3.76.58 \(\int \frac {-2+4 x^2+8 x^2 \log (3 x)}{x} \, dx\)

Optimal. Leaf size=18 \[ \left (2+x \left (-\frac {4}{x}+4 x\right )\right ) \log (3 x) \]

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Rubi [A]  time = 0.02, antiderivative size = 14, normalized size of antiderivative = 0.78, number of steps used = 5, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {14, 2304} \begin {gather*} 4 x^2 \log (3 x)-2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2 + 4*x^2 + 8*x^2*Log[3*x])/x,x]

[Out]

-2*Log[x] + 4*x^2*Log[3*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]]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2 \left (-1+2 x^2\right )}{x}+8 x \log (3 x)\right ) \, dx\\ &=2 \int \frac {-1+2 x^2}{x} \, dx+8 \int x \log (3 x) \, dx\\ &=-2 x^2+4 x^2 \log (3 x)+2 \int \left (-\frac {1}{x}+2 x\right ) \, dx\\ &=-2 \log (x)+4 x^2 \log (3 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 14, normalized size = 0.78 \begin {gather*} -2 \log (x)+4 x^2 \log (3 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 + 4*x^2 + 8*x^2*Log[3*x])/x,x]

[Out]

-2*Log[x] + 4*x^2*Log[3*x]

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fricas [A]  time = 0.79, size = 13, normalized size = 0.72 \begin {gather*} 2 \, {\left (2 \, x^{2} - 1\right )} \log \left (3 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^2*log(3*x)+4*x^2-2)/x,x, algorithm="fricas")

[Out]

2*(2*x^2 - 1)*log(3*x)

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giac [A]  time = 0.24, size = 14, normalized size = 0.78 \begin {gather*} 4 \, x^{2} \log \left (3 \, x\right ) - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^2*log(3*x)+4*x^2-2)/x,x, algorithm="giac")

[Out]

4*x^2*log(3*x) - 2*log(x)

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maple [A]  time = 0.02, size = 15, normalized size = 0.83

method	result	size
risch	\(4 x^{2} \ln \left (3 x \right )-2 \ln \relax (x )\)	\(15\)
derivativedivides	\(4 x^{2} \ln \left (3 x \right )-2 \ln \left (3 x \right )\)	\(17\)
default	\(4 x^{2} \ln \left (3 x \right )-2 \ln \left (3 x \right )\)	\(17\)
norman	\(4 x^{2} \ln \left (3 x \right )-2 \ln \left (3 x \right )\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x^2*ln(3*x)+4*x^2-2)/x,x,method=_RETURNVERBOSE)

[Out]

4*x^2*ln(3*x)-2*ln(x)

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maxima [A]  time = 0.35, size = 14, normalized size = 0.78 \begin {gather*} 4 \, x^{2} \log \left (3 \, x\right ) - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x^2*log(3*x)+4*x^2-2)/x,x, algorithm="maxima")

[Out]

4*x^2*log(3*x) - 2*log(x)

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mupad [B]  time = 5.10, size = 19, normalized size = 1.06 \begin {gather*} 4\,x^2\,\ln \relax (x)-2\,\ln \relax (x)+4\,x^2\,\ln \relax (3) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x^2*log(3*x) + 4*x^2 - 2)/x,x)

[Out]

4*x^2*log(x) - 2*log(x) + 4*x^2*log(3)

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sympy [A]  time = 0.10, size = 14, normalized size = 0.78 \begin {gather*} 4 x^{2} \log {\left (3 x \right )} - 2 \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x**2*ln(3*x)+4*x**2-2)/x,x)

[Out]

4*x**2*log(3*x) - 2*log(x)

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