\(\int \frac {e^{\frac {-3 x+3 e^{20} x+4 x^2-2 x^3-x^2 \log (5)}{-3+3 e^{20}+3 x}} (3+3 e^{40}-8 x+10 x^2-4 x^3+e^{20} (-6+8 x-6 x^2)+(2 x-2 e^{20} x-x^2) \log (5))}{3+3 e^{40}-6 x+3 x^2+e^{20} (-6+6 x)} \, dx\) [1648]

Optimal result

Integrand size = 121, antiderivative size = 32 \[ \int \frac {e^{\frac {-3 x+3 e^{20} x+4 x^2-2 x^3-x^2 \log (5)}{-3+3 e^{20}+3 x}} \left (3+3 e^{40}-8 x+10 x^2-4 x^3+e^{20} \left (-6+8 x-6 x^2\right )+\left (2 x-2 e^{20} x-x^2\right ) \log (5)\right )}{3+3 e^{40}-6 x+3 x^2+e^{20} (-6+6 x)} \, dx=e^{x-\frac {x^2 (1-2 x-\log (5))}{3 \left (1-e^{20}-x\right )}} \] Output:

exp(x-1/3*x^2*(1-ln(5)-2*x)/(1-x-exp(20)))
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.50 \[ \int \frac {e^{\frac {-3 x+3 e^{20} x+4 x^2-2 x^3-x^2 \log (5)}{-3+3 e^{20}+3 x}} \left (3+3 e^{40}-8 x+10 x^2-4 x^3+e^{20} \left (-6+8 x-6 x^2\right )+\left (2 x-2 e^{20} x-x^2\right ) \log (5)\right )}{3+3 e^{40}-6 x+3 x^2+e^{20} (-6+6 x)} \, dx=5^{-\frac {x^2}{3 \left (-1+e^{20}+x\right )}} e^{-\frac {x \left (3-3 e^{20}-4 x+2 x^2\right )}{3 \left (-1+e^{20}+x\right )}} \] Input:

Integrate[(E^((-3*x + 3*E^20*x + 4*x^2 - 2*x^3 - x^2*Log[5])/(-3 + 3*E^20 
+ 3*x))*(3 + 3*E^40 - 8*x + 10*x^2 - 4*x^3 + E^20*(-6 + 8*x - 6*x^2) + (2* 
x - 2*E^20*x - x^2)*Log[5]))/(3 + 3*E^40 - 6*x + 3*x^2 + E^20*(-6 + 6*x)), 
x]
 

Output:

1/(5^(x^2/(3*(-1 + E^20 + x)))*E^((x*(3 - 3*E^20 - 4*x + 2*x^2))/(3*(-1 + 
E^20 + x))))
 

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.

\(\displaystyle \int \frac {\left (-4 x^3+10 x^2+e^{20} \left (-6 x^2+8 x-6\right )+\left (-x^2-2 e^{20} x+2 x\right ) \log (5)-8 x+3 e^{40}+3\right ) \exp \left (\frac {-2 x^3+4 x^2-x^2 \log (5)+3 e^{20} x-3 x}{3 x+3 e^{20}-3}\right )}{3 x^2-6 x+e^{20} (6 x-6)+3 e^{40}+3} \, dx\)

\(\Big \downarrow \) 2007

\(\displaystyle \int \frac {\left (-4 x^3+10 x^2+e^{20} \left (-6 x^2+8 x-6\right )+\left (-x^2-2 e^{20} x+2 x\right ) \log (5)-8 x+3 e^{40}+3\right ) \exp \left (\frac {-2 x^3+4 x^2-x^2 \log (5)+3 e^{20} x-3 x}{3 x+3 e^{20}-3}\right )}{\left (\sqrt {3} x+\sqrt {3} \left (e^{20}-1\right )\right )^2}dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {\left (-4 x^3+x^2 \left (10-6 e^{20}-\log (5)\right )-2 \left (1-e^{20}\right ) x (4-\log (5))+3 \left (1-e^{20}\right )^2\right ) \exp \left (\frac {x \left (-2 x^2+x (4-\log (5))-3 \left (1-e^{20}\right )\right )}{3 x+3 e^{20}-3}\right )}{\left (\sqrt {3} x+\sqrt {3} \left (e^{20}-1\right )\right )^2}dx\)

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

Int[(E^((-3*x + 3*E^20*x + 4*x^2 - 2*x^3 - x^2*Log[5])/(-3 + 3*E^20 + 3*x) 
)*(3 + 3*E^40 - 8*x + 10*x^2 - 4*x^3 + E^20*(-6 + 8*x - 6*x^2) + (2*x - 2* 
E^20*x - x^2)*Log[5]))/(3 + 3*E^40 - 6*x + 3*x^2 + E^20*(-6 + 6*x)),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.85 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94

method	result	size
gosper	\({\mathrm e}^{-\frac {x \left (x \ln \left (5\right )+2 x^{2}-3 \,{\mathrm e}^{20}-4 x +3\right )}{3 \left ({\mathrm e}^{20}+x -1\right )}}\)	\(30\)
parallelrisch	\({\mathrm e}^{-\frac {x \left (x \ln \left (5\right )+2 x^{2}-3 \,{\mathrm e}^{20}-4 x +3\right )}{3 \left ({\mathrm e}^{20}+x -1\right )}}\)	\(30\)
risch	\({\mathrm e}^{\frac {\left (-x \ln \left (5\right )-2 x^{2}+3 \,{\mathrm e}^{20}+4 x -3\right ) x}{3 \,{\mathrm e}^{20}+3 x -3}}\)	\(31\)
norman	\(\frac {x \,{\mathrm e}^{\frac {-x^{2} \ln \left (5\right )+3 x \,{\mathrm e}^{20}-2 x^{3}+4 x^{2}-3 x}{3 \,{\mathrm e}^{20}+3 x -3}}+\left ({\mathrm e}^{20}-1\right ) {\mathrm e}^{\frac {-x^{2} \ln \left (5\right )+3 x \,{\mathrm e}^{20}-2 x^{3}+4 x^{2}-3 x}{3 \,{\mathrm e}^{20}+3 x -3}}}{{\mathrm e}^{20}+x -1}\)	\(95\)

Input:

int(((-2*x*exp(20)-x^2+2*x)*ln(5)+3*exp(20)^2+(-6*x^2+8*x-6)*exp(20)-4*x^3 
+10*x^2-8*x+3)*exp((-x^2*ln(5)+3*x*exp(20)-2*x^3+4*x^2-3*x)/(3*exp(20)+3*x 
-3))/(3*exp(20)^2+(6*x-6)*exp(20)+3*x^2-6*x+3),x,method=_RETURNVERBOSE)
 

Output:

exp(-1/3*x*(x*ln(5)+2*x^2-3*exp(20)-4*x+3)/(exp(20)+x-1))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09 \[ \int \frac {e^{\frac {-3 x+3 e^{20} x+4 x^2-2 x^3-x^2 \log (5)}{-3+3 e^{20}+3 x}} \left (3+3 e^{40}-8 x+10 x^2-4 x^3+e^{20} \left (-6+8 x-6 x^2\right )+\left (2 x-2 e^{20} x-x^2\right ) \log (5)\right )}{3+3 e^{40}-6 x+3 x^2+e^{20} (-6+6 x)} \, dx=e^{\left (-\frac {2 \, x^{3} + x^{2} \log \left (5\right ) - 4 \, x^{2} - 3 \, x e^{20} + 3 \, x}{3 \, {\left (x + e^{20} - 1\right )}}\right )} \] Input:

integrate(((-2*x*exp(20)-x^2+2*x)*log(5)+3*exp(20)^2+(-6*x^2+8*x-6)*exp(20 
)-4*x^3+10*x^2-8*x+3)*exp((-x^2*log(5)+3*x*exp(20)-2*x^3+4*x^2-3*x)/(3*exp 
(20)+3*x-3))/(3*exp(20)^2+(6*x-6)*exp(20)+3*x^2-6*x+3),x, algorithm="frica 
s")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16 \[ \int \frac {e^{\frac {-3 x+3 e^{20} x+4 x^2-2 x^3-x^2 \log (5)}{-3+3 e^{20}+3 x}} \left (3+3 e^{40}-8 x+10 x^2-4 x^3+e^{20} \left (-6+8 x-6 x^2\right )+\left (2 x-2 e^{20} x-x^2\right ) \log (5)\right )}{3+3 e^{40}-6 x+3 x^2+e^{20} (-6+6 x)} \, dx=e^{\frac {- 2 x^{3} - x^{2} \log {\left (5 \right )} + 4 x^{2} - 3 x + 3 x e^{20}}{3 x - 3 + 3 e^{20}}} \] Input:

integrate(((-2*x*exp(20)-x**2+2*x)*ln(5)+3*exp(20)**2+(-6*x**2+8*x-6)*exp( 
20)-4*x**3+10*x**2-8*x+3)*exp((-x**2*ln(5)+3*x*exp(20)-2*x**3+4*x**2-3*x)/ 
(3*exp(20)+3*x-3))/(3*exp(20)**2+(6*x-6)*exp(20)+3*x**2-6*x+3),x)
 

Output:

exp((-2*x**3 - x**2*log(5) + 4*x**2 - 3*x + 3*x*exp(20))/(3*x - 3 + 3*exp( 
20)))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 115 vs. \(2 (22) = 44\).

Time = 0.54 (sec) , antiderivative size = 115, normalized size of antiderivative = 3.59 \[ \int \frac {e^{\frac {-3 x+3 e^{20} x+4 x^2-2 x^3-x^2 \log (5)}{-3+3 e^{20}+3 x}} \left (3+3 e^{40}-8 x+10 x^2-4 x^3+e^{20} \left (-6+8 x-6 x^2\right )+\left (2 x-2 e^{20} x-x^2\right ) \log (5)\right )}{3+3 e^{40}-6 x+3 x^2+e^{20} (-6+6 x)} \, dx=5^{\frac {1}{3} \, e^{20} - \frac {1}{3}} e^{\left (-\frac {2}{3} \, x^{2} + \frac {2}{3} \, x e^{20} - \frac {1}{3} \, x \log \left (5\right ) + \frac {2}{3} \, x - \frac {e^{40} \log \left (5\right )}{3 \, {\left (x + e^{20} - 1\right )}} + \frac {2 \, e^{20} \log \left (5\right )}{3 \, {\left (x + e^{20} - 1\right )}} + \frac {2 \, e^{60}}{3 \, {\left (x + e^{20} - 1\right )}} - \frac {5 \, e^{40}}{3 \, {\left (x + e^{20} - 1\right )}} + \frac {4 \, e^{20}}{3 \, {\left (x + e^{20} - 1\right )}} - \frac {\log \left (5\right )}{3 \, {\left (x + e^{20} - 1\right )}} - \frac {1}{3 \, {\left (x + e^{20} - 1\right )}} - \frac {2}{3} \, e^{40} + e^{20} - \frac {1}{3}\right )} \] Input:

integrate(((-2*x*exp(20)-x^2+2*x)*log(5)+3*exp(20)^2+(-6*x^2+8*x-6)*exp(20 
)-4*x^3+10*x^2-8*x+3)*exp((-x^2*log(5)+3*x*exp(20)-2*x^3+4*x^2-3*x)/(3*exp 
(20)+3*x-3))/(3*exp(20)^2+(6*x-6)*exp(20)+3*x^2-6*x+3),x, algorithm="maxim 
a")
 

Output:

5^(1/3*e^20 - 1/3)*e^(-2/3*x^2 + 2/3*x*e^20 - 1/3*x*log(5) + 2/3*x - 1/3*e 
^40*log(5)/(x + e^20 - 1) + 2/3*e^20*log(5)/(x + e^20 - 1) + 2/3*e^60/(x + 
 e^20 - 1) - 5/3*e^40/(x + e^20 - 1) + 4/3*e^20/(x + e^20 - 1) - 1/3*log(5 
)/(x + e^20 - 1) - 1/3/(x + e^20 - 1) - 2/3*e^40 + e^20 - 1/3)
 

Giac [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09 \[ \int \frac {e^{\frac {-3 x+3 e^{20} x+4 x^2-2 x^3-x^2 \log (5)}{-3+3 e^{20}+3 x}} \left (3+3 e^{40}-8 x+10 x^2-4 x^3+e^{20} \left (-6+8 x-6 x^2\right )+\left (2 x-2 e^{20} x-x^2\right ) \log (5)\right )}{3+3 e^{40}-6 x+3 x^2+e^{20} (-6+6 x)} \, dx=e^{\left (-\frac {2 \, x^{3} + x^{2} \log \left (5\right ) - 4 \, x^{2} - 3 \, x e^{20} + 3 \, x}{3 \, {\left (x + e^{20} - 1\right )}}\right )} \] Input:

integrate(((-2*x*exp(20)-x^2+2*x)*log(5)+3*exp(20)^2+(-6*x^2+8*x-6)*exp(20 
)-4*x^3+10*x^2-8*x+3)*exp((-x^2*log(5)+3*x*exp(20)-2*x^3+4*x^2-3*x)/(3*exp 
(20)+3*x-3))/(3*exp(20)^2+(6*x-6)*exp(20)+3*x^2-6*x+3),x, algorithm="giac" 
)
 

Output:

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

Mupad [B] (verification not implemented)

Time = 3.50 (sec) , antiderivative size = 86, normalized size of antiderivative = 2.69 \[ \int \frac {e^{\frac {-3 x+3 e^{20} x+4 x^2-2 x^3-x^2 \log (5)}{-3+3 e^{20}+3 x}} \left (3+3 e^{40}-8 x+10 x^2-4 x^3+e^{20} \left (-6+8 x-6 x^2\right )+\left (2 x-2 e^{20} x-x^2\right ) \log (5)\right )}{3+3 e^{40}-6 x+3 x^2+e^{20} (-6+6 x)} \, dx=\frac {{\mathrm {e}}^{-\frac {3\,x}{3\,x+3\,{\mathrm {e}}^{20}-3}}\,{\mathrm {e}}^{-\frac {2\,x^3}{3\,x+3\,{\mathrm {e}}^{20}-3}}\,{\mathrm {e}}^{\frac {4\,x^2}{3\,x+3\,{\mathrm {e}}^{20}-3}}\,{\mathrm {e}}^{\frac {3\,x\,{\mathrm {e}}^{20}}{3\,x+3\,{\mathrm {e}}^{20}-3}}}{5^{\frac {x^2}{3\,x+3\,{\mathrm {e}}^{20}-3}}} \] Input:

int(-(exp(-(3*x - 3*x*exp(20) + x^2*log(5) - 4*x^2 + 2*x^3)/(3*x + 3*exp(2 
0) - 3))*(8*x - 3*exp(40) + log(5)*(2*x*exp(20) - 2*x + x^2) + exp(20)*(6* 
x^2 - 8*x + 6) - 10*x^2 + 4*x^3 - 3))/(3*exp(40) - 6*x + 3*x^2 + exp(20)*( 
6*x - 6) + 3),x)
 

Output:

(exp(-(3*x)/(3*x + 3*exp(20) - 3))*exp(-(2*x^3)/(3*x + 3*exp(20) - 3))*exp 
((4*x^2)/(3*x + 3*exp(20) - 3))*exp((3*x*exp(20))/(3*x + 3*exp(20) - 3)))/ 
5^(x^2/(3*x + 3*exp(20) - 3))
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.88 \[ \int \frac {e^{\frac {-3 x+3 e^{20} x+4 x^2-2 x^3-x^2 \log (5)}{-3+3 e^{20}+3 x}} \left (3+3 e^{40}-8 x+10 x^2-4 x^3+e^{20} \left (-6+8 x-6 x^2\right )+\left (2 x-2 e^{20} x-x^2\right ) \log (5)\right )}{3+3 e^{40}-6 x+3 x^2+e^{20} (-6+6 x)} \, dx=\frac {e^{\frac {3 e^{20} x +4 x^{2}-3 x}{3 e^{20}+3 x -3}}}{e^{\frac {\mathrm {log}\left (5\right ) x^{2}+2 x^{3}}{3 e^{20}+3 x -3}}} \] Input:

int(((-2*x*exp(20)-x^2+2*x)*log(5)+3*exp(20)^2+(-6*x^2+8*x-6)*exp(20)-4*x^ 
3+10*x^2-8*x+3)*exp((-x^2*log(5)+3*x*exp(20)-2*x^3+4*x^2-3*x)/(3*exp(20)+3 
*x-3))/(3*exp(20)^2+(6*x-6)*exp(20)+3*x^2-6*x+3),x)
 

Output:

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