3.88.14 \(\int \frac {12-16 x}{(-x+x^2) \log (\frac {-x^3+x^4}{9 \log (\log (4))})+(-2 x+2 x^2) \log (\frac {-x^3+x^4}{9 \log (\log (4))}) \log (\log (\frac {-x^3+x^4}{9 \log (\log (4))}))+(-x+x^2) \log (\frac {-x^3+x^4}{9 \log (\log (4))}) \log ^2(\log (\frac {-x^3+x^4}{9 \log (\log (4))}))} \, dx\)

Optimal. Leaf size=27 \[ \frac {4}{1+\log \left (\log \left (\frac {x^2 \left (-x+x^2\right )}{9 \log (\log (4))}\right )\right )} \]

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Rubi [A]  time = 0.24, antiderivative size = 25, normalized size of antiderivative = 0.93, number of steps used = 3, number of rules used = 3, integrand size = 134, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.022, Rules used = {6688, 12, 6686} \begin {gather*} \frac {4}{\log \left (\log \left (-\frac {(1-x) x^3}{9 \log (\log (4))}\right )\right )+1} \end {gather*}

Antiderivative was successfully verified.

[In]

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]

[Out]

4/(1 + Log[Log[-1/9*((1 - x)*x^3)/Log[Log[4]]]])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6686

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 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 (-3+4 x)}{(1-x) x \log \left (\frac {(-1+x) x^3}{9 \log (\log (4))}\right ) \left (1+\log \left (\log \left (\frac {(-1+x) x^3}{9 \log (\log (4))}\right )\right )\right )^2} \, dx\\ &=4 \int \frac {-3+4 x}{(1-x) x \log \left (\frac {(-1+x) x^3}{9 \log (\log (4))}\right ) \left (1+\log \left (\log \left (\frac {(-1+x) x^3}{9 \log (\log (4))}\right )\right )\right )^2} \, dx\\ &=\frac {4}{1+\log \left (\log \left (-\frac {(1-x) x^3}{9 \log (\log (4))}\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 23, normalized size = 0.85 \begin {gather*} \frac {4}{1+\log \left (\log \left (\frac {(-1+x) x^3}{9 \log (\log (4))}\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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[L
og[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]

[Out]

4/(1 + Log[Log[((-1 + x)*x^3)/(9*Log[Log[4]])]])

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fricas [A]  time = 0.66, size = 26, normalized size = 0.96 \begin {gather*} \frac {4}{\log \left (\log \left (\frac {x^{4} - x^{3}}{9 \, \log \left (2 \, \log \relax (2)\right )}\right )\right ) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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*lo
g(2)))),x, algorithm="fricas")

[Out]

4/(log(log(1/9*(x^4 - x^3)/log(2*log(2)))) + 1)

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giac [A]  time = 0.87, size = 31, normalized size = 1.15 \begin {gather*} \frac {4}{\log \left (\log \left (x^{4} - x^{3}\right ) - \log \left (9 \, \log \relax (2) + 9 \, \log \left (\log \relax (2)\right )\right )\right ) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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*lo
g(2)))),x, algorithm="giac")

[Out]

4/(log(log(x^4 - x^3) - log(9*log(2) + 9*log(log(2)))) + 1)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {-16 x +12}{\left (x^{2}-x \right ) \ln \left (\frac {x^{4}-x^{3}}{9 \ln \left (2 \ln \relax (2)\right )}\right ) \ln \left (\ln \left (\frac {x^{4}-x^{3}}{9 \ln \left (2 \ln \relax (2)\right )}\right )\right )^{2}+\left (2 x^{2}-2 x \right ) \ln \left (\frac {x^{4}-x^{3}}{9 \ln \left (2 \ln \relax (2)\right )}\right ) \ln \left (\ln \left (\frac {x^{4}-x^{3}}{9 \ln \left (2 \ln \relax (2)\right )}\right )\right )+\left (x^{2}-x \right ) \ln \left (\frac {x^{4}-x^{3}}{9 \ln \left (2 \ln \relax (2)\right )}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

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)

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maxima [A]  time = 0.51, size = 29, normalized size = 1.07 \begin {gather*} \frac {4}{\log \left (-2 \, \log \relax (3) + \log \left (x - 1\right ) + 3 \, \log \relax (x) - \log \left (\log \relax (2) + \log \left (\log \relax (2)\right )\right )\right ) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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*lo
g(2)))),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 6.78, size = 31, normalized size = 1.15 \begin {gather*} \frac {4}{\ln \left (\ln \left (x^4-x^3\right )-\ln \left (9\,\ln \relax (2)+9\,\ln \left (\ln \relax (2)\right )\right )\right )+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*x - 12)/(log(-(x^3/9 - x^4/9)/log(2*log(2)))*(x - x^2) + log(-(x^3/9 - x^4/9)/log(2*log(2)))*log(log(-
(x^3/9 - x^4/9)/log(2*log(2))))*(2*x - 2*x^2) + log(-(x^3/9 - x^4/9)/log(2*log(2)))*log(log(-(x^3/9 - x^4/9)/l
og(2*log(2))))^2*(x - x^2)),x)

[Out]

4/(log(log(x^4 - x^3) - log(9*log(2) + 9*log(log(2)))) + 1)

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sympy [A]  time = 0.38, size = 22, normalized size = 0.81 \begin {gather*} \frac {4}{\log {\left (\log {\left (\frac {\frac {x^{4}}{9} - \frac {x^{3}}{9}}{\log {\left (2 \log {\relax (2 )} \right )}} \right )} \right )} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)))*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)

[Out]

4/(log(log((x**4/9 - x**3/9)/log(2*log(2)))) + 1)

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