3.99.5 \(\int \frac {-20+5 x}{(-2 x+x^2) \log ^2(\frac {2-x}{x^2})} \, dx\)

Optimal. Leaf size=16 \[ \frac {5}{\log \left (\frac {2}{x^2}-\frac {1}{x}\right )} \]

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Rubi [F]  time = 0.09, 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 {-20+5 x}{\left (-2 x+x^2\right ) \log ^2\left (\frac {2-x}{x^2}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-20 + 5*x)/((-2*x + x^2)*Log[(2 - x)/x^2]^2),x]

[Out]

Defer[Int][(-20 + 5*x)/((-2 + x)*x*Log[(2 - x)/x^2]^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-20+5 x}{(-2+x) x \log ^2\left (\frac {2-x}{x^2}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 14, normalized size = 0.88 \begin {gather*} \frac {5}{\log \left (\frac {2-x}{x^2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20 + 5*x)/((-2*x + x^2)*Log[(2 - x)/x^2]^2),x]

[Out]

5/Log[(2 - x)/x^2]

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fricas [A]  time = 0.75, size = 13, normalized size = 0.81 \begin {gather*} \frac {5}{\log \left (-\frac {x - 2}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x-20)/(x^2-2*x)/log((2-x)/x^2)^2,x, algorithm="fricas")

[Out]

5/log(-(x - 2)/x^2)

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giac [A]  time = 0.17, size = 13, normalized size = 0.81 \begin {gather*} \frac {5}{\log \left (-\frac {x - 2}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x-20)/(x^2-2*x)/log((2-x)/x^2)^2,x, algorithm="giac")

[Out]

5/log(-(x - 2)/x^2)

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maple [A]  time = 0.10, size = 15, normalized size = 0.94

method	result	size
norman	\(\frac {5}{\ln \left (\frac {2-x}{x^{2}}\right )}\)	\(15\)
risch	\(\frac {5}{\ln \left (\frac {2-x}{x^{2}}\right )}\)	\(15\)
derivativedivides	\(\frac {5}{\ln \left (\frac {\frac {2}{x}-1}{x}\right )}\)	\(17\)
default	\(\frac {5}{\ln \left (\frac {\frac {2}{x}-1}{x}\right )}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x-20)/(x^2-2*x)/ln((2-x)/x^2)^2,x,method=_RETURNVERBOSE)

[Out]

5/ln((2-x)/x^2)

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maxima [A]  time = 0.39, size = 17, normalized size = 1.06 \begin {gather*} -\frac {5}{2 \, \log \relax (x) - \log \left (-x + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x-20)/(x^2-2*x)/log((2-x)/x^2)^2,x, algorithm="maxima")

[Out]

-5/(2*log(x) - log(-x + 2))

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mupad [B]  time = 5.73, size = 13, normalized size = 0.81 \begin {gather*} \frac {5}{\ln \left (-\frac {x-2}{x^2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*x - 20)/(log(-(x - 2)/x^2)^2*(2*x - x^2)),x)

[Out]

5/log(-(x - 2)/x^2)

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sympy [A]  time = 0.12, size = 8, normalized size = 0.50 \begin {gather*} \frac {5}{\log {\left (\frac {2 - x}{x^{2}} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x-20)/(x**2-2*x)/ln((2-x)/x**2)**2,x)

[Out]

5/log((2 - x)/x**2)

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