac {x^{2} \ln \left (5\right )}{2}-\frac {{\mathrm e}^{x}}{2}\right ) {\mathrm e}^{\frac {3 x -15}{x +\ln \left (x \right )}}\]

Input:

int(((-exp(x)*x-2*x^2*ln(15)+3*x^3)*ln(x)^2+((-2*x^2-3*x)*exp(x)-7*x^3*ln( 
15)+9*x^4)*ln(x)+(-x^3-12*x-15)*exp(x)+(-2*x^4-12*x^3-15*x^2)*ln(15)+3*x^5 
+12*x^4+15*x^3)*exp((3*x-15)/(x+ln(x)))/(2*x*ln(x)^2+4*x^2*ln(x)+2*x^3),x)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {-15+3 x}{x+\log (x)}} \left (15 x^3+12 x^4+3 x^5+e^x \left (-15-12 x-x^3\right )+\left (-15 x^2-12 x^3-2 x^4\right ) \log (15)+\left (9 x^4+e^x \left (-3 x-2 x^2\right )-7 x^3 \log (15)\right ) \log (x)+\left (-e^x x+3 x^3-2 x^2 \log (15)\right ) \log ^2(x)\right )}{2 x^3+4 x^2 \log (x)+2 x \log ^2(x)} \, dx=\frac {1}{2} \, {\left (x^{3} - x^{2} \log \left (15\right ) - e^{x}\right )} e^{\left (\frac {3 \, {\left (x - 5\right )}}{x + \log \left (x\right )}\right )} \] Input:

integrate(((-exp(x)*x-2*x^2*log(15)+3*x^3)*log(x)^2+((-2*x^2-3*x)*exp(x)-7 
*x^3*log(15)+9*x^4)*log(x)+(-x^3-12*x-15)*exp(x)+(-2*x^4-12*x^3-15*x^2)*lo 
g(15)+3*x^5+12*x^4+15*x^3)*exp((3*x-15)/(x+log(x)))/(2*x*log(x)^2+4*x^2*lo 
g(x)+2*x^3),x, algorithm="fricas")
 

Output:

1/2*(x^3 - x^2*log(15) - e^x)*e^(3*(x - 5)/(x + log(x)))
 

Sympy [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {-15+3 x}{x+\log (x)}} \left (15 x^3+12 x^4+3 x^5+e^x \left (-15-12 x-x^3\right )+\left (-15 x^2-12 x^3-2 x^4\right ) \log (15)+\left (9 x^4+e^x \left (-3 x-2 x^2\right )-7 x^3 \log (15)\right ) \log (x)+\left (-e^x x+3 x^3-2 x^2 \log (15)\right ) \log ^2(x)\right )}{2 x^3+4 x^2 \log (x)+2 x \log ^2(x)} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(((-exp(x)*x-2*x**2*ln(15)+3*x**3)*ln(x)**2+((-2*x**2-3*x)*exp(x) 
-7*x**3*ln(15)+9*x**4)*ln(x)+(-x**3-12*x-15)*exp(x)+(-2*x**4-12*x**3-15*x* 
*2)*ln(15)+3*x**5+12*x**4+15*x**3)*exp((3*x-15)/(x+ln(x)))/(2*x*ln(x)**2+4 
*x**2*ln(x)+2*x**3),x)
 

Output:

Exception raised: TypeError >> '>' not supported between instances of 'Pol 
y' and 'int'
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {-15+3 x}{x+\log (x)}} \left (15 x^3+12 x^4+3 x^5+e^x \left (-15-12 x-x^3\right )+\left (-15 x^2-12 x^3-2 x^4\right ) \log (15)+\left (9 x^4+e^x \left (-3 x-2 x^2\right )-7 x^3 \log (15)\right ) \log (x)+\left (-e^x x+3 x^3-2 x^2 \log (15)\right ) \log ^2(x)\right )}{2 x^3+4 x^2 \log (x)+2 x \log ^2(x)} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(((-exp(x)*x-2*x^2*log(15)+3*x^3)*log(x)^2+((-2*x^2-3*x)*exp(x)-7 
*x^3*log(15)+9*x^4)*log(x)+(-x^3-12*x-15)*exp(x)+(-2*x^4-12*x^3-15*x^2)*lo 
g(15)+3*x^5+12*x^4+15*x^3)*exp((3*x-15)/(x+log(x)))/(2*x*log(x)^2+4*x^2*lo 
g(x)+2*x^3),x, algorithm="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: In function CAR, the value of 
the first argument is  0which is not of the expected type LIST
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (29) = 58\).

Time = 0.18 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.79 \[ \int \frac {e^{\frac {-15+3 x}{x+\log (x)}} \left (15 x^3+12 x^4+3 x^5+e^x \left (-15-12 x-x^3\right )+\left (-15 x^2-12 x^3-2 x^4\right ) \log (15)+\left (9 x^4+e^x \left (-3 x-2 x^2\right )-7 x^3 \log (15)\right ) \log (x)+\left (-e^x x+3 x^3-2 x^2 \log (15)\right ) \log ^2(x)\right )}{2 x^3+4 x^2 \log (x)+2 x \log ^2(x)} \, dx=\frac {1}{2} \, x^{3} e^{\left (\frac {3 \, {\left (x - 5\right )}}{x + \log \left (x\right )}\right )} - \frac {1}{2} \, x^{2} e^{\left (\frac {3 \, {\left (x - 5\right )}}{x + \log \left (x\right )}\right )} \log \left (15\right ) - \frac {1}{2} \, e^{\left (\frac {x^{2} + x \log \left (x\right ) + 3 \, x - 15}{x + \log \left (x\right )}\right )} \] Input:

integrate(((-exp(x)*x-2*x^2*log(15)+3*x^3)*log(x)^2+((-2*x^2-3*x)*exp(x)-7 
*x^3*log(15)+9*x^4)*log(x)+(-x^3-12*x-15)*exp(x)+(-2*x^4-12*x^3-15*x^2)*lo 
g(15)+3*x^5+12*x^4+15*x^3)*exp((3*x-15)/(x+log(x)))/(2*x*log(x)^2+4*x^2*lo 
g(x)+2*x^3),x, algorithm="giac")
 

Output:

1/2*x^3*e^(3*(x - 5)/(x + log(x))) - 1/2*x^2*e^(3*(x - 5)/(x + log(x)))*lo 
g(15) - 1/2*e^((x^2 + x*log(x) + 3*x - 15)/(x + log(x)))
 

Mupad [B] (verification not implemented)

Time = 3.89 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.70 \[ \int \frac {e^{\frac {-15+3 x}{x+\log (x)}} \left (15 x^3+12 x^4+3 x^5+e^x \left (-15-12 x-x^3\right )+\left (-15 x^2-12 x^3-2 x^4\right ) \log (15)+\left (9 x^4+e^x \left (-3 x-2 x^2\right )-7 x^3 \log (15)\right ) \log (x)+\left (-e^x x+3 x^3-2 x^2 \log (15)\right ) \log ^2(x)\right )}{2 x^3+4 x^2 \log (x)+2 x \log ^2(x)} \, dx=\frac {x^3\,{\mathrm {e}}^{\frac {3\,x-15}{x+\ln \left (x\right )}}}{2}-\frac {{\mathrm {e}}^{\frac {3\,x-15}{x+\ln \left (x\right )}}\,{\mathrm {e}}^x}{2}-\frac {x^2\,{\mathrm {e}}^{\frac {3\,x-15}{x+\ln \left (x\right )}}\,\ln \left (15\right )}{2} \] Input:

int(-(exp((3*x - 15)/(x + log(x)))*(log(x)*(7*x^3*log(15) + exp(x)*(3*x + 
2*x^2) - 9*x^4) + exp(x)*(12*x + x^3 + 15) + log(x)^2*(2*x^2*log(15) + x*e 
xp(x) - 3*x^3) + log(15)*(15*x^2 + 12*x^3 + 2*x^4) - 15*x^3 - 12*x^4 - 3*x 
^5))/(2*x*log(x)^2 + 4*x^2*log(x) + 2*x^3),x)
                                                                                    
                                                                                    
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.24 \[ \int \frac {e^{\frac {-15+3 x}{x+\log (x)}} \left (15 x^3+12 x^4+3 x^5+e^x \left (-15-12 x-x^3\right )+\left (-15 x^2-12 x^3-2 x^4\right ) \log (15)+\left (9 x^4+e^x \left (-3 x-2 x^2\right )-7 x^3 \log (15)\right ) \log (x)+\left (-e^x x+3 x^3-2 x^2 \log (15)\right ) \log ^2(x)\right )}{2 x^3+4 x^2 \log (x)+2 x \log ^2(x)} \, dx=\frac {e^{\frac {3 x}{\mathrm {log}\left (x \right )+x}} \left (-e^{x}-\mathrm {log}\left (15\right ) x^{2}+x^{3}\right )}{2 e^{\frac {15}{\mathrm {log}\left (x \right )+x}}} \] Input:

int(((-exp(x)*x-2*x^2*log(15)+3*x^3)*log(x)^2+((-2*x^2-3*x)*exp(x)-7*x^3*l 
og(15)+9*x^4)*log(x)+(-x^3-12*x-15)*exp(x)+(-2*x^4-12*x^3-15*x^2)*log(15)+ 
3*x^5+12*x^4+15*x^3)*exp((3*x-15)/(x+log(x)))/(2*x*log(x)^2+4*x^2*log(x)+2 
*x^3),x)
 

Output:

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