3.103.27 \(\int (-8 x+16 x^2) \log (3) \, dx\)

Optimal. Leaf size=23 \[ \frac {2}{3} (4+2 (-2+x)) \left (4-\frac {3}{x}\right ) x^2 \log (3) \]

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Rubi [A]  time = 0.00, antiderivative size = 17, normalized size of antiderivative = 0.74, number of steps used = 2, number of rules used = 1, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {12} \begin {gather*} \frac {16}{3} x^3 \log (3)-4 x^2 \log (3) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-8*x + 16*x^2)*Log[3],x]

[Out]

-4*x^2*Log[3] + (16*x^3*Log[3])/3

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log (3) \int \left (-8 x+16 x^2\right ) \, dx\\ &=-4 x^2 \log (3)+\frac {16}{3} x^3 \log (3)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 19, normalized size = 0.83 \begin {gather*} 8 \left (-\frac {x^2}{2}+\frac {2 x^3}{3}\right ) \log (3) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8*x + 16*x^2)*Log[3],x]

[Out]

8*(-1/2*x^2 + (2*x^3)/3)*Log[3]

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fricas [A]  time = 0.92, size = 15, normalized size = 0.65 \begin {gather*} \frac {4}{3} \, {\left (4 \, x^{3} - 3 \, x^{2}\right )} \log \relax (3) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*x^2-8*x)*log(3),x, algorithm="fricas")

[Out]

4/3*(4*x^3 - 3*x^2)*log(3)

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giac [A]  time = 0.14, size = 15, normalized size = 0.65 \begin {gather*} \frac {4}{3} \, {\left (4 \, x^{3} - 3 \, x^{2}\right )} \log \relax (3) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*x^2-8*x)*log(3),x, algorithm="giac")

[Out]

4/3*(4*x^3 - 3*x^2)*log(3)

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maple [A]  time = 0.02, size = 13, normalized size = 0.57

method	result	size
gosper	\(\frac {4 \ln \relax (3) \left (4 x -3\right ) x^{2}}{3}\)	\(13\)
default	\(\ln \relax (3) \left (\frac {16}{3} x^{3}-4 x^{2}\right )\)	\(15\)
norman	\(-4 x^{2} \ln \relax (3)+\frac {16 x^{3} \ln \relax (3)}{3}\)	\(16\)
risch	\(-4 x^{2} \ln \relax (3)+\frac {16 x^{3} \ln \relax (3)}{3}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*x^2-8*x)*ln(3),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.36, size = 15, normalized size = 0.65 \begin {gather*} \frac {4}{3} \, {\left (4 \, x^{3} - 3 \, x^{2}\right )} \log \relax (3) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*x^2-8*x)*log(3),x, algorithm="maxima")

[Out]

4/3*(4*x^3 - 3*x^2)*log(3)

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mupad [B]  time = 0.06, size = 12, normalized size = 0.52 \begin {gather*} \frac {4\,x^2\,\ln \relax (3)\,\left (4\,x-3\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-log(3)*(8*x - 16*x^2),x)

[Out]

(4*x^2*log(3)*(4*x - 3))/3

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sympy [A]  time = 0.06, size = 17, normalized size = 0.74 \begin {gather*} \frac {16 x^{3} \log {\relax (3 )}}{3} - 4 x^{2} \log {\relax (3 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((16*x**2-8*x)*ln(3),x)

[Out]

16*x**3*log(3)/3 - 4*x**2*log(3)

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