3.62.77 \(\int \frac {-8+8 x}{(-x+2 x^2) \log (\frac {50-100 x}{3 x^2})} \, dx\)

Optimal. Leaf size=20 \[ 4 \left (2-\log \left (\log \left (\frac {50 (1-2 x)}{3 x^2}\right )\right )\right ) \]

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Rubi [F]  time = 0.10, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-8+8 x}{\left (-x+2 x^2\right ) \log \left (\frac {50-100 x}{3 x^2}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-8 + 8*x)/((-x + 2*x^2)*Log[(50 - 100*x)/(3*x^2)]),x]

[Out]

Defer[Int][(-8 + 8*x)/(x*(-1 + 2*x)*Log[(50 - 100*x)/(3*x^2)]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-8+8 x}{x (-1+2 x) \log \left (\frac {50-100 x}{3 x^2}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 16, normalized size = 0.80 \begin {gather*} -4 \log \left (\log \left (-\frac {50 (-1+2 x)}{3 x^2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8 + 8*x)/((-x + 2*x^2)*Log[(50 - 100*x)/(3*x^2)]),x]

[Out]

-4*Log[Log[(-50*(-1 + 2*x))/(3*x^2)]]

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fricas [A]  time = 0.72, size = 14, normalized size = 0.70 \begin {gather*} -4 \, \log \left (\log \left (-\frac {50 \, {\left (2 \, x - 1\right )}}{3 \, x^{2}}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x-8)/(2*x^2-x)/log(1/3*(-100*x+50)/x^2),x, algorithm="fricas")

[Out]

-4*log(log(-50/3*(2*x - 1)/x^2))

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giac [A]  time = 0.26, size = 14, normalized size = 0.70 \begin {gather*} -4 \, \log \left (\log \left (-\frac {50 \, {\left (2 \, x - 1\right )}}{3 \, x^{2}}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x-8)/(2*x^2-x)/log(1/3*(-100*x+50)/x^2),x, algorithm="giac")

[Out]

-4*log(log(-50/3*(2*x - 1)/x^2))

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maple [A]  time = 0.35, size = 15, normalized size = 0.75

method	result	size
norman	\(-4 \ln \left (\ln \left (\frac {-100 x +50}{3 x^{2}}\right )\right )\)	\(15\)
risch	\(-4 \ln \left (\ln \left (\frac {-100 x +50}{3 x^{2}}\right )\right )\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x-8)/(2*x^2-x)/ln(1/3*(-100*x+50)/x^2),x,method=_RETURNVERBOSE)

[Out]

-4*ln(ln(1/3*(-100*x+50)/x^2))

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maxima [C]  time = 0.46, size = 27, normalized size = 1.35 \begin {gather*} -4 \, \log \left (i \, \pi + 2 \, \log \relax (5) - \log \relax (3) + \log \relax (2) + \log \left (2 \, x - 1\right ) - 2 \, \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x-8)/(2*x^2-x)/log(1/3*(-100*x+50)/x^2),x, algorithm="maxima")

[Out]

-4*log(I*pi + 2*log(5) - log(3) + log(2) + log(2*x - 1) - 2*log(x))

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mupad [B]  time = 4.69, size = 14, normalized size = 0.70 \begin {gather*} -4\,\ln \left (\ln \left (-\frac {100\,x-50}{3\,x^2}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(8*x - 8)/(log(-((100*x)/3 - 50/3)/x^2)*(x - 2*x^2)),x)

[Out]

-4*log(log(-(100*x - 50)/(3*x^2)))

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sympy [A]  time = 0.18, size = 17, normalized size = 0.85 \begin {gather*} - 4 \log {\left (\log {\left (\frac {\frac {50}{3} - \frac {100 x}{3}}{x^{2}} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x-8)/(2*x**2-x)/ln(1/3*(-100*x+50)/x**2),x)

[Out]

-4*log(log((50/3 - 100*x/3)/x**2))

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