\(\int -\frac {216 \log (\log (1-x))}{(-81+81 x) \log (1-x)+(-18+18 x) \log (1-x) \log ^2(\log (1-x))+(-1+x) \log (1-x) \log ^4(\log (1-x))} \, dx\) [7804]

Optimal result

Integrand size = 64, antiderivative size = 20 \[ \int -\frac {216 \log (\log (1-x))}{(-81+81 x) \log (1-x)+(-18+18 x) \log (1-x) \log ^2(\log (1-x))+(-1+x) \log (1-x) \log ^4(\log (1-x))} \, dx=\frac {108 x}{x+x \left (8+\log ^2(\log (1-x))\right )} \]

[Out]

12/(1/9*x+1/9*(ln(ln(1-x))^2+8)*x)*x

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {12, 6820, 6818} \[ \int -\frac {216 \log (\log (1-x))}{(-81+81 x) \log (1-x)+(-18+18 x) \log (1-x) \log ^2(\log (1-x))+(-1+x) \log (1-x) \log ^4(\log (1-x))} \, dx=\frac {108}{\log ^2(\log (1-x))+9} \]

[In]

Int[(-216*Log[Log[1 - x]])/((-81 + 81*x)*Log[1 - x] + (-18 + 18*x)*Log[1 - x]*Log[Log[1 - x]]^2 + (-1 + x)*Log
[1 - x]*Log[Log[1 - x]]^4),x]

[Out]

108/(9 + Log[Log[1 - x]]^2)

Rule 12

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

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}& = -\left (216 \int \frac {\log (\log (1-x))}{(-81+81 x) \log (1-x)+(-18+18 x) \log (1-x) \log ^2(\log (1-x))+(-1+x) \log (1-x) \log ^4(\log (1-x))} \, dx\right ) \\ & = -\left (216 \int \frac {\log (\log (1-x))}{(-1+x) \log (1-x) \left (9+\log ^2(\log (1-x))\right )^2} \, dx\right ) \\ & = \frac {108}{9+\log ^2(\log (1-x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int -\frac {216 \log (\log (1-x))}{(-81+81 x) \log (1-x)+(-18+18 x) \log (1-x) \log ^2(\log (1-x))+(-1+x) \log (1-x) \log ^4(\log (1-x))} \, dx=\frac {108}{9+\log ^2(\log (1-x))} \]

[In]

Integrate[(-216*Log[Log[1 - x]])/((-81 + 81*x)*Log[1 - x] + (-18 + 18*x)*Log[1 - x]*Log[Log[1 - x]]^2 + (-1 +
x)*Log[1 - x]*Log[Log[1 - x]]^4),x]

[Out]

108/(9 + Log[Log[1 - x]]^2)

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80

method	result	size
default	\(\frac {108}{\ln \left (\ln \left (1-x \right )\right )^{2}+9}\)	\(16\)
risch	\(\frac {108}{\ln \left (\ln \left (1-x \right )\right )^{2}+9}\)	\(16\)
parallelrisch	\(\frac {108}{\ln \left (\ln \left (1-x \right )\right )^{2}+9}\)	\(16\)

[In]

int(-216*ln(ln(1-x))/((-1+x)*ln(1-x)*ln(ln(1-x))^4+(18*x-18)*ln(1-x)*ln(ln(1-x))^2+(81*x-81)*ln(1-x)),x,method
=_RETURNVERBOSE)

[Out]

108/(ln(ln(1-x))^2+9)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int -\frac {216 \log (\log (1-x))}{(-81+81 x) \log (1-x)+(-18+18 x) \log (1-x) \log ^2(\log (1-x))+(-1+x) \log (1-x) \log ^4(\log (1-x))} \, dx=\frac {108}{\log \left (\log \left (-x + 1\right )\right )^{2} + 9} \]

[In]

integrate(-216*log(log(1-x))/((-1+x)*log(1-x)*log(log(1-x))^4+(18*x-18)*log(1-x)*log(log(1-x))^2+(81*x-81)*log
(1-x)),x, algorithm="fricas")

[Out]

108/(log(log(-x + 1))^2 + 9)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.50 \[ \int -\frac {216 \log (\log (1-x))}{(-81+81 x) \log (1-x)+(-18+18 x) \log (1-x) \log ^2(\log (1-x))+(-1+x) \log (1-x) \log ^4(\log (1-x))} \, dx=\frac {108}{\log {\left (\log {\left (1 - x \right )} \right )}^{2} + 9} \]

[In]

integrate(-216*ln(ln(1-x))/((-1+x)*ln(1-x)*ln(ln(1-x))**4+(18*x-18)*ln(1-x)*ln(ln(1-x))**2+(81*x-81)*ln(1-x)),
x)

[Out]

108/(log(log(1 - x))**2 + 9)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int -\frac {216 \log (\log (1-x))}{(-81+81 x) \log (1-x)+(-18+18 x) \log (1-x) \log ^2(\log (1-x))+(-1+x) \log (1-x) \log ^4(\log (1-x))} \, dx=\frac {108}{\log \left (\log \left (-x + 1\right )\right )^{2} + 9} \]

[In]

integrate(-216*log(log(1-x))/((-1+x)*log(1-x)*log(log(1-x))^4+(18*x-18)*log(1-x)*log(log(1-x))^2+(81*x-81)*log
(1-x)),x, algorithm="maxima")

[Out]

108/(log(log(-x + 1))^2 + 9)

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int -\frac {216 \log (\log (1-x))}{(-81+81 x) \log (1-x)+(-18+18 x) \log (1-x) \log ^2(\log (1-x))+(-1+x) \log (1-x) \log ^4(\log (1-x))} \, dx=\frac {108}{\log \left (\log \left (-x + 1\right )\right )^{2} + 9} \]

[In]

integrate(-216*log(log(1-x))/((-1+x)*log(1-x)*log(log(1-x))^4+(18*x-18)*log(1-x)*log(log(1-x))^2+(81*x-81)*log
(1-x)),x, algorithm="giac")

[Out]

108/(log(log(-x + 1))^2 + 9)

Mupad [B] (verification not implemented)

Time = 13.84 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int -\frac {216 \log (\log (1-x))}{(-81+81 x) \log (1-x)+(-18+18 x) \log (1-x) \log ^2(\log (1-x))+(-1+x) \log (1-x) \log ^4(\log (1-x))} \, dx=\frac {108}{{\ln \left (\ln \left (1-x\right )\right )}^2+9} \]

[In]

int(-(216*log(log(1 - x)))/(log(1 - x)*(81*x - 81) + log(log(1 - x))^4*log(1 - x)*(x - 1) + log(log(1 - x))^2*
log(1 - x)*(18*x - 18)),x)

[Out]

108/(log(log(1 - x))^2 + 9)