\(\int \frac {e^{\frac {3 x^3-e^{x^2} x^3+x^4-x^3 \log (x)}{-3+3 x}} (-8 x^2+x^3+3 x^4+e^{x^2} (3 x^2-2 x^3+2 x^4-2 x^5)+(3 x^2-2 x^3) \log (x))}{15-30 x+15 x^2} \, dx\) [1411]

Optimal result

Integrand size = 104, antiderivative size = 33 \[ \int \frac {e^{\frac {3 x^3-e^{x^2} x^3+x^4-x^3 \log (x)}{-3+3 x}} \left (-8 x^2+x^3+3 x^4+e^{x^2} \left (3 x^2-2 x^3+2 x^4-2 x^5\right )+\left (3 x^2-2 x^3\right ) \log (x)\right )}{15-30 x+15 x^2} \, dx=\frac {1}{5} e^{\frac {x^2 \left (3-e^{x^2}+x-\log (x)\right )}{3-\frac {3}{x}}} \] Output:

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.24 \[ \int \frac {e^{\frac {3 x^3-e^{x^2} x^3+x^4-x^3 \log (x)}{-3+3 x}} \left (-8 x^2+x^3+3 x^4+e^{x^2} \left (3 x^2-2 x^3+2 x^4-2 x^5\right )+\left (3 x^2-2 x^3\right ) \log (x)\right )}{15-30 x+15 x^2} \, dx=\frac {1}{5} e^{\frac {x^3 \left (3-e^{x^2}+x\right )}{3 (-1+x)}} x^{\frac {x^3}{3-3 x}} \] Input:

Integrate[(E^((3*x^3 - E^x^2*x^3 + x^4 - x^3*Log[x])/(-3 + 3*x))*(-8*x^2 + 
 x^3 + 3*x^4 + E^x^2*(3*x^2 - 2*x^3 + 2*x^4 - 2*x^5) + (3*x^2 - 2*x^3)*Log 
[x]))/(15 - 30*x + 15*x^2),x]
 

Output:

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

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.

\(\displaystyle \int \frac {e^{\frac {x^4+3 x^3-x^3 \log (x)-e^{x^2} x^3}{3 x-3}} \left (3 x^4+x^3-8 x^2+\left (3 x^2-2 x^3\right ) \log (x)+e^{x^2} \left (-2 x^5+2 x^4-2 x^3+3 x^2\right )\right )}{15 x^2-30 x+15} \, dx\)

\(\Big \downarrow \) 7277

\(\displaystyle 60 \int -\frac {e^{-\frac {x^4-e^{x^2} x^3+3 x^3}{3 (1-x)}} x^{\frac {x^3}{3-3 x}} \left (-3 x^4-x^3+8 x^2-e^{x^2} \left (-2 x^5+2 x^4-2 x^3+3 x^2\right )-\left (3 x^2-2 x^3\right ) \log (x)\right )}{900 (1-x)^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {1}{15} \int \frac {e^{-\frac {x^4-e^{x^2} x^3+3 x^3}{3 (1-x)}} x^{\frac {x^3}{3 (1-x)}} \left (-3 x^4-x^3+8 x^2-e^{x^2} \left (-2 x^5+2 x^4-2 x^3+3 x^2\right )-\left (3 x^2-2 x^3\right ) \log (x)\right )}{(1-x)^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle -\frac {1}{15} \int \left (\frac {e^{x^2-\frac {x^4-e^{x^2} x^3+3 x^3}{3 (1-x)}} \left (2 x^3-2 x^2+2 x-3\right ) x^{\frac {x^3}{3 (1-x)}+2}}{(x-1)^2}+\frac {e^{-\frac {x^4-e^{x^2} x^3+3 x^3}{3 (1-x)}} (2 x-3) \log (x) x^{\frac {x^3}{3 (1-x)}+2}}{(x-1)^2}+\frac {8 e^{-\frac {x^4-e^{x^2} x^3+3 x^3}{3 (1-x)}} x^{\frac {x^3}{3 (1-x)}+2}}{(x-1)^2}-\frac {e^{-\frac {x^4-e^{x^2} x^3+3 x^3}{3 (1-x)}} x^{\frac {x^3}{3 (1-x)}+3}}{(x-1)^2}-\frac {3 e^{-\frac {x^4-e^{x^2} x^3+3 x^3}{3 (1-x)}} x^{\frac {x^3}{3 (1-x)}+4}}{(x-1)^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{15} \left (-2 \int e^{\frac {x^2 \left (x^2-e^{x^2} x+6 x-3\right )}{3 (x-1)}} x^{\frac {x^3}{3 (1-x)}+2}dx+\int \frac {e^{\frac {x^2 \left (x^2-e^{x^2} x+6 x-3\right )}{3 (x-1)}} x^{\frac {x^3}{3 (1-x)}+2}}{(x-1)^2}dx-4 \int \frac {e^{\frac {x^2 \left (x^2-e^{x^2} x+6 x-3\right )}{3 (x-1)}} x^{\frac {x^3}{3 (1-x)}+2}}{x-1}dx-2 \int e^{\frac {x^2 \left (x^2-e^{x^2} x+6 x-3\right )}{3 (x-1)}} x^{\frac {x^3}{3 (1-x)}+3}dx-\int \frac {\int \frac {e^{\frac {x^3 \left (x-e^{x^2}+3\right )}{3 (x-1)}} x^{\frac {x^3}{3-3 x}+2}}{(x-1)^2}dx}{x}dx+2 \int \frac {\int \frac {e^{\frac {x^3 \left (x-e^{x^2}+3\right )}{3 (x-1)}} x^{\frac {x^3}{3-3 x}+2}}{x-1}dx}{x}dx-8 \int \frac {e^{-\frac {x^4-e^{x^2} x^3+3 x^3}{3 (1-x)}} x^{\frac {x^3}{3 (1-x)}+2}}{(x-1)^2}dx+\int \frac {e^{-\frac {x^4-e^{x^2} x^3+3 x^3}{3 (1-x)}} x^{\frac {x^3}{3 (1-x)}+3}}{(x-1)^2}dx+3 \int \frac {e^{-\frac {x^4-e^{x^2} x^3+3 x^3}{3 (1-x)}} x^{\frac {x^3}{3 (1-x)}+4}}{(x-1)^2}dx+\log (x) \int \frac {e^{-\frac {x^4-e^{x^2} x^3+3 x^3}{3 (1-x)}} x^{\frac {x^3}{3 (1-x)}+2}}{(x-1)^2}dx-2 \log (x) \int \frac {e^{-\frac {x^4-e^{x^2} x^3+3 x^3}{3 (1-x)}} x^{\frac {x^3}{3 (1-x)}+2}}{x-1}dx\right )\)

Input:

Int[(E^((3*x^3 - E^x^2*x^3 + x^4 - x^3*Log[x])/(-3 + 3*x))*(-8*x^2 + x^3 + 
 3*x^4 + E^x^2*(3*x^2 - 2*x^3 + 2*x^4 - 2*x^5) + (3*x^2 - 2*x^3)*Log[x]))/ 
(15 - 30*x + 15*x^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 6.74 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76

method	result	size
risch	\(\frac {{\mathrm e}^{-\frac {x^{3} \left (\ln \left (x \right )+{\mathrm e}^{x^{2}}-x -3\right )}{3 \left (-1+x \right )}}}{5}\)	\(25\)
parallelrisch	\(\frac {{\mathrm e}^{-\frac {x^{3} \left (\ln \left (x \right )+{\mathrm e}^{x^{2}}-x -3\right )}{3 \left (-1+x \right )}}}{5}\)	\(25\)

Input:

int(((-2*x^3+3*x^2)*ln(x)+(-2*x^5+2*x^4-2*x^3+3*x^2)*exp(x^2)+3*x^4+x^3-8* 
x^2)*exp((-x^3*ln(x)-x^3*exp(x^2)+x^4+3*x^3)/(-3+3*x))/(15*x^2-30*x+15),x, 
method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int \frac {e^{\frac {3 x^3-e^{x^2} x^3+x^4-x^3 \log (x)}{-3+3 x}} \left (-8 x^2+x^3+3 x^4+e^{x^2} \left (3 x^2-2 x^3+2 x^4-2 x^5\right )+\left (3 x^2-2 x^3\right ) \log (x)\right )}{15-30 x+15 x^2} \, dx=\frac {1}{5} \, e^{\left (\frac {x^{4} - x^{3} e^{\left (x^{2}\right )} - x^{3} \log \left (x\right ) + 3 \, x^{3}}{3 \, {\left (x - 1\right )}}\right )} \] Input:

integrate(((-2*x^3+3*x^2)*log(x)+(-2*x^5+2*x^4-2*x^3+3*x^2)*exp(x^2)+3*x^4 
+x^3-8*x^2)*exp((-x^3*log(x)-x^3*exp(x^2)+x^4+3*x^3)/(-3+3*x))/(15*x^2-30* 
x+15),x, algorithm="fricas")
 

Output:

1/5*e^(1/3*(x^4 - x^3*e^(x^2) - x^3*log(x) + 3*x^3)/(x - 1))
 

Sympy [A] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \frac {e^{\frac {3 x^3-e^{x^2} x^3+x^4-x^3 \log (x)}{-3+3 x}} \left (-8 x^2+x^3+3 x^4+e^{x^2} \left (3 x^2-2 x^3+2 x^4-2 x^5\right )+\left (3 x^2-2 x^3\right ) \log (x)\right )}{15-30 x+15 x^2} \, dx=\frac {e^{\frac {x^{4} - x^{3} e^{x^{2}} - x^{3} \log {\left (x \right )} + 3 x^{3}}{3 x - 3}}}{5} \] Input:

integrate(((-2*x**3+3*x**2)*ln(x)+(-2*x**5+2*x**4-2*x**3+3*x**2)*exp(x**2) 
+3*x**4+x**3-8*x**2)*exp((-x**3*ln(x)-exp(x**2)*x**3+x**4+3*x**3)/(-3+3*x) 
)/(15*x**2-30*x+15),x)
 

Output:

exp((x**4 - x**3*exp(x**2) - x**3*log(x) + 3*x**3)/(3*x - 3))/5
 

Maxima [F]

\[ \int \frac {e^{\frac {3 x^3-e^{x^2} x^3+x^4-x^3 \log (x)}{-3+3 x}} \left (-8 x^2+x^3+3 x^4+e^{x^2} \left (3 x^2-2 x^3+2 x^4-2 x^5\right )+\left (3 x^2-2 x^3\right ) \log (x)\right )}{15-30 x+15 x^2} \, dx=\int { \frac {{\left (3 \, x^{4} + x^{3} - 8 \, x^{2} - {\left (2 \, x^{5} - 2 \, x^{4} + 2 \, x^{3} - 3 \, x^{2}\right )} e^{\left (x^{2}\right )} - {\left (2 \, x^{3} - 3 \, x^{2}\right )} \log \left (x\right )\right )} e^{\left (\frac {x^{4} - x^{3} e^{\left (x^{2}\right )} - x^{3} \log \left (x\right ) + 3 \, x^{3}}{3 \, {\left (x - 1\right )}}\right )}}{15 \, {\left (x^{2} - 2 \, x + 1\right )}} \,d x } \] Input:

integrate(((-2*x^3+3*x^2)*log(x)+(-2*x^5+2*x^4-2*x^3+3*x^2)*exp(x^2)+3*x^4 
+x^3-8*x^2)*exp((-x^3*log(x)-x^3*exp(x^2)+x^4+3*x^3)/(-3+3*x))/(15*x^2-30* 
x+15),x, algorithm="maxima")
 

Output:

1/15*integrate((3*x^4 + x^3 - 8*x^2 - (2*x^5 - 2*x^4 + 2*x^3 - 3*x^2)*e^(x 
^2) - (2*x^3 - 3*x^2)*log(x))*e^(1/3*(x^4 - x^3*e^(x^2) - x^3*log(x) + 3*x 
^3)/(x - 1))/(x^2 - 2*x + 1), x)
 

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.48 \[ \int \frac {e^{\frac {3 x^3-e^{x^2} x^3+x^4-x^3 \log (x)}{-3+3 x}} \left (-8 x^2+x^3+3 x^4+e^{x^2} \left (3 x^2-2 x^3+2 x^4-2 x^5\right )+\left (3 x^2-2 x^3\right ) \log (x)\right )}{15-30 x+15 x^2} \, dx=\frac {1}{5} \, e^{\left (\frac {x^{4}}{3 \, {\left (x - 1\right )}} - \frac {x^{3} e^{\left (x^{2}\right )}}{3 \, {\left (x - 1\right )}} - \frac {x^{3} \log \left (x\right )}{3 \, {\left (x - 1\right )}} + \frac {x^{3}}{x - 1}\right )} \] Input:

integrate(((-2*x^3+3*x^2)*log(x)+(-2*x^5+2*x^4-2*x^3+3*x^2)*exp(x^2)+3*x^4 
+x^3-8*x^2)*exp((-x^3*log(x)-x^3*exp(x^2)+x^4+3*x^3)/(-3+3*x))/(15*x^2-30* 
x+15),x, algorithm="giac")
 

Output:

1/5*e^(1/3*x^4/(x - 1) - 1/3*x^3*e^(x^2)/(x - 1) - 1/3*x^3*log(x)/(x - 1) 
+ x^3/(x - 1))
 

Mupad [B] (verification not implemented)

Time = 3.15 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.70 \[ \int \frac {e^{\frac {3 x^3-e^{x^2} x^3+x^4-x^3 \log (x)}{-3+3 x}} \left (-8 x^2+x^3+3 x^4+e^{x^2} \left (3 x^2-2 x^3+2 x^4-2 x^5\right )+\left (3 x^2-2 x^3\right ) \log (x)\right )}{15-30 x+15 x^2} \, dx=\frac {{\mathrm {e}}^{\frac {x^4}{3\,x-3}}\,{\mathrm {e}}^{-\frac {x^3\,{\mathrm {e}}^{x^2}}{3\,x-3}}\,{\mathrm {e}}^{\frac {x^3}{x-1}}}{5\,x^{\frac {x^3}{3\,x-3}}} \] Input:

int((exp(-(x^3*log(x) + x^3*exp(x^2) - 3*x^3 - x^4)/(3*x - 3))*(log(x)*(3* 
x^2 - 2*x^3) + exp(x^2)*(3*x^2 - 2*x^3 + 2*x^4 - 2*x^5) - 8*x^2 + x^3 + 3* 
x^4))/(15*x^2 - 30*x + 15),x)
 

Output:

(exp(x^4/(3*x - 3))*exp(-(x^3*exp(x^2))/(3*x - 3))*exp(x^3/(x - 1)))/(5*x^ 
(x^3/(3*x - 3)))
 

Reduce [B] (verification not implemented)

Time = 187.49 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.64 \[ \int \frac {e^{\frac {3 x^3-e^{x^2} x^3+x^4-x^3 \log (x)}{-3+3 x}} \left (-8 x^2+x^3+3 x^4+e^{x^2} \left (3 x^2-2 x^3+2 x^4-2 x^5\right )+\left (3 x^2-2 x^3\right ) \log (x)\right )}{15-30 x+15 x^2} \, dx=\frac {e^{\frac {x^{4}+3 x^{3}-3 x +3}{3 x -3}} e}{5 e^{\frac {e^{x^{2}} x^{3}+\mathrm {log}\left (x \right ) x^{3}}{3 x -3}}} \] Input:

int(((-2*x^3+3*x^2)*log(x)+(-2*x^5+2*x^4-2*x^3+3*x^2)*exp(x^2)+3*x^4+x^3-8 
*x^2)*exp((-x^3*log(x)-exp(x^2)*x^3+x^4+3*x^3)/(-3+3*x))/(15*x^2-30*x+15), 
x)
 

Output:

(e**((x**4 + 3*x**3 - 3*x + 3)/(3*x - 3))*e)/(5*e**((e**(x**2)*x**3 + log( 
x)*x**3)/(3*x - 3)))