\(\int \frac {e^{e^{25 x^2 \log ^2(\frac {e^3-x}{1+e^3-x})}+25 x^2 \log ^2(\frac {e^3-x}{1+e^3-x})} (-25 x^2 \log (\frac {e^3-x}{1+e^3-x})+(25 e^6 x-25 x^2+25 x^3+e^3 (25 x-50 x^2)) \log ^2(\frac {e^3-x}{1+e^3-x}))}{8 e^6+e^3 (8-16 x)-8 x+8 x^2} \, dx\) [1809]

Optimal result

Integrand size = 160, antiderivative size = 33 \[ \int \frac {e^{e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )} \left (-25 x^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )+\left (25 e^6 x-25 x^2+25 x^3+e^3 \left (25 x-50 x^2\right )\right ) \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right )}{8 e^6+e^3 (8-16 x)-8 x+8 x^2} \, dx=\frac {1}{16} e^{e^{25 x^2 \log ^2\left (\frac {x}{x+\frac {x}{e^3-x}}\right )}} \] Output:

1/16*exp(exp(25*x^2*ln(x/(x+x/(-x+exp(3))))^2))
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {e^{e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )} \left (-25 x^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )+\left (25 e^6 x-25 x^2+25 x^3+e^3 \left (25 x-50 x^2\right )\right ) \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right )}{8 e^6+e^3 (8-16 x)-8 x+8 x^2} \, dx=\frac {1}{16} e^{e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}} \] Input:

Integrate[(E^(E^(25*x^2*Log[(E^3 - x)/(1 + E^3 - x)]^2) + 25*x^2*Log[(E^3 
- x)/(1 + E^3 - x)]^2)*(-25*x^2*Log[(E^3 - x)/(1 + E^3 - x)] + (25*E^6*x - 
 25*x^2 + 25*x^3 + E^3*(25*x - 50*x^2))*Log[(E^3 - x)/(1 + E^3 - x)]^2))/( 
8*E^6 + E^3*(8 - 16*x) - 8*x + 8*x^2),x]
 

Output:

E^E^(25*x^2*Log[(E^3 - x)/(1 + E^3 - x)]^2)/16
 

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.

\(\displaystyle \int \frac {\left (\left (25 x^3-25 x^2+e^3 \left (25 x-50 x^2\right )+25 e^6 x\right ) \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )-25 x^2 \log \left (\frac {e^3-x}{-x+e^3+1}\right )\right ) \exp \left (25 x^2 \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )+e^{25 x^2 \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )}\right )}{8 x^2-8 x+e^3 (8-16 x)+8 e^6} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {\left (\left (25 x^3-25 x^2+e^3 \left (25 x-50 x^2\right )+25 e^6 x\right ) \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )-25 x^2 \log \left (\frac {e^3-x}{-x+e^3+1}\right )\right ) \exp \left (25 x^2 \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )+e^{25 x^2 \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )}\right )}{8 x^2-8 \left (1+2 e^3\right ) x+8 e^3 \left (1+e^3\right )}dx\)

\(\Big \downarrow \) 7279

\(\displaystyle \int \left (\frac {25}{8} x \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right ) \exp \left (25 x^2 \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )+e^{25 x^2 \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )}\right )-\frac {25 x^2 \log \left (\frac {e^3-x}{-x+e^3+1}\right ) \exp \left (25 x^2 \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )+e^{25 x^2 \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )}\right )}{8 \left (e^3-x\right ) \left (-x+e^3+1\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {25}{8} \int \exp \left (25 x^2 \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )+e^{25 x^2 \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )}\right ) \log \left (\frac {e^3-x}{-x+e^3+1}\right )dx-\frac {25}{8} \int \frac {\exp \left (25 x^2 \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )+e^{25 x^2 \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )}+6\right ) \log \left (\frac {e^3-x}{-x+e^3+1}\right )}{e^3-x}dx+\frac {25}{8} \left (1+e^3\right )^2 \int \frac {\exp \left (25 x^2 \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )+e^{25 x^2 \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )}\right ) \log \left (\frac {e^3-x}{-x+e^3+1}\right )}{-x+e^3+1}dx+\frac {25}{8} \int \exp \left (25 x^2 \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )+e^{25 x^2 \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )}\right ) x \log ^2\left (\frac {e^3-x}{-x+e^3+1}\right )dx\)

Input:

Int[(E^(E^(25*x^2*Log[(E^3 - x)/(1 + E^3 - x)]^2) + 25*x^2*Log[(E^3 - x)/( 
1 + E^3 - x)]^2)*(-25*x^2*Log[(E^3 - x)/(1 + E^3 - x)] + (25*E^6*x - 25*x^ 
2 + 25*x^3 + E^3*(25*x - 50*x^2))*Log[(E^3 - x)/(1 + E^3 - x)]^2))/(8*E^6 
+ E^3*(8 - 16*x) - 8*x + 8*x^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 28.51 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88

method	result	size
risch	\(\frac {{\mathrm e}^{{\mathrm e}^{25 x^{2} \ln \left (\frac {-x +{\mathrm e}^{3}}{{\mathrm e}^{3}-x +1}\right )^{2}}}}{16}\)	\(29\)
parallelrisch	\(\frac {{\mathrm e}^{{\mathrm e}^{25 x^{2} \ln \left (\frac {-x +{\mathrm e}^{3}}{{\mathrm e}^{3}-x +1}\right )^{2}}}}{16}\)	\(29\)

Input:

int(((25*x*exp(3)^2+(-50*x^2+25*x)*exp(3)+25*x^3-25*x^2)*ln((-x+exp(3))/(e 
xp(3)-x+1))^2-25*x^2*ln((-x+exp(3))/(exp(3)-x+1)))*exp(25*x^2*ln((-x+exp(3 
))/(exp(3)-x+1))^2)*exp(exp(25*x^2*ln((-x+exp(3))/(exp(3)-x+1))^2))/(8*exp 
(3)^2+(-16*x+8)*exp(3)+8*x^2-8*x),x,method=_RETURNVERBOSE)
 

Output:

1/16*exp(exp(25*x^2*ln((-x+exp(3))/(exp(3)-x+1))^2))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {e^{e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )} \left (-25 x^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )+\left (25 e^6 x-25 x^2+25 x^3+e^3 \left (25 x-50 x^2\right )\right ) \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right )}{8 e^6+e^3 (8-16 x)-8 x+8 x^2} \, dx=\frac {1}{16} \, e^{\left (e^{\left (25 \, x^{2} \log \left (\frac {x - e^{3}}{x - e^{3} - 1}\right )^{2}\right )}\right )} \] Input:

integrate(((25*x*exp(3)^2+(-50*x^2+25*x)*exp(3)+25*x^3-25*x^2)*log((-x+exp 
(3))/(exp(3)-x+1))^2-25*x^2*log((-x+exp(3))/(exp(3)-x+1)))*exp(25*x^2*log( 
(-x+exp(3))/(exp(3)-x+1))^2)*exp(exp(25*x^2*log((-x+exp(3))/(exp(3)-x+1))^ 
2))/(8*exp(3)^2+(-16*x+8)*exp(3)+8*x^2-8*x),x, algorithm="fricas")
 

Output:

1/16*e^(e^(25*x^2*log((x - e^3)/(x - e^3 - 1))^2))
 

Sympy [A] (verification not implemented)

Time = 3.17 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73 \[ \int \frac {e^{e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )} \left (-25 x^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )+\left (25 e^6 x-25 x^2+25 x^3+e^3 \left (25 x-50 x^2\right )\right ) \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right )}{8 e^6+e^3 (8-16 x)-8 x+8 x^2} \, dx=\frac {e^{e^{25 x^{2} \log {\left (\frac {- x + e^{3}}{- x + 1 + e^{3}} \right )}^{2}}}}{16} \] Input:

integrate(((25*x*exp(3)**2+(-50*x**2+25*x)*exp(3)+25*x**3-25*x**2)*ln((-x+ 
exp(3))/(exp(3)-x+1))**2-25*x**2*ln((-x+exp(3))/(exp(3)-x+1)))*exp(25*x**2 
*ln((-x+exp(3))/(exp(3)-x+1))**2)*exp(exp(25*x**2*ln((-x+exp(3))/(exp(3)-x 
+1))**2))/(8*exp(3)**2+(-16*x+8)*exp(3)+8*x**2-8*x),x)
 

Output:

exp(exp(25*x**2*log((-x + exp(3))/(-x + 1 + exp(3)))**2))/16
 

Maxima [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.64 \[ \int \frac {e^{e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )} \left (-25 x^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )+\left (25 e^6 x-25 x^2+25 x^3+e^3 \left (25 x-50 x^2\right )\right ) \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right )}{8 e^6+e^3 (8-16 x)-8 x+8 x^2} \, dx=\frac {1}{16} \, e^{\left (e^{\left (25 \, x^{2} \log \left (x - e^{3}\right )^{2} - 50 \, x^{2} \log \left (x - e^{3}\right ) \log \left (x - e^{3} - 1\right ) + 25 \, x^{2} \log \left (x - e^{3} - 1\right )^{2}\right )}\right )} \] Input:

integrate(((25*x*exp(3)^2+(-50*x^2+25*x)*exp(3)+25*x^3-25*x^2)*log((-x+exp 
(3))/(exp(3)-x+1))^2-25*x^2*log((-x+exp(3))/(exp(3)-x+1)))*exp(25*x^2*log( 
(-x+exp(3))/(exp(3)-x+1))^2)*exp(exp(25*x^2*log((-x+exp(3))/(exp(3)-x+1))^ 
2))/(8*exp(3)^2+(-16*x+8)*exp(3)+8*x^2-8*x),x, algorithm="maxima")
 

Output:

1/16*e^(e^(25*x^2*log(x - e^3)^2 - 50*x^2*log(x - e^3)*log(x - e^3 - 1) + 
25*x^2*log(x - e^3 - 1)^2))
 

Giac [F]

\[ \int \frac {e^{e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )} \left (-25 x^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )+\left (25 e^6 x-25 x^2+25 x^3+e^3 \left (25 x-50 x^2\right )\right ) \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right )}{8 e^6+e^3 (8-16 x)-8 x+8 x^2} \, dx=\int { -\frac {25 \, {\left (x^{2} \log \left (\frac {x - e^{3}}{x - e^{3} - 1}\right ) - {\left (x^{3} - x^{2} + x e^{6} - {\left (2 \, x^{2} - x\right )} e^{3}\right )} \log \left (\frac {x - e^{3}}{x - e^{3} - 1}\right )^{2}\right )} e^{\left (25 \, x^{2} \log \left (\frac {x - e^{3}}{x - e^{3} - 1}\right )^{2} + e^{\left (25 \, x^{2} \log \left (\frac {x - e^{3}}{x - e^{3} - 1}\right )^{2}\right )}\right )}}{8 \, {\left (x^{2} - {\left (2 \, x - 1\right )} e^{3} - x + e^{6}\right )}} \,d x } \] Input:

integrate(((25*x*exp(3)^2+(-50*x^2+25*x)*exp(3)+25*x^3-25*x^2)*log((-x+exp 
(3))/(exp(3)-x+1))^2-25*x^2*log((-x+exp(3))/(exp(3)-x+1)))*exp(25*x^2*log( 
(-x+exp(3))/(exp(3)-x+1))^2)*exp(exp(25*x^2*log((-x+exp(3))/(exp(3)-x+1))^ 
2))/(8*exp(3)^2+(-16*x+8)*exp(3)+8*x^2-8*x),x, algorithm="giac")
 

Output:

integrate(-25/8*(x^2*log((x - e^3)/(x - e^3 - 1)) - (x^3 - x^2 + x*e^6 - ( 
2*x^2 - x)*e^3)*log((x - e^3)/(x - e^3 - 1))^2)*e^(25*x^2*log((x - e^3)/(x 
 - e^3 - 1))^2 + e^(25*x^2*log((x - e^3)/(x - e^3 - 1))^2))/(x^2 - (2*x - 
1)*e^3 - x + e^6), x)
 

Mupad [B] (verification not implemented)

Time = 3.60 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {e^{e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )} \left (-25 x^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )+\left (25 e^6 x-25 x^2+25 x^3+e^3 \left (25 x-50 x^2\right )\right ) \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right )}{8 e^6+e^3 (8-16 x)-8 x+8 x^2} \, dx=\frac {{\mathrm {e}}^{{\mathrm {e}}^{25\,x^2\,{\ln \left (-\frac {x-{\mathrm {e}}^3}{{\mathrm {e}}^3-x+1}\right )}^2}}}{16} \] Input:

int(-(exp(25*x^2*log(-(x - exp(3))/(exp(3) - x + 1))^2)*exp(exp(25*x^2*log 
(-(x - exp(3))/(exp(3) - x + 1))^2))*(log(-(x - exp(3))/(exp(3) - x + 1))^ 
2*(exp(3)*(25*x - 50*x^2) + 25*x*exp(6) - 25*x^2 + 25*x^3) - 25*x^2*log(-( 
x - exp(3))/(exp(3) - x + 1))))/(8*x - 8*exp(6) - 8*x^2 + exp(3)*(16*x - 8 
)),x)
 

Output:

exp(exp(25*x^2*log(-(x - exp(3))/(exp(3) - x + 1))^2))/16
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {e^{e^{25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )}+25 x^2 \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )} \left (-25 x^2 \log \left (\frac {e^3-x}{1+e^3-x}\right )+\left (25 e^6 x-25 x^2+25 x^3+e^3 \left (25 x-50 x^2\right )\right ) \log ^2\left (\frac {e^3-x}{1+e^3-x}\right )\right )}{8 e^6+e^3 (8-16 x)-8 x+8 x^2} \, dx=\frac {e^{e^{25 \mathrm {log}\left (\frac {e^{3}-x}{e^{3}-x +1}\right )^{2} x^{2}}}}{16} \] Input:

int(((25*x*exp(3)^2+(-50*x^2+25*x)*exp(3)+25*x^3-25*x^2)*log((-x+exp(3))/( 
exp(3)-x+1))^2-25*x^2*log((-x+exp(3))/(exp(3)-x+1)))*exp(25*x^2*log((-x+ex 
p(3))/(exp(3)-x+1))^2)*exp(exp(25*x^2*log((-x+exp(3))/(exp(3)-x+1))^2))/(8 
*exp(3)^2+(-16*x+8)*exp(3)+8*x^2-8*x),x)
 

Output:

e**(e**(25*log((e**3 - x)/(e**3 - x + 1))**2*x**2))/16