Integrand size = 134, antiderivative size = 27 \[ \int \frac {12-16 x}{\left (-x+x^2\right ) \log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right )+\left (-2 x+2 x^2\right ) \log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right ) \log \left (\log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right )\right )+\left (-x+x^2\right ) \log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right ) \log ^2\left (\log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right )\right )} \, dx=\frac {4}{1+\log \left (\log \left (\frac {x^2 \left (-x+x^2\right )}{9 \log (\log (4))}\right )\right )} \]
Time = 0.04 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {12-16 x}{\left (-x+x^2\right ) \log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right )+\left (-2 x+2 x^2\right ) \log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right ) \log \left (\log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right )\right )+\left (-x+x^2\right ) \log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right ) \log ^2\left (\log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right )\right )} \, dx=\frac {4}{1+\log \left (\log \left (\frac {(-1+x) x^3}{9 \log (\log (4))}\right )\right )} \]
Integrate[(12 - 16*x)/((-x + x^2)*Log[(-x^3 + x^4)/(9*Log[Log[4]])] + (-2* x + 2*x^2)*Log[(-x^3 + x^4)/(9*Log[Log[4]])]*Log[Log[(-x^3 + x^4)/(9*Log[L og[4]])]] + (-x + x^2)*Log[(-x^3 + x^4)/(9*Log[Log[4]])]*Log[Log[(-x^3 + x ^4)/(9*Log[Log[4]])]]^2),x]
Time = 0.50 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {7239, 27, 25, 7237}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {12-16 x}{\left (x^2-x\right ) \log \left (\frac {x^4-x^3}{9 \log (\log (4))}\right ) \log ^2\left (\log \left (\frac {x^4-x^3}{9 \log (\log (4))}\right )\right )+\left (2 x^2-2 x\right ) \log \left (\frac {x^4-x^3}{9 \log (\log (4))}\right ) \log \left (\log \left (\frac {x^4-x^3}{9 \log (\log (4))}\right )\right )+\left (x^2-x\right ) \log \left (\frac {x^4-x^3}{9 \log (\log (4))}\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {4 (4 x-3)}{(1-x) x \log \left (\frac {(x-1) x^3}{9 \log (\log (4))}\right ) \left (\log \left (\log \left (\frac {(x-1) x^3}{9 \log (\log (4))}\right )\right )+1\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \int -\frac {3-4 x}{(1-x) x \log \left (-\frac {(1-x) x^3}{9 \log (\log (4))}\right ) \left (\log \left (\log \left (-\frac {(1-x) x^3}{9 \log (\log (4))}\right )\right )+1\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -4 \int \frac {3-4 x}{(1-x) x \log \left (-\frac {(1-x) x^3}{9 \log (\log (4))}\right ) \left (\log \left (\log \left (-\frac {(1-x) x^3}{9 \log (\log (4))}\right )\right )+1\right )^2}dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle \frac {4}{\log \left (\log \left (-\frac {(1-x) x^3}{9 \log (\log (4))}\right )\right )+1}\) |
Int[(12 - 16*x)/((-x + x^2)*Log[(-x^3 + x^4)/(9*Log[Log[4]])] + (-2*x + 2* x^2)*Log[(-x^3 + x^4)/(9*Log[Log[4]])]*Log[Log[(-x^3 + x^4)/(9*Log[Log[4]] )]] + (-x + x^2)*Log[(-x^3 + x^4)/(9*Log[Log[4]])]*Log[Log[(-x^3 + x^4)/(9 *Log[Log[4]])]]^2),x]
3.28.14.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.73 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89
method | result | size |
parallelrisch | \(\frac {4}{\ln \left (\ln \left (\frac {x^{3} \left (-1+x \right )}{9 \ln \left (2 \ln \left (2\right )\right )}\right )\right )+1}\) | \(24\) |
default | \(\frac {4}{\ln \left (-2 \ln \left (3\right )-\ln \left (\ln \left (2\right )+\ln \left (\ln \left (2\right )\right )\right )+\ln \left (x^{3} \left (-1+x \right )\right )\right )+1}\) | \(30\) |
int((-16*x+12)/((x^2-x)*ln(1/9*(x^4-x^3)/ln(2*ln(2)))*ln(ln(1/9*(x^4-x^3)/ ln(2*ln(2))))^2+(2*x^2-2*x)*ln(1/9*(x^4-x^3)/ln(2*ln(2)))*ln(ln(1/9*(x^4-x ^3)/ln(2*ln(2))))+(x^2-x)*ln(1/9*(x^4-x^3)/ln(2*ln(2)))),x,method=_RETURNV ERBOSE)
Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {12-16 x}{\left (-x+x^2\right ) \log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right )+\left (-2 x+2 x^2\right ) \log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right ) \log \left (\log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right )\right )+\left (-x+x^2\right ) \log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right ) \log ^2\left (\log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right )\right )} \, dx=\frac {4}{\log \left (\log \left (\frac {x^{4} - x^{3}}{9 \, \log \left (2 \, \log \left (2\right )\right )}\right )\right ) + 1} \]
integrate((-16*x+12)/((x^2-x)*log(1/9*(x^4-x^3)/log(2*log(2)))*log(log(1/9 *(x^4-x^3)/log(2*log(2))))^2+(2*x^2-2*x)*log(1/9*(x^4-x^3)/log(2*log(2)))* log(log(1/9*(x^4-x^3)/log(2*log(2))))+(x^2-x)*log(1/9*(x^4-x^3)/log(2*log( 2)))),x, algorithm=\
Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {12-16 x}{\left (-x+x^2\right ) \log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right )+\left (-2 x+2 x^2\right ) \log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right ) \log \left (\log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right )\right )+\left (-x+x^2\right ) \log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right ) \log ^2\left (\log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right )\right )} \, dx=\frac {4}{\log {\left (\log {\left (\frac {\frac {x^{4}}{9} - \frac {x^{3}}{9}}{\log {\left (2 \log {\left (2 \right )} \right )}} \right )} \right )} + 1} \]
integrate((-16*x+12)/((x**2-x)*ln(1/9*(x**4-x**3)/ln(2*ln(2)))*ln(ln(1/9*( x**4-x**3)/ln(2*ln(2))))**2+(2*x**2-2*x)*ln(1/9*(x**4-x**3)/ln(2*ln(2)))*l n(ln(1/9*(x**4-x**3)/ln(2*ln(2))))+(x**2-x)*ln(1/9*(x**4-x**3)/ln(2*ln(2)) )),x)
Time = 0.33 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {12-16 x}{\left (-x+x^2\right ) \log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right )+\left (-2 x+2 x^2\right ) \log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right ) \log \left (\log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right )\right )+\left (-x+x^2\right ) \log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right ) \log ^2\left (\log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right )\right )} \, dx=\frac {4}{\log \left (-2 \, \log \left (3\right ) + \log \left (x - 1\right ) + 3 \, \log \left (x\right ) - \log \left (\log \left (2\right ) + \log \left (\log \left (2\right )\right )\right )\right ) + 1} \]
integrate((-16*x+12)/((x^2-x)*log(1/9*(x^4-x^3)/log(2*log(2)))*log(log(1/9 *(x^4-x^3)/log(2*log(2))))^2+(2*x^2-2*x)*log(1/9*(x^4-x^3)/log(2*log(2)))* log(log(1/9*(x^4-x^3)/log(2*log(2))))+(x^2-x)*log(1/9*(x^4-x^3)/log(2*log( 2)))),x, algorithm=\
Time = 0.64 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {12-16 x}{\left (-x+x^2\right ) \log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right )+\left (-2 x+2 x^2\right ) \log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right ) \log \left (\log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right )\right )+\left (-x+x^2\right ) \log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right ) \log ^2\left (\log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right )\right )} \, dx=\frac {4}{\log \left (\log \left (x^{4} - x^{3}\right ) - \log \left (9 \, \log \left (2\right ) + 9 \, \log \left (\log \left (2\right )\right )\right )\right ) + 1} \]
integrate((-16*x+12)/((x^2-x)*log(1/9*(x^4-x^3)/log(2*log(2)))*log(log(1/9 *(x^4-x^3)/log(2*log(2))))^2+(2*x^2-2*x)*log(1/9*(x^4-x^3)/log(2*log(2)))* log(log(1/9*(x^4-x^3)/log(2*log(2))))+(x^2-x)*log(1/9*(x^4-x^3)/log(2*log( 2)))),x, algorithm=\
Time = 15.40 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {12-16 x}{\left (-x+x^2\right ) \log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right )+\left (-2 x+2 x^2\right ) \log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right ) \log \left (\log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right )\right )+\left (-x+x^2\right ) \log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right ) \log ^2\left (\log \left (\frac {-x^3+x^4}{9 \log (\log (4))}\right )\right )} \, dx=\frac {4}{\ln \left (\ln \left (x^4-x^3\right )-\ln \left (9\,\ln \left (2\right )+9\,\ln \left (\ln \left (2\right )\right )\right )\right )+1} \]