3.99.22 \(\int \frac {e^{-e^{6-2 e^{2 x}} x^2-2 e^{3-e^{2 x}} x^2 \log (\log (4))-x^2 \log ^2(\log (4))} (-4+e^{6-2 e^{2 x}} (-8 x^2+16 e^{2 x} x^3)+e^{3-e^{2 x}} (-16 x^2+16 e^{2 x} x^3) \log (\log (4))-8 x^2 \log ^2(\log (4)))}{x^2} \, dx\)

Optimal. Leaf size=29 \[ \frac {4 e^{-x^2 \left (e^{3-e^{2 x}}+\log (\log (4))\right )^2}}{x} \]

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Rubi [B]  time = 1.65, antiderivative size = 200, normalized size of antiderivative = 6.90, number of steps used = 1, number of rules used = 1, integrand size = 123, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.008, Rules used = {2288} \begin {gather*} \frac {4 e^{-e^{6-2 e^{2 x}} x^2-x^2 \log ^2(\log (4))} \log ^{-2 e^{3-e^{2 x}} x^2}(4) \left (x^2 \log ^2(\log (4))+e^{6-2 e^{2 x}} \left (x^2-2 e^{2 x} x^3\right )+2 e^{3-e^{2 x}} \left (x^2-e^{2 x} x^3\right ) \log (\log (4))\right )}{x^2 \left (-2 e^{2 x-2 e^{2 x}+6} x^2-2 e^{2 x-e^{2 x}+3} x^2 \log (\log (4))+e^{6-2 e^{2 x}} x+x \log ^2(\log (4))+2 e^{3-e^{2 x}} x \log (\log (4))\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-(E^(6 - 2*E^(2*x))*x^2) - 2*E^(3 - E^(2*x))*x^2*Log[Log[4]] - x^2*Log[Log[4]]^2)*(-4 + E^(6 - 2*E^(2*
x))*(-8*x^2 + 16*E^(2*x)*x^3) + E^(3 - E^(2*x))*(-16*x^2 + 16*E^(2*x)*x^3)*Log[Log[4]] - 8*x^2*Log[Log[4]]^2))
/x^2,x]

[Out]

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

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {4 e^{-e^{6-2 e^{2 x}} x^2-x^2 \log ^2(\log (4))} \log ^{-2 e^{3-e^{2 x}} x^2}(4) \left (e^{6-2 e^{2 x}} \left (x^2-2 e^{2 x} x^3\right )+2 e^{3-e^{2 x}} \left (x^2-e^{2 x} x^3\right ) \log (\log (4))+x^2 \log ^2(\log (4))\right )}{x^2 \left (e^{6-2 e^{2 x}} x-2 e^{6-2 e^{2 x}+2 x} x^2+2 e^{3-e^{2 x}} x \log (\log (4))-2 e^{3-e^{2 x}+2 x} x^2 \log (\log (4))+x \log ^2(\log (4))\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.26, size = 48, normalized size = 1.66 \begin {gather*} \frac {4 e^{-x^2 \left (e^{6-2 e^{2 x}}+\log ^2(\log (4))\right )} \log ^{-2 e^{3-e^{2 x}} x^2}(4)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-(E^(6 - 2*E^(2*x))*x^2) - 2*E^(3 - E^(2*x))*x^2*Log[Log[4]] - x^2*Log[Log[4]]^2)*(-4 + E^(6 - 2
*E^(2*x))*(-8*x^2 + 16*E^(2*x)*x^3) + E^(3 - E^(2*x))*(-16*x^2 + 16*E^(2*x)*x^3)*Log[Log[4]] - 8*x^2*Log[Log[4
]]^2))/x^2,x]

[Out]

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

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fricas [A]  time = 0.69, size = 52, normalized size = 1.79 \begin {gather*} \frac {4 \, e^{\left (-2 \, x^{2} e^{\left (-e^{\left (2 \, x\right )} + 3\right )} \log \left (2 \, \log \relax (2)\right ) - x^{2} \log \left (2 \, \log \relax (2)\right )^{2} - x^{2} e^{\left (-2 \, e^{\left (2 \, x\right )} + 6\right )}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*exp(x)^2*x^3-8*x^2)*exp(-exp(x)^2+3)^2+(16*exp(x)^2*x^3-16*x^2)*log(2*log(2))*exp(-exp(x)^2+3)-
8*x^2*log(2*log(2))^2-4)/x^2/exp(x^2*exp(-exp(x)^2+3)^2+2*x^2*log(2*log(2))*exp(-exp(x)^2+3)+x^2*log(2*log(2))
^2),x, algorithm="fricas")

[Out]

4*e^(-2*x^2*e^(-e^(2*x) + 3)*log(2*log(2)) - x^2*log(2*log(2))^2 - x^2*e^(-2*e^(2*x) + 6))/x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*exp(x)^2*x^3-8*x^2)*exp(-exp(x)^2+3)^2+(16*exp(x)^2*x^3-16*x^2)*log(2*log(2))*exp(-exp(x)^2+3)-
8*x^2*log(2*log(2))^2-4)/x^2/exp(x^2*exp(-exp(x)^2+3)^2+2*x^2*log(2*log(2))*exp(-exp(x)^2+3)+x^2*log(2*log(2))
^2),x, algorithm="giac")

[Out]

undef

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maple [B]  time = 0.18, size = 65, normalized size = 2.24

method	result	size
risch	\(\frac {4 \,{\mathrm e}^{-x^{2} \left (\ln \relax (2)^{2}+2 \,{\mathrm e}^{3-{\mathrm e}^{2 x}} \ln \relax (2)+2 \ln \relax (2) \ln \left (\ln \relax (2)\right )+2 \,{\mathrm e}^{3-{\mathrm e}^{2 x}} \ln \left (\ln \relax (2)\right )+\ln \left (\ln \relax (2)\right )^{2}+{\mathrm e}^{6-2 \,{\mathrm e}^{2 x}}\right )}}{x}\)	\(65\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((16*exp(x)^2*x^3-8*x^2)*exp(-exp(x)^2+3)^2+(16*exp(x)^2*x^3-16*x^2)*ln(2*ln(2))*exp(-exp(x)^2+3)-8*x^2*ln
(2*ln(2))^2-4)/x^2/exp(x^2*exp(-exp(x)^2+3)^2+2*x^2*ln(2*ln(2))*exp(-exp(x)^2+3)+x^2*ln(2*ln(2))^2),x,method=_
RETURNVERBOSE)

[Out]

4/x*exp(-x^2*(ln(2)^2+2*exp(3-exp(2*x))*ln(2)+2*ln(2)*ln(ln(2))+2*exp(3-exp(2*x))*ln(ln(2))+ln(ln(2))^2+exp(6-
2*exp(2*x))))

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maxima [B]  time = 0.63, size = 83, normalized size = 2.86 \begin {gather*} \frac {4 \, e^{\left (-2 \, x^{2} e^{\left (-e^{\left (2 \, x\right )} + 3\right )} \log \relax (2) - x^{2} \log \relax (2)^{2} - 2 \, x^{2} e^{\left (-e^{\left (2 \, x\right )} + 3\right )} \log \left (\log \relax (2)\right ) - 2 \, x^{2} \log \relax (2) \log \left (\log \relax (2)\right ) - x^{2} \log \left (\log \relax (2)\right )^{2} - x^{2} e^{\left (-2 \, e^{\left (2 \, x\right )} + 6\right )}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*exp(x)^2*x^3-8*x^2)*exp(-exp(x)^2+3)^2+(16*exp(x)^2*x^3-16*x^2)*log(2*log(2))*exp(-exp(x)^2+3)-
8*x^2*log(2*log(2))^2-4)/x^2/exp(x^2*exp(-exp(x)^2+3)^2+2*x^2*log(2*log(2))*exp(-exp(x)^2+3)+x^2*log(2*log(2))
^2),x, algorithm="maxima")

[Out]

4*e^(-2*x^2*e^(-e^(2*x) + 3)*log(2) - x^2*log(2)^2 - 2*x^2*e^(-e^(2*x) + 3)*log(log(2)) - 2*x^2*log(2)*log(log
(2)) - x^2*log(log(2))^2 - x^2*e^(-2*e^(2*x) + 6))/x

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mupad [B]  time = 5.97, size = 90, normalized size = 3.10 \begin {gather*} \frac {4\,{\mathrm {e}}^{-x^2\,{\mathrm {e}}^{-2\,{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^6}\,{\mathrm {e}}^{-x^2\,{\ln \relax (2)}^2}\,{\mathrm {e}}^{-x^2\,{\ln \left (\ln \relax (2)\right )}^2}}{2^{2\,x^2\,\ln \left (\ln \relax (2)\right )}\,2^{2\,x^2\,{\mathrm {e}}^{-{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^3}\,x\,{\ln \relax (2)}^{2\,x^2\,{\mathrm {e}}^{-{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(- x^2*exp(6 - 2*exp(2*x)) - x^2*log(2*log(2))^2 - 2*x^2*log(2*log(2))*exp(3 - exp(2*x)))*(exp(6 - 2*e
xp(2*x))*(16*x^3*exp(2*x) - 8*x^2) - 8*x^2*log(2*log(2))^2 + log(2*log(2))*exp(3 - exp(2*x))*(16*x^3*exp(2*x)
- 16*x^2) - 4))/x^2,x)

[Out]

(4*exp(-x^2*exp(-2*exp(2*x))*exp(6))*exp(-x^2*log(2)^2)*exp(-x^2*log(log(2))^2))/(2^(2*x^2*log(log(2)))*2^(2*x
^2*exp(-exp(2*x))*exp(3))*x*log(2)^(2*x^2*exp(-exp(2*x))*exp(3)))

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sympy [B]  time = 0.61, size = 51, normalized size = 1.76 \begin {gather*} \frac {4 e^{- 2 x^{2} e^{3 - e^{2 x}} \log {\left (2 \log {\relax (2 )} \right )} - x^{2} e^{6 - 2 e^{2 x}} - x^{2} \log {\left (2 \log {\relax (2 )} \right )}^{2}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*exp(x)**2*x**3-8*x**2)*exp(-exp(x)**2+3)**2+(16*exp(x)**2*x**3-16*x**2)*ln(2*ln(2))*exp(-exp(x)
**2+3)-8*x**2*ln(2*ln(2))**2-4)/x**2/exp(x**2*exp(-exp(x)**2+3)**2+2*x**2*ln(2*ln(2))*exp(-exp(x)**2+3)+x**2*l
n(2*ln(2))**2),x)

[Out]

4*exp(-2*x**2*exp(3 - exp(2*x))*log(2*log(2)) - x**2*exp(6 - 2*exp(2*x)) - x**2*log(2*log(2))**2)/x

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