3.6.68 \(\int \frac {-16-5 \log (5) \log (32)}{5 x^2 \log (5)} \, dx\)

Optimal. Leaf size=19 \[ 5+\frac {16}{5 x \log (5)}+\frac {\log (32)}{x} \]

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Rubi [A]  time = 0.01, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {12, 30} \begin {gather*} \frac {16+5 \log (5) \log (32)}{5 x \log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-16 - 5*Log[5]*Log[32])/(5*x^2*Log[5]),x]

[Out]

(16 + 5*Log[5]*Log[32])/(5*x*Log[5])

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {(-16-5 \log (5) \log (32)) \int \frac {1}{x^2} \, dx}{5 \log (5)}\\ &=\frac {16+5 \log (5) \log (32)}{5 x \log (5)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 19, normalized size = 1.00 \begin {gather*} \frac {16+5 \log (5) \log (32)}{5 x \log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16 - 5*Log[5]*Log[32])/(5*x^2*Log[5]),x]

[Out]

(16 + 5*Log[5]*Log[32])/(5*x*Log[5])

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fricas [A]  time = 0.66, size = 17, normalized size = 0.89 \begin {gather*} \frac {25 \, \log \relax (5) \log \relax (2) + 16}{5 \, x \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-25*log(2)*log(5)-16)/x^2/log(5),x, algorithm="fricas")

[Out]

1/5*(25*log(5)*log(2) + 16)/(x*log(5))

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giac [A]  time = 0.81, size = 17, normalized size = 0.89 \begin {gather*} \frac {25 \, \log \relax (5) \log \relax (2) + 16}{5 \, x \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-25*log(2)*log(5)-16)/x^2/log(5),x, algorithm="giac")

[Out]

1/5*(25*log(5)*log(2) + 16)/(x*log(5))

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maple [A]  time = 0.02, size = 18, normalized size = 0.95

method	result	size
gosper	\(\frac {25 \ln \relax (2) \ln \relax (5)+16}{5 x \ln \relax (5)}\)	\(18\)
default	\(-\frac {-25 \ln \relax (2) \ln \relax (5)-16}{5 x \ln \relax (5)}\)	\(18\)
norman	\(\frac {25 \ln \relax (2) \ln \relax (5)+16}{5 x \ln \relax (5)}\)	\(18\)
risch	\(\frac {5 \ln \relax (2)}{x}+\frac {16}{5 x \ln \relax (5)}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(-25*ln(2)*ln(5)-16)/x^2/ln(5),x,method=_RETURNVERBOSE)

[Out]

1/5*(25*ln(2)*ln(5)+16)/x/ln(5)

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maxima [A]  time = 0.41, size = 17, normalized size = 0.89 \begin {gather*} \frac {25 \, \log \relax (5) \log \relax (2) + 16}{5 \, x \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-25*log(2)*log(5)-16)/x^2/log(5),x, algorithm="maxima")

[Out]

1/5*(25*log(5)*log(2) + 16)/(x*log(5))

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mupad [B]  time = 0.04, size = 16, normalized size = 0.84 \begin {gather*} \frac {5\,\ln \relax (2)\,\ln \relax (5)+\frac {16}{5}}{x\,\ln \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*log(2)*log(5) + 16/5)/(x^2*log(5)),x)

[Out]

(5*log(2)*log(5) + 16/5)/(x*log(5))

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sympy [A]  time = 0.06, size = 19, normalized size = 1.00 \begin {gather*} - \frac {- 25 \log {\relax (2 )} \log {\relax (5 )} - 16}{5 x \log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-25*ln(2)*ln(5)-16)/x**2/ln(5),x)

[Out]

-(-25*log(2)*log(5) - 16)/(5*x*log(5))

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