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

Optimal. Leaf size=20 \[ \frac {108 x}{x+x \left (8+\log ^2(\log (1-x))\right )} \]

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Rubi [A]  time = 0.12, antiderivative size = 15, normalized size of antiderivative = 0.75, number of steps used = 3, number of rules used = 3, integrand size = 64, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {12, 6688, 6686} \begin {gather*} \frac {108}{\log ^2(\log (1-x))+9} \end {gather*}

Antiderivative was successfully verified.

[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 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} &=-\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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 15, normalized size = 0.75 \begin {gather*} \frac {108}{9+\log ^2(\log (1-x))} \end {gather*}

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.85, size = 15, normalized size = 0.75 \begin {gather*} \frac {108}{\log \left (\log \left (-x + 1\right )\right )^{2} + 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

108/(log(log(-x + 1))^2 + 9)

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giac [A]  time = 0.25, size = 15, normalized size = 0.75 \begin {gather*} \frac {108}{\log \left (\log \left (-x + 1\right )\right )^{2} + 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

108/(log(log(-x + 1))^2 + 9)

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maple [A]  time = 0.02, size = 16, normalized size = 0.80

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-216*ln(ln(1-x))/((x-1)*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)

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maxima [A]  time = 0.41, size = 15, normalized size = 0.75 \begin {gather*} \frac {108}{\log \left (\log \left (-x + 1\right )\right )^{2} + 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

108/(log(log(-x + 1))^2 + 9)

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mupad [B]  time = 6.07, size = 15, normalized size = 0.75 \begin {gather*} \frac {108}{{\ln \left (\ln \left (1-x\right )\right )}^2+9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.26, size = 10, normalized size = 0.50 \begin {gather*} \frac {108}{\log {\left (\log {\left (1 - x \right )} \right )}^{2} + 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

108/(log(log(1 - x))**2 + 9)

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