3.93.97 \(\int \frac {8}{(8 x-108 x^2) \log (\frac {2-27 x}{x})+(-2 x+27 x^2) \log (\frac {2-27 x}{x}) \log (\log (\frac {2-27 x}{x}))} \, dx\)

Optimal. Leaf size=18 \[ 4 \log \left (4-\log \left (\log \left (5+2 \left (-16+\frac {1}{x}\right )\right )\right )\right ) \]

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Rubi [A]  time = 0.16, antiderivative size = 16, normalized size of antiderivative = 0.89, number of steps used = 3, number of rules used = 3, integrand size = 56, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {12, 6688, 6684} \begin {gather*} 4 \log \left (4-\log \left (\log \left (\frac {2}{x}-27\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[8/((8*x - 108*x^2)*Log[(2 - 27*x)/x] + (-2*x + 27*x^2)*Log[(2 - 27*x)/x]*Log[Log[(2 - 27*x)/x]]),x]

[Out]

4*Log[4 - Log[Log[-27 + 2/x]]]

Rule 12

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

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

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} &=8 \int \frac {1}{\left (8 x-108 x^2\right ) \log \left (\frac {2-27 x}{x}\right )+\left (-2 x+27 x^2\right ) \log \left (\frac {2-27 x}{x}\right ) \log \left (\log \left (\frac {2-27 x}{x}\right )\right )} \, dx\\ &=8 \int \frac {1}{(2-27 x) x \log \left (-27+\frac {2}{x}\right ) \left (4-\log \left (\log \left (-27+\frac {2}{x}\right )\right )\right )} \, dx\\ &=4 \log \left (4-\log \left (\log \left (-27+\frac {2}{x}\right )\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 5.02, size = 14, normalized size = 0.78 \begin {gather*} 4 \log \left (-4+\log \left (\log \left (-27+\frac {2}{x}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[8/((8*x - 108*x^2)*Log[(2 - 27*x)/x] + (-2*x + 27*x^2)*Log[(2 - 27*x)/x]*Log[Log[(2 - 27*x)/x]]),x]

[Out]

4*Log[-4 + Log[Log[-27 + 2/x]]]

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fricas [A]  time = 0.65, size = 17, normalized size = 0.94 \begin {gather*} 4 \, \log \left (\log \left (\log \left (-\frac {27 \, x - 2}{x}\right )\right ) - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(8/((27*x^2-2*x)*log((-27*x+2)/x)*log(log((-27*x+2)/x))+(-108*x^2+8*x)*log((-27*x+2)/x)),x, algorithm
="fricas")

[Out]

4*log(log(log(-(27*x - 2)/x)) - 4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {8}{{\left (27 \, x^{2} - 2 \, x\right )} \log \left (-\frac {27 \, x - 2}{x}\right ) \log \left (\log \left (-\frac {27 \, x - 2}{x}\right )\right ) - 4 \, {\left (27 \, x^{2} - 2 \, x\right )} \log \left (-\frac {27 \, x - 2}{x}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(8/((27*x^2-2*x)*log((-27*x+2)/x)*log(log((-27*x+2)/x))+(-108*x^2+8*x)*log((-27*x+2)/x)),x, algorithm
="giac")

[Out]

integrate(8/((27*x^2 - 2*x)*log(-(27*x - 2)/x)*log(log(-(27*x - 2)/x)) - 4*(27*x^2 - 2*x)*log(-(27*x - 2)/x)),
x)

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maple [A]  time = 0.04, size = 17, normalized size = 0.94

method	result	size
norman	\(4 \ln \left (\ln \left (\ln \left (\frac {-27 x +2}{x}\right )\right )-4\right )\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(8/((27*x^2-2*x)*ln((-27*x+2)/x)*ln(ln((-27*x+2)/x))+(-108*x^2+8*x)*ln((-27*x+2)/x)),x,method=_RETURNVERBOS
E)

[Out]

4*ln(ln(ln((-27*x+2)/x))-4)

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maxima [A]  time = 0.35, size = 17, normalized size = 0.94 \begin {gather*} 4 \, \log \left (\log \left (-\log \relax (x) + \log \left (-27 \, x + 2\right )\right ) - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(8/((27*x^2-2*x)*log((-27*x+2)/x)*log(log((-27*x+2)/x))+(-108*x^2+8*x)*log((-27*x+2)/x)),x, algorithm
="maxima")

[Out]

4*log(log(-log(x) + log(-27*x + 2)) - 4)

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mupad [B]  time = 7.82, size = 14, normalized size = 0.78 \begin {gather*} 4\,\ln \left (\ln \left (\ln \left (\frac {2}{x}-27\right )\right )-4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(8/(log(-(27*x - 2)/x)*(8*x - 108*x^2) - log(log(-(27*x - 2)/x))*log(-(27*x - 2)/x)*(2*x - 27*x^2)),x)

[Out]

4*log(log(log(2/x - 27)) - 4)

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sympy [A]  time = 0.37, size = 14, normalized size = 0.78 \begin {gather*} 4 \log {\left (\log {\left (\log {\left (\frac {2 - 27 x}{x} \right )} \right )} - 4 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(8/((27*x**2-2*x)*ln((-27*x+2)/x)*ln(ln((-27*x+2)/x))+(-108*x**2+8*x)*ln((-27*x+2)/x)),x)

[Out]

4*log(log(log((2 - 27*x)/x)) - 4)

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