3.8.77 \(\int \frac {1-6 x^2 \log (5)}{5 x} \, dx\)

Optimal. Leaf size=15 \[ \frac {1}{5} \left (3-3 x^2 \log (5)+\log (x)\right ) \]

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Rubi [A]  time = 0.00, antiderivative size = 16, normalized size of antiderivative = 1.07, number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 14} \begin {gather*} \frac {\log (x)}{5}-\frac {3}{5} x^2 \log (5) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - 6*x^2*Log[5])/(5*x),x]

[Out]

(-3*x^2*Log[5])/5 + Log[x]/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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {1-6 x^2 \log (5)}{x} \, dx\\ &=\frac {1}{5} \int \left (\frac {1}{x}-6 x \log (5)\right ) \, dx\\ &=-\frac {3}{5} x^2 \log (5)+\frac {\log (x)}{5}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 16, normalized size = 1.07 \begin {gather*} -\frac {3}{5} x^2 \log (5)+\frac {\log (x)}{5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - 6*x^2*Log[5])/(5*x),x]

[Out]

(-3*x^2*Log[5])/5 + Log[x]/5

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fricas [A]  time = 1.01, size = 12, normalized size = 0.80 \begin {gather*} -\frac {3}{5} \, x^{2} \log \relax (5) + \frac {1}{5} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/5*x^2*log(5) + 1/5*log(x)

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giac [A]  time = 0.45, size = 14, normalized size = 0.93 \begin {gather*} -\frac {3}{5} \, x^{2} \log \relax (5) + \frac {1}{10} \, \log \left (x^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/5*x^2*log(5) + 1/10*log(x^2)

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maple [A]  time = 0.04, size = 13, normalized size = 0.87

method	result	size
default	\(-\frac {3 x^{2} \ln \relax (5)}{5}+\frac {\ln \relax (x )}{5}\)	\(13\)
norman	\(-\frac {3 x^{2} \ln \relax (5)}{5}+\frac {\ln \relax (x )}{5}\)	\(13\)
risch	\(-\frac {3 x^{2} \ln \relax (5)}{5}+\frac {\ln \relax (x )}{5}\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/5*x^2*ln(5)+1/5*ln(x)

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maxima [A]  time = 0.48, size = 14, normalized size = 0.93 \begin {gather*} -\frac {3}{5} \, x^{2} \log \relax (5) + \frac {1}{10} \, \log \left (x^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/5*x^2*log(5) + 1/10*log(x^2)

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mupad [B]  time = 0.61, size = 12, normalized size = 0.80 \begin {gather*} \frac {\ln \relax (x)}{5}-\frac {3\,x^2\,\ln \relax (5)}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((6*x^2*log(5))/5 - 1/5)/x,x)

[Out]

log(x)/5 - (3*x^2*log(5))/5

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sympy [A]  time = 0.07, size = 14, normalized size = 0.93 \begin {gather*} - \frac {3 x^{2} \log {\relax (5 )}}{5} + \frac {\log {\relax (x )}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*x**2*log(5)/5 + log(x)/5

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