\(\int \frac {(-4-2 x+e^{4 e^{2 x}} x)^{\frac {1}{x}} (-2 x+e^{4 e^{2 x}} (x+8 e^{2 x} x^2)+(4+2 x-e^{4 e^{2 x}} x) \log (-4-2 x+e^{4 e^{2 x}} x))}{-4 x^2-2 x^3+e^{4 e^{2 x}} x^3} \, dx\) [231]

Optimal result

Integrand size = 108, antiderivative size = 22 \[ \int \frac {\left (-4-2 x+e^{4 e^{2 x}} x\right )^{\frac {1}{x}} \left (-2 x+e^{4 e^{2 x}} \left (x+8 e^{2 x} x^2\right )+\left (4+2 x-e^{4 e^{2 x}} x\right ) \log \left (-4-2 x+e^{4 e^{2 x}} x\right )\right )}{-4 x^2-2 x^3+e^{4 e^{2 x}} x^3} \, dx=\left (\left (-2+e^{4 e^{2 x}}-\frac {4}{x}\right ) x\right )^{\frac {1}{x}} \] Output:

exp(ln((exp(4*exp(2*x))-2-4/x)*x)/x)
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {\left (-4-2 x+e^{4 e^{2 x}} x\right )^{\frac {1}{x}} \left (-2 x+e^{4 e^{2 x}} \left (x+8 e^{2 x} x^2\right )+\left (4+2 x-e^{4 e^{2 x}} x\right ) \log \left (-4-2 x+e^{4 e^{2 x}} x\right )\right )}{-4 x^2-2 x^3+e^{4 e^{2 x}} x^3} \, dx=\left (-4+\left (-2+e^{4 e^{2 x}}\right ) x\right )^{\frac {1}{x}} \] Input:

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

Output:

(-4 + (-2 + E^(4*E^(2*x)))*x)^x^(-1)
 

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.

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

\(\displaystyle -4 \int \frac {\int \frac {\left (\left (-2+e^{4 e^{2 x}}\right ) x-4\right )^{\frac {1}{x}-1}}{x^2}dx}{x}dx-16 \int \frac {\int \frac {\left (\left (-2+e^{4 e^{2 x}}\right ) x-4\right )^{\frac {1}{x}-1}}{x^2}dx}{x \left (e^{4 e^{2 x}} x-2 x-4\right )}dx-32 \int \frac {e^{2 \left (x+2 e^{2 x}\right )} x \int \frac {\left (\left (-2+e^{4 e^{2 x}}\right ) x-4\right )^{\frac {1}{x}-1}}{x^2}dx}{e^{4 e^{2 x}} x-2 x-4}dx+4 \log \left (-\left (\left (2-e^{4 e^{2 x}}\right ) x\right )-4\right ) \int \frac {\left (e^{4 e^{2 x}} x-2 x-4\right )^{\frac {1}{x}-1}}{x^2}dx+8 \int e^{2 \left (x+2 e^{2 x}\right )} \left (e^{4 e^{2 x}} x-2 x-4\right )^{\frac {1}{x}-1}dx-2 \int \frac {\left (e^{4 e^{2 x}} x-2 x-4\right )^{\frac {1}{x}-1}}{x}dx+\int \frac {e^{4 e^{2 x}} \left (e^{4 e^{2 x}} x-2 x-4\right )^{\frac {1}{x}-1}}{x}dx-2 \int \frac {\int \frac {\left (\left (-2+e^{4 e^{2 x}}\right ) x-4\right )^{\frac {1}{x}-1}}{x}dx}{x}dx-8 \int \frac {\int \frac {\left (\left (-2+e^{4 e^{2 x}}\right ) x-4\right )^{\frac {1}{x}-1}}{x}dx}{x \left (e^{4 e^{2 x}} x-2 x-4\right )}dx-16 \int \frac {e^{2 \left (x+2 e^{2 x}\right )} x \int \frac {\left (\left (-2+e^{4 e^{2 x}}\right ) x-4\right )^{\frac {1}{x}-1}}{x}dx}{e^{4 e^{2 x}} x-2 x-4}dx+\int \frac {\int \frac {e^{4 e^{2 x}} \left (\left (-2+e^{4 e^{2 x}}\right ) x-4\right )^{\frac {1}{x}-1}}{x}dx}{x}dx+4 \int \frac {\int \frac {e^{4 e^{2 x}} \left (\left (-2+e^{4 e^{2 x}}\right ) x-4\right )^{\frac {1}{x}-1}}{x}dx}{x \left (e^{4 e^{2 x}} x-2 x-4\right )}dx+8 \int \frac {e^{2 \left (x+2 e^{2 x}\right )} x \int \frac {e^{4 e^{2 x}} \left (\left (-2+e^{4 e^{2 x}}\right ) x-4\right )^{\frac {1}{x}-1}}{x}dx}{e^{4 e^{2 x}} x-2 x-4}dx+2 \log \left (-\left (\left (2-e^{4 e^{2 x}}\right ) x\right )-4\right ) \int \frac {\left (e^{4 e^{2 x}} x-2 x-4\right )^{\frac {1}{x}-1}}{x}dx-\log \left (-\left (\left (2-e^{4 e^{2 x}}\right ) x\right )-4\right ) \int \frac {e^{4 e^{2 x}} \left (e^{4 e^{2 x}} x-2 x-4\right )^{\frac {1}{x}-1}}{x}dx\)

Input:

Int[((-4 - 2*x + E^(4*E^(2*x))*x)^x^(-1)*(-2*x + E^(4*E^(2*x))*(x + 8*E^(2 
*x)*x^2) + (4 + 2*x - E^(4*E^(2*x))*x)*Log[-4 - 2*x + E^(4*E^(2*x))*x]))/( 
-4*x^2 - 2*x^3 + E^(4*E^(2*x))*x^3),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 15.73 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86

method	result	size
risch	\(\left (x \,{\mathrm e}^{4 \,{\mathrm e}^{2 x}}-2 x -4\right )^{\frac {1}{x}}\)	\(19\)
parallelrisch	\({\mathrm e}^{\frac {\ln \left (x \,{\mathrm e}^{4 \,{\mathrm e}^{2 x}}-2 x -4\right )}{x}}\)	\(21\)

Input:

int(((-x*exp(4*exp(2*x))+2*x+4)*ln(x*exp(4*exp(2*x))-2*x-4)+(8*exp(2*x)*x^ 
2+x)*exp(4*exp(2*x))-2*x)*exp(ln(x*exp(4*exp(2*x))-2*x-4)/x)/(x^3*exp(4*ex 
p(2*x))-2*x^3-4*x^2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {\left (-4-2 x+e^{4 e^{2 x}} x\right )^{\frac {1}{x}} \left (-2 x+e^{4 e^{2 x}} \left (x+8 e^{2 x} x^2\right )+\left (4+2 x-e^{4 e^{2 x}} x\right ) \log \left (-4-2 x+e^{4 e^{2 x}} x\right )\right )}{-4 x^2-2 x^3+e^{4 e^{2 x}} x^3} \, dx={\left (x e^{\left (4 \, e^{\left (2 \, x\right )}\right )} - 2 \, x - 4\right )}^{\left (\frac {1}{x}\right )} \] Input:

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

Output:

(x*e^(4*e^(2*x)) - 2*x - 4)^(1/x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (-4-2 x+e^{4 e^{2 x}} x\right )^{\frac {1}{x}} \left (-2 x+e^{4 e^{2 x}} \left (x+8 e^{2 x} x^2\right )+\left (4+2 x-e^{4 e^{2 x}} x\right ) \log \left (-4-2 x+e^{4 e^{2 x}} x\right )\right )}{-4 x^2-2 x^3+e^{4 e^{2 x}} x^3} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {\left (-4-2 x+e^{4 e^{2 x}} x\right )^{\frac {1}{x}} \left (-2 x+e^{4 e^{2 x}} \left (x+8 e^{2 x} x^2\right )+\left (4+2 x-e^{4 e^{2 x}} x\right ) \log \left (-4-2 x+e^{4 e^{2 x}} x\right )\right )}{-4 x^2-2 x^3+e^{4 e^{2 x}} x^3} \, dx={\left (x e^{\left (4 \, e^{\left (2 \, x\right )}\right )} - 2 \, x - 4\right )}^{\left (\frac {1}{x}\right )} \] Input:

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

Output:

(x*e^(4*e^(2*x)) - 2*x - 4)^(1/x)
 

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 4.39 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {\left (-4-2 x+e^{4 e^{2 x}} x\right )^{\frac {1}{x}} \left (-2 x+e^{4 e^{2 x}} \left (x+8 e^{2 x} x^2\right )+\left (4+2 x-e^{4 e^{2 x}} x\right ) \log \left (-4-2 x+e^{4 e^{2 x}} x\right )\right )}{-4 x^2-2 x^3+e^{4 e^{2 x}} x^3} \, dx={\left (x\,{\mathrm {e}}^{4\,{\mathrm {e}}^{2\,x}}-2\,x-4\right )}^{1/x} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {\left (-4-2 x+e^{4 e^{2 x}} x\right )^{\frac {1}{x}} \left (-2 x+e^{4 e^{2 x}} \left (x+8 e^{2 x} x^2\right )+\left (4+2 x-e^{4 e^{2 x}} x\right ) \log \left (-4-2 x+e^{4 e^{2 x}} x\right )\right )}{-4 x^2-2 x^3+e^{4 e^{2 x}} x^3} \, dx=e^{\frac {\mathrm {log}\left (e^{4 e^{2 x}} x -2 x -4\right )}{x}} \] Input:

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

Output:

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