\(\int \frac {e^{-x^2} x (2 x^2+e^{2+2 e^{e^3}} (-2+2 x^2)+e^2 (-8+6 x^2+2 x^4)+e^{e^{e^3}} (-2 x+e^2 (4 x-4 x^3)))+e^{-x^2} x (8-6 x^2-2 x^4+e^{2 e^{e^3}} (2-2 x^2)+e^{e^{e^3}} (-4 x+4 x^3)) \log (4+e^{2 e^{e^3}}-2 e^{e^{e^3}} x+x^2)}{4+e^{2 e^{e^3}}-2 e^{e^{e^3}} x+x^2} \, dx\) [3539]

Optimal result

Integrand size = 183, antiderivative size = 33 \[ \int \frac {e^{-x^2} x \left (2 x^2+e^{2+2 e^{e^3}} \left (-2+2 x^2\right )+e^2 \left (-8+6 x^2+2 x^4\right )+e^{e^{e^3}} \left (-2 x+e^2 \left (4 x-4 x^3\right )\right )\right )+e^{-x^2} x \left (8-6 x^2-2 x^4+e^{2 e^{e^3}} \left (2-2 x^2\right )+e^{e^{e^3}} \left (-4 x+4 x^3\right )\right ) \log \left (4+e^{2 e^{e^3}}-2 e^{e^{e^3}} x+x^2\right )}{4+e^{2 e^{e^3}}-2 e^{e^{e^3}} x+x^2} \, dx=e^{-x^2} x^2 \left (-e^2+\log \left (4+\left (e^{e^{e^3}}-x\right )^2\right )\right ) \]

[Out]

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

Rubi [F(-1)]

Timed out. \[ \int \frac {e^{-x^2} x \left (2 x^2+e^{2+2 e^{e^3}} \left (-2+2 x^2\right )+e^2 \left (-8+6 x^2+2 x^4\right )+e^{e^{e^3}} \left (-2 x+e^2 \left (4 x-4 x^3\right )\right )\right )+e^{-x^2} x \left (8-6 x^2-2 x^4+e^{2 e^{e^3}} \left (2-2 x^2\right )+e^{e^{e^3}} \left (-4 x+4 x^3\right )\right ) \log \left (4+e^{2 e^{e^3}}-2 e^{e^{e^3}} x+x^2\right )}{4+e^{2 e^{e^3}}-2 e^{e^{e^3}} x+x^2} \, dx=\text {\$Aborted} \]

[In]

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

[Out]

$Aborted

Rubi steps Aborted

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.30 \[ \int \frac {e^{-x^2} x \left (2 x^2+e^{2+2 e^{e^3}} \left (-2+2 x^2\right )+e^2 \left (-8+6 x^2+2 x^4\right )+e^{e^{e^3}} \left (-2 x+e^2 \left (4 x-4 x^3\right )\right )\right )+e^{-x^2} x \left (8-6 x^2-2 x^4+e^{2 e^{e^3}} \left (2-2 x^2\right )+e^{e^{e^3}} \left (-4 x+4 x^3\right )\right ) \log \left (4+e^{2 e^{e^3}}-2 e^{e^{e^3}} x+x^2\right )}{4+e^{2 e^{e^3}}-2 e^{e^{e^3}} x+x^2} \, dx=-e^{-x^2} x^2 \left (e^2-\log \left (4+e^{2 e^{e^3}}-2 e^{e^{e^3}} x+x^2\right )\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 45.05 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09

method	result	size
risch	\(\left (-{\mathrm e}^{2} x +x \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{{\mathrm e}^{3}}}-2 x \,{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{3}}}+x^{2}+4\right )\right ) x \,{\mathrm e}^{-x^{2}}\)	\(36\)
parallelrisch	\(-{\mathrm e}^{2} {\mathrm e}^{\ln \left (x \right )-x^{2}} x +x \,{\mathrm e}^{\ln \left (x \right )-x^{2}} \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{{\mathrm e}^{3}}}-2 x \,{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{3}}}+x^{2}+4\right )\)	\(46\)

[In]

int((((-2*x^2+2)*exp(exp(exp(3)))^2+(4*x^3-4*x)*exp(exp(exp(3)))-2*x^4-6*x^2+8)*exp(ln(x)-x^2)*ln(exp(exp(exp(
3)))^2-2*x*exp(exp(exp(3)))+x^2+4)+((2*x^2-2)*exp(2)*exp(exp(exp(3)))^2+((-4*x^3+4*x)*exp(2)-2*x)*exp(exp(exp(
3)))+(2*x^4+6*x^2-8)*exp(2)+2*x^2)*exp(ln(x)-x^2))/(exp(exp(exp(3)))^2-2*x*exp(exp(exp(3)))+x^2+4),x,method=_R
ETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.33 \[ \int \frac {e^{-x^2} x \left (2 x^2+e^{2+2 e^{e^3}} \left (-2+2 x^2\right )+e^2 \left (-8+6 x^2+2 x^4\right )+e^{e^{e^3}} \left (-2 x+e^2 \left (4 x-4 x^3\right )\right )\right )+e^{-x^2} x \left (8-6 x^2-2 x^4+e^{2 e^{e^3}} \left (2-2 x^2\right )+e^{e^{e^3}} \left (-4 x+4 x^3\right )\right ) \log \left (4+e^{2 e^{e^3}}-2 e^{e^{e^3}} x+x^2\right )}{4+e^{2 e^{e^3}}-2 e^{e^{e^3}} x+x^2} \, dx=x e^{\left (-x^{2} + \log \left (x\right )\right )} \log \left (x^{2} - 2 \, x e^{\left (e^{\left (e^{3}\right )}\right )} + e^{\left (2 \, e^{\left (e^{3}\right )}\right )} + 4\right ) - x e^{\left (-x^{2} + \log \left (x\right ) + 2\right )} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.18 \[ \int \frac {e^{-x^2} x \left (2 x^2+e^{2+2 e^{e^3}} \left (-2+2 x^2\right )+e^2 \left (-8+6 x^2+2 x^4\right )+e^{e^{e^3}} \left (-2 x+e^2 \left (4 x-4 x^3\right )\right )\right )+e^{-x^2} x \left (8-6 x^2-2 x^4+e^{2 e^{e^3}} \left (2-2 x^2\right )+e^{e^{e^3}} \left (-4 x+4 x^3\right )\right ) \log \left (4+e^{2 e^{e^3}}-2 e^{e^{e^3}} x+x^2\right )}{4+e^{2 e^{e^3}}-2 e^{e^{e^3}} x+x^2} \, dx=\left (x^{2} \log {\left (x^{2} - 2 x e^{e^{e^{3}}} + 4 + e^{2 e^{e^{3}}} \right )} - x^{2} e^{2}\right ) e^{- x^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.30 \[ \int \frac {e^{-x^2} x \left (2 x^2+e^{2+2 e^{e^3}} \left (-2+2 x^2\right )+e^2 \left (-8+6 x^2+2 x^4\right )+e^{e^{e^3}} \left (-2 x+e^2 \left (4 x-4 x^3\right )\right )\right )+e^{-x^2} x \left (8-6 x^2-2 x^4+e^{2 e^{e^3}} \left (2-2 x^2\right )+e^{e^{e^3}} \left (-4 x+4 x^3\right )\right ) \log \left (4+e^{2 e^{e^3}}-2 e^{e^{e^3}} x+x^2\right )}{4+e^{2 e^{e^3}}-2 e^{e^{e^3}} x+x^2} \, dx=x^{2} e^{\left (-x^{2}\right )} \log \left (x^{2} - 2 \, x e^{\left (e^{\left (e^{3}\right )}\right )} + e^{\left (2 \, e^{\left (e^{3}\right )}\right )} + 4\right ) - x^{2} e^{\left (-x^{2} + 2\right )} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.18 \[ \int \frac {e^{-x^2} x \left (2 x^2+e^{2+2 e^{e^3}} \left (-2+2 x^2\right )+e^2 \left (-8+6 x^2+2 x^4\right )+e^{e^{e^3}} \left (-2 x+e^2 \left (4 x-4 x^3\right )\right )\right )+e^{-x^2} x \left (8-6 x^2-2 x^4+e^{2 e^{e^3}} \left (2-2 x^2\right )+e^{e^{e^3}} \left (-4 x+4 x^3\right )\right ) \log \left (4+e^{2 e^{e^3}}-2 e^{e^{e^3}} x+x^2\right )}{4+e^{2 e^{e^3}}-2 e^{e^{e^3}} x+x^2} \, dx=-{\left (x^{2} e^{2} - x^{2} \log \left (x^{2} - 2 \, x e^{\left (e^{\left (e^{3}\right )}\right )} + e^{\left (2 \, e^{\left (e^{3}\right )}\right )} + 4\right )\right )} e^{\left (-x^{2}\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.64 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {e^{-x^2} x \left (2 x^2+e^{2+2 e^{e^3}} \left (-2+2 x^2\right )+e^2 \left (-8+6 x^2+2 x^4\right )+e^{e^{e^3}} \left (-2 x+e^2 \left (4 x-4 x^3\right )\right )\right )+e^{-x^2} x \left (8-6 x^2-2 x^4+e^{2 e^{e^3}} \left (2-2 x^2\right )+e^{e^{e^3}} \left (-4 x+4 x^3\right )\right ) \log \left (4+e^{2 e^{e^3}}-2 e^{e^{e^3}} x+x^2\right )}{4+e^{2 e^{e^3}}-2 e^{e^{e^3}} x+x^2} \, dx=x^2\,{\mathrm {e}}^{-x^2}\,\left (\ln \left (x^2-2\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^3}}\,x+{\mathrm {e}}^{2\,{\mathrm {e}}^{{\mathrm {e}}^3}}+4\right )-{\mathrm {e}}^2\right ) \]

[In]

int((exp(log(x) - x^2)*(exp(2)*(6*x^2 + 2*x^4 - 8) + 2*x^2 - exp(exp(exp(3)))*(2*x - exp(2)*(4*x - 4*x^3)) + e
xp(2)*exp(2*exp(exp(3)))*(2*x^2 - 2)) - log(exp(2*exp(exp(3))) - 2*x*exp(exp(exp(3))) + x^2 + 4)*exp(log(x) -
x^2)*(exp(2*exp(exp(3)))*(2*x^2 - 2) + 6*x^2 + 2*x^4 + exp(exp(exp(3)))*(4*x - 4*x^3) - 8))/(exp(2*exp(exp(3))
) - 2*x*exp(exp(exp(3))) + x^2 + 4),x)

[Out]

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