∫ e−x2 x(2x2+e2+2ee3 (−2+2x2)+e2(−8+6x2+2x4)+eee3 (−2x+e2(4x−4x3)))+e−x2 x(8−6x2−2x4+e2ee3 (2−2x2)+eee3 (−4x+4x3)) log(4+e2ee3 −2eee3 x+x2)______________________________________________________________________________________________________________

3.36.39 $\int \frac {e^{-x^2} x (2 x^2+e^{2+2 e^{e^3}} (-2+2 x^2)+e^2 (-8+6 x^2+2 x^4)+e^{e^{e^3}} (-2 x+e^2 (4 x-4 x^3)))+e^{-x^2} x (8-6 x^2-2 x^4+e^{2 e^{e^3}} (2-2 x^2)+e^{e^{e^3}} (-4 x+4 x^3)) \log (4+e^{2 e^{e^3}}-2 e^{e^{e^3}} x+x^2)}{4+e^{2 e^{e^3}}-2 e^{e^{e^3}} x+x^2} \, dx$

Optimal. Leaf size=33 \[ e^{-x^2} x^2 \left (-e^2+\log \left (4+\left (e^{e^{e^3}}-x\right )^2\right )\right ) \]

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Rubi [F] time = 180.10, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, $\frac {\text {number of rules}}{\text {integrand size}}$ = 0.000, Rules used = {} \begin {gather*} \text {\$Aborted} \end {gather*}

Veriﬁcation is not applicable to the result.

[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

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Mathematica [A] time = 0.15, size = 43, normalized size = 1.30 \begin {gather*} -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 ) \end {gather*}

Antiderivative was successfully veriﬁed.

[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)

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fricas [A] time = 0.67, size = 44, normalized size = 1.33 \begin {gather*} x e^{\left (-x^{2} + \log \relax (x)\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 \relax (x) + 2\right )} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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giac [A] time = 0.18, size = 39, normalized size = 1.18 \begin {gather*} -{\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 )} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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maple [F(-1)] time = 180.00, size = 0, normalized size = 0.00 hanged

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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]

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)

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maxima [A] time = 0.58, size = 43, normalized size = 1.30 \begin {gather*} 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 )} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B] time = 2.60, size = 34, normalized size = 1.03 \begin {gather*} 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 ) \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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))

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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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]

Timed out

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