3.73.11 \(\int \frac {e^{-2 e^{3-x^2}} (-54 x-104 e^{3-x^2} x^3+(2 x+4 e^{3-x^2} x^3) \log (x))}{-17576+2028 \log (x)-78 \log ^2(x)+\log ^3(x)} \, dx\)

Optimal. Leaf size=25 \[ \frac {e^{-2 e^{3-x^2}} x^2}{(26-\log (x))^2} \]

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Rubi [F]  time = 2.05, 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*} \int \frac {e^{-2 e^{3-x^2}} \left (-54 x-104 e^{3-x^2} x^3+\left (2 x+4 e^{3-x^2} x^3\right ) \log (x)\right )}{-17576+2028 \log (x)-78 \log ^2(x)+\log ^3(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-54*x - 104*E^(3 - x^2)*x^3 + (2*x + 4*E^(3 - x^2)*x^3)*Log[x])/(E^(2*E^(3 - x^2))*(-17576 + 2028*Log[x]
- 78*Log[x]^2 + Log[x]^3)),x]

[Out]

-2*Defer[Int][x/(E^(2*E^(3 - x^2))*(-26 + Log[x])^3), x] + 2*Defer[Int][x/(E^(2*E^(3 - x^2))*(-26 + Log[x])^2)
, x] + 4*Defer[Int][(E^(3 - 2*E^(3 - x^2) - x^2)*x^3)/(-26 + Log[x])^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-2 e^{3-x^2}} \left (54 x+104 e^{3-x^2} x^3-\left (2 x+4 e^{3-x^2} x^3\right ) \log (x)\right )}{(26-\log (x))^3} \, dx\\ &=\int \left (\frac {2 e^{-2 e^{3-x^2}} x (-27+\log (x))}{(-26+\log (x))^3}+\frac {4 e^{3-2 e^{3-x^2}-x^2} x^3}{(-26+\log (x))^2}\right ) \, dx\\ &=2 \int \frac {e^{-2 e^{3-x^2}} x (-27+\log (x))}{(-26+\log (x))^3} \, dx+4 \int \frac {e^{3-2 e^{3-x^2}-x^2} x^3}{(-26+\log (x))^2} \, dx\\ &=2 \int \left (-\frac {e^{-2 e^{3-x^2}} x}{(-26+\log (x))^3}+\frac {e^{-2 e^{3-x^2}} x}{(-26+\log (x))^2}\right ) \, dx+4 \int \frac {e^{3-2 e^{3-x^2}-x^2} x^3}{(-26+\log (x))^2} \, dx\\ &=-\left (2 \int \frac {e^{-2 e^{3-x^2}} x}{(-26+\log (x))^3} \, dx\right )+2 \int \frac {e^{-2 e^{3-x^2}} x}{(-26+\log (x))^2} \, dx+4 \int \frac {e^{3-2 e^{3-x^2}-x^2} x^3}{(-26+\log (x))^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.51, size = 23, normalized size = 0.92 \begin {gather*} \frac {e^{-2 e^{3-x^2}} x^2}{(-26+\log (x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-54*x - 104*E^(3 - x^2)*x^3 + (2*x + 4*E^(3 - x^2)*x^3)*Log[x])/(E^(2*E^(3 - x^2))*(-17576 + 2028*L
og[x] - 78*Log[x]^2 + Log[x]^3)),x]

[Out]

x^2/(E^(2*E^(3 - x^2))*(-26 + Log[x])^2)

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fricas [A]  time = 0.81, size = 27, normalized size = 1.08 \begin {gather*} \frac {x^{2} e^{\left (-2 \, e^{\left (-x^{2} + 3\right )}\right )}}{\log \relax (x)^{2} - 52 \, \log \relax (x) + 676} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3*exp(-x^2+3)+2*x)*log(x)-104*x^3*exp(-x^2+3)-54*x)/(log(x)^3-78*log(x)^2+2028*log(x)-17576)/e
xp(exp(-x^2+3))^2,x, algorithm="fricas")

[Out]

x^2*e^(-2*e^(-x^2 + 3))/(log(x)^2 - 52*log(x) + 676)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {2 \, {\left (52 \, x^{3} e^{\left (-x^{2} + 3\right )} - {\left (2 \, x^{3} e^{\left (-x^{2} + 3\right )} + x\right )} \log \relax (x) + 27 \, x\right )} e^{\left (-2 \, e^{\left (-x^{2} + 3\right )}\right )}}{\log \relax (x)^{3} - 78 \, \log \relax (x)^{2} + 2028 \, \log \relax (x) - 17576}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3*exp(-x^2+3)+2*x)*log(x)-104*x^3*exp(-x^2+3)-54*x)/(log(x)^3-78*log(x)^2+2028*log(x)-17576)/e
xp(exp(-x^2+3))^2,x, algorithm="giac")

[Out]

integrate(-2*(52*x^3*e^(-x^2 + 3) - (2*x^3*e^(-x^2 + 3) + x)*log(x) + 27*x)*e^(-2*e^(-x^2 + 3))/(log(x)^3 - 78
*log(x)^2 + 2028*log(x) - 17576), x)

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maple [A]  time = 0.04, size = 22, normalized size = 0.88

method	result	size
risch	\(\frac {x^{2} {\mathrm e}^{-2 \,{\mathrm e}^{-x^{2}+3}}}{\left (-26+\ln \relax (x )\right )^{2}}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^3*exp(-x^2+3)+2*x)*ln(x)-104*x^3*exp(-x^2+3)-54*x)/(ln(x)^3-78*ln(x)^2+2028*ln(x)-17576)/exp(exp(-x^
2+3))^2,x,method=_RETURNVERBOSE)

[Out]

x^2/(-26+ln(x))^2*exp(-2*exp(-x^2+3))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -2 \, \int \frac {{\left (52 \, x^{3} e^{\left (-x^{2} + 3\right )} - {\left (2 \, x^{3} e^{\left (-x^{2} + 3\right )} + x\right )} \log \relax (x) + 27 \, x\right )} e^{\left (-2 \, e^{\left (-x^{2} + 3\right )}\right )}}{\log \relax (x)^{3} - 78 \, \log \relax (x)^{2} + 2028 \, \log \relax (x) - 17576}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3*exp(-x^2+3)+2*x)*log(x)-104*x^3*exp(-x^2+3)-54*x)/(log(x)^3-78*log(x)^2+2028*log(x)-17576)/e
xp(exp(-x^2+3))^2,x, algorithm="maxima")

[Out]

-2*integrate((52*x^3*e^(-x^2 + 3) - (2*x^3*e^(-x^2 + 3) + x)*log(x) + 27*x)*e^(-2*e^(-x^2 + 3))/(log(x)^3 - 78
*log(x)^2 + 2028*log(x) - 17576), x)

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mupad [B]  time = 4.43, size = 21, normalized size = 0.84 \begin {gather*} \frac {x^2\,{\mathrm {e}}^{-2\,{\mathrm {e}}^3\,{\mathrm {e}}^{-x^2}}}{{\left (\ln \relax (x)-26\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-2*exp(3 - x^2))*(54*x + 104*x^3*exp(3 - x^2) - log(x)*(2*x + 4*x^3*exp(3 - x^2))))/(2028*log(x) - 7
8*log(x)^2 + log(x)^3 - 17576),x)

[Out]

(x^2*exp(-2*exp(3)*exp(-x^2)))/(log(x) - 26)^2

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sympy [A]  time = 0.45, size = 24, normalized size = 0.96 \begin {gather*} \frac {x^{2} e^{- 2 e^{3 - x^{2}}}}{\log {\relax (x )}^{2} - 52 \log {\relax (x )} + 676} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**3*exp(-x**2+3)+2*x)*ln(x)-104*x**3*exp(-x**2+3)-54*x)/(ln(x)**3-78*ln(x)**2+2028*ln(x)-17576)
/exp(exp(-x**2+3))**2,x)

[Out]

x**2*exp(-2*exp(3 - x**2))/(log(x)**2 - 52*log(x) + 676)

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