3.92.77 \(\int \frac {e^{-2+\frac {16}{x^2 \log ^2(x)}} (-32-32 \log (x))}{x^3 \log ^3(x)} \, dx\)

Optimal. Leaf size=13 \[ e^{-2+\frac {16}{x^2 \log ^2(x)}} \]

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Rubi [A]  time = 0.31, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {6706} \begin {gather*} e^{\frac {16}{x^2 \log ^2(x)}-2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-2 + 16/(x^2*Log[x]^2))*(-32 - 32*Log[x]))/(x^3*Log[x]^3),x]

[Out]

E^(-2 + 16/(x^2*Log[x]^2))

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=e^{-2+\frac {16}{x^2 \log ^2(x)}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 13, normalized size = 1.00 \begin {gather*} e^{-2+\frac {16}{x^2 \log ^2(x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-2 + 16/(x^2*Log[x]^2))*(-32 - 32*Log[x]))/(x^3*Log[x]^3),x]

[Out]

E^(-2 + 16/(x^2*Log[x]^2))

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fricas [A]  time = 0.57, size = 20, normalized size = 1.54 \begin {gather*} e^{\left (-\frac {2 \, {\left (x^{2} \log \relax (x)^{2} - 8\right )}}{x^{2} \log \relax (x)^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*log(x)-32)*exp(8/x^2/log(x)^2)^2/x^3/exp(1)^2/log(x)^3,x, algorithm="fricas")

[Out]

e^(-2*(x^2*log(x)^2 - 8)/(x^2*log(x)^2))

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giac [A]  time = 0.14, size = 12, normalized size = 0.92 \begin {gather*} e^{\left (\frac {16}{x^{2} \log \relax (x)^{2}} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*log(x)-32)*exp(8/x^2/log(x)^2)^2/x^3/exp(1)^2/log(x)^3,x, algorithm="giac")

[Out]

e^(16/(x^2*log(x)^2) - 2)

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maple [A]  time = 0.03, size = 21, normalized size = 1.62

method	result	size
risch	\({\mathrm e}^{-\frac {2 \left (x^{2} \ln \relax (x )^{2}-8\right )}{\ln \relax (x )^{2} x^{2}}}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-32*ln(x)-32)*exp(8/x^2/ln(x)^2)^2/x^3/exp(1)^2/ln(x)^3,x,method=_RETURNVERBOSE)

[Out]

exp(-2*(x^2*ln(x)^2-8)/ln(x)^2/x^2)

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maxima [A]  time = 0.46, size = 12, normalized size = 0.92 \begin {gather*} e^{\left (\frac {16}{x^{2} \log \relax (x)^{2}} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*log(x)-32)*exp(8/x^2/log(x)^2)^2/x^3/exp(1)^2/log(x)^3,x, algorithm="maxima")

[Out]

e^(16/(x^2*log(x)^2) - 2)

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mupad [B]  time = 5.43, size = 13, normalized size = 1.00 \begin {gather*} {\mathrm {e}}^{-2}\,{\mathrm {e}}^{\frac {16}{x^2\,{\ln \relax (x)}^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-2)*exp(16/(x^2*log(x)^2))*(32*log(x) + 32))/(x^3*log(x)^3),x)

[Out]

exp(-2)*exp(16/(x^2*log(x)^2))

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sympy [A]  time = 0.39, size = 14, normalized size = 1.08 \begin {gather*} \frac {e^{\frac {16}{x^{2} \log {\relax (x )}^{2}}}}{e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*ln(x)-32)*exp(8/x**2/ln(x)**2)**2/x**3/exp(1)**2/ln(x)**3,x)

[Out]

exp(-2)*exp(16/(x**2*log(x)**2))

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