3.24.69 \(\int -\frac {2 \log (\log (4))}{x \log ^3(-x)} \, dx\)

Optimal. Leaf size=10 \[ \frac {\log (\log (4))}{\log ^2(-x)} \]

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Rubi [A]  time = 0.01, antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {12, 2302, 30} \begin {gather*} \frac {\log (\log (4))}{\log ^2(-x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*Log[Log[4]])/(x*Log[-x]^3),x]

[Out]

Log[Log[4]]/Log[-x]^2

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} &=-\left ((2 \log (\log (4))) \int \frac {1}{x \log ^3(-x)} \, dx\right )\\ &=-\left ((2 \log (\log (4))) \operatorname {Subst}\left (\int \frac {1}{x^3} \, dx,x,\log (-x)\right )\right )\\ &=\frac {\log (\log (4))}{\log ^2(-x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 10, normalized size = 1.00 \begin {gather*} \frac {\log (\log (4))}{\log ^2(-x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*Log[Log[4]])/(x*Log[-x]^3),x]

[Out]

Log[Log[4]]/Log[-x]^2

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fricas [A]  time = 0.51, size = 12, normalized size = 1.20 \begin {gather*} \frac {\log \left (2 \, \log \relax (2)\right )}{\log \left (-x\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.35, size = 12, normalized size = 1.20 \begin {gather*} \frac {\log \left (2 \, \log \relax (2)\right )}{\log \left (-x\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 13, normalized size = 1.30

method	result	size
derivativedivides	\(\frac {\ln \left (2 \ln \relax (2)\right )}{\ln \left (-x \right )^{2}}\)	\(13\)
default	\(\frac {\ln \left (2 \ln \relax (2)\right )}{\ln \left (-x \right )^{2}}\)	\(13\)
norman	\(\frac {\ln \relax (2)+\ln \left (\ln \relax (2)\right )}{\ln \left (-x \right )^{2}}\)	\(14\)
risch	\(\frac {\ln \relax (2)}{\ln \left (-x \right )^{2}}+\frac {\ln \left (\ln \relax (2)\right )}{\ln \left (-x \right )^{2}}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-2*ln(2*ln(2))/x/ln(-x)^3,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.40, size = 12, normalized size = 1.20 \begin {gather*} \frac {\log \left (2 \, \log \relax (2)\right )}{\log \left (-x\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.46, size = 10, normalized size = 1.00 \begin {gather*} \frac {\ln \left (\ln \relax (4)\right )}{{\ln \left (-x\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*log(2*log(2)))/(x*log(-x)^3),x)

[Out]

log(log(4))/log(-x)^2

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sympy [A]  time = 0.09, size = 14, normalized size = 1.40 \begin {gather*} \frac {\log {\left (\log {\relax (2 )} \right )} + \log {\relax (2 )}}{\log {\left (- x \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-2*ln(2*ln(2))/x/ln(-x)**3,x)

[Out]

(log(log(2)) + log(2))/log(-x)**2

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