3.69.68 \(\int \frac {10-5 \log (\log (6))}{x \log ^2(x)} \, dx\)

Optimal. Leaf size=11 \[ \frac {5 (-2+\log (\log (6)))}{\log (x)} \]

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Rubi [A]  time = 0.02, antiderivative size = 13, normalized size of antiderivative = 1.18, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {12, 2302, 30} \begin {gather*} -\frac {5 (2-\log (\log (6)))}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10 - 5*Log[Log[6]])/(x*Log[x]^2),x]

[Out]

(-5*(2 - Log[Log[6]]))/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} &=(5 (2-\log (\log (6)))) \int \frac {1}{x \log ^2(x)} \, dx\\ &=(5 (2-\log (\log (6)))) \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right )\\ &=-\frac {5 (2-\log (\log (6)))}{\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 16, normalized size = 1.45 \begin {gather*} -\frac {10}{\log (x)}+\frac {5 \log (\log (6))}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10 - 5*Log[Log[6]])/(x*Log[x]^2),x]

[Out]

-10/Log[x] + (5*Log[Log[6]])/Log[x]

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fricas [A]  time = 1.44, size = 11, normalized size = 1.00 \begin {gather*} \frac {5 \, {\left (\log \left (\log \relax (6)\right ) - 2\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*log(log(6))+10)/x/log(x)^2,x, algorithm="fricas")

[Out]

5*(log(log(6)) - 2)/log(x)

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giac [A]  time = 0.18, size = 11, normalized size = 1.00 \begin {gather*} \frac {5 \, {\left (\log \left (\log \relax (6)\right ) - 2\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*log(log(6))+10)/x/log(x)^2,x, algorithm="giac")

[Out]

5*(log(log(6)) - 2)/log(x)

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maple [A]  time = 0.03, size = 13, normalized size = 1.18

method	result	size
norman	\(\frac {5 \ln \left (\ln \relax (6)\right )-10}{\ln \relax (x )}\)	\(13\)
derivativedivides	\(-\frac {-5 \ln \left (\ln \relax (6)\right )+10}{\ln \relax (x )}\)	\(14\)
default	\(-\frac {-5 \ln \left (\ln \relax (6)\right )+10}{\ln \relax (x )}\)	\(14\)
risch	\(\frac {5 \ln \left (\ln \relax (2)+\ln \relax (3)\right )}{\ln \relax (x )}-\frac {10}{\ln \relax (x )}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-5*ln(ln(6))+10)/x/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

(5*ln(ln(6))-10)/ln(x)

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maxima [A]  time = 0.35, size = 11, normalized size = 1.00 \begin {gather*} \frac {5 \, {\left (\log \left (\log \relax (6)\right ) - 2\right )}}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*log(log(6))+10)/x/log(x)^2,x, algorithm="maxima")

[Out]

5*(log(log(6)) - 2)/log(x)

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mupad [B]  time = 0.03, size = 12, normalized size = 1.09 \begin {gather*} \frac {5\,\ln \left (\ln \relax (6)\right )-10}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*log(log(6)) - 10)/(x*log(x)^2),x)

[Out]

(5*log(log(6)) - 10)/log(x)

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sympy [A]  time = 0.09, size = 10, normalized size = 0.91 \begin {gather*} \frac {-10 + 5 \log {\left (\log {\relax (6 )} \right )}}{\log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*ln(ln(6))+10)/x/ln(x)**2,x)

[Out]

(-10 + 5*log(log(6)))/log(x)

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