\(\int \frac {-4+6 x+(-2+2 x+(2-2 x) \log (2 x^2-2 x^3)) \log (1-\log (2 x^2-2 x^3))}{x^2-x^3+(-x^2+x^3) \log (2 x^2-2 x^3)} \, dx\) [6238]

Optimal result

Integrand size = 80, antiderivative size = 23 \[ \int \frac {-4+6 x+\left (-2+2 x+(2-2 x) \log \left (2 x^2-2 x^3\right )\right ) \log \left (1-\log \left (2 x^2-2 x^3\right )\right )}{x^2-x^3+\left (-x^2+x^3\right ) \log \left (2 x^2-2 x^3\right )} \, dx=\frac {2 \left (x+\log \left (1-\log \left (2 (1-x) x^2\right )\right )\right )}{x} \]

[Out]

2*(ln(1-ln(x^2*(2-2*x)))+x)/x

Rubi [F]

\[ \int \frac {-4+6 x+\left (-2+2 x+(2-2 x) \log \left (2 x^2-2 x^3\right )\right ) \log \left (1-\log \left (2 x^2-2 x^3\right )\right )}{x^2-x^3+\left (-x^2+x^3\right ) \log \left (2 x^2-2 x^3\right )} \, dx=\int \frac {-4+6 x+\left (-2+2 x+(2-2 x) \log \left (2 x^2-2 x^3\right )\right ) \log \left (1-\log \left (2 x^2-2 x^3\right )\right )}{x^2-x^3+\left (-x^2+x^3\right ) \log \left (2 x^2-2 x^3\right )} \, dx \]

[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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {-4+6 x+\left (-2+2 x+(2-2 x) \log \left (2 x^2-2 x^3\right )\right ) \log \left (1-\log \left (2 x^2-2 x^3\right )\right )}{x^2-x^3+\left (-x^2+x^3\right ) \log \left (2 x^2-2 x^3\right )} \, dx=\frac {2 \log \left (1-\log \left (-2 (-1+x) x^2\right )\right )}{x} \]

[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

Maple [A] (verified)

Time = 1.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00

method	result	size
parallelrisch	\(\frac {2 \ln \left (-\ln \left (-2 x^{3}+2 x^{2}\right )+1\right )}{x}\)	\(23\)

[In]

int((((2-2*x)*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,me
thod=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {-4+6 x+\left (-2+2 x+(2-2 x) \log \left (2 x^2-2 x^3\right )\right ) \log \left (1-\log \left (2 x^2-2 x^3\right )\right )}{x^2-x^3+\left (-x^2+x^3\right ) \log \left (2 x^2-2 x^3\right )} \, dx=\frac {2 \, \log \left (-\log \left (-2 \, x^{3} + 2 \, x^{2}\right ) + 1\right )}{x} \]

[In]

integrate((((2-2*x)*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

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {-4+6 x+\left (-2+2 x+(2-2 x) \log \left (2 x^2-2 x^3\right )\right ) \log \left (1-\log \left (2 x^2-2 x^3\right )\right )}{x^2-x^3+\left (-x^2+x^3\right ) \log \left (2 x^2-2 x^3\right )} \, dx=\frac {2 \log {\left (1 - \log {\left (- 2 x^{3} + 2 x^{2} \right )} \right )}}{x} \]

[In]

integrate((((2-2*x)*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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {-4+6 x+\left (-2+2 x+(2-2 x) \log \left (2 x^2-2 x^3\right )\right ) \log \left (1-\log \left (2 x^2-2 x^3\right )\right )}{x^2-x^3+\left (-x^2+x^3\right ) \log \left (2 x^2-2 x^3\right )} \, dx=\frac {2 \, \log \left (-i \, \pi - \log \left (2\right ) - \log \left (x - 1\right ) - 2 \, \log \left (x\right ) + 1\right )}{x} \]

[In]

integrate((((2-2*x)*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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {-4+6 x+\left (-2+2 x+(2-2 x) \log \left (2 x^2-2 x^3\right )\right ) \log \left (1-\log \left (2 x^2-2 x^3\right )\right )}{x^2-x^3+\left (-x^2+x^3\right ) \log \left (2 x^2-2 x^3\right )} \, dx=\frac {2 \, \log \left (-\log \left (-2 \, x^{3} + 2 \, x^{2}\right ) + 1\right )}{x} \]

[In]

integrate((((2-2*x)*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

Mupad [B] (verification not implemented)

Time = 12.99 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {-4+6 x+\left (-2+2 x+(2-2 x) \log \left (2 x^2-2 x^3\right )\right ) \log \left (1-\log \left (2 x^2-2 x^3\right )\right )}{x^2-x^3+\left (-x^2+x^3\right ) \log \left (2 x^2-2 x^3\right )} \, dx=\frac {2\,\ln \left (1-\ln \left (2\,x^2-2\,x^3\right )\right )}{x} \]

[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