3.2.81 \(\int -\frac {1}{e^{253} x \log ^2(x)} \, dx\)

Optimal. Leaf size=8 \[ \frac {1}{e^{253} \log (x)} \]

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Rubi [A]  time = 0.02, antiderivative size = 8, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {12, 2302, 30} \begin {gather*} \frac {1}{e^{253} \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-(1/(E^253*x*Log[x]^2)),x]

[Out]

1/(E^253*Log[x])

Rule 12

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\frac {\int \frac {1}{x \log ^2(x)} \, dx}{e^{253}}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right )}{e^{253}}\\ &=\frac {1}{e^{253} \log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 8, normalized size = 1.00 \begin {gather*} \frac {1}{e^{253} \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-(1/(E^253*x*Log[x]^2)),x]

[Out]

1/(E^253*Log[x])

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fricas [A]  time = 1.02, size = 7, normalized size = 0.88 \begin {gather*} \frac {e^{\left (-253\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(-log(log(x))-253)/x/log(x),x, algorithm="fricas")

[Out]

e^(-253)/log(x)

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giac [A]  time = 0.28, size = 7, normalized size = 0.88 \begin {gather*} \frac {e^{\left (-253\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(-log(log(x))-253)/x/log(x),x, algorithm="giac")

[Out]

e^(-253)/log(x)

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maple [A]  time = 0.04, size = 8, normalized size = 1.00

method	result	size
risch	\(\frac {{\mathrm e}^{-253}}{\ln \relax (x )}\)	\(8\)
derivativedivides	\({\mathrm e}^{-\ln \left (\ln \relax (x )\right )-253}\)	\(9\)
default	\({\mathrm e}^{-\ln \left (\ln \relax (x )\right )-253}\)	\(9\)
norman	\(\frac {{\mathrm e}^{-253}}{\ln \relax (x )}\)	\(10\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(-ln(ln(x))-253)/x/ln(x),x,method=_RETURNVERBOSE)

[Out]

exp(-253)/ln(x)

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maxima [A]  time = 0.41, size = 7, normalized size = 0.88 \begin {gather*} \frac {e^{\left (-253\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(-log(log(x))-253)/x/log(x),x, algorithm="maxima")

[Out]

e^(-253)/log(x)

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mupad [B]  time = 0.34, size = 7, normalized size = 0.88 \begin {gather*} \frac {{\mathrm {e}}^{-253}}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(- log(log(x)) - 253)/(x*log(x)),x)

[Out]

exp(-253)/log(x)

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sympy [A]  time = 0.08, size = 7, normalized size = 0.88 \begin {gather*} \frac {1}{e^{253} \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-exp(-ln(ln(x))-253)/x/ln(x),x)

[Out]

exp(-253)/log(x)

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