3.67.52 \(\int \frac {-20+(10 x-35 x^2+30 x^3) \log (\frac {2-3 x}{2 x})}{(-2 x+3 x^2) \log (\frac {2-3 x}{2 x})} \, dx\)

Optimal. Leaf size=22 \[ x \left (-5+5 \left (x-\frac {2 \log \left (\log \left (-\frac {3}{2}+\frac {1}{x}\right )\right )}{x}\right )\right ) \]

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Rubi [F]  time = 0.39, 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+\left (10 x-35 x^2+30 x^3\right ) \log \left (\frac {2-3 x}{2 x}\right )}{\left (-2 x+3 x^2\right ) \log \left (\frac {2-3 x}{2 x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-20 + (10*x - 35*x^2 + 30*x^3)*Log[(2 - 3*x)/(2*x)])/((-2*x + 3*x^2)*Log[(2 - 3*x)/(2*x)]),x]

[Out]

-5*x + 5*x^2 + 10*Defer[Int][1/(x*Log[-3/2 + x^(-1)]), x] - 30*Defer[Int][1/((-2 + 3*x)*Log[-3/2 + x^(-1)]), x
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-20+\left (10 x-35 x^2+30 x^3\right ) \log \left (\frac {2-3 x}{2 x}\right )}{x (-2+3 x) \log \left (\frac {2-3 x}{2 x}\right )} \, dx\\ &=\int \frac {5 \left (-2+7 x-6 x^2+\frac {4}{x \log \left (-\frac {3}{2}+\frac {1}{x}\right )}\right )}{2-3 x} \, dx\\ &=5 \int \frac {-2+7 x-6 x^2+\frac {4}{x \log \left (-\frac {3}{2}+\frac {1}{x}\right )}}{2-3 x} \, dx\\ &=5 \int \left (-1+2 x-\frac {4}{x (-2+3 x) \log \left (-\frac {3}{2}+\frac {1}{x}\right )}\right ) \, dx\\ &=-5 x+5 x^2-20 \int \frac {1}{x (-2+3 x) \log \left (-\frac {3}{2}+\frac {1}{x}\right )} \, dx\\ &=-5 x+5 x^2-20 \int \left (-\frac {1}{2 x \log \left (-\frac {3}{2}+\frac {1}{x}\right )}+\frac {3}{2 (-2+3 x) \log \left (-\frac {3}{2}+\frac {1}{x}\right )}\right ) \, dx\\ &=-5 x+5 x^2+10 \int \frac {1}{x \log \left (-\frac {3}{2}+\frac {1}{x}\right )} \, dx-30 \int \frac {1}{(-2+3 x) \log \left (-\frac {3}{2}+\frac {1}{x}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 5.02, size = 20, normalized size = 0.91 \begin {gather*} 5 \left (-x+x^2-2 \log \left (\log \left (-\frac {3}{2}+\frac {1}{x}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20 + (10*x - 35*x^2 + 30*x^3)*Log[(2 - 3*x)/(2*x)])/((-2*x + 3*x^2)*Log[(2 - 3*x)/(2*x)]),x]

[Out]

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

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fricas [A]  time = 0.60, size = 23, normalized size = 1.05 \begin {gather*} 5 \, x^{2} - 5 \, x - 10 \, \log \left (\log \left (-\frac {3 \, x - 2}{2 \, x}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((30*x^3-35*x^2+10*x)*log(1/2*(-3*x+2)/x)-20)/(3*x^2-2*x)/log(1/2*(-3*x+2)/x),x, algorithm="fricas")

[Out]

5*x^2 - 5*x - 10*log(log(-1/2*(3*x - 2)/x))

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giac [B]  time = 0.25, size = 53, normalized size = 2.41 \begin {gather*} \frac {10 \, {\left (\frac {3 \, x - 2}{x} - 1\right )}}{\frac {{\left (3 \, x - 2\right )}^{2}}{x^{2}} - \frac {6 \, {\left (3 \, x - 2\right )}}{x} + 9} - 10 \, \log \left (\log \left (-\frac {3 \, x - 2}{2 \, x}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((30*x^3-35*x^2+10*x)*log(1/2*(-3*x+2)/x)-20)/(3*x^2-2*x)/log(1/2*(-3*x+2)/x),x, algorithm="giac")

[Out]

10*((3*x - 2)/x - 1)/((3*x - 2)^2/x^2 - 6*(3*x - 2)/x + 9) - 10*log(log(-1/2*(3*x - 2)/x))

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maple [A]  time = 0.08, size = 19, normalized size = 0.86




method result size



derivativedivides \(-10 \ln \left (\ln \left (\frac {1}{x}-\frac {3}{2}\right )\right )-5 x +5 x^{2}\) \(19\)
default \(-10 \ln \left (\ln \left (\frac {1}{x}-\frac {3}{2}\right )\right )-5 x +5 x^{2}\) \(19\)
norman \(-5 x +5 x^{2}-10 \ln \left (\ln \left (\frac {-3 x +2}{2 x}\right )\right )\) \(24\)
risch \(-5 x +5 x^{2}-10 \ln \left (\ln \left (\frac {-3 x +2}{2 x}\right )\right )\) \(24\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((30*x^3-35*x^2+10*x)*ln(1/2*(-3*x+2)/x)-20)/(3*x^2-2*x)/ln(1/2*(-3*x+2)/x),x,method=_RETURNVERBOSE)

[Out]

-10*ln(ln(1/x-3/2))-5*x+5*x^2

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maxima [A]  time = 0.53, size = 27, normalized size = 1.23 \begin {gather*} 5 \, x^{2} - 5 \, x - 10 \, \log \left (-\log \relax (2) - \log \relax (x) + \log \left (-3 \, x + 2\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((30*x^3-35*x^2+10*x)*log(1/2*(-3*x+2)/x)-20)/(3*x^2-2*x)/log(1/2*(-3*x+2)/x),x, algorithm="maxima")

[Out]

5*x^2 - 5*x - 10*log(-log(2) - log(x) + log(-3*x + 2))

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mupad [B]  time = 4.49, size = 23, normalized size = 1.05 \begin {gather*} 5\,x^2-10\,\ln \left (\ln \left (-\frac {3\,x-2}{2\,x}\right )\right )-5\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(-((3*x)/2 - 1)/x)*(10*x - 35*x^2 + 30*x^3) - 20)/(log(-((3*x)/2 - 1)/x)*(2*x - 3*x^2)),x)

[Out]

5*x^2 - 10*log(log(-(3*x - 2)/(2*x))) - 5*x

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sympy [A]  time = 0.15, size = 20, normalized size = 0.91 \begin {gather*} 5 x^{2} - 5 x - 10 \log {\left (\log {\left (\frac {1 - \frac {3 x}{2}}{x} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((30*x**3-35*x**2+10*x)*ln(1/2*(-3*x+2)/x)-20)/(3*x**2-2*x)/ln(1/2*(-3*x+2)/x),x)

[Out]

5*x**2 - 5*x - 10*log(log((1 - 3*x/2)/x))

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