3.95.79 \(\int \frac {\log (2)-\log (2) \log (5)}{x \log (64)} \, dx\)

Optimal. Leaf size=17 \[ \frac {\log (2) (-1+\log (5)) \log \left (\frac {15}{x}\right )}{\log (64)} \]

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Rubi [A]  time = 0.01, antiderivative size = 15, normalized size of antiderivative = 0.88, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 29} \begin {gather*} \frac {\log (2) (1-\log (5)) \log (x)}{\log (64)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Log[2] - Log[2]*Log[5])/(x*Log[64]),x]

[Out]

(Log[2]*(1 - Log[5])*Log[x])/Log[64]

Rule 12

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {(\log (2) (1-\log (5))) \int \frac {1}{x} \, dx}{\log (64)}\\ &=\frac {\log (2) (1-\log (5)) \log (x)}{\log (64)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 14, normalized size = 0.82 \begin {gather*} -\frac {\log (2) (-1+\log (5)) \log (x)}{\log (64)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Log[2] - Log[2]*Log[5])/(x*Log[64]),x]

[Out]

-((Log[2]*(-1 + Log[5])*Log[x])/Log[64])

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fricas [A]  time = 0.49, size = 8, normalized size = 0.47 \begin {gather*} -\frac {1}{6} \, {\left (\log \relax (5) - 1\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*(-log(2)*log(5)+log(2))/x/log(2),x, algorithm="fricas")

[Out]

-1/6*(log(5) - 1)*log(x)

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giac [A]  time = 0.12, size = 19, normalized size = 1.12 \begin {gather*} -\frac {{\left (\log \relax (5) \log \relax (2) - \log \relax (2)\right )} \log \left ({\left | x \right |}\right )}{6 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*(-log(2)*log(5)+log(2))/x/log(2),x, algorithm="giac")

[Out]

-1/6*(log(5)*log(2) - log(2))*log(abs(x))/log(2)

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maple [A]  time = 0.04, size = 10, normalized size = 0.59

method	result	size
norman	\(\left (-\frac {\ln \relax (5)}{6}+\frac {1}{6}\right ) \ln \relax (x )\)	\(10\)
risch	\(-\frac {\ln \relax (5) \ln \relax (x )}{6}+\frac {\ln \relax (x )}{6}\)	\(12\)
default	\(\frac {\left (-\ln \relax (2) \ln \relax (5)+\ln \relax (2)\right ) \ln \relax (x )}{6 \ln \relax (2)}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/6*(-ln(2)*ln(5)+ln(2))/x/ln(2),x,method=_RETURNVERBOSE)

[Out]

(-1/6*ln(5)+1/6)*ln(x)

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maxima [A]  time = 0.34, size = 18, normalized size = 1.06 \begin {gather*} -\frac {{\left (\log \relax (5) \log \relax (2) - \log \relax (2)\right )} \log \relax (x)}{6 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*(-log(2)*log(5)+log(2))/x/log(2),x, algorithm="maxima")

[Out]

-1/6*(log(5)*log(2) - log(2))*log(x)/log(2)

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mupad [B]  time = 0.05, size = 10, normalized size = 0.59 \begin {gather*} -\ln \relax (x)\,\left (\frac {\ln \relax (5)}{6}-\frac {1}{6}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(x)*(log(5)/6 - 1/6)

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sympy [A]  time = 0.06, size = 10, normalized size = 0.59 \begin {gather*} \left (\frac {1}{6} - \frac {\log {\relax (5 )}}{6}\right ) \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*(-ln(2)*ln(5)+ln(2))/x/ln(2),x)

[Out]

(1/6 - log(5)/6)*log(x)

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