3.63.38 \(\int \frac {-4+6 x+(-2+2 x+(2-2 x) \log (2 x^2-2 x^3)) \log (1-\log (2 x^2-2 x^3))}{x^2-x^3+(-x^2+x^3) \log (2 x^2-2 x^3)} \, dx\)

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

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Rubi [F]  time = 0.85, 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+6 x+\left (-2+2 x+(2-2 x) \log \left (2 x^2-2 x^3\right )\right ) \log \left (1-\log \left (2 x^2-2 x^3\right )\right )}{x^2-x^3+\left (-x^2+x^3\right ) \log \left (2 x^2-2 x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

2*Defer[Int][1/((-1 + x)*(-1 + Log[-2*(-1 + x)*x^2])), x] + 4*Defer[Int][1/(x^2*(-1 + Log[-2*(-1 + x)*x^2])),
x] - 2*Defer[Int][1/(x*(-1 + Log[-2*(-1 + x)*x^2])), x] - 2*Defer[Int][Log[1 - Log[-2*(-1 + x)*x^2]]/x^2, x]

Rubi steps

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

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Mathematica [A]  time = 0.17, size = 19, normalized size = 0.83 \begin {gather*} \frac {2 \log \left (1-\log \left (-2 (-1+x) x^2\right )\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Log[1 - Log[-2*(-1 + x)*x^2]])/x

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fricas [A]  time = 0.55, size = 22, normalized size = 0.96 \begin {gather*} \frac {2 \, \log \left (-\log \left (-2 \, x^{3} + 2 \, x^{2}\right ) + 1\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*log(-log(-2*x^3 + 2*x^2) + 1)/x

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giac [A]  time = 0.24, size = 22, normalized size = 0.96 \begin {gather*} \frac {2 \, \log \left (-\log \left (-2 \, x^{3} + 2 \, x^{2}\right ) + 1\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*log(-log(-2*x^3 + 2*x^2) + 1)/x

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (-2 x +2\right ) \ln \left (-2 x^{3}+2 x^{2}\right )+2 x -2\right ) \ln \left (-\ln \left (-2 x^{3}+2 x^{2}\right )+1\right )+6 x -4}{\left (x^{3}-x^{2}\right ) \ln \left (-2 x^{3}+2 x^{2}\right )-x^{3}+x^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-2*x+2)*ln(-2*x^3+2*x^2)+2*x-2)*ln(-ln(-2*x^3+2*x^2)+1)+6*x-4)/((x^3-x^2)*ln(-2*x^3+2*x^2)-x^3+x^2),x)

[Out]

int((((-2*x+2)*ln(-2*x^3+2*x^2)+2*x-2)*ln(-ln(-2*x^3+2*x^2)+1)+6*x-4)/((x^3-x^2)*ln(-2*x^3+2*x^2)-x^3+x^2),x)

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maxima [C]  time = 0.47, size = 25, normalized size = 1.09 \begin {gather*} \frac {2 \, \log \left (-i \, \pi - \log \relax (2) - \log \left (x - 1\right ) - 2 \, \log \relax (x) + 1\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*log(-I*pi - log(2) - log(x - 1) - 2*log(x) + 1)/x

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mupad [B]  time = 4.57, size = 22, normalized size = 0.96 \begin {gather*} \frac {2\,\ln \left (1-\ln \left (2\,x^2-2\,x^3\right )\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*log(1 - log(2*x^2 - 2*x^3)))/x

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sympy [A]  time = 0.43, size = 17, normalized size = 0.74 \begin {gather*} \frac {2 \log {\left (1 - \log {\left (- 2 x^{3} + 2 x^{2} \right )} \right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*log(1 - log(-2*x**3 + 2*x**2))/x

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