\(\int \frac {e^x (-15 x^2+5 x^3+x^4-x^5)+e^{x^2} (60 x^2-44 x^4+8 x^6)+(e^x (4 x^3-2 x^4)+e^{x^2} (-16 x^3+16 x^5)) \log (\log (3))}{e^{x+x^2} (-1000+400 x^2-40 x^4)+e^{2 x} (125-50 x^2+5 x^4)+e^{2 x^2} (2000-800 x^2+80 x^4)+(e^{x+x^2} (800 x-160 x^3)+e^{2 x} (-100 x+20 x^3)+e^{2 x^2} (-1600 x+320 x^3)) \log (\log (3))+(20 e^{2 x} x^2+320 e^{2 x^2} x^2-160 e^{x+x^2} x^2) \log ^2(\log (3))} \, dx\) [666]

Optimal result

Integrand size = 240, antiderivative size = 34 \[ \int \frac {e^x \left (-15 x^2+5 x^3+x^4-x^5\right )+e^{x^2} \left (60 x^2-44 x^4+8 x^6\right )+\left (e^x \left (4 x^3-2 x^4\right )+e^{x^2} \left (-16 x^3+16 x^5\right )\right ) \log (\log (3))}{e^{x+x^2} \left (-1000+400 x^2-40 x^4\right )+e^{2 x} \left (125-50 x^2+5 x^4\right )+e^{2 x^2} \left (2000-800 x^2+80 x^4\right )+\left (e^{x+x^2} \left (800 x-160 x^3\right )+e^{2 x} \left (-100 x+20 x^3\right )+e^{2 x^2} \left (-1600 x+320 x^3\right )\right ) \log (\log (3))+\left (20 e^{2 x} x^2+320 e^{2 x^2} x^2-160 e^{x+x^2} x^2\right ) \log ^2(\log (3))} \, dx=\frac {x^2}{5 \left (e^x-4 e^{x^2}\right ) \left (-\frac {5}{x}+x+2 \log (\log (3))\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97 \[ \int \frac {e^x \left (-15 x^2+5 x^3+x^4-x^5\right )+e^{x^2} \left (60 x^2-44 x^4+8 x^6\right )+\left (e^x \left (4 x^3-2 x^4\right )+e^{x^2} \left (-16 x^3+16 x^5\right )\right ) \log (\log (3))}{e^{x+x^2} \left (-1000+400 x^2-40 x^4\right )+e^{2 x} \left (125-50 x^2+5 x^4\right )+e^{2 x^2} \left (2000-800 x^2+80 x^4\right )+\left (e^{x+x^2} \left (800 x-160 x^3\right )+e^{2 x} \left (-100 x+20 x^3\right )+e^{2 x^2} \left (-1600 x+320 x^3\right )\right ) \log (\log (3))+\left (20 e^{2 x} x^2+320 e^{2 x^2} x^2-160 e^{x+x^2} x^2\right ) \log ^2(\log (3))} \, dx=\frac {x^3}{5 \left (e^x-4 e^{x^2}\right ) \left (-5+x^2+2 x \log (\log (3))\right )} \] Input:

Integrate[(E^x*(-15*x^2 + 5*x^3 + x^4 - x^5) + E^x^2*(60*x^2 - 44*x^4 + 8* 
x^6) + (E^x*(4*x^3 - 2*x^4) + E^x^2*(-16*x^3 + 16*x^5))*Log[Log[3]])/(E^(x 
 + x^2)*(-1000 + 400*x^2 - 40*x^4) + E^(2*x)*(125 - 50*x^2 + 5*x^4) + E^(2 
*x^2)*(2000 - 800*x^2 + 80*x^4) + (E^(x + x^2)*(800*x - 160*x^3) + E^(2*x) 
*(-100*x + 20*x^3) + E^(2*x^2)*(-1600*x + 320*x^3))*Log[Log[3]] + (20*E^(2 
*x)*x^2 + 320*E^(2*x^2)*x^2 - 160*E^(x + x^2)*x^2)*Log[Log[3]]^2),x]
 

Output:

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

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.

\(\displaystyle \int \frac {e^{x^2} \left (8 x^6-44 x^4+60 x^2\right )+e^x \left (-x^5+x^4+5 x^3-15 x^2\right )+\left (e^x \left (4 x^3-2 x^4\right )+e^{x^2} \left (16 x^5-16 x^3\right )\right ) \log (\log (3))}{\left (20 e^{2 x} x^2+320 e^{2 x^2} x^2-160 e^{x^2+x} x^2\right ) \log ^2(\log (3))+e^{x^2+x} \left (-40 x^4+400 x^2-1000\right )+e^{2 x} \left (5 x^4-50 x^2+125\right )+e^{2 x^2} \left (80 x^4-800 x^2+2000\right )+\left (e^{2 x} \left (20 x^3-100 x\right )+e^{x^2+x} \left (800 x-160 x^3\right )+e^{2 x^2} \left (320 x^3-1600 x\right )\right ) \log (\log (3))} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {x^2 \left (e^x \left (-x^3+x^2 (1-2 \log (\log (3)))+x (5+4 \log (\log (3)))-15\right )+4 e^{x^2} \left (2 x^4+4 x^3 \log (\log (3))-11 x^2-4 x \log (\log (3))+15\right )\right )}{5 \left (e^x-4 e^{x^2}\right )^2 \left (-x^2-2 x \log (\log (3))+5\right )^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \int \frac {x^2 \left (4 e^{x^2} \left (2 x^4+4 \log (\log (3)) x^3-11 x^2-4 \log (\log (3)) x+15\right )-e^x \left (x^3-(1-2 \log (\log (3))) x^2-(5+4 \log (\log (3))) x+15\right )\right )}{\left (e^x-4 e^{x^2}\right )^2 \left (-x^2-2 \log (\log (3)) x+5\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \frac {1}{5} \int \left (\frac {e^x x^3 (2 x-1)}{\left (e^x-4 e^{x^2}\right )^2 \left (x^2+2 \log (\log (3)) x-5\right )}-\frac {x^2 \left (2 x^4+4 \log (\log (3)) x^3-11 x^2-4 \log (\log (3)) x+15\right )}{\left (e^x-4 e^{x^2}\right ) \left (x^2+2 \log (\log (3)) x-5\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{5} \left (2 \left (5+\log (\log (3))+4 \log ^2(\log (3))\right ) \int \frac {e^x}{\left (e^x-4 e^{x^2}\right )^2}dx-\left (9+8 \log ^2(\log (3))\right ) \int \frac {1}{e^x-4 e^{x^2}}dx-(1+4 \log (\log (3))) \int \frac {e^x x}{\left (e^x-4 e^{x^2}\right )^2}dx+4 \log (\log (3)) \int \frac {x}{e^x-4 e^{x^2}}dx+2 \int \frac {e^x x^2}{\left (e^x-4 e^{x^2}\right )^2}dx-2 \int \frac {x^2}{e^x-4 e^{x^2}}dx-10 \left (5+2 \log ^2(\log (3))\right ) \int \frac {1}{\left (e^x-4 e^{x^2}\right ) \left (x^2+2 \log (\log (3)) x-5\right )^2}dx+2 \log (\log (3)) \left (15+4 \log ^2(\log (3))\right ) \int \frac {x}{\left (e^x-4 e^{x^2}\right ) \left (x^2+2 \log (\log (3)) x-5\right )^2}dx-\left (5+40 \log (\log (3))+4 \log ^2(\log (3))+16 \log ^3(\log (3))\right ) \left (1-\frac {\log (\log (3))}{\sqrt {5+\log ^2(\log (3))}}\right ) \int \frac {e^x}{\left (e^x-4 e^{x^2}\right )^2 \left (2 x-2 \sqrt {5+\log ^2(\log (3))}+2 \log (\log (3))\right )}dx+8 \log (\log (3)) \left (5+2 \log ^2(\log (3))\right ) \left (1-\frac {\log (\log (3))}{\sqrt {5+\log ^2(\log (3))}}\right ) \int \frac {1}{\left (e^x-4 e^{x^2}\right ) \left (2 x-2 \sqrt {5+\log ^2(\log (3))}+2 \log (\log (3))\right )}dx-\frac {10 \left (5+\log (\log (3))+4 \log ^2(\log (3))\right ) \int \frac {e^x}{\left (e^x-4 e^{x^2}\right )^2 \left (-2 x+2 \sqrt {5+\log ^2(\log (3))}-2 \log (\log (3))\right )}dx}{\sqrt {5+\log ^2(\log (3))}}+\frac {11 \left (5+4 \log ^2(\log (3))\right ) \int \frac {1}{\left (e^x-4 e^{x^2}\right ) \left (-2 x+2 \sqrt {5+\log ^2(\log (3))}-2 \log (\log (3))\right )}dx}{\sqrt {5+\log ^2(\log (3))}}-\left (5+40 \log (\log (3))+4 \log ^2(\log (3))+16 \log ^3(\log (3))\right ) \left (1+\frac {\log (\log (3))}{\sqrt {5+\log ^2(\log (3))}}\right ) \int \frac {e^x}{\left (e^x-4 e^{x^2}\right )^2 \left (2 x+2 \sqrt {5+\log ^2(\log (3))}+2 \log (\log (3))\right )}dx-\frac {10 \left (5+\log (\log (3))+4 \log ^2(\log (3))\right ) \int \frac {e^x}{\left (e^x-4 e^{x^2}\right )^2 \left (2 x+2 \sqrt {5+\log ^2(\log (3))}+2 \log (\log (3))\right )}dx}{\sqrt {5+\log ^2(\log (3))}}+8 \log (\log (3)) \left (5+2 \log ^2(\log (3))\right ) \left (1+\frac {\log (\log (3))}{\sqrt {5+\log ^2(\log (3))}}\right ) \int \frac {1}{\left (e^x-4 e^{x^2}\right ) \left (2 x+2 \sqrt {5+\log ^2(\log (3))}+2 \log (\log (3))\right )}dx+\frac {11 \left (5+4 \log ^2(\log (3))\right ) \int \frac {1}{\left (e^x-4 e^{x^2}\right ) \left (2 x+2 \sqrt {5+\log ^2(\log (3))}+2 \log (\log (3))\right )}dx}{\sqrt {5+\log ^2(\log (3))}}\right )\)

Input:

Int[(E^x*(-15*x^2 + 5*x^3 + x^4 - x^5) + E^x^2*(60*x^2 - 44*x^4 + 8*x^6) + 
 (E^x*(4*x^3 - 2*x^4) + E^x^2*(-16*x^3 + 16*x^5))*Log[Log[3]])/(E^(x + x^2 
)*(-1000 + 400*x^2 - 40*x^4) + E^(2*x)*(125 - 50*x^2 + 5*x^4) + E^(2*x^2)* 
(2000 - 800*x^2 + 80*x^4) + (E^(x + x^2)*(800*x - 160*x^3) + E^(2*x)*(-100 
*x + 20*x^3) + E^(2*x^2)*(-1600*x + 320*x^3))*Log[Log[3]] + (20*E^(2*x)*x^ 
2 + 320*E^(2*x^2)*x^2 - 160*E^(x + x^2)*x^2)*Log[Log[3]]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 2.61 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88

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

Input:

int((((16*x^5-16*x^3)*exp(x^2)+(-2*x^4+4*x^3)*exp(x))*ln(ln(3))+(8*x^6-44* 
x^4+60*x^2)*exp(x^2)+(-x^5+x^4+5*x^3-15*x^2)*exp(x))/((320*x^2*exp(x^2)^2- 
160*x^2*exp(x)*exp(x^2)+20*exp(x)^2*x^2)*ln(ln(3))^2+((320*x^3-1600*x)*exp 
(x^2)^2+(-160*x^3+800*x)*exp(x)*exp(x^2)+(20*x^3-100*x)*exp(x)^2)*ln(ln(3) 
)+(80*x^4-800*x^2+2000)*exp(x^2)^2+(-40*x^4+400*x^2-1000)*exp(x)*exp(x^2)+ 
(5*x^4-50*x^2+125)*exp(x)^2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.82 \[ \int \frac {e^x \left (-15 x^2+5 x^3+x^4-x^5\right )+e^{x^2} \left (60 x^2-44 x^4+8 x^6\right )+\left (e^x \left (4 x^3-2 x^4\right )+e^{x^2} \left (-16 x^3+16 x^5\right )\right ) \log (\log (3))}{e^{x+x^2} \left (-1000+400 x^2-40 x^4\right )+e^{2 x} \left (125-50 x^2+5 x^4\right )+e^{2 x^2} \left (2000-800 x^2+80 x^4\right )+\left (e^{x+x^2} \left (800 x-160 x^3\right )+e^{2 x} \left (-100 x+20 x^3\right )+e^{2 x^2} \left (-1600 x+320 x^3\right )\right ) \log (\log (3))+\left (20 e^{2 x} x^2+320 e^{2 x^2} x^2-160 e^{x+x^2} x^2\right ) \log ^2(\log (3))} \, dx=-\frac {x^{3} e^{\left (x^{2}\right )}}{5 \, {\left (4 \, {\left (x^{2} - 5\right )} e^{\left (2 \, x^{2}\right )} - {\left (x^{2} - 5\right )} e^{\left (x^{2} + x\right )} + 2 \, {\left (4 \, x e^{\left (2 \, x^{2}\right )} - x e^{\left (x^{2} + x\right )}\right )} \log \left (\log \left (3\right )\right )\right )}} \] Input:

integrate((((16*x^5-16*x^3)*exp(x^2)+(-2*x^4+4*x^3)*exp(x))*log(log(3))+(8 
*x^6-44*x^4+60*x^2)*exp(x^2)+(-x^5+x^4+5*x^3-15*x^2)*exp(x))/((320*x^2*exp 
(x^2)^2-160*x^2*exp(x)*exp(x^2)+20*exp(x)^2*x^2)*log(log(3))^2+((320*x^3-1 
600*x)*exp(x^2)^2+(-160*x^3+800*x)*exp(x)*exp(x^2)+(20*x^3-100*x)*exp(x)^2 
)*log(log(3))+(80*x^4-800*x^2+2000)*exp(x^2)^2+(-40*x^4+400*x^2-1000)*exp( 
x)*exp(x^2)+(5*x^4-50*x^2+125)*exp(x)^2),x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.44 \[ \int \frac {e^x \left (-15 x^2+5 x^3+x^4-x^5\right )+e^{x^2} \left (60 x^2-44 x^4+8 x^6\right )+\left (e^x \left (4 x^3-2 x^4\right )+e^{x^2} \left (-16 x^3+16 x^5\right )\right ) \log (\log (3))}{e^{x+x^2} \left (-1000+400 x^2-40 x^4\right )+e^{2 x} \left (125-50 x^2+5 x^4\right )+e^{2 x^2} \left (2000-800 x^2+80 x^4\right )+\left (e^{x+x^2} \left (800 x-160 x^3\right )+e^{2 x} \left (-100 x+20 x^3\right )+e^{2 x^2} \left (-1600 x+320 x^3\right )\right ) \log (\log (3))+\left (20 e^{2 x} x^2+320 e^{2 x^2} x^2-160 e^{x+x^2} x^2\right ) \log ^2(\log (3))} \, dx=- \frac {x^{3}}{- 5 x^{2} e^{x} - 10 x e^{x} \log {\left (\log {\left (3 \right )} \right )} + \left (20 x^{2} + 40 x \log {\left (\log {\left (3 \right )} \right )} - 100\right ) e^{x^{2}} + 25 e^{x}} \] Input:

integrate((((16*x**5-16*x**3)*exp(x**2)+(-2*x**4+4*x**3)*exp(x))*ln(ln(3)) 
+(8*x**6-44*x**4+60*x**2)*exp(x**2)+(-x**5+x**4+5*x**3-15*x**2)*exp(x))/(( 
320*x**2*exp(x**2)**2-160*x**2*exp(x)*exp(x**2)+20*exp(x)**2*x**2)*ln(ln(3 
))**2+((320*x**3-1600*x)*exp(x**2)**2+(-160*x**3+800*x)*exp(x)*exp(x**2)+( 
20*x**3-100*x)*exp(x)**2)*ln(ln(3))+(80*x**4-800*x**2+2000)*exp(x**2)**2+( 
-40*x**4+400*x**2-1000)*exp(x)*exp(x**2)+(5*x**4-50*x**2+125)*exp(x)**2),x 
)
 

Output:

-x**3/(-5*x**2*exp(x) - 10*x*exp(x)*log(log(3)) + (20*x**2 + 40*x*log(log( 
3)) - 100)*exp(x**2) + 25*exp(x))
 

Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.18 \[ \int \frac {e^x \left (-15 x^2+5 x^3+x^4-x^5\right )+e^{x^2} \left (60 x^2-44 x^4+8 x^6\right )+\left (e^x \left (4 x^3-2 x^4\right )+e^{x^2} \left (-16 x^3+16 x^5\right )\right ) \log (\log (3))}{e^{x+x^2} \left (-1000+400 x^2-40 x^4\right )+e^{2 x} \left (125-50 x^2+5 x^4\right )+e^{2 x^2} \left (2000-800 x^2+80 x^4\right )+\left (e^{x+x^2} \left (800 x-160 x^3\right )+e^{2 x} \left (-100 x+20 x^3\right )+e^{2 x^2} \left (-1600 x+320 x^3\right )\right ) \log (\log (3))+\left (20 e^{2 x} x^2+320 e^{2 x^2} x^2-160 e^{x+x^2} x^2\right ) \log ^2(\log (3))} \, dx=-\frac {x^{3}}{5 \, {\left (4 \, {\left (x^{2} + 2 \, x \log \left (\log \left (3\right )\right ) - 5\right )} e^{\left (x^{2}\right )} - {\left (x^{2} + 2 \, x \log \left (\log \left (3\right )\right ) - 5\right )} e^{x}\right )}} \] Input:

integrate((((16*x^5-16*x^3)*exp(x^2)+(-2*x^4+4*x^3)*exp(x))*log(log(3))+(8 
*x^6-44*x^4+60*x^2)*exp(x^2)+(-x^5+x^4+5*x^3-15*x^2)*exp(x))/((320*x^2*exp 
(x^2)^2-160*x^2*exp(x)*exp(x^2)+20*exp(x)^2*x^2)*log(log(3))^2+((320*x^3-1 
600*x)*exp(x^2)^2+(-160*x^3+800*x)*exp(x)*exp(x^2)+(20*x^3-100*x)*exp(x)^2 
)*log(log(3))+(80*x^4-800*x^2+2000)*exp(x^2)^2+(-40*x^4+400*x^2-1000)*exp( 
x)*exp(x^2)+(5*x^4-50*x^2+125)*exp(x)^2),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.53 \[ \int \frac {e^x \left (-15 x^2+5 x^3+x^4-x^5\right )+e^{x^2} \left (60 x^2-44 x^4+8 x^6\right )+\left (e^x \left (4 x^3-2 x^4\right )+e^{x^2} \left (-16 x^3+16 x^5\right )\right ) \log (\log (3))}{e^{x+x^2} \left (-1000+400 x^2-40 x^4\right )+e^{2 x} \left (125-50 x^2+5 x^4\right )+e^{2 x^2} \left (2000-800 x^2+80 x^4\right )+\left (e^{x+x^2} \left (800 x-160 x^3\right )+e^{2 x} \left (-100 x+20 x^3\right )+e^{2 x^2} \left (-1600 x+320 x^3\right )\right ) \log (\log (3))+\left (20 e^{2 x} x^2+320 e^{2 x^2} x^2-160 e^{x+x^2} x^2\right ) \log ^2(\log (3))} \, dx=-\frac {x^{3}}{5 \, {\left (4 \, x^{2} e^{\left (x^{2}\right )} - x^{2} e^{x} + 8 \, x e^{\left (x^{2}\right )} \log \left (\log \left (3\right )\right ) - 2 \, x e^{x} \log \left (\log \left (3\right )\right ) - 20 \, e^{\left (x^{2}\right )} + 5 \, e^{x}\right )}} \] Input:

integrate((((16*x^5-16*x^3)*exp(x^2)+(-2*x^4+4*x^3)*exp(x))*log(log(3))+(8 
*x^6-44*x^4+60*x^2)*exp(x^2)+(-x^5+x^4+5*x^3-15*x^2)*exp(x))/((320*x^2*exp 
(x^2)^2-160*x^2*exp(x)*exp(x^2)+20*exp(x)^2*x^2)*log(log(3))^2+((320*x^3-1 
600*x)*exp(x^2)^2+(-160*x^3+800*x)*exp(x)*exp(x^2)+(20*x^3-100*x)*exp(x)^2 
)*log(log(3))+(80*x^4-800*x^2+2000)*exp(x^2)^2+(-40*x^4+400*x^2-1000)*exp( 
x)*exp(x^2)+(5*x^4-50*x^2+125)*exp(x)^2),x, algorithm="giac")
 

Output:

-1/5*x^3/(4*x^2*e^(x^2) - x^2*e^x + 8*x*e^(x^2)*log(log(3)) - 2*x*e^x*log( 
log(3)) - 20*e^(x^2) + 5*e^x)
 

Mupad [B] (verification not implemented)

Time = 4.37 (sec) , antiderivative size = 100, normalized size of antiderivative = 2.94 \[ \int \frac {e^x \left (-15 x^2+5 x^3+x^4-x^5\right )+e^{x^2} \left (60 x^2-44 x^4+8 x^6\right )+\left (e^x \left (4 x^3-2 x^4\right )+e^{x^2} \left (-16 x^3+16 x^5\right )\right ) \log (\log (3))}{e^{x+x^2} \left (-1000+400 x^2-40 x^4\right )+e^{2 x} \left (125-50 x^2+5 x^4\right )+e^{2 x^2} \left (2000-800 x^2+80 x^4\right )+\left (e^{x+x^2} \left (800 x-160 x^3\right )+e^{2 x} \left (-100 x+20 x^3\right )+e^{2 x^2} \left (-1600 x+320 x^3\right )\right ) \log (\log (3))+\left (20 e^{2 x} x^2+320 e^{2 x^2} x^2-160 e^{x+x^2} x^2\right ) \log ^2(\log (3))} \, dx=-\frac {{\mathrm {e}}^x\,\left (\frac {2\,x^4\,\ln \left (\ln \left (3\right )\right )}{5}-\frac {4\,x^5\,\ln \left (\ln \left (3\right )\right )}{5}-x^3+2\,x^4+\frac {x^5}{5}-\frac {2\,x^6}{5}\right )}{\left (4\,{\mathrm {e}}^{x^2}-{\mathrm {e}}^x\right )\,\left ({\mathrm {e}}^x-2\,x\,{\mathrm {e}}^x\right )\,\left (4\,x^3\,\ln \left (\ln \left (3\right )\right )+4\,x^2\,{\ln \left (\ln \left (3\right )\right )}^2-20\,x\,\ln \left (\ln \left (3\right )\right )-10\,x^2+x^4+25\right )} \] Input:

int((log(log(3))*(exp(x)*(4*x^3 - 2*x^4) - exp(x^2)*(16*x^3 - 16*x^5)) - e 
xp(x)*(15*x^2 - 5*x^3 - x^4 + x^5) + exp(x^2)*(60*x^2 - 44*x^4 + 8*x^6))/( 
exp(2*x)*(5*x^4 - 50*x^2 + 125) + log(log(3))^2*(20*x^2*exp(2*x) + 320*x^2 
*exp(2*x^2) - 160*x^2*exp(x^2)*exp(x)) + exp(2*x^2)*(80*x^4 - 800*x^2 + 20 
00) - log(log(3))*(exp(2*x)*(100*x - 20*x^3) + exp(2*x^2)*(1600*x - 320*x^ 
3) - exp(x^2)*exp(x)*(800*x - 160*x^3)) - exp(x^2)*exp(x)*(40*x^4 - 400*x^ 
2 + 1000)),x)
 

Output:

-(exp(x)*((2*x^4*log(log(3)))/5 - (4*x^5*log(log(3)))/5 - x^3 + 2*x^4 + x^ 
5/5 - (2*x^6)/5))/((4*exp(x^2) - exp(x))*(exp(x) - 2*x*exp(x))*(4*x^3*log( 
log(3)) + 4*x^2*log(log(3))^2 - 20*x*log(log(3)) - 10*x^2 + x^4 + 25))
 

Reduce [B] (verification not implemented)

Time = 1.74 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.71 \[ \int \frac {e^x \left (-15 x^2+5 x^3+x^4-x^5\right )+e^{x^2} \left (60 x^2-44 x^4+8 x^6\right )+\left (e^x \left (4 x^3-2 x^4\right )+e^{x^2} \left (-16 x^3+16 x^5\right )\right ) \log (\log (3))}{e^{x+x^2} \left (-1000+400 x^2-40 x^4\right )+e^{2 x} \left (125-50 x^2+5 x^4\right )+e^{2 x^2} \left (2000-800 x^2+80 x^4\right )+\left (e^{x+x^2} \left (800 x-160 x^3\right )+e^{2 x} \left (-100 x+20 x^3\right )+e^{2 x^2} \left (-1600 x+320 x^3\right )\right ) \log (\log (3))+\left (20 e^{2 x} x^2+320 e^{2 x^2} x^2-160 e^{x+x^2} x^2\right ) \log ^2(\log (3))} \, dx=-\frac {x^{3}}{40 e^{x^{2}} \mathrm {log}\left (\mathrm {log}\left (3\right )\right ) x +20 e^{x^{2}} x^{2}-100 e^{x^{2}}-10 e^{x} \mathrm {log}\left (\mathrm {log}\left (3\right )\right ) x -5 e^{x} x^{2}+25 e^{x}} \] Input:

int((((16*x^5-16*x^3)*exp(x^2)+(-2*x^4+4*x^3)*exp(x))*log(log(3))+(8*x^6-4 
4*x^4+60*x^2)*exp(x^2)+(-x^5+x^4+5*x^3-15*x^2)*exp(x))/((320*x^2*exp(x^2)^ 
2-160*x^2*exp(x)*exp(x^2)+20*exp(x)^2*x^2)*log(log(3))^2+((320*x^3-1600*x) 
*exp(x^2)^2+(-160*x^3+800*x)*exp(x)*exp(x^2)+(20*x^3-100*x)*exp(x)^2)*log( 
log(3))+(80*x^4-800*x^2+2000)*exp(x^2)^2+(-40*x^4+400*x^2-1000)*exp(x)*exp 
(x^2)+(5*x^4-50*x^2+125)*exp(x)^2),x)
 

Output:

( - x**3)/(5*(8*e**(x**2)*log(log(3))*x + 4*e**(x**2)*x**2 - 20*e**(x**2) 
- 2*e**x*log(log(3))*x - e**x*x**2 + 5*e**x))