\(\int \frac {(6-3 x) \log (-4+2 x)+(2-4 x+2 x^2) \log (\log (-4+2 x))}{(-6+15 x-12 x^2+3 x^3) \log (-4+2 x)} \, dx\) [3264]

Optimal result

Integrand size = 57, antiderivative size = 27 \[ \int \frac {(6-3 x) \log (-4+2 x)+\left (2-4 x+2 x^2\right ) \log (\log (-4+2 x))}{\left (-6+15 x-12 x^2+3 x^3\right ) \log (-4+2 x)} \, dx=1-\frac {x}{x-x^2}+\frac {1}{3} \log ^2(\log (-4+2 x)) \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {6820, 6818} \[ \int \frac {(6-3 x) \log (-4+2 x)+\left (2-4 x+2 x^2\right ) \log (\log (-4+2 x))}{\left (-6+15 x-12 x^2+3 x^3\right ) \log (-4+2 x)} \, dx=\frac {1}{x-1}+\frac {1}{3} \log ^2(\log (2 x-4)) \]

[In]

Int[((6 - 3*x)*Log[-4 + 2*x] + (2 - 4*x + 2*x^2)*Log[Log[-4 + 2*x]])/((-6 + 15*x - 12*x^2 + 3*x^3)*Log[-4 + 2*
x]),x]

[Out]

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

Rule 6818

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*(y^(m + 1)/(m + 1)), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {(6-3 x) \log (-4+2 x)+\left (2-4 x+2 x^2\right ) \log (\log (-4+2 x))}{\left (-6+15 x-12 x^2+3 x^3\right ) \log (-4+2 x)} \, dx=\frac {1}{-1+x}+\frac {1}{3} \log ^2(\log (2 (-2+x))) \]

[In]

Integrate[((6 - 3*x)*Log[-4 + 2*x] + (2 - 4*x + 2*x^2)*Log[Log[-4 + 2*x]])/((-6 + 15*x - 12*x^2 + 3*x^3)*Log[-
4 + 2*x]),x]

[Out]

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

Maple [A] (verified)

Time = 0.85 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67

method	result	size
risch	\(\frac {\ln \left (\ln \left (2 x -4\right )\right )^{2}}{3}+\frac {1}{-1+x}\)	\(18\)
parts	\(\frac {\ln \left (\ln \left (2 x -4\right )\right )^{2}}{3}+\frac {1}{-1+x}\)	\(18\)
default	\(\frac {1}{-1+x}+\frac {\ln \left (\ln \left (2\right )+\ln \left (-2+x \right )\right )^{2}}{3}\)	\(19\)
parallelrisch	\(\frac {12+5 x \ln \left (\ln \left (2 x -4\right )\right )^{2}-5 \ln \left (\ln \left (2 x -4\right )\right )^{2}+3 x}{15 x -15}\)	\(36\)

[In]

int(((2*x^2-4*x+2)*ln(ln(2*x-4))+(-3*x+6)*ln(2*x-4))/(3*x^3-12*x^2+15*x-6)/ln(2*x-4),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {(6-3 x) \log (-4+2 x)+\left (2-4 x+2 x^2\right ) \log (\log (-4+2 x))}{\left (-6+15 x-12 x^2+3 x^3\right ) \log (-4+2 x)} \, dx=\frac {{\left (x - 1\right )} \log \left (\log \left (2 \, x - 4\right )\right )^{2} + 3}{3 \, {\left (x - 1\right )}} \]

[In]

integrate(((2*x^2-4*x+2)*log(log(2*x-4))+(-3*x+6)*log(2*x-4))/(3*x^3-12*x^2+15*x-6)/log(2*x-4),x, algorithm="f
ricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.56 \[ \int \frac {(6-3 x) \log (-4+2 x)+\left (2-4 x+2 x^2\right ) \log (\log (-4+2 x))}{\left (-6+15 x-12 x^2+3 x^3\right ) \log (-4+2 x)} \, dx=\frac {\log {\left (\log {\left (2 x - 4 \right )} \right )}^{2}}{3} + \frac {1}{x - 1} \]

[In]

integrate(((2*x**2-4*x+2)*ln(ln(2*x-4))+(-3*x+6)*ln(2*x-4))/(3*x**3-12*x**2+15*x-6)/ln(2*x-4),x)

[Out]

log(log(2*x - 4))**2/3 + 1/(x - 1)

Maxima [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.67 \[ \int \frac {(6-3 x) \log (-4+2 x)+\left (2-4 x+2 x^2\right ) \log (\log (-4+2 x))}{\left (-6+15 x-12 x^2+3 x^3\right ) \log (-4+2 x)} \, dx=\frac {1}{3} \, \log \left (\log \left (2\right ) + \log \left (x - 2\right )\right )^{2} + \frac {1}{x - 1} \]

[In]

integrate(((2*x^2-4*x+2)*log(log(2*x-4))+(-3*x+6)*log(2*x-4))/(3*x^3-12*x^2+15*x-6)/log(2*x-4),x, algorithm="m
axima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.63 \[ \int \frac {(6-3 x) \log (-4+2 x)+\left (2-4 x+2 x^2\right ) \log (\log (-4+2 x))}{\left (-6+15 x-12 x^2+3 x^3\right ) \log (-4+2 x)} \, dx=\frac {1}{3} \, \log \left (\log \left (2 \, x - 4\right )\right )^{2} + \frac {1}{x - 1} \]

[In]

integrate(((2*x^2-4*x+2)*log(log(2*x-4))+(-3*x+6)*log(2*x-4))/(3*x^3-12*x^2+15*x-6)/log(2*x-4),x, algorithm="g
iac")

[Out]

1/3*log(log(2*x - 4))^2 + 1/(x - 1)

Mupad [B] (verification not implemented)

Time = 9.68 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.63 \[ \int \frac {(6-3 x) \log (-4+2 x)+\left (2-4 x+2 x^2\right ) \log (\log (-4+2 x))}{\left (-6+15 x-12 x^2+3 x^3\right ) \log (-4+2 x)} \, dx=\frac {1}{x-1}+\frac {{\ln \left (\ln \left (2\,x-4\right )\right )}^2}{3} \]

[In]

int(-(log(2*x - 4)*(3*x - 6) - log(log(2*x - 4))*(2*x^2 - 4*x + 2))/(log(2*x - 4)*(15*x - 12*x^2 + 3*x^3 - 6))
,x)

[Out]

1/(x - 1) + log(log(2*x - 4))^2/3