3.88.39 \(\int \frac {-1+2 x+(12 x-4 x^2) \log (x)+(17 x-12 x^2+2 x^3) \log ^2(x)}{x+(6 x-2 x^2) \log (x)+(9 x-6 x^2+x^3) \log ^2(x)} \, dx\)

Optimal. Leaf size=17 \[ 2 x-\frac {1}{3-x+\frac {1}{\log (x)}} \]

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Rubi [F]  time = 0.66, 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 {-1+2 x+\left (12 x-4 x^2\right ) \log (x)+\left (17 x-12 x^2+2 x^3\right ) \log ^2(x)}{x+\left (6 x-2 x^2\right ) \log (x)+\left (9 x-6 x^2+x^3\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-1 + 2*x + (12*x - 4*x^2)*Log[x] + (17*x - 12*x^2 + 2*x^3)*Log[x]^2)/(x + (6*x - 2*x^2)*Log[x] + (9*x - 6
*x^2 + x^3)*Log[x]^2),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-1+2 x-4 (-3+x) x \log (x)+x \left (17-12 x+2 x^2\right ) \log ^2(x)}{x (1-(-3+x) \log (x))^2} \, dx\\ &=\int \left (\frac {17-12 x+2 x^2}{(-3+x)^2}+\frac {-9+5 x-x^2}{(-3+x)^2 x (-1-3 \log (x)+x \log (x))^2}-\frac {2}{(-3+x)^2 (-1-3 \log (x)+x \log (x))}\right ) \, dx\\ &=-\left (2 \int \frac {1}{(-3+x)^2 (-1-3 \log (x)+x \log (x))} \, dx\right )+\int \frac {17-12 x+2 x^2}{(-3+x)^2} \, dx+\int \frac {-9+5 x-x^2}{(-3+x)^2 x (-1-3 \log (x)+x \log (x))^2} \, dx\\ &=-\left (2 \int \frac {1}{(-3+x)^2 (-1-3 \log (x)+x \log (x))} \, dx\right )+\int \left (2-\frac {1}{(-3+x)^2}\right ) \, dx+\int \left (-\frac {1}{(-3+x)^2 (-1-3 \log (x)+x \log (x))^2}-\frac {1}{x (-1-3 \log (x)+x \log (x))^2}\right ) \, dx\\ &=\frac {1}{-3+x}+2 x-2 \int \frac {1}{(-3+x)^2 (-1-3 \log (x)+x \log (x))} \, dx-\int \frac {1}{(-3+x)^2 (-1-3 \log (x)+x \log (x))^2} \, dx-\int \frac {1}{x (-1-3 \log (x)+x \log (x))^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 27, normalized size = 1.59 \begin {gather*} \frac {1}{-3+x}+2 x+\frac {1}{(-3+x) (-1-3 \log (x)+x \log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + 2*x + (12*x - 4*x^2)*Log[x] + (17*x - 12*x^2 + 2*x^3)*Log[x]^2)/(x + (6*x - 2*x^2)*Log[x] + (9
*x - 6*x^2 + x^3)*Log[x]^2),x]

[Out]

(-3 + x)^(-1) + 2*x + 1/((-3 + x)*(-1 - 3*Log[x] + x*Log[x]))

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fricas [A]  time = 0.60, size = 28, normalized size = 1.65 \begin {gather*} \frac {{\left (2 \, x^{2} - 6 \, x + 1\right )} \log \relax (x) - 2 \, x}{{\left (x - 3\right )} \log \relax (x) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3-12*x^2+17*x)*log(x)^2+(-4*x^2+12*x)*log(x)+2*x-1)/((x^3-6*x^2+9*x)*log(x)^2+(-2*x^2+6*x)*log
(x)+x),x, algorithm="fricas")

[Out]

((2*x^2 - 6*x + 1)*log(x) - 2*x)/((x - 3)*log(x) - 1)

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giac [B]  time = 0.23, size = 31, normalized size = 1.82 \begin {gather*} 2 \, x + \frac {1}{x^{2} \log \relax (x) - 6 \, x \log \relax (x) - x + 9 \, \log \relax (x) + 3} + \frac {1}{x - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3-12*x^2+17*x)*log(x)^2+(-4*x^2+12*x)*log(x)+2*x-1)/((x^3-6*x^2+9*x)*log(x)^2+(-2*x^2+6*x)*log
(x)+x),x, algorithm="giac")

[Out]

2*x + 1/(x^2*log(x) - 6*x*log(x) - x + 9*log(x) + 3) + 1/(x - 3)

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maple [A]  time = 0.05, size = 30, normalized size = 1.76

method	result	size
norman	\(\frac {-17 \ln \relax (x )-2 x +2 x^{2} \ln \relax (x )-6}{x \ln \relax (x )-3 \ln \relax (x )-1}\)	\(30\)
risch	\(\frac {2 x^{2}-6 x +1}{x -3}+\frac {1}{\left (x -3\right ) \left (x \ln \relax (x )-3 \ln \relax (x )-1\right )}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^3-12*x^2+17*x)*ln(x)^2+(-4*x^2+12*x)*ln(x)+2*x-1)/((x^3-6*x^2+9*x)*ln(x)^2+(-2*x^2+6*x)*ln(x)+x),x,m
ethod=_RETURNVERBOSE)

[Out]

(-17*ln(x)-2*x+2*x^2*ln(x)-6)/(x*ln(x)-3*ln(x)-1)

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maxima [A]  time = 0.39, size = 28, normalized size = 1.65 \begin {gather*} \frac {{\left (2 \, x^{2} - 6 \, x + 1\right )} \log \relax (x) - 2 \, x}{{\left (x - 3\right )} \log \relax (x) - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3-12*x^2+17*x)*log(x)^2+(-4*x^2+12*x)*log(x)+2*x-1)/((x^3-6*x^2+9*x)*log(x)^2+(-2*x^2+6*x)*log
(x)+x),x, algorithm="maxima")

[Out]

((2*x^2 - 6*x + 1)*log(x) - 2*x)/((x - 3)*log(x) - 1)

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mupad [B]  time = 5.54, size = 21, normalized size = 1.24 \begin {gather*} 2\,x-\frac {\ln \relax (x)}{3\,\ln \relax (x)-x\,\ln \relax (x)+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x + log(x)^2*(17*x - 12*x^2 + 2*x^3) + log(x)*(12*x - 4*x^2) - 1)/(x + log(x)*(6*x - 2*x^2) + log(x)^2*
(9*x - 6*x^2 + x^3)),x)

[Out]

2*x - log(x)/(3*log(x) - x*log(x) + 1)

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sympy [A]  time = 0.16, size = 24, normalized size = 1.41 \begin {gather*} 2 x + \frac {1}{- x + \left (x^{2} - 6 x + 9\right ) \log {\relax (x )} + 3} + \frac {1}{x - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**3-12*x**2+17*x)*ln(x)**2+(-4*x**2+12*x)*ln(x)+2*x-1)/((x**3-6*x**2+9*x)*ln(x)**2+(-2*x**2+6*x
)*ln(x)+x),x)

[Out]

2*x + 1/(-x + (x**2 - 6*x + 9)*log(x) + 3) + 1/(x - 3)

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