\(\int \frac {-4+4 e^3-4 e^{e^x} x^4+e^{e^x} (40 x^3+10 e^x x^4) \log (1-e^3+e^{e^x} x^4)}{5-5 e^3+5 e^{e^x} x^4} \, dx\) [961]

Optimal result

Integrand size = 74, antiderivative size = 26 \[ \int \frac {-4+4 e^3-4 e^{e^x} x^4+e^{e^x} \left (40 x^3+10 e^x x^4\right ) \log \left (1-e^3+e^{e^x} x^4\right )}{5-5 e^3+5 e^{e^x} x^4} \, dx=-4-\frac {4 x}{5}+\log ^2\left (1-e^3+e^{e^x} x^4\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {-4+4 e^3-4 e^{e^x} x^4+e^{e^x} \left (40 x^3+10 e^x x^4\right ) \log \left (1-e^3+e^{e^x} x^4\right )}{5-5 e^3+5 e^{e^x} x^4} \, dx=\frac {1}{5} \left (-4 x+5 \log ^2\left (1-e^3+e^{e^x} x^4\right )\right ) \] Input:

Integrate[(-4 + 4*E^3 - 4*E^E^x*x^4 + E^E^x*(40*x^3 + 10*E^x*x^4)*Log[1 - 
E^3 + E^E^x*x^4])/(5 - 5*E^3 + 5*E^E^x*x^4),x]
 

Output:

(-4*x + 5*Log[1 - E^3 + E^E^x*x^4]^2)/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.

\(\displaystyle \int \frac {-4 e^{e^x} x^4+e^{e^x} \left (10 e^x x^4+40 x^3\right ) \log \left (e^{e^x} x^4-e^3+1\right )+4 e^3-4}{5 e^{e^x} x^4-5 e^3+5} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {-4 e^{e^x} x^4+e^{e^x} \left (10 e^x x^4+40 x^3\right ) \log \left (e^{e^x} x^4-e^3+1\right )-4 \left (1-e^3\right )}{5 \left (e^{e^x} x^4-e^3+1\right )}dx\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 7293

\(\displaystyle -\frac {2}{5} \int \left (\frac {2 \left (-e^{e^x} x^4+10 e^{e^x} \log \left (e^{e^x} x^4-e^3+1\right ) x^3+e^3-1\right )}{-e^{e^x} x^4+e^3-1}-\frac {5 e^{x+e^x} x^4 \log \left (e^{e^x} x^4-e^3+1\right )}{e^{e^x} x^4-e^3+1}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2}{5} \left (5 \int \frac {e^{x+e^x} x^4 \int \frac {e^{x+e^x} x^4}{e^{e^x} x^4-e^3+1}dx}{e^{e^x} x^4-e^3+1}dx-20 \left (1-e^3\right ) \log \left (e^{e^x} x^4-e^3+1\right ) \int \frac {1}{x \left (-e^{e^x} x^4+e^3-1\right )}dx-5 \log \left (e^{e^x} x^4-e^3+1\right ) \int \frac {e^{x+e^x} x^4}{e^{e^x} x^4-e^3+1}dx-20 \int \frac {\log \left (e^{e^x} x^4-e^3+1\right )}{x}dx+20 \left (1-e^3\right ) \int \frac {e^{x+e^x} x^4 \int -\frac {1}{e^{e^x} x^5+\left (1-e^3\right ) x}dx}{e^{e^x} x^4-e^3+1}dx+20 \int \frac {e^{e^x} x^3 \int \frac {e^{x+e^x} x^4}{e^{e^x} x^4-e^3+1}dx}{e^{e^x} x^4-e^3+1}dx+80 \left (1-e^3\right ) \int \frac {e^{e^x} x^3 \int -\frac {1}{e^{e^x} x^5+\left (1-e^3\right ) x}dx}{e^{e^x} x^4-e^3+1}dx+2 x\right )\)

Input:

Int[(-4 + 4*E^3 - 4*E^E^x*x^4 + E^E^x*(40*x^3 + 10*E^x*x^4)*Log[1 - E^3 + 
E^E^x*x^4])/(5 - 5*E^3 + 5*E^E^x*x^4),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 106.09 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81

method	result	size
parallelrisch	\(\ln \left (x^{4} {\mathrm e}^{{\mathrm e}^{x}}-{\mathrm e}^{3}+1\right )^{2}-\frac {4 x}{5}\)	\(21\)

Input:

int(((10*exp(x)*x^4+40*x^3)*exp(exp(x))*ln(x^4*exp(exp(x))-exp(3)+1)-4*x^4 
*exp(exp(x))+4*exp(3)-4)/(5*x^4*exp(exp(x))-5*exp(3)+5),x,method=_RETURNVE 
RBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {-4+4 e^3-4 e^{e^x} x^4+e^{e^x} \left (40 x^3+10 e^x x^4\right ) \log \left (1-e^3+e^{e^x} x^4\right )}{5-5 e^3+5 e^{e^x} x^4} \, dx=\log \left (x^{4} e^{\left (e^{x}\right )} - e^{3} + 1\right )^{2} - \frac {4}{5} \, x \] Input:

integrate(((10*exp(x)*x^4+40*x^3)*exp(exp(x))*log(x^4*exp(exp(x))-exp(3)+1 
)-4*x^4*exp(exp(x))+4*exp(3)-4)/(5*x^4*exp(exp(x))-5*exp(3)+5),x, algorith 
m="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {-4+4 e^3-4 e^{e^x} x^4+e^{e^x} \left (40 x^3+10 e^x x^4\right ) \log \left (1-e^3+e^{e^x} x^4\right )}{5-5 e^3+5 e^{e^x} x^4} \, dx=- \frac {4 x}{5} + \log {\left (x^{4} e^{e^{x}} - e^{3} + 1 \right )}^{2} \] Input:

integrate(((10*exp(x)*x**4+40*x**3)*exp(exp(x))*ln(x**4*exp(exp(x))-exp(3) 
+1)-4*x**4*exp(exp(x))+4*exp(3)-4)/(5*x**4*exp(exp(x))-5*exp(3)+5),x)
 

Output:

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

Maxima [F]

\[ \int \frac {-4+4 e^3-4 e^{e^x} x^4+e^{e^x} \left (40 x^3+10 e^x x^4\right ) \log \left (1-e^3+e^{e^x} x^4\right )}{5-5 e^3+5 e^{e^x} x^4} \, dx=\int { -\frac {2 \, {\left (2 \, x^{4} e^{\left (e^{x}\right )} - 5 \, {\left (x^{4} e^{x} + 4 \, x^{3}\right )} e^{\left (e^{x}\right )} \log \left (x^{4} e^{\left (e^{x}\right )} - e^{3} + 1\right ) - 2 \, e^{3} + 2\right )}}{5 \, {\left (x^{4} e^{\left (e^{x}\right )} - e^{3} + 1\right )}} \,d x } \] Input:

integrate(((10*exp(x)*x^4+40*x^3)*exp(exp(x))*log(x^4*exp(exp(x))-exp(3)+1 
)-4*x^4*exp(exp(x))+4*exp(3)-4)/(5*x^4*exp(exp(x))-5*exp(3)+5),x, algorith 
m="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {-4+4 e^3-4 e^{e^x} x^4+e^{e^x} \left (40 x^3+10 e^x x^4\right ) \log \left (1-e^3+e^{e^x} x^4\right )}{5-5 e^3+5 e^{e^x} x^4} \, dx=\int { -\frac {2 \, {\left (2 \, x^{4} e^{\left (e^{x}\right )} - 5 \, {\left (x^{4} e^{x} + 4 \, x^{3}\right )} e^{\left (e^{x}\right )} \log \left (x^{4} e^{\left (e^{x}\right )} - e^{3} + 1\right ) - 2 \, e^{3} + 2\right )}}{5 \, {\left (x^{4} e^{\left (e^{x}\right )} - e^{3} + 1\right )}} \,d x } \] Input:

integrate(((10*exp(x)*x^4+40*x^3)*exp(exp(x))*log(x^4*exp(exp(x))-exp(3)+1 
)-4*x^4*exp(exp(x))+4*exp(3)-4)/(5*x^4*exp(exp(x))-5*exp(3)+5),x, algorith 
m="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {-4+4 e^3-4 e^{e^x} x^4+e^{e^x} \left (40 x^3+10 e^x x^4\right ) \log \left (1-e^3+e^{e^x} x^4\right )}{5-5 e^3+5 e^{e^x} x^4} \, dx={\ln \left (x^4\,{\mathrm {e}}^{{\mathrm {e}}^x}-{\mathrm {e}}^3+1\right )}^2-\frac {4\,x}{5} \] Input:

int((4*exp(3) - 4*x^4*exp(exp(x)) + log(x^4*exp(exp(x)) - exp(3) + 1)*exp( 
exp(x))*(10*x^4*exp(x) + 40*x^3) - 4)/(5*x^4*exp(exp(x)) - 5*exp(3) + 5),x 
)
 

Output:

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

Reduce [F]

\[ \int \frac {-4+4 e^3-4 e^{e^x} x^4+e^{e^x} \left (40 x^3+10 e^x x^4\right ) \log \left (1-e^3+e^{e^x} x^4\right )}{5-5 e^3+5 e^{e^x} x^4} \, dx=-\frac {4 \left (\int \frac {e^{e^{x}} x^{4}}{e^{e^{x}} x^{4}-e^{3}+1}d x \right )}{5}+8 \left (\int \frac {e^{e^{x}} \mathrm {log}\left (e^{e^{x}} x^{4}-e^{3}+1\right ) x^{3}}{e^{e^{x}} x^{4}-e^{3}+1}d x \right )+2 \left (\int \frac {e^{e^{x}+x} \mathrm {log}\left (e^{e^{x}} x^{4}-e^{3}+1\right ) x^{4}}{e^{e^{x}} x^{4}-e^{3}+1}d x \right )+\frac {4 \left (\int \frac {1}{e^{e^{x}} x^{4}-e^{3}+1}d x \right ) e^{3}}{5}-\frac {4 \left (\int \frac {1}{e^{e^{x}} x^{4}-e^{3}+1}d x \right )}{5} \] Input:

int(((10*exp(x)*x^4+40*x^3)*exp(exp(x))*log(x^4*exp(exp(x))-exp(3)+1)-4*x^ 
4*exp(exp(x))+4*exp(3)-4)/(5*x^4*exp(exp(x))-5*exp(3)+5),x)
 

Output:

(2*( - 2*int((e**(e**x)*x**4)/(e**(e**x)*x**4 - e**3 + 1),x) + 20*int((e** 
(e**x)*log(e**(e**x)*x**4 - e**3 + 1)*x**3)/(e**(e**x)*x**4 - e**3 + 1),x) 
 + 5*int((e**(e**x + x)*log(e**(e**x)*x**4 - e**3 + 1)*x**4)/(e**(e**x)*x* 
*4 - e**3 + 1),x) + 2*int(1/(e**(e**x)*x**4 - e**3 + 1),x)*e**3 - 2*int(1/ 
(e**(e**x)*x**4 - e**3 + 1),x)))/5