\(\int \frac {e^{e^{\frac {-5+x^2+x^3}{x}} (4 x-x^2)+8 e^{\frac {-5+x^2+x^3}{x}} x^3 \log (x)} (e^{\frac {-5+x^2+x^3}{x}} (20-x+2 x^2+15 x^3-2 x^4)+e^{\frac {-5+x^2+x^3}{x}} (40 x^2+24 x^3+8 x^4+16 x^5) \log (x))}{x} \, dx\) [776]

Optimal result

Integrand size = 126, antiderivative size = 28 \[ \int \frac {e^{e^{\frac {-5+x^2+x^3}{x}} \left (4 x-x^2\right )+8 e^{\frac {-5+x^2+x^3}{x}} x^3 \log (x)} \left (e^{\frac {-5+x^2+x^3}{x}} \left (20-x+2 x^2+15 x^3-2 x^4\right )+e^{\frac {-5+x^2+x^3}{x}} \left (40 x^2+24 x^3+8 x^4+16 x^5\right ) \log (x)\right )}{x} \, dx=e^{e^{-\frac {5}{x}+x+x^2} x \left (4-x+8 x^2 \log (x)\right )} \] Output:

exp(x*exp(x+x^2-5/x)*(4-x+8*x^2*ln(x)))
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.43 \[ \int \frac {e^{e^{\frac {-5+x^2+x^3}{x}} \left (4 x-x^2\right )+8 e^{\frac {-5+x^2+x^3}{x}} x^3 \log (x)} \left (e^{\frac {-5+x^2+x^3}{x}} \left (20-x+2 x^2+15 x^3-2 x^4\right )+e^{\frac {-5+x^2+x^3}{x}} \left (40 x^2+24 x^3+8 x^4+16 x^5\right ) \log (x)\right )}{x} \, dx=e^{-e^{-\frac {5}{x}+x+x^2} (-4+x) x} x^{8 e^{-\frac {5}{x}+x+x^2} x^3} \] Input:

Integrate[(E^(E^((-5 + x^2 + x^3)/x)*(4*x - x^2) + 8*E^((-5 + x^2 + x^3)/x 
)*x^3*Log[x])*(E^((-5 + x^2 + x^3)/x)*(20 - x + 2*x^2 + 15*x^3 - 2*x^4) + 
E^((-5 + x^2 + x^3)/x)*(40*x^2 + 24*x^3 + 8*x^4 + 16*x^5)*Log[x]))/x,x]
 

Output:

x^(8*E^(-5/x + x + x^2)*x^3)/E^(E^(-5/x + x + x^2)*(-4 + x)*x)
 

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.

\(\displaystyle \int \frac {\left (e^{\frac {x^3+x^2-5}{x}} \left (-2 x^4+15 x^3+2 x^2-x+20\right )+e^{\frac {x^3+x^2-5}{x}} \left (16 x^5+8 x^4+24 x^3+40 x^2\right ) \log (x)\right ) \exp \left (e^{\frac {x^3+x^2-5}{x}} \left (4 x-x^2\right )+8 e^{\frac {x^3+x^2-5}{x}} x^3 \log (x)\right )}{x} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {\left (16 x^5 \log (x)-2 x^4+8 x^4 \log (x)+15 x^3+24 x^3 \log (x)+2 x^2+40 x^2 \log (x)-x+20\right ) \exp \left (x^2+e^{x^2+x-\frac {5}{x}} x \left (8 x^2 \log (x)-x+4\right )+x-\frac {5}{x}\right )}{x}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (8 x \left (2 x^3+x^2+3 x+5\right ) \log (x) \exp \left (x^2+e^{x^2+x-\frac {5}{x}} x \left (8 x^2 \log (x)-x+4\right )+x-\frac {5}{x}\right )+\frac {\left (-2 x^4+15 x^3+2 x^2-x+20\right ) \exp \left (x^2+e^{x^2+x-\frac {5}{x}} x \left (8 x^2 \log (x)-x+4\right )+x-\frac {5}{x}\right )}{x}\right )dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (8 x \left (2 x^3+x^2+3 x+5\right ) \log (x) \exp \left (x^2+e^{x^2+x-\frac {5}{x}} x \left (8 x^2 \log (x)-x+4\right )+x-\frac {5}{x}\right )-\frac {\left (2 x^4-15 x^3-2 x^2+x-20\right ) \exp \left (x^2+e^{x^2+x-\frac {5}{x}} x \left (8 x^2 \log (x)-x+4\right )+x-\frac {5}{x}\right )}{x}\right )dx\)

\(\Big \downarrow \) 7299

\(\displaystyle \int \left (8 x \left (2 x^3+x^2+3 x+5\right ) \log (x) \exp \left (x^2+e^{x^2+x-\frac {5}{x}} x \left (8 x^2 \log (x)-x+4\right )+x-\frac {5}{x}\right )-\frac {\left (2 x^4-15 x^3-2 x^2+x-20\right ) \exp \left (x^2+e^{x^2+x-\frac {5}{x}} x \left (8 x^2 \log (x)-x+4\right )+x-\frac {5}{x}\right )}{x}\right )dx\)

Input:

Int[(E^(E^((-5 + x^2 + x^3)/x)*(4*x - x^2) + 8*E^((-5 + x^2 + x^3)/x)*x^3* 
Log[x])*(E^((-5 + x^2 + x^3)/x)*(20 - x + 2*x^2 + 15*x^3 - 2*x^4) + E^((-5 
 + x^2 + x^3)/x)*(40*x^2 + 24*x^3 + 8*x^4 + 16*x^5)*Log[x]))/x,x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 1.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04

method	result	size
parallelrisch	\({\mathrm e}^{{\mathrm e}^{\frac {x^{3}+x^{2}-5}{x}} x \left (4-x +8 x^{2} \ln \left (x \right )\right )}\)	\(29\)
risch	\(x^{8 \,{\mathrm e}^{\frac {x^{3}+x^{2}-5}{x}} x^{3}} {\mathrm e}^{-\left (x -4\right ) x \,{\mathrm e}^{\frac {x^{3}+x^{2}-5}{x}}}\)	\(42\)

Input:

int(((16*x^5+8*x^4+24*x^3+40*x^2)*exp((x^3+x^2-5)/x)*ln(x)+(-2*x^4+15*x^3+ 
2*x^2-x+20)*exp((x^3+x^2-5)/x))*exp(8*x^3*exp((x^3+x^2-5)/x)*ln(x)+(-x^2+4 
*x)*exp((x^3+x^2-5)/x))/x,x,method=_RETURNVERBOSE)
 

Output:

exp(exp((x^3+x^2-5)/x)*x*(4-x+8*x^2*ln(x)))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.57 \[ \int \frac {e^{e^{\frac {-5+x^2+x^3}{x}} \left (4 x-x^2\right )+8 e^{\frac {-5+x^2+x^3}{x}} x^3 \log (x)} \left (e^{\frac {-5+x^2+x^3}{x}} \left (20-x+2 x^2+15 x^3-2 x^4\right )+e^{\frac {-5+x^2+x^3}{x}} \left (40 x^2+24 x^3+8 x^4+16 x^5\right ) \log (x)\right )}{x} \, dx=e^{\left (8 \, x^{3} e^{\left (\frac {x^{3} + x^{2} - 5}{x}\right )} \log \left (x\right ) - {\left (x^{2} - 4 \, x\right )} e^{\left (\frac {x^{3} + x^{2} - 5}{x}\right )}\right )} \] Input:

integrate(((16*x^5+8*x^4+24*x^3+40*x^2)*exp((x^3+x^2-5)/x)*log(x)+(-2*x^4+ 
15*x^3+2*x^2-x+20)*exp((x^3+x^2-5)/x))*exp(8*x^3*exp((x^3+x^2-5)/x)*log(x) 
+(-x^2+4*x)*exp((x^3+x^2-5)/x))/x,x, algorithm="fricas")
 

Output:

e^(8*x^3*e^((x^3 + x^2 - 5)/x)*log(x) - (x^2 - 4*x)*e^((x^3 + x^2 - 5)/x))
 

Sympy [A] (verification not implemented)

Time = 0.54 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {e^{e^{\frac {-5+x^2+x^3}{x}} \left (4 x-x^2\right )+8 e^{\frac {-5+x^2+x^3}{x}} x^3 \log (x)} \left (e^{\frac {-5+x^2+x^3}{x}} \left (20-x+2 x^2+15 x^3-2 x^4\right )+e^{\frac {-5+x^2+x^3}{x}} \left (40 x^2+24 x^3+8 x^4+16 x^5\right ) \log (x)\right )}{x} \, dx=e^{8 x^{3} e^{\frac {x^{3} + x^{2} - 5}{x}} \log {\left (x \right )} + \left (- x^{2} + 4 x\right ) e^{\frac {x^{3} + x^{2} - 5}{x}}} \] Input:

integrate(((16*x**5+8*x**4+24*x**3+40*x**2)*exp((x**3+x**2-5)/x)*ln(x)+(-2 
*x**4+15*x**3+2*x**2-x+20)*exp((x**3+x**2-5)/x))*exp(8*x**3*exp((x**3+x**2 
-5)/x)*ln(x)+(-x**2+4*x)*exp((x**3+x**2-5)/x))/x,x)
 

Output:

exp(8*x**3*exp((x**3 + x**2 - 5)/x)*log(x) + (-x**2 + 4*x)*exp((x**3 + x** 
2 - 5)/x))
 

Maxima [A] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.79 \[ \int \frac {e^{e^{\frac {-5+x^2+x^3}{x}} \left (4 x-x^2\right )+8 e^{\frac {-5+x^2+x^3}{x}} x^3 \log (x)} \left (e^{\frac {-5+x^2+x^3}{x}} \left (20-x+2 x^2+15 x^3-2 x^4\right )+e^{\frac {-5+x^2+x^3}{x}} \left (40 x^2+24 x^3+8 x^4+16 x^5\right ) \log (x)\right )}{x} \, dx=e^{\left (8 \, x^{3} e^{\left (x^{2} + x - \frac {5}{x}\right )} \log \left (x\right ) - x^{2} e^{\left (x^{2} + x - \frac {5}{x}\right )} + 4 \, x e^{\left (x^{2} + x - \frac {5}{x}\right )}\right )} \] Input:

integrate(((16*x^5+8*x^4+24*x^3+40*x^2)*exp((x^3+x^2-5)/x)*log(x)+(-2*x^4+ 
15*x^3+2*x^2-x+20)*exp((x^3+x^2-5)/x))*exp(8*x^3*exp((x^3+x^2-5)/x)*log(x) 
+(-x^2+4*x)*exp((x^3+x^2-5)/x))/x,x, algorithm="maxima")
 

Output:

e^(8*x^3*e^(x^2 + x - 5/x)*log(x) - x^2*e^(x^2 + x - 5/x) + 4*x*e^(x^2 + x 
 - 5/x))
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.79 \[ \int \frac {e^{e^{\frac {-5+x^2+x^3}{x}} \left (4 x-x^2\right )+8 e^{\frac {-5+x^2+x^3}{x}} x^3 \log (x)} \left (e^{\frac {-5+x^2+x^3}{x}} \left (20-x+2 x^2+15 x^3-2 x^4\right )+e^{\frac {-5+x^2+x^3}{x}} \left (40 x^2+24 x^3+8 x^4+16 x^5\right ) \log (x)\right )}{x} \, dx=e^{\left (8 \, x^{3} e^{\left (x^{2} + x - \frac {5}{x}\right )} \log \left (x\right ) - x^{2} e^{\left (x^{2} + x - \frac {5}{x}\right )} + 4 \, x e^{\left (x^{2} + x - \frac {5}{x}\right )}\right )} \] Input:

integrate(((16*x^5+8*x^4+24*x^3+40*x^2)*exp((x^3+x^2-5)/x)*log(x)+(-2*x^4+ 
15*x^3+2*x^2-x+20)*exp((x^3+x^2-5)/x))*exp(8*x^3*exp((x^3+x^2-5)/x)*log(x) 
+(-x^2+4*x)*exp((x^3+x^2-5)/x))/x,x, algorithm="giac")
 

Output:

e^(8*x^3*e^(x^2 + x - 5/x)*log(x) - x^2*e^(x^2 + x - 5/x) + 4*x*e^(x^2 + x 
 - 5/x))
 

Mupad [B] (verification not implemented)

Time = 0.74 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.93 \[ \int \frac {e^{e^{\frac {-5+x^2+x^3}{x}} \left (4 x-x^2\right )+8 e^{\frac {-5+x^2+x^3}{x}} x^3 \log (x)} \left (e^{\frac {-5+x^2+x^3}{x}} \left (20-x+2 x^2+15 x^3-2 x^4\right )+e^{\frac {-5+x^2+x^3}{x}} \left (40 x^2+24 x^3+8 x^4+16 x^5\right ) \log (x)\right )}{x} \, dx=x^{8\,x^3\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{-\frac {5}{x}}\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-x^2\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{-\frac {5}{x}}\,{\mathrm {e}}^x}\,{\mathrm {e}}^{4\,x\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{-\frac {5}{x}}\,{\mathrm {e}}^x} \] Input:

int((exp(exp((x^2 + x^3 - 5)/x)*(4*x - x^2) + 8*x^3*exp((x^2 + x^3 - 5)/x) 
*log(x))*(exp((x^2 + x^3 - 5)/x)*(2*x^2 - x + 15*x^3 - 2*x^4 + 20) + exp(( 
x^2 + x^3 - 5)/x)*log(x)*(40*x^2 + 24*x^3 + 8*x^4 + 16*x^5)))/x,x)
 

Output:

x^(8*x^3*exp(x^2)*exp(-5/x)*exp(x))*exp(-x^2*exp(x^2)*exp(-5/x)*exp(x))*ex 
p(4*x*exp(x^2)*exp(-5/x)*exp(x))
 

Reduce [F]

\[ \int \frac {e^{e^{\frac {-5+x^2+x^3}{x}} \left (4 x-x^2\right )+8 e^{\frac {-5+x^2+x^3}{x}} x^3 \log (x)} \left (e^{\frac {-5+x^2+x^3}{x}} \left (20-x+2 x^2+15 x^3-2 x^4\right )+e^{\frac {-5+x^2+x^3}{x}} \left (40 x^2+24 x^3+8 x^4+16 x^5\right ) \log (x)\right )}{x} \, dx=\int \frac {\left (\left (16 x^{5}+8 x^{4}+24 x^{3}+40 x^{2}\right ) {\mathrm e}^{\frac {x^{3}+x^{2}-5}{x}} \mathrm {log}\left (x \right )+\left (-2 x^{4}+15 x^{3}+2 x^{2}-x +20\right ) {\mathrm e}^{\frac {x^{3}+x^{2}-5}{x}}\right ) {\mathrm e}^{8 x^{3} {\mathrm e}^{\frac {x^{3}+x^{2}-5}{x}} \mathrm {log}\left (x \right )+\left (-x^{2}+4 x \right ) {\mathrm e}^{\frac {x^{3}+x^{2}-5}{x}}}}{x}d x \] Input:

int(((16*x^5+8*x^4+24*x^3+40*x^2)*exp((x^3+x^2-5)/x)*log(x)+(-2*x^4+15*x^3 
+2*x^2-x+20)*exp((x^3+x^2-5)/x))*exp(8*x^3*exp((x^3+x^2-5)/x)*log(x)+(-x^2 
+4*x)*exp((x^3+x^2-5)/x))/x,x)
 

Output:

int(((16*x^5+8*x^4+24*x^3+40*x^2)*exp((x^3+x^2-5)/x)*log(x)+(-2*x^4+15*x^3 
+2*x^2-x+20)*exp((x^3+x^2-5)/x))*exp(8*x^3*exp((x^3+x^2-5)/x)*log(x)+(-x^2 
+4*x)*exp((x^3+x^2-5)/x))/x,x)