3.81.61 \(\int \frac {8-12 x}{-x+2 x^2} \, dx\)

Optimal. Leaf size=15 \[ \log \left (\frac {(1-2 x)^2}{e^3 x^8}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 13, normalized size of antiderivative = 0.87, number of steps used = 2, number of rules used = 1, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {631} \begin {gather*} 2 \log (1-2 x)-8 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(8 - 12*x)/(-x + 2*x^2),x]

[Out]

2*Log[1 - 2*x] - 8*Log[x]

Rule 631

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)
*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0]
|| EqQ[a, 0])

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 13, normalized size = 0.87 \begin {gather*} 2 \log (1-2 x)-8 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8 - 12*x)/(-x + 2*x^2),x]

[Out]

2*Log[1 - 2*x] - 8*Log[x]

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fricas [A]  time = 0.61, size = 13, normalized size = 0.87 \begin {gather*} 2 \, \log \left (2 \, x - 1\right ) - 8 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*x+8)/(2*x^2-x),x, algorithm="fricas")

[Out]

2*log(2*x - 1) - 8*log(x)

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giac [A]  time = 0.32, size = 15, normalized size = 1.00 \begin {gather*} 2 \, \log \left ({\left | 2 \, x - 1 \right |}\right ) - 8 \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*x+8)/(2*x^2-x),x, algorithm="giac")

[Out]

2*log(abs(2*x - 1)) - 8*log(abs(x))

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maple [A]  time = 0.31, size = 14, normalized size = 0.93

method	result	size
default	\(2 \ln \left (2 x -1\right )-8 \ln \relax (x )\)	\(14\)
norman	\(2 \ln \left (2 x -1\right )-8 \ln \relax (x )\)	\(14\)
risch	\(2 \ln \left (2 x -1\right )-8 \ln \relax (x )\)	\(14\)
meijerg	\(-8 \ln \relax (x )-8 \ln \relax (2)-8 i \pi +2 \ln \left (1-2 x \right )\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-12*x+8)/(2*x^2-x),x,method=_RETURNVERBOSE)

[Out]

2*ln(2*x-1)-8*ln(x)

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maxima [A]  time = 0.36, size = 13, normalized size = 0.87 \begin {gather*} 2 \, \log \left (2 \, x - 1\right ) - 8 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*x+8)/(2*x^2-x),x, algorithm="maxima")

[Out]

2*log(2*x - 1) - 8*log(x)

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mupad [B]  time = 0.08, size = 11, normalized size = 0.73 \begin {gather*} 2\,\ln \left (x-\frac {1}{2}\right )-8\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((12*x - 8)/(x - 2*x^2),x)

[Out]

2*log(x - 1/2) - 8*log(x)

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sympy [A]  time = 0.11, size = 12, normalized size = 0.80 \begin {gather*} - 8 \log {\relax (x )} + 2 \log {\left (x - \frac {1}{2} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*x+8)/(2*x**2-x),x)

[Out]

-8*log(x) + 2*log(x - 1/2)

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