\(\int \frac {e^{\frac {-3 x^3+(x-4 x^2) \log (3)}{3 x+4 \log (3)}} (90 x^2-180 x^4+(240 x-480 x^3) \log (3)+(160+40 x-320 x^2) \log ^2(3))}{9 e^2 x^2+24 e^2 x \log (3)+16 e^2 \log ^2(3)} \, dx\) [2536]

Optimal result

Integrand size = 96, antiderivative size = 25 \[ \int \frac {e^{\frac {-3 x^3+\left (x-4 x^2\right ) \log (3)}{3 x+4 \log (3)}} \left (90 x^2-180 x^4+\left (240 x-480 x^3\right ) \log (3)+\left (160+40 x-320 x^2\right ) \log ^2(3)\right )}{9 e^2 x^2+24 e^2 x \log (3)+16 e^2 \log ^2(3)} \, dx=10 e^{-2-x^2+\frac {x}{4+\frac {3 x}{\log (3)}}} x \] Output:

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

Mathematica [F]

\[ \int \frac {e^{\frac {-3 x^3+\left (x-4 x^2\right ) \log (3)}{3 x+4 \log (3)}} \left (90 x^2-180 x^4+\left (240 x-480 x^3\right ) \log (3)+\left (160+40 x-320 x^2\right ) \log ^2(3)\right )}{9 e^2 x^2+24 e^2 x \log (3)+16 e^2 \log ^2(3)} \, dx=\int \frac {e^{\frac {-3 x^3+\left (x-4 x^2\right ) \log (3)}{3 x+4 \log (3)}} \left (90 x^2-180 x^4+\left (240 x-480 x^3\right ) \log (3)+\left (160+40 x-320 x^2\right ) \log ^2(3)\right )}{9 e^2 x^2+24 e^2 x \log (3)+16 e^2 \log ^2(3)} \, dx \] Input:

Integrate[(E^((-3*x^3 + (x - 4*x^2)*Log[3])/(3*x + 4*Log[3]))*(90*x^2 - 18 
0*x^4 + (240*x - 480*x^3)*Log[3] + (160 + 40*x - 320*x^2)*Log[3]^2))/(9*E^ 
2*x^2 + 24*E^2*x*Log[3] + 16*E^2*Log[3]^2),x]
 

Output:

Integrate[(E^((-3*x^3 + (x - 4*x^2)*Log[3])/(3*x + 4*Log[3]))*(90*x^2 - 18 
0*x^4 + (240*x - 480*x^3)*Log[3] + (160 + 40*x - 320*x^2)*Log[3]^2))/(9*E^ 
2*x^2 + 24*E^2*x*Log[3] + 16*E^2*Log[3]^2), 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.

\(\displaystyle \int \frac {\left (-180 x^4+\left (240 x-480 x^3\right ) \log (3)+90 x^2+\left (-320 x^2+40 x+160\right ) \log ^2(3)\right ) \exp \left (\frac {\left (x-4 x^2\right ) \log (3)-3 x^3}{3 x+4 \log (3)}\right )}{9 e^2 x^2+24 e^2 x \log (3)+16 e^2 \log ^2(3)} \, dx\)

\(\Big \downarrow \) 2007

\(\displaystyle \int \frac {\left (-180 x^4+\left (240 x-480 x^3\right ) \log (3)+90 x^2+\left (-320 x^2+40 x+160\right ) \log ^2(3)\right ) \exp \left (\frac {\left (x-4 x^2\right ) \log (3)-3 x^3}{3 x+4 \log (3)}\right )}{(3 e x+4 e \log (3))^2}dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {\left (-180 x^4-480 x^3 \log (3)+10 x^2 \left (9-32 \log ^2(3)\right )+40 x \log (3) (6+\log (3))+160 \log ^2(3)\right ) \exp \left (\frac {x \left (-3 x^2-4 x \log (3)+\log (3)\right )}{3 x+4 \log (3)}\right )}{(3 e x+4 e \log (3))^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (-\frac {160 \log ^3(3) \exp \left (\frac {x \left (-3 x^2-4 x \log (3)+\log (3)\right )}{3 x+4 \log (3)}-2\right )}{3 (3 x+\log (81))^2}+\frac {40 \log ^2(3) \exp \left (\frac {x \left (-3 x^2-4 x \log (3)+\log (3)\right )}{3 x+4 \log (3)}-2\right )}{3 (3 x+\log (81))}-20 x^2 \exp \left (\frac {x \left (-3 x^2-4 x \log (3)+\log (3)\right )}{3 x+4 \log (3)}-2\right )+10 \exp \left (\frac {x \left (-3 x^2-4 x \log (3)+\log (3)\right )}{3 x+4 \log (3)}-2\right )\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {160}{3} \log ^3(3) \int \frac {\exp \left (\frac {x \left (-3 x^2-4 \log (3) x+\log (3)\right )}{3 x+4 \log (3)}-2\right )}{(3 x+\log (81))^2}dx+\frac {40}{3} \log ^2(3) \int \frac {\exp \left (\frac {x \left (-3 x^2-4 \log (3) x+\log (3)\right )}{3 x+4 \log (3)}-2\right )}{3 x+\log (81)}dx+10 \int \exp \left (\frac {x \left (-3 x^2-4 \log (3) x+\log (3)\right )}{3 x+4 \log (3)}-2\right )dx-20 \int \exp \left (\frac {x \left (-3 x^2-4 \log (3) x+\log (3)\right )}{3 x+4 \log (3)}-2\right ) x^2dx\)

Input:

Int[(E^((-3*x^3 + (x - 4*x^2)*Log[3])/(3*x + 4*Log[3]))*(90*x^2 - 180*x^4 
+ (240*x - 480*x^3)*Log[3] + (160 + 40*x - 320*x^2)*Log[3]^2))/(9*E^2*x^2 
+ 24*E^2*x*Log[3] + 16*E^2*Log[3]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44

method	result	size
parallelrisch	\(10 \,{\mathrm e}^{-2} x \,{\mathrm e}^{\frac {\left (-4 x^{2}+x \right ) \ln \left (3\right )-3 x^{3}}{4 \ln \left (3\right )+3 x}}\)	\(36\)
gosper	\(10 \,{\mathrm e}^{-2} x \,{\mathrm e}^{-\frac {x \left (4 x \ln \left (3\right )+3 x^{2}-\ln \left (3\right )\right )}{4 \ln \left (3\right )+3 x}}\)	\(37\)
risch	\(10 x \,{\mathrm e}^{-\frac {4 x^{2} \ln \left (3\right )+3 x^{3}-x \ln \left (3\right )+8 \ln \left (3\right )+6 x}{4 \ln \left (3\right )+3 x}}\)	\(42\)
norman	\(\frac {30 x^{2} {\mathrm e}^{-2} {\mathrm e}^{\frac {\left (-4 x^{2}+x \right ) \ln \left (3\right )-3 x^{3}}{4 \ln \left (3\right )+3 x}}+40 \,{\mathrm e}^{-2} \ln \left (3\right ) x \,{\mathrm e}^{\frac {\left (-4 x^{2}+x \right ) \ln \left (3\right )-3 x^{3}}{4 \ln \left (3\right )+3 x}}}{4 \ln \left (3\right )+3 x}\)	\(87\)

Input:

int(((-320*x^2+40*x+160)*ln(3)^2+(-480*x^3+240*x)*ln(3)-180*x^4+90*x^2)*ex 
p(((-4*x^2+x)*ln(3)-3*x^3)/(4*ln(3)+3*x))/(16*exp(2)*ln(3)^2+24*x*exp(2)*l 
n(3)+9*x^2*exp(2)),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44 \[ \int \frac {e^{\frac {-3 x^3+\left (x-4 x^2\right ) \log (3)}{3 x+4 \log (3)}} \left (90 x^2-180 x^4+\left (240 x-480 x^3\right ) \log (3)+\left (160+40 x-320 x^2\right ) \log ^2(3)\right )}{9 e^2 x^2+24 e^2 x \log (3)+16 e^2 \log ^2(3)} \, dx=10 \, x e^{\left (-\frac {3 \, x^{3} + {\left (4 \, x^{2} - x\right )} \log \left (3\right )}{3 \, x + 4 \, \log \left (3\right )} - 2\right )} \] Input:

integrate(((-320*x^2+40*x+160)*log(3)^2+(-480*x^3+240*x)*log(3)-180*x^4+90 
*x^2)*exp(((-4*x^2+x)*log(3)-3*x^3)/(4*log(3)+3*x))/(16*exp(2)*log(3)^2+24 
*x*exp(2)*log(3)+9*x^2*exp(2)),x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {e^{\frac {-3 x^3+\left (x-4 x^2\right ) \log (3)}{3 x+4 \log (3)}} \left (90 x^2-180 x^4+\left (240 x-480 x^3\right ) \log (3)+\left (160+40 x-320 x^2\right ) \log ^2(3)\right )}{9 e^2 x^2+24 e^2 x \log (3)+16 e^2 \log ^2(3)} \, dx=\frac {10 x e^{\frac {- 3 x^{3} + \left (- 4 x^{2} + x\right ) \log {\left (3 \right )}}{3 x + 4 \log {\left (3 \right )}}}}{e^{2}} \] Input:

integrate(((-320*x**2+40*x+160)*ln(3)**2+(-480*x**3+240*x)*ln(3)-180*x**4+ 
90*x**2)*exp(((-4*x**2+x)*ln(3)-3*x**3)/(4*ln(3)+3*x))/(16*exp(2)*ln(3)**2 
+24*x*exp(2)*ln(3)+9*x**2*exp(2)),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {e^{\frac {-3 x^3+\left (x-4 x^2\right ) \log (3)}{3 x+4 \log (3)}} \left (90 x^2-180 x^4+\left (240 x-480 x^3\right ) \log (3)+\left (160+40 x-320 x^2\right ) \log ^2(3)\right )}{9 e^2 x^2+24 e^2 x \log (3)+16 e^2 \log ^2(3)} \, dx=10 \cdot 3^{\frac {1}{3}} x e^{\left (-x^{2} - \frac {4 \, \log \left (3\right )^{2}}{3 \, {\left (3 \, x + 4 \, \log \left (3\right )\right )}} - 2\right )} \] Input:

integrate(((-320*x^2+40*x+160)*log(3)^2+(-480*x^3+240*x)*log(3)-180*x^4+90 
*x^2)*exp(((-4*x^2+x)*log(3)-3*x^3)/(4*log(3)+3*x))/(16*exp(2)*log(3)^2+24 
*x*exp(2)*log(3)+9*x^2*exp(2)),x, algorithm="maxima")
 

Output:

10*3^(1/3)*x*e^(-x^2 - 4/3*log(3)^2/(3*x + 4*log(3)) - 2)
 

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.60 \[ \int \frac {e^{\frac {-3 x^3+\left (x-4 x^2\right ) \log (3)}{3 x+4 \log (3)}} \left (90 x^2-180 x^4+\left (240 x-480 x^3\right ) \log (3)+\left (160+40 x-320 x^2\right ) \log ^2(3)\right )}{9 e^2 x^2+24 e^2 x \log (3)+16 e^2 \log ^2(3)} \, dx=10 \cdot 3^{\frac {1}{3}} x e^{\left (-\frac {9 \, x^{3} + 12 \, x^{2} \log \left (3\right ) + 4 \, \log \left (3\right )^{2}}{3 \, {\left (3 \, x + 4 \, \log \left (3\right )\right )}} - 2\right )} \] Input:

integrate(((-320*x^2+40*x+160)*log(3)^2+(-480*x^3+240*x)*log(3)-180*x^4+90 
*x^2)*exp(((-4*x^2+x)*log(3)-3*x^3)/(4*log(3)+3*x))/(16*exp(2)*log(3)^2+24 
*x*exp(2)*log(3)+9*x^2*exp(2)),x, algorithm="giac")
 

Output:

10*3^(1/3)*x*e^(-1/3*(9*x^3 + 12*x^2*log(3) + 4*log(3)^2)/(3*x + 4*log(3)) 
 - 2)
 

Mupad [B] (verification not implemented)

Time = 3.64 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {e^{\frac {-3 x^3+\left (x-4 x^2\right ) \log (3)}{3 x+4 \log (3)}} \left (90 x^2-180 x^4+\left (240 x-480 x^3\right ) \log (3)+\left (160+40 x-320 x^2\right ) \log ^2(3)\right )}{9 e^2 x^2+24 e^2 x \log (3)+16 e^2 \log ^2(3)} \, dx=10\,3^{\frac {x-4\,x^2}{3\,x+\ln \left (81\right )}}\,x\,{\mathrm {e}}^{-\frac {3\,x^3}{3\,x+\ln \left (81\right )}-2} \] Input:

int((exp(-(3*x^3 - log(3)*(x - 4*x^2))/(3*x + 4*log(3)))*(log(3)*(240*x - 
480*x^3) + log(3)^2*(40*x - 320*x^2 + 160) + 90*x^2 - 180*x^4))/(16*exp(2) 
*log(3)^2 + 9*x^2*exp(2) + 24*x*exp(2)*log(3)),x)
 

Output:

10*3^((x - 4*x^2)/(3*x + log(81)))*x*exp(- (3*x^3)/(3*x + log(81)) - 2)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {e^{\frac {-3 x^3+\left (x-4 x^2\right ) \log (3)}{3 x+4 \log (3)}} \left (90 x^2-180 x^4+\left (240 x-480 x^3\right ) \log (3)+\left (160+40 x-320 x^2\right ) \log ^2(3)\right )}{9 e^2 x^2+24 e^2 x \log (3)+16 e^2 \log ^2(3)} \, dx=\frac {10 e^{\frac {\mathrm {log}\left (3\right ) x}{4 \,\mathrm {log}\left (3\right )+3 x}} x}{e^{x^{2}} e^{2}} \] Input:

int(((-320*x^2+40*x+160)*log(3)^2+(-480*x^3+240*x)*log(3)-180*x^4+90*x^2)* 
exp(((-4*x^2+x)*log(3)-3*x^3)/(4*log(3)+3*x))/(16*exp(2)*log(3)^2+24*x*exp 
(2)*log(3)+9*x^2*exp(2)),x)
 

Output:

(10*e**((log(3)*x)/(4*log(3) + 3*x))*x)/(e**(x**2)*e**2)