3.61.46 \(\int \frac {-4-\log (x)}{25-10 x+x^2+(10-2 x) \log (x)+\log ^2(x)} \, dx\)

Optimal. Leaf size=14 \[ 4-\frac {x}{5-x+\log (x)} \]

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Rubi [F]  time = 0.16, 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 {-4-\log (x)}{25-10 x+x^2+(10-2 x) \log (x)+\log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-4 - Log[x])/(25 - 10*x + x^2 + (10 - 2*x)*Log[x] + Log[x]^2),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-4-\log (x)}{(5-x+\log (x))^2} \, dx\\ &=\int \left (\frac {1-x}{(-5+x-\log (x))^2}+\frac {1}{-5+x-\log (x)}\right ) \, dx\\ &=\int \frac {1-x}{(-5+x-\log (x))^2} \, dx+\int \frac {1}{-5+x-\log (x)} \, dx\\ &=\int \left (\frac {1}{(-5+x-\log (x))^2}-\frac {x}{(-5+x-\log (x))^2}\right ) \, dx+\int \frac {1}{-5+x-\log (x)} \, dx\\ &=\int \frac {1}{(-5+x-\log (x))^2} \, dx-\int \frac {x}{(-5+x-\log (x))^2} \, dx+\int \frac {1}{-5+x-\log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 12, normalized size = 0.86 \begin {gather*} -\frac {x}{5-x+\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 - Log[x])/(25 - 10*x + x^2 + (10 - 2*x)*Log[x] + Log[x]^2),x]

[Out]

-(x/(5 - x + Log[x]))

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fricas [A]  time = 0.64, size = 11, normalized size = 0.79 \begin {gather*} \frac {x}{x - \log \relax (x) - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)-4)/(log(x)^2+(-2*x+10)*log(x)+x^2-10*x+25),x, algorithm="fricas")

[Out]

x/(x - log(x) - 5)

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giac [A]  time = 0.17, size = 11, normalized size = 0.79 \begin {gather*} \frac {x}{x - \log \relax (x) - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)-4)/(log(x)^2+(-2*x+10)*log(x)+x^2-10*x+25),x, algorithm="giac")

[Out]

x/(x - log(x) - 5)

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maple [A]  time = 0.02, size = 12, normalized size = 0.86

method	result	size
risch	\(\frac {x}{-\ln \relax (x )+x -5}\)	\(12\)
norman	\(\frac {5+\ln \relax (x )}{-\ln \relax (x )+x -5}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-ln(x)-4)/(ln(x)^2+(-2*x+10)*ln(x)+x^2-10*x+25),x,method=_RETURNVERBOSE)

[Out]

x/(-ln(x)+x-5)

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maxima [A]  time = 0.37, size = 11, normalized size = 0.79 \begin {gather*} \frac {x}{x - \log \relax (x) - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)-4)/(log(x)^2+(-2*x+10)*log(x)+x^2-10*x+25),x, algorithm="maxima")

[Out]

x/(x - log(x) - 5)

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mupad [B]  time = 4.63, size = 12, normalized size = 0.86 \begin {gather*} -\frac {x}{\ln \relax (x)-x+5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x) + 4)/(log(x)^2 - 10*x - log(x)*(2*x - 10) + x^2 + 25),x)

[Out]

-x/(log(x) - x + 5)

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sympy [A]  time = 0.09, size = 8, normalized size = 0.57 \begin {gather*} - \frac {x}{- x + \log {\relax (x )} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-ln(x)-4)/(ln(x)**2+(-2*x+10)*ln(x)+x**2-10*x+25),x)

[Out]

-x/(-x + log(x) + 5)

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