3.83.50 \(\int \frac {(-6 x^2-48 \log (2)) \log (3)}{x^2} \, dx\)

Optimal. Leaf size=24 \[ \frac {\left (x+x^2-(6 (-8+x)+x) (x+\log (2))\right ) \log (3)}{x} \]

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Rubi [A]  time = 0.01, antiderivative size = 15, normalized size of antiderivative = 0.62, 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 {48 \log (2) \log (3)}{x}-6 x \log (3) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-6*x^2 - 48*Log[2])*Log[3])/x^2,x]

[Out]

-6*x*Log[3] + (48*Log[2]*Log[3])/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 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} &=\log (3) \int \frac {-6 x^2-48 \log (2)}{x^2} \, dx\\ &=\log (3) \int \left (-6-\frac {48 \log (2)}{x^2}\right ) \, dx\\ &=-6 x \log (3)+\frac {48 \log (2) \log (3)}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 13, normalized size = 0.54 \begin {gather*} -6 \log (3) \left (x-\frac {\log (256)}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-6*x^2 - 48*Log[2])*Log[3])/x^2,x]

[Out]

-6*Log[3]*(x - Log[256]/x)

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fricas [A]  time = 1.11, size = 15, normalized size = 0.62 \begin {gather*} -\frac {6 \, {\left (x^{2} - 8 \, \log \relax (2)\right )} \log \relax (3)}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-48*log(2)-6*x^2)*log(3)/x^2,x, algorithm="fricas")

[Out]

-6*(x^2 - 8*log(2))*log(3)/x

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giac [A]  time = 0.11, size = 13, normalized size = 0.54 \begin {gather*} -6 \, {\left (x - \frac {8 \, \log \relax (2)}{x}\right )} \log \relax (3) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-48*log(2)-6*x^2)*log(3)/x^2,x, algorithm="giac")

[Out]

-6*(x - 8*log(2)/x)*log(3)

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maple [A]  time = 0.03, size = 16, normalized size = 0.67

method	result	size
default	\(6 \ln \relax (3) \left (-x +\frac {8 \ln \relax (2)}{x}\right )\)	\(16\)
risch	\(-6 x \ln \relax (3)+\frac {48 \ln \relax (3) \ln \relax (2)}{x}\)	\(16\)
gosper	\(\frac {6 \ln \relax (3) \left (-x^{2}+8 \ln \relax (2)\right )}{x}\)	\(18\)
norman	\(\frac {-6 x^{2} \ln \relax (3)+48 \ln \relax (2) \ln \relax (3)}{x}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-48*ln(2)-6*x^2)*ln(3)/x^2,x,method=_RETURNVERBOSE)

[Out]

6*ln(3)*(-x+8*ln(2)/x)

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maxima [A]  time = 0.37, size = 13, normalized size = 0.54 \begin {gather*} -6 \, {\left (x - \frac {8 \, \log \relax (2)}{x}\right )} \log \relax (3) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-48*log(2)-6*x^2)*log(3)/x^2,x, algorithm="maxima")

[Out]

-6*(x - 8*log(2)/x)*log(3)

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mupad [B]  time = 0.04, size = 17, normalized size = 0.71 \begin {gather*} \frac {6\,\ln \relax (3)\,\left (8\,\ln \relax (2)-x^2\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(3)*(48*log(2) + 6*x^2))/x^2,x)

[Out]

(6*log(3)*(8*log(2) - x^2))/x

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sympy [A]  time = 0.09, size = 15, normalized size = 0.62 \begin {gather*} - 6 x \log {\relax (3 )} + \frac {48 \log {\relax (2 )} \log {\relax (3 )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-48*ln(2)-6*x**2)*ln(3)/x**2,x)

[Out]

-6*x*log(3) + 48*log(2)*log(3)/x

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