3.21.28 \(\int -\frac {2 \log (x)}{e^8 x} \, dx\)

Optimal. Leaf size=19 \[ -\frac {e^{10}}{9}+\log (3)-\frac {\log ^2(x)}{e^8} \]

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Rubi [A]  time = 0.01, antiderivative size = 9, normalized size of antiderivative = 0.47, number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {12, 2301} \begin {gather*} -\frac {\log ^2(x)}{e^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*Log[x])/(E^8*x),x]

[Out]

-(Log[x]^2/E^8)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\frac {2 \int \frac {\log (x)}{x} \, dx}{e^8}\\ &=-\frac {\log ^2(x)}{e^8}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 9, normalized size = 0.47 \begin {gather*} -\frac {\log ^2(x)}{e^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*Log[x])/(E^8*x),x]

[Out]

-(Log[x]^2/E^8)

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fricas [A]  time = 0.85, size = 8, normalized size = 0.42 \begin {gather*} -e^{\left (-8\right )} \log \relax (x)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.25, size = 8, normalized size = 0.42 \begin {gather*} -e^{\left (-8\right )} \log \relax (x)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 9, normalized size = 0.47

method	result	size
risch	\(-\ln \relax (x )^{2} {\mathrm e}^{-8}\)	\(9\)
derivativedivides	\(-\ln \relax (x )^{2} {\mathrm e}^{-8}\)	\(11\)
default	\(-\ln \relax (x )^{2} {\mathrm e}^{-8}\)	\(11\)
norman	\(-\ln \relax (x )^{2} {\mathrm e}^{-8}\)	\(11\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-2*ln(x)/x/exp(8),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.41, size = 8, normalized size = 0.42 \begin {gather*} -e^{\left (-8\right )} \log \relax (x)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.02, size = 8, normalized size = 0.42 \begin {gather*} -{\mathrm {e}}^{-8}\,{\ln \relax (x)}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*exp(-8)*log(x))/x,x)

[Out]

-exp(-8)*log(x)^2

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sympy [A]  time = 0.10, size = 8, normalized size = 0.42 \begin {gather*} - \frac {\log {\relax (x )}^{2}}{e^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-exp(-8)*log(x)**2

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