3.28.63 \(\int \frac {12 x^2+(3+24 x^2) \log (3)}{4 x^2} \, dx\)

Optimal. Leaf size=24 \[ x \left (3+\frac {\left (x-\left (-5+\frac {3}{4 x^2}\right ) x\right ) \log (3)}{x}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 17, normalized size of antiderivative = 0.71, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {12, 14} \begin {gather*} 3 x (1+\log (9))-\frac {\log (27)}{4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(12*x^2 + (3 + 24*x^2)*Log[3])/(4*x^2),x]

[Out]

3*x*(1 + Log[9]) - Log[27]/(4*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} &=\frac {1}{4} \int \frac {12 x^2+\left (3+24 x^2\right ) \log (3)}{x^2} \, dx\\ &=\frac {1}{4} \int \left (12 (1+\log (9))+\frac {\log (27)}{x^2}\right ) \, dx\\ &=3 x (1+\log (9))-\frac {\log (27)}{4 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 17, normalized size = 0.71 \begin {gather*} 3 x-\frac {3 \log (3)}{4 x}+x \log (729) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(12*x^2 + (3 + 24*x^2)*Log[3])/(4*x^2),x]

[Out]

3*x - (3*Log[3])/(4*x) + x*Log[729]

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fricas [A]  time = 0.49, size = 21, normalized size = 0.88 \begin {gather*} \frac {3 \, {\left (4 \, x^{2} + {\left (8 \, x^{2} - 1\right )} \log \relax (3)\right )}}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((24*x^2+3)*log(3)+12*x^2)/x^2,x, algorithm="fricas")

[Out]

3/4*(4*x^2 + (8*x^2 - 1)*log(3))/x

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giac [A]  time = 0.23, size = 16, normalized size = 0.67 \begin {gather*} 6 \, x \log \relax (3) + 3 \, x - \frac {3 \, \log \relax (3)}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((24*x^2+3)*log(3)+12*x^2)/x^2,x, algorithm="giac")

[Out]

6*x*log(3) + 3*x - 3/4*log(3)/x

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maple [A]  time = 0.03, size = 17, normalized size = 0.71

method	result	size
default	\(6 x \ln \relax (3)+3 x -\frac {3 \ln \relax (3)}{4 x}\)	\(17\)
risch	\(6 x \ln \relax (3)+3 x -\frac {3 \ln \relax (3)}{4 x}\)	\(17\)
norman	\(\frac {\left (6 \ln \relax (3)+3\right ) x^{2}-\frac {3 \ln \relax (3)}{4}}{x}\)	\(20\)
gosper	\(\frac {6 x^{2} \ln \relax (3)+3 x^{2}-\frac {3 \ln \relax (3)}{4}}{x}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*((24*x^2+3)*ln(3)+12*x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

6*x*ln(3)+3*x-3/4*ln(3)/x

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maxima [A]  time = 0.35, size = 17, normalized size = 0.71 \begin {gather*} 3 \, x {\left (2 \, \log \relax (3) + 1\right )} - \frac {3 \, \log \relax (3)}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((24*x^2+3)*log(3)+12*x^2)/x^2,x, algorithm="maxima")

[Out]

3*x*(2*log(3) + 1) - 3/4*log(3)/x

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mupad [B]  time = 0.07, size = 16, normalized size = 0.67 \begin {gather*} x\,\left (6\,\ln \relax (3)+3\right )-\frac {3\,\ln \relax (3)}{4\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((log(3)*(24*x^2 + 3))/4 + 3*x^2)/x^2,x)

[Out]

x*(6*log(3) + 3) - (3*log(3))/(4*x)

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sympy [A]  time = 0.08, size = 17, normalized size = 0.71 \begin {gather*} \frac {x \left (12 + 24 \log {\relax (3 )}\right )}{4} - \frac {3 \log {\relax (3 )}}{4 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((24*x**2+3)*ln(3)+12*x**2)/x**2,x)

[Out]

x*(12 + 24*log(3))/4 - 3*log(3)/(4*x)

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