3.17.64 \(\int \frac {-6+16 x-12 x^2}{-21+3 x-4 x^2+2 x^3} \, dx\) [1664]

Optimal. Leaf size=24 \[ \log \left (\frac {1}{\left (-1-x+2 \left (4+\frac {1}{3} (2-x) x^2\right )\right )^2}\right ) \]

[Out]

ln(1/(2/3*(2-x)*x^2+7-x)^2)

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Rubi [A]
time = 0.01, antiderivative size = 18, normalized size of antiderivative = 0.75, number of steps used = 1, number of rules used = 1, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {1601} \begin {gather*} -2 \log \left (-2 x^3+4 x^2-3 x+21\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-6 + 16*x - 12*x^2)/(-21 + 3*x - 4*x^2 + 2*x^3),x]

[Out]

-2*Log[21 - 3*x + 4*x^2 - 2*x^3]

Rule 1601

Int[(Pp_)/(Qq_), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*(Log[RemoveConte
nt[Qq, x]]/(q*Coeff[Qq, x, q])), x] /; EqQ[p, q - 1] && EqQ[Pp, Simplify[(Coeff[Pp, x, p]/(q*Coeff[Qq, x, q]))
*D[Qq, x]]]] /; PolyQ[Pp, x] && PolyQ[Qq, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-2 \log \left (21-3 x+4 x^2-2 x^3\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.00, size = 18, normalized size = 0.75 \begin {gather*} -2 \log \left (21-3 x+4 x^2-2 x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-6 + 16*x - 12*x^2)/(-21 + 3*x - 4*x^2 + 2*x^3),x]

[Out]

-2*Log[21 - 3*x + 4*x^2 - 2*x^3]

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Maple [A]
time = 0.03, size = 19, normalized size = 0.79

method	result	size
default	\(-2 \ln \left (2 x^{3}-4 x^{2}+3 x -21\right )\)	\(19\)
norman	\(-2 \ln \left (2 x^{3}-4 x^{2}+3 x -21\right )\)	\(19\)
risch	\(-2 \ln \left (2 x^{3}-4 x^{2}+3 x -21\right )\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-12*x^2+16*x-6)/(2*x^3-4*x^2+3*x-21),x,method=_RETURNVERBOSE)

[Out]

-2*ln(2*x^3-4*x^2+3*x-21)

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Maxima [A]
time = 0.26, size = 18, normalized size = 0.75 \begin {gather*} -2 \, \log \left (2 \, x^{3} - 4 \, x^{2} + 3 \, x - 21\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*x^2+16*x-6)/(2*x^3-4*x^2+3*x-21),x, algorithm="maxima")

[Out]

-2*log(2*x^3 - 4*x^2 + 3*x - 21)

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Fricas [A]
time = 0.34, size = 18, normalized size = 0.75 \begin {gather*} -2 \, \log \left (2 \, x^{3} - 4 \, x^{2} + 3 \, x - 21\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*x^2+16*x-6)/(2*x^3-4*x^2+3*x-21),x, algorithm="fricas")

[Out]

-2*log(2*x^3 - 4*x^2 + 3*x - 21)

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Sympy [A]
time = 0.03, size = 19, normalized size = 0.79 \begin {gather*} - 2 \log {\left (2 x^{3} - 4 x^{2} + 3 x - 21 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*x**2+16*x-6)/(2*x**3-4*x**2+3*x-21),x)

[Out]

-2*log(2*x**3 - 4*x**2 + 3*x - 21)

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Giac [A]
time = 0.37, size = 19, normalized size = 0.79 \begin {gather*} -2 \, \log \left ({\left | 2 \, x^{3} - 4 \, x^{2} + 3 \, x - 21 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*x^2+16*x-6)/(2*x^3-4*x^2+3*x-21),x, algorithm="giac")

[Out]

-2*log(abs(2*x^3 - 4*x^2 + 3*x - 21))

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Mupad [B]
time = 0.07, size = 16, normalized size = 0.67 \begin {gather*} -2\,\ln \left (x^3-2\,x^2+\frac {3\,x}{2}-\frac {21}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(12*x^2 - 16*x + 6)/(3*x - 4*x^2 + 2*x^3 - 21),x)

[Out]

-2*log((3*x)/2 - 2*x^2 + x^3 - 21/2)

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