\(\int \frac {x^{\frac {4}{-2 x+\log ^4(3+e^{3-x})}} (-24 x-8 e^{3-x} x+(12+4 e^{3-x}) \log ^4(3+e^{3-x})+(24 x+8 e^{3-x} x) \log (x)+16 e^{3-x} x \log ^3(3+e^{3-x}) \log (x))}{12 x^3+4 e^{3-x} x^3+(-12 x^2-4 e^{3-x} x^2) \log ^4(3+e^{3-x})+(3 x+e^{3-x} x) \log ^8(3+e^{3-x})} \, dx\) [1407]

Optimal result

Integrand size = 179, antiderivative size = 26 \[ \int \frac {x^{\frac {4}{-2 x+\log ^4\left (3+e^{3-x}\right )}} \left (-24 x-8 e^{3-x} x+\left (12+4 e^{3-x}\right ) \log ^4\left (3+e^{3-x}\right )+\left (24 x+8 e^{3-x} x\right ) \log (x)+16 e^{3-x} x \log ^3\left (3+e^{3-x}\right ) \log (x)\right )}{12 x^3+4 e^{3-x} x^3+\left (-12 x^2-4 e^{3-x} x^2\right ) \log ^4\left (3+e^{3-x}\right )+\left (3 x+e^{3-x} x\right ) \log ^8\left (3+e^{3-x}\right )} \, dx=x^{\frac {2}{-x+\frac {1}{2} \log ^4\left (3+e^{3-x}\right )}} \] Output:

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {x^{\frac {4}{-2 x+\log ^4\left (3+e^{3-x}\right )}} \left (-24 x-8 e^{3-x} x+\left (12+4 e^{3-x}\right ) \log ^4\left (3+e^{3-x}\right )+\left (24 x+8 e^{3-x} x\right ) \log (x)+16 e^{3-x} x \log ^3\left (3+e^{3-x}\right ) \log (x)\right )}{12 x^3+4 e^{3-x} x^3+\left (-12 x^2-4 e^{3-x} x^2\right ) \log ^4\left (3+e^{3-x}\right )+\left (3 x+e^{3-x} x\right ) \log ^8\left (3+e^{3-x}\right )} \, dx=x^{-\frac {4}{2 x-\log ^4\left (3+e^{3-x}\right )}} \] Input:

Integrate[(x^(4/(-2*x + Log[3 + E^(3 - x)]^4))*(-24*x - 8*E^(3 - x)*x + (1 
2 + 4*E^(3 - x))*Log[3 + E^(3 - x)]^4 + (24*x + 8*E^(3 - x)*x)*Log[x] + 16 
*E^(3 - x)*x*Log[3 + E^(3 - x)]^3*Log[x]))/(12*x^3 + 4*E^(3 - x)*x^3 + (-1 
2*x^2 - 4*E^(3 - x)*x^2)*Log[3 + E^(3 - x)]^4 + (3*x + E^(3 - x)*x)*Log[3 
+ E^(3 - x)]^8),x]
 

Output:

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

\(\displaystyle \int \frac {x^{\frac {4}{\log ^4\left (e^{3-x}+3\right )-2 x}} \left (-8 e^{3-x} x-24 x+\left (4 e^{3-x}+12\right ) \log ^4\left (e^{3-x}+3\right )+16 e^{3-x} x \log (x) \log ^3\left (e^{3-x}+3\right )+\left (8 e^{3-x} x+24 x\right ) \log (x)\right )}{4 e^{3-x} x^3+12 x^3+\left (-4 e^{3-x} x^2-12 x^2\right ) \log ^4\left (e^{3-x}+3\right )+\left (e^{3-x} x+3 x\right ) \log ^8\left (e^{3-x}+3\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {e^x x^{\frac {4}{\log ^4\left (e^{3-x}+3\right )-2 x}-1} \left (-8 e^{3-x} x-24 x+\left (4 e^{3-x}+12\right ) \log ^4\left (e^{3-x}+3\right )+16 e^{3-x} x \log (x) \log ^3\left (e^{3-x}+3\right )+\left (8 e^{3-x} x+24 x\right ) \log (x)\right )}{\left (3 e^x+e^3\right ) \left (2 x-\log ^4\left (e^{3-x}+3\right )\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {4 x^{\frac {4}{\log ^4\left (e^{3-x}+3\right )-2 x}-1} \left (-2 x+\log ^4\left (e^{3-x}+3\right )+4 x \log (x) \log ^3\left (e^{3-x}+3\right )+2 x \log (x)\right )}{\left (2 x-\log ^4\left (e^{3-x}+3\right )\right )^2}-\frac {48 e^x x^{\frac {4}{\log ^4\left (e^{3-x}+3\right )-2 x}} \log ^3\left (e^{3-x}+3\right ) \log (x)}{\left (3 e^x+e^3\right ) \left (2 x-\log ^4\left (e^{3-x}+3\right )\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 4 \int \frac {x^{\frac {4}{\log ^4\left (3+e^{3-x}\right )-2 x}-1}}{\log ^4\left (3+e^{3-x}\right )-2 x}dx+8 \int \frac {x^{\frac {4}{\log ^4\left (3+e^{3-x}\right )-2 x}} \log (x)}{\left (2 x-\log ^4\left (3+e^{3-x}\right )\right )^2}dx-48 \int \frac {e^x x^{\frac {4}{\log ^4\left (3+e^{3-x}\right )-2 x}} \log ^3\left (3+e^{3-x}\right ) \log (x)}{\left (e^3+3 e^x\right ) \left (2 x-\log ^4\left (3+e^{3-x}\right )\right )^2}dx+16 \int \frac {x^{\frac {4}{\log ^4\left (3+e^{3-x}\right )-2 x}} \log ^3\left (3+e^{3-x}\right ) \log (x)}{\left (\log ^4\left (3+e^{3-x}\right )-2 x\right )^2}dx\)

Input:

Int[(x^(4/(-2*x + Log[3 + E^(3 - x)]^4))*(-24*x - 8*E^(3 - x)*x + (12 + 4* 
E^(3 - x))*Log[3 + E^(3 - x)]^4 + (24*x + 8*E^(3 - x)*x)*Log[x] + 16*E^(3 
- x)*x*Log[3 + E^(3 - x)]^3*Log[x]))/(12*x^3 + 4*E^(3 - x)*x^3 + (-12*x^2 
- 4*E^(3 - x)*x^2)*Log[3 + E^(3 - x)]^4 + (3*x + E^(3 - x)*x)*Log[3 + E^(3 
 - x)]^8),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85

\[x^{\frac {4}{\ln \left ({\mathrm e}^{-x +3}+3\right )^{4}-2 x}}\]

Input:

int(((4*exp(-x+3)+12)*ln(exp(-x+3)+3)^4+16*x*exp(-x+3)*ln(x)*ln(exp(-x+3)+ 
3)^3+(8*x*exp(-x+3)+24*x)*ln(x)-8*x*exp(-x+3)-24*x)*exp(4*ln(x)/(ln(exp(-x 
+3)+3)^4-2*x))/((x*exp(-x+3)+3*x)*ln(exp(-x+3)+3)^8+(-4*x^2*exp(-x+3)-12*x 
^2)*ln(exp(-x+3)+3)^4+4*x^3*exp(-x+3)+12*x^3),x)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {x^{\frac {4}{-2 x+\log ^4\left (3+e^{3-x}\right )}} \left (-24 x-8 e^{3-x} x+\left (12+4 e^{3-x}\right ) \log ^4\left (3+e^{3-x}\right )+\left (24 x+8 e^{3-x} x\right ) \log (x)+16 e^{3-x} x \log ^3\left (3+e^{3-x}\right ) \log (x)\right )}{12 x^3+4 e^{3-x} x^3+\left (-12 x^2-4 e^{3-x} x^2\right ) \log ^4\left (3+e^{3-x}\right )+\left (3 x+e^{3-x} x\right ) \log ^8\left (3+e^{3-x}\right )} \, dx=x^{\frac {4}{\log \left (e^{\left (-x + 3\right )} + 3\right )^{4} - 2 \, x}} \] Input:

integrate(((4*exp(3-x)+12)*log(exp(3-x)+3)^4+16*x*exp(3-x)*log(x)*log(exp( 
3-x)+3)^3+(8*x*exp(3-x)+24*x)*log(x)-8*x*exp(3-x)-24*x)*exp(4*log(x)/(log( 
exp(3-x)+3)^4-2*x))/((x*exp(3-x)+3*x)*log(exp(3-x)+3)^8+(-4*x^2*exp(3-x)-1 
2*x^2)*log(exp(3-x)+3)^4+4*x^3*exp(3-x)+12*x^3),x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{\frac {4}{-2 x+\log ^4\left (3+e^{3-x}\right )}} \left (-24 x-8 e^{3-x} x+\left (12+4 e^{3-x}\right ) \log ^4\left (3+e^{3-x}\right )+\left (24 x+8 e^{3-x} x\right ) \log (x)+16 e^{3-x} x \log ^3\left (3+e^{3-x}\right ) \log (x)\right )}{12 x^3+4 e^{3-x} x^3+\left (-12 x^2-4 e^{3-x} x^2\right ) \log ^4\left (3+e^{3-x}\right )+\left (3 x+e^{3-x} x\right ) \log ^8\left (3+e^{3-x}\right )} \, dx=\text {Timed out} \] Input:

integrate(((4*exp(3-x)+12)*ln(exp(3-x)+3)**4+16*x*exp(3-x)*ln(x)*ln(exp(3- 
x)+3)**3+(8*x*exp(3-x)+24*x)*ln(x)-8*x*exp(3-x)-24*x)*exp(4*ln(x)/(ln(exp( 
3-x)+3)**4-2*x))/((x*exp(3-x)+3*x)*ln(exp(3-x)+3)**8+(-4*x**2*exp(3-x)-12* 
x**2)*ln(exp(3-x)+3)**4+4*x**3*exp(3-x)+12*x**3),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (21) = 42\).

Time = 0.21 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.50 \[ \int \frac {x^{\frac {4}{-2 x+\log ^4\left (3+e^{3-x}\right )}} \left (-24 x-8 e^{3-x} x+\left (12+4 e^{3-x}\right ) \log ^4\left (3+e^{3-x}\right )+\left (24 x+8 e^{3-x} x\right ) \log (x)+16 e^{3-x} x \log ^3\left (3+e^{3-x}\right ) \log (x)\right )}{12 x^3+4 e^{3-x} x^3+\left (-12 x^2-4 e^{3-x} x^2\right ) \log ^4\left (3+e^{3-x}\right )+\left (3 x+e^{3-x} x\right ) \log ^8\left (3+e^{3-x}\right )} \, dx=x^{\frac {4}{x^{4} - 4 \, x^{3} \log \left (e^{3} + 3 \, e^{x}\right ) + 6 \, x^{2} \log \left (e^{3} + 3 \, e^{x}\right )^{2} + \log \left (e^{3} + 3 \, e^{x}\right )^{4} - 2 \, {\left (2 \, \log \left (e^{3} + 3 \, e^{x}\right )^{3} + 1\right )} x}} \] Input:

integrate(((4*exp(3-x)+12)*log(exp(3-x)+3)^4+16*x*exp(3-x)*log(x)*log(exp( 
3-x)+3)^3+(8*x*exp(3-x)+24*x)*log(x)-8*x*exp(3-x)-24*x)*exp(4*log(x)/(log( 
exp(3-x)+3)^4-2*x))/((x*exp(3-x)+3*x)*log(exp(3-x)+3)^8+(-4*x^2*exp(3-x)-1 
2*x^2)*log(exp(3-x)+3)^4+4*x^3*exp(3-x)+12*x^3),x, algorithm="maxima")
 

Output:

x^(4/(x^4 - 4*x^3*log(e^3 + 3*e^x) + 6*x^2*log(e^3 + 3*e^x)^2 + log(e^3 + 
3*e^x)^4 - 2*(2*log(e^3 + 3*e^x)^3 + 1)*x))
 

Giac [F(-1)]

Timed out. \[ \int \frac {x^{\frac {4}{-2 x+\log ^4\left (3+e^{3-x}\right )}} \left (-24 x-8 e^{3-x} x+\left (12+4 e^{3-x}\right ) \log ^4\left (3+e^{3-x}\right )+\left (24 x+8 e^{3-x} x\right ) \log (x)+16 e^{3-x} x \log ^3\left (3+e^{3-x}\right ) \log (x)\right )}{12 x^3+4 e^{3-x} x^3+\left (-12 x^2-4 e^{3-x} x^2\right ) \log ^4\left (3+e^{3-x}\right )+\left (3 x+e^{3-x} x\right ) \log ^8\left (3+e^{3-x}\right )} \, dx=\text {Timed out} \] Input:

integrate(((4*exp(3-x)+12)*log(exp(3-x)+3)^4+16*x*exp(3-x)*log(x)*log(exp( 
3-x)+3)^3+(8*x*exp(3-x)+24*x)*log(x)-8*x*exp(3-x)-24*x)*exp(4*log(x)/(log( 
exp(3-x)+3)^4-2*x))/((x*exp(3-x)+3*x)*log(exp(3-x)+3)^8+(-4*x^2*exp(3-x)-1 
2*x^2)*log(exp(3-x)+3)^4+4*x^3*exp(3-x)+12*x^3),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 3.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {x^{\frac {4}{-2 x+\log ^4\left (3+e^{3-x}\right )}} \left (-24 x-8 e^{3-x} x+\left (12+4 e^{3-x}\right ) \log ^4\left (3+e^{3-x}\right )+\left (24 x+8 e^{3-x} x\right ) \log (x)+16 e^{3-x} x \log ^3\left (3+e^{3-x}\right ) \log (x)\right )}{12 x^3+4 e^{3-x} x^3+\left (-12 x^2-4 e^{3-x} x^2\right ) \log ^4\left (3+e^{3-x}\right )+\left (3 x+e^{3-x} x\right ) \log ^8\left (3+e^{3-x}\right )} \, dx=\frac {1}{x^{\frac {4}{2\,x-{\ln \left ({\mathrm {e}}^{-x}\,{\mathrm {e}}^3+3\right )}^4}}} \] Input:

int((exp(-(4*log(x))/(2*x - log(exp(3 - x) + 3)^4))*(log(exp(3 - x) + 3)^4 
*(4*exp(3 - x) + 12) - 8*x*exp(3 - x) - 24*x + log(x)*(24*x + 8*x*exp(3 - 
x)) + 16*x*exp(3 - x)*log(exp(3 - x) + 3)^3*log(x)))/(log(exp(3 - x) + 3)^ 
8*(3*x + x*exp(3 - x)) - log(exp(3 - x) + 3)^4*(4*x^2*exp(3 - x) + 12*x^2) 
 + 4*x^3*exp(3 - x) + 12*x^3),x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {x^{\frac {4}{-2 x+\log ^4\left (3+e^{3-x}\right )}} \left (-24 x-8 e^{3-x} x+\left (12+4 e^{3-x}\right ) \log ^4\left (3+e^{3-x}\right )+\left (24 x+8 e^{3-x} x\right ) \log (x)+16 e^{3-x} x \log ^3\left (3+e^{3-x}\right ) \log (x)\right )}{12 x^3+4 e^{3-x} x^3+\left (-12 x^2-4 e^{3-x} x^2\right ) \log ^4\left (3+e^{3-x}\right )+\left (3 x+e^{3-x} x\right ) \log ^8\left (3+e^{3-x}\right )} \, dx=e^{\frac {4 \,\mathrm {log}\left (x \right )}{\mathrm {log}\left (\frac {3 e^{x}+e^{3}}{e^{x}}\right )^{4}-2 x}} \] Input:

int(((4*exp(3-x)+12)*log(exp(3-x)+3)^4+16*x*exp(3-x)*log(x)*log(exp(3-x)+3 
)^3+(8*x*exp(3-x)+24*x)*log(x)-8*x*exp(3-x)-24*x)*exp(4*log(x)/(log(exp(3- 
x)+3)^4-2*x))/((x*exp(3-x)+3*x)*log(exp(3-x)+3)^8+(-4*x^2*exp(3-x)-12*x^2) 
*log(exp(3-x)+3)^4+4*x^3*exp(3-x)+12*x^3),x)
 

Output:

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