\(\int \frac {e^{\frac {e^{e^3} (25-5 x)}{-x-5 e^{e^3} x+5 \log ^2(x)}} (25 e^{e^3} x+125 e^{2 e^3} x+e^{e^3} (-250+50 x) \log (x)-25 e^{e^3} x \log ^2(x))}{x^3+10 e^{e^3} x^3+25 e^{2 e^3} x^3+(-10 x^2-50 e^{e^3} x^2) \log ^2(x)+25 x \log ^4(x)} \, dx\) [168]

Optimal result

Integrand size = 134, antiderivative size = 30 \[ \int \frac {e^{\frac {e^{e^3} (25-5 x)}{-x-5 e^{e^3} x+5 \log ^2(x)}} \left (25 e^{e^3} x+125 e^{2 e^3} x+e^{e^3} (-250+50 x) \log (x)-25 e^{e^3} x \log ^2(x)\right )}{x^3+10 e^{e^3} x^3+25 e^{2 e^3} x^3+\left (-10 x^2-50 e^{e^3} x^2\right ) \log ^2(x)+25 x \log ^4(x)} \, dx=e^{\frac {-5+x}{x+e^{-e^3} \left (\frac {x}{5}-\log ^2(x)\right )}} \] Output:

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {e^{\frac {e^{e^3} (25-5 x)}{-x-5 e^{e^3} x+5 \log ^2(x)}} \left (25 e^{e^3} x+125 e^{2 e^3} x+e^{e^3} (-250+50 x) \log (x)-25 e^{e^3} x \log ^2(x)\right )}{x^3+10 e^{e^3} x^3+25 e^{2 e^3} x^3+\left (-10 x^2-50 e^{e^3} x^2\right ) \log ^2(x)+25 x \log ^4(x)} \, dx=e^{-\frac {5 e^{e^3} (-5+x)}{-x-5 e^{e^3} x+5 \log ^2(x)}} \] Input:

Integrate[(E^((E^E^3*(25 - 5*x))/(-x - 5*E^E^3*x + 5*Log[x]^2))*(25*E^E^3* 
x + 125*E^(2*E^3)*x + E^E^3*(-250 + 50*x)*Log[x] - 25*E^E^3*x*Log[x]^2))/( 
x^3 + 10*E^E^3*x^3 + 25*E^(2*E^3)*x^3 + (-10*x^2 - 50*E^E^3*x^2)*Log[x]^2 
+ 25*x*Log[x]^4),x]
 

Output:

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

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.

\(\displaystyle \int \frac {\left (125 e^{2 e^3} x+25 e^{e^3} x-25 e^{e^3} x \log ^2(x)+e^{e^3} (50 x-250) \log (x)\right ) \exp \left (\frac {e^{e^3} (25-5 x)}{-5 e^{e^3} x-x+5 \log ^2(x)}\right )}{25 e^{2 e^3} x^3+10 e^{e^3} x^3+x^3+\left (-50 e^{e^3} x^2-10 x^2\right ) \log ^2(x)+25 x \log ^4(x)} \, dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {\left (\left (25 e^{e^3}+125 e^{2 e^3}\right ) x-25 e^{e^3} x \log ^2(x)+e^{e^3} (50 x-250) \log (x)\right ) \exp \left (\frac {e^{e^3} (25-5 x)}{-5 e^{e^3} x-x+5 \log ^2(x)}\right )}{25 e^{2 e^3} x^3+10 e^{e^3} x^3+x^3+\left (-50 e^{e^3} x^2-10 x^2\right ) \log ^2(x)+25 x \log ^4(x)}dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {\left (\left (25 e^{e^3}+125 e^{2 e^3}\right ) x-25 e^{e^3} x \log ^2(x)+e^{e^3} (50 x-250) \log (x)\right ) \exp \left (\frac {e^{e^3} (25-5 x)}{-5 e^{e^3} x-x+5 \log ^2(x)}\right )}{\left (1+10 e^{e^3}\right ) x^3+25 e^{2 e^3} x^3+\left (-50 e^{e^3} x^2-10 x^2\right ) \log ^2(x)+25 x \log ^4(x)}dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {\left (\left (25 e^{e^3}+125 e^{2 e^3}\right ) x-25 e^{e^3} x \log ^2(x)+e^{e^3} (50 x-250) \log (x)\right ) \exp \left (\frac {e^{e^3} (25-5 x)}{-5 e^{e^3} x-x+5 \log ^2(x)}\right )}{\left (1+10 e^{e^3}+25 e^{2 e^3}\right ) x^3+\left (-50 e^{e^3} x^2-10 x^2\right ) \log ^2(x)+25 x \log ^4(x)}dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {25 \left (\left (1+5 e^{e^3}\right ) x-x \log ^2(x)+2 x \log (x)-10 \log (x)\right ) \exp \left (\frac {e^{e^3} (25-5 x)}{5 \log ^2(x)-\left (1+5 e^{e^3}\right ) x}+e^3\right )}{x \left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 25 \int \frac {\exp \left (e^3-\frac {5 e^{e^3} (5-x)}{\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)}\right ) \left (-x \log ^2(x)+2 x \log (x)-10 \log (x)+\left (1+5 e^{e^3}\right ) x\right )}{x \left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle 25 \int \left (\frac {\exp \left (e^3-\frac {5 e^{e^3} (5-x)}{\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)}\right ) (5-x) \left (\left (1+5 e^{e^3}\right ) x-10 \log (x)\right )}{5 x \left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2}+\frac {\exp \left (e^3-\frac {5 e^{e^3} (5-x)}{\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)}\right )}{5 \left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 25 \left (\left (1+5 e^{e^3}\right ) \int \frac {\exp \left (e^3-\frac {5 e^{e^3} (5-x)}{\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)}\right )}{\left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2}dx-\frac {1}{5} \left (1+5 e^{e^3}\right ) \int \frac {\exp \left (e^3-\frac {5 e^{e^3} (5-x)}{\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)}\right ) x}{\left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2}dx+2 \int \frac {\exp \left (e^3-\frac {5 e^{e^3} (5-x)}{\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)}\right ) \log (x)}{\left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2}dx-10 \int \frac {\exp \left (e^3-\frac {5 e^{e^3} (5-x)}{\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)}\right ) \log (x)}{x \left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2}dx+\frac {1}{5} \int \frac {\exp \left (e^3-\frac {5 e^{e^3} (5-x)}{\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)}\right )}{\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)}dx\right )\)

Input:

Int[(E^((E^E^3*(25 - 5*x))/(-x - 5*E^E^3*x + 5*Log[x]^2))*(25*E^E^3*x + 12 
5*E^(2*E^3)*x + E^E^3*(-250 + 50*x)*Log[x] - 25*E^E^3*x*Log[x]^2))/(x^3 + 
10*E^E^3*x^3 + 25*E^(2*E^3)*x^3 + (-10*x^2 - 50*E^E^3*x^2)*Log[x]^2 + 25*x 
*Log[x]^4),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 3.47 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87

method	result	size
parallelrisch	\({\mathrm e}^{\frac {5 \left (-5+x \right ) {\mathrm e}^{{\mathrm e}^{3}}}{5 x \,{\mathrm e}^{{\mathrm e}^{3}}-5 \ln \left (x \right )^{2}+x}}\)	\(26\)
risch	\({\mathrm e}^{-\frac {5 \left (-5+x \right ) {\mathrm e}^{{\mathrm e}^{3}}}{5 \ln \left (x \right )^{2}-5 x \,{\mathrm e}^{{\mathrm e}^{3}}-x}}\)	\(28\)

Input:

int((-25*x*exp(exp(3))*ln(x)^2+(50*x-250)*exp(exp(3))*ln(x)+125*x*exp(exp( 
3))^2+25*x*exp(exp(3)))*exp((-5*x+25)*exp(exp(3))/(5*ln(x)^2-5*x*exp(exp(3 
))-x))/(25*x*ln(x)^4+(-50*x^2*exp(exp(3))-10*x^2)*ln(x)^2+25*x^3*exp(exp(3 
))^2+10*x^3*exp(exp(3))+x^3),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {e^{\frac {e^{e^3} (25-5 x)}{-x-5 e^{e^3} x+5 \log ^2(x)}} \left (25 e^{e^3} x+125 e^{2 e^3} x+e^{e^3} (-250+50 x) \log (x)-25 e^{e^3} x \log ^2(x)\right )}{x^3+10 e^{e^3} x^3+25 e^{2 e^3} x^3+\left (-10 x^2-50 e^{e^3} x^2\right ) \log ^2(x)+25 x \log ^4(x)} \, dx=e^{\left (\frac {5 \, {\left (x - 5\right )} e^{\left (e^{3}\right )}}{5 \, x e^{\left (e^{3}\right )} - 5 \, \log \left (x\right )^{2} + x}\right )} \] Input:

integrate((-25*x*exp(exp(3))*log(x)^2+(50*x-250)*exp(exp(3))*log(x)+125*x* 
exp(exp(3))^2+25*x*exp(exp(3)))*exp((-5*x+25)*exp(exp(3))/(5*log(x)^2-5*x* 
exp(exp(3))-x))/(25*x*log(x)^4+(-50*x^2*exp(exp(3))-10*x^2)*log(x)^2+25*x^ 
3*exp(exp(3))^2+10*x^3*exp(exp(3))+x^3),x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\frac {e^{e^3} (25-5 x)}{-x-5 e^{e^3} x+5 \log ^2(x)}} \left (25 e^{e^3} x+125 e^{2 e^3} x+e^{e^3} (-250+50 x) \log (x)-25 e^{e^3} x \log ^2(x)\right )}{x^3+10 e^{e^3} x^3+25 e^{2 e^3} x^3+\left (-10 x^2-50 e^{e^3} x^2\right ) \log ^2(x)+25 x \log ^4(x)} \, dx=e^{\frac {\left (25 - 5 x\right ) e^{e^{3}}}{- 5 x e^{e^{3}} - x + 5 \log {\left (x \right )}^{2}}} \] Input:

integrate((-25*x*exp(exp(3))*ln(x)**2+(50*x-250)*exp(exp(3))*ln(x)+125*x*e 
xp(exp(3))**2+25*x*exp(exp(3)))*exp((-5*x+25)*exp(exp(3))/(5*ln(x)**2-5*x* 
exp(exp(3))-x))/(25*x*ln(x)**4+(-50*x**2*exp(exp(3))-10*x**2)*ln(x)**2+25* 
x**3*exp(exp(3))**2+10*x**3*exp(exp(3))+x**3),x)
 

Output:

exp((25 - 5*x)*exp(exp(3))/(-5*x*exp(exp(3)) - x + 5*log(x)**2))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 81 vs. \(2 (28) = 56\).

Time = 0.26 (sec) , antiderivative size = 81, normalized size of antiderivative = 2.70 \[ \int \frac {e^{\frac {e^{e^3} (25-5 x)}{-x-5 e^{e^3} x+5 \log ^2(x)}} \left (25 e^{e^3} x+125 e^{2 e^3} x+e^{e^3} (-250+50 x) \log (x)-25 e^{e^3} x \log ^2(x)\right )}{x^3+10 e^{e^3} x^3+25 e^{2 e^3} x^3+\left (-10 x^2-50 e^{e^3} x^2\right ) \log ^2(x)+25 x \log ^4(x)} \, dx=e^{\left (-\frac {25 \, e^{\left (e^{3}\right )} \log \left (x\right )^{2}}{5 \, {\left (5 \, e^{\left (e^{3}\right )} + 1\right )} \log \left (x\right )^{2} - x {\left (25 \, e^{\left (2 \, e^{3}\right )} + 10 \, e^{\left (e^{3}\right )} + 1\right )}} - \frac {25 \, e^{\left (e^{3}\right )}}{x {\left (5 \, e^{\left (e^{3}\right )} + 1\right )} - 5 \, \log \left (x\right )^{2}} + \frac {5 \, e^{\left (e^{3}\right )}}{5 \, e^{\left (e^{3}\right )} + 1}\right )} \] Input:

integrate((-25*x*exp(exp(3))*log(x)^2+(50*x-250)*exp(exp(3))*log(x)+125*x* 
exp(exp(3))^2+25*x*exp(exp(3)))*exp((-5*x+25)*exp(exp(3))/(5*log(x)^2-5*x* 
exp(exp(3))-x))/(25*x*log(x)^4+(-50*x^2*exp(exp(3))-10*x^2)*log(x)^2+25*x^ 
3*exp(exp(3))^2+10*x^3*exp(exp(3))+x^3),x, algorithm="maxima")
 

Output:

e^(-25*e^(e^3)*log(x)^2/(5*(5*e^(e^3) + 1)*log(x)^2 - x*(25*e^(2*e^3) + 10 
*e^(e^3) + 1)) - 25*e^(e^3)/(x*(5*e^(e^3) + 1) - 5*log(x)^2) + 5*e^(e^3)/( 
5*e^(e^3) + 1))
 

Giac [F(-1)]

Timed out. \[ \int \frac {e^{\frac {e^{e^3} (25-5 x)}{-x-5 e^{e^3} x+5 \log ^2(x)}} \left (25 e^{e^3} x+125 e^{2 e^3} x+e^{e^3} (-250+50 x) \log (x)-25 e^{e^3} x \log ^2(x)\right )}{x^3+10 e^{e^3} x^3+25 e^{2 e^3} x^3+\left (-10 x^2-50 e^{e^3} x^2\right ) \log ^2(x)+25 x \log ^4(x)} \, dx=\text {Timed out} \] Input:

integrate((-25*x*exp(exp(3))*log(x)^2+(50*x-250)*exp(exp(3))*log(x)+125*x* 
exp(exp(3))^2+25*x*exp(exp(3)))*exp((-5*x+25)*exp(exp(3))/(5*log(x)^2-5*x* 
exp(exp(3))-x))/(25*x*log(x)^4+(-50*x^2*exp(exp(3))-10*x^2)*log(x)^2+25*x^ 
3*exp(exp(3))^2+10*x^3*exp(exp(3))+x^3),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 3.12 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {e^{\frac {e^{e^3} (25-5 x)}{-x-5 e^{e^3} x+5 \log ^2(x)}} \left (25 e^{e^3} x+125 e^{2 e^3} x+e^{e^3} (-250+50 x) \log (x)-25 e^{e^3} x \log ^2(x)\right )}{x^3+10 e^{e^3} x^3+25 e^{2 e^3} x^3+\left (-10 x^2-50 e^{e^3} x^2\right ) \log ^2(x)+25 x \log ^4(x)} \, dx={\mathrm {e}}^{-\frac {25\,{\mathrm {e}}^{{\mathrm {e}}^3}-5\,x\,{\mathrm {e}}^{{\mathrm {e}}^3}}{-5\,{\ln \left (x\right )}^2+x+5\,x\,{\mathrm {e}}^{{\mathrm {e}}^3}}} \] Input:

int((exp((exp(exp(3))*(5*x - 25))/(x - 5*log(x)^2 + 5*x*exp(exp(3))))*(125 
*x*exp(2*exp(3)) + 25*x*exp(exp(3)) + exp(exp(3))*log(x)*(50*x - 250) - 25 
*x*exp(exp(3))*log(x)^2))/(10*x^3*exp(exp(3)) + 25*x*log(x)^4 - log(x)^2*( 
50*x^2*exp(exp(3)) + 10*x^2) + 25*x^3*exp(2*exp(3)) + x^3),x)
 

Output:

exp(-(25*exp(exp(3)) - 5*x*exp(exp(3)))/(x - 5*log(x)^2 + 5*x*exp(exp(3))) 
)
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.00 \[ \int \frac {e^{\frac {e^{e^3} (25-5 x)}{-x-5 e^{e^3} x+5 \log ^2(x)}} \left (25 e^{e^3} x+125 e^{2 e^3} x+e^{e^3} (-250+50 x) \log (x)-25 e^{e^3} x \log ^2(x)\right )}{x^3+10 e^{e^3} x^3+25 e^{2 e^3} x^3+\left (-10 x^2-50 e^{e^3} x^2\right ) \log ^2(x)+25 x \log ^4(x)} \, dx=\frac {e^{\frac {5 \mathrm {log}\left (x \right )^{2}}{5 e^{e^{3}} x -5 \mathrm {log}\left (x \right )^{2}+x}} e}{e^{\frac {25 e^{e^{3}}+x}{5 e^{e^{3}} x -5 \mathrm {log}\left (x \right )^{2}+x}}} \] Input:

int((-25*x*exp(exp(3))*log(x)^2+(50*x-250)*exp(exp(3))*log(x)+125*x*exp(ex 
p(3))^2+25*x*exp(exp(3)))*exp((-5*x+25)*exp(exp(3))/(5*log(x)^2-5*x*exp(ex 
p(3))-x))/(25*x*log(x)^4+(-50*x^2*exp(exp(3))-10*x^2)*log(x)^2+25*x^3*exp( 
exp(3))^2+10*x^3*exp(exp(3))+x^3),x)
 

Output:

(e**((5*log(x)**2)/(5*e**(e**3)*x - 5*log(x)**2 + x))*e)/e**((25*e**(e**3) 
 + x)/(5*e**(e**3)*x - 5*log(x)**2 + x))