3.10.11 \(\int \frac {-2+\log (x)}{1-2 x+x^2+(-2+2 x) \log (x)+\log ^2(x)} \, dx\)

Optimal. Leaf size=23 \[ \frac {3}{4} \left (-1+5 e^3\right )^2+\frac {x}{-1+x+\log (x)} \]

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Rubi [F]  time = 0.17, 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 {-2+\log (x)}{1-2 x+x^2+(-2+2 x) \log (x)+\log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-2 + Log[x])/(1 - 2*x + x^2 + (-2 + 2*x)*Log[x] + Log[x]^2),x]

[Out]

-Defer[Int][(-1 + x + Log[x])^(-2), x] - Defer[Int][x/(-1 + x + Log[x])^2, x] + Defer[Int][(-1 + x + Log[x])^(
-1), x]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 9, normalized size = 0.39 \begin {gather*} \frac {x}{-1+x+\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 + Log[x])/(1 - 2*x + x^2 + (-2 + 2*x)*Log[x] + Log[x]^2),x]

[Out]

x/(-1 + x + Log[x])

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fricas [A]  time = 0.86, size = 9, normalized size = 0.39 \begin {gather*} \frac {x}{x + \log \relax (x) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(x + log(x) - 1)

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giac [A]  time = 0.31, size = 9, normalized size = 0.39 \begin {gather*} \frac {x}{x + \log \relax (x) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(x + log(x) - 1)

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maple [A]  time = 0.02, size = 10, normalized size = 0.43

method	result	size
norman	\(\frac {x}{-1+\ln \relax (x )+x}\)	\(10\)
risch	\(\frac {x}{-1+\ln \relax (x )+x}\)	\(10\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(-1+ln(x)+x)

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maxima [A]  time = 0.56, size = 9, normalized size = 0.39 \begin {gather*} \frac {x}{x + \log \relax (x) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/(x + log(x) - 1)

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mupad [B]  time = 0.69, size = 9, normalized size = 0.39 \begin {gather*} \frac {x}{x+\ln \relax (x)-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x) - 2)/(log(x)^2 - 2*x + log(x)*(2*x - 2) + x^2 + 1),x)

[Out]

x/(x + log(x) - 1)

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sympy [A]  time = 0.09, size = 7, normalized size = 0.30 \begin {gather*} \frac {x}{x + \log {\relax (x )} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((ln(x)-2)/(ln(x)**2+(2*x-2)*ln(x)+x**2-2*x+1),x)

[Out]

x/(x + log(x) - 1)

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