3.69.40 \(\int \frac {18+6 \log (\log (\frac {-4+16 x^3}{9 x^3}))}{(-x+4 x^4) \log (\frac {-4+16 x^3}{9 x^3})} \, dx\)

Optimal. Leaf size=20 \[ \left (3+\log \left (\log \left (\frac {4 \left (-1+4 x^3\right )}{9 x^3}\right )\right )\right )^2 \]

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Rubi [A]  time = 0.13, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {1593, 6686} \begin {gather*} \left (\log \left (\log \left (-\frac {4 \left (1-4 x^3\right )}{9 x^3}\right )\right )+3\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(18 + 6*Log[Log[(-4 + 16*x^3)/(9*x^3)]])/((-x + 4*x^4)*Log[(-4 + 16*x^3)/(9*x^3)]),x]

[Out]

(3 + Log[Log[(-4*(1 - 4*x^3))/(9*x^3)]])^2

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {18+6 \log \left (\log \left (\frac {-4+16 x^3}{9 x^3}\right )\right )}{x \left (-1+4 x^3\right ) \log \left (\frac {-4+16 x^3}{9 x^3}\right )} \, dx\\ &=\left (3+\log \left (\log \left (-\frac {4 \left (1-4 x^3\right )}{9 x^3}\right )\right )\right )^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 17, normalized size = 0.85 \begin {gather*} \left (3+\log \left (\log \left (\frac {4}{9} \left (4-\frac {1}{x^3}\right )\right )\right )\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(18 + 6*Log[Log[(-4 + 16*x^3)/(9*x^3)]])/((-x + 4*x^4)*Log[(-4 + 16*x^3)/(9*x^3)]),x]

[Out]

(3 + Log[Log[(4*(4 - x^(-3)))/9]])^2

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fricas [A]  time = 1.01, size = 33, normalized size = 1.65 \begin {gather*} \log \left (\log \left (\frac {4 \, {\left (4 \, x^{3} - 1\right )}}{9 \, x^{3}}\right )\right )^{2} + 6 \, \log \left (\log \left (\frac {4 \, {\left (4 \, x^{3} - 1\right )}}{9 \, x^{3}}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*log(log(1/9*(16*x^3-4)/x^3))+18)/(4*x^4-x)/log(1/9*(16*x^3-4)/x^3),x, algorithm="fricas")

[Out]

log(log(4/9*(4*x^3 - 1)/x^3))^2 + 6*log(log(4/9*(4*x^3 - 1)/x^3))

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giac [B]  time = 0.21, size = 77, normalized size = 3.85 \begin {gather*} -\log \left (-\log \left (9 \, x^{3}\right ) + \log \left (16 \, x^{3} - 4\right )\right )^{2} + 2 \, \log \left (-\log \left (9 \, x^{3}\right ) + \log \left (16 \, x^{3} - 4\right )\right ) \log \left (\log \left (\frac {4 \, {\left (4 \, x^{3} - 1\right )}}{9 \, x^{3}}\right )\right ) + 6 \, \log \left (-\log \left (9 \, x^{3}\right ) + \log \left (16 \, x^{3} - 4\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*log(log(1/9*(16*x^3-4)/x^3))+18)/(4*x^4-x)/log(1/9*(16*x^3-4)/x^3),x, algorithm="giac")

[Out]

-log(-log(9*x^3) + log(16*x^3 - 4))^2 + 2*log(-log(9*x^3) + log(16*x^3 - 4))*log(log(4/9*(4*x^3 - 1)/x^3)) + 6
*log(-log(9*x^3) + log(16*x^3 - 4))

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {6 \ln \left (\ln \left (\frac {16 x^{3}-4}{9 x^{3}}\right )\right )+18}{\left (4 x^{4}-x \right ) \ln \left (\frac {16 x^{3}-4}{9 x^{3}}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*ln(ln(1/9*(16*x^3-4)/x^3))+18)/(4*x^4-x)/ln(1/9*(16*x^3-4)/x^3),x)

[Out]

int((6*ln(ln(1/9*(16*x^3-4)/x^3))+18)/(4*x^4-x)/ln(1/9*(16*x^3-4)/x^3),x)

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maxima [B]  time = 0.45, size = 84, normalized size = 4.20 \begin {gather*} -\log \left (-2 \, \log \relax (3) + 2 \, \log \relax (2) + \log \left (4 \, x^{3} - 1\right ) - 3 \, \log \relax (x)\right )^{2} + 2 \, \log \left (-2 \, \log \relax (3) + 2 \, \log \relax (2) + \log \left (4 \, x^{3} - 1\right ) - 3 \, \log \relax (x)\right ) \log \left (\log \left (-\frac {4}{9 \, x^{3}} + \frac {16}{9}\right )\right ) + 6 \, \log \left (-2 \, \log \relax (3) + 2 \, \log \relax (2) + \log \left (4 \, x^{3} - 1\right ) - 3 \, \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*log(log(1/9*(16*x^3-4)/x^3))+18)/(4*x^4-x)/log(1/9*(16*x^3-4)/x^3),x, algorithm="maxima")

[Out]

-log(-2*log(3) + 2*log(2) + log(4*x^3 - 1) - 3*log(x))^2 + 2*log(-2*log(3) + 2*log(2) + log(4*x^3 - 1) - 3*log
(x))*log(log(-4/9/x^3 + 16/9)) + 6*log(-2*log(3) + 2*log(2) + log(4*x^3 - 1) - 3*log(x))

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mupad [B]  time = 4.52, size = 29, normalized size = 1.45 \begin {gather*} \ln \left (\ln \left (\frac {\frac {16\,x^3}{9}-\frac {4}{9}}{x^3}\right )\right )\,\left (\ln \left (\ln \left (\frac {\frac {16\,x^3}{9}-\frac {4}{9}}{x^3}\right )\right )+6\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(6*log(log(((16*x^3)/9 - 4/9)/x^3)) + 18)/(log(((16*x^3)/9 - 4/9)/x^3)*(x - 4*x^4)),x)

[Out]

log(log(((16*x^3)/9 - 4/9)/x^3))*(log(log(((16*x^3)/9 - 4/9)/x^3)) + 6)

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sympy [A]  time = 0.52, size = 36, normalized size = 1.80 \begin {gather*} \log {\left (\log {\left (\frac {\frac {16 x^{3}}{9} - \frac {4}{9}}{x^{3}} \right )} \right )}^{2} + 6 \log {\left (\log {\left (\frac {\frac {16 x^{3}}{9} - \frac {4}{9}}{x^{3}} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*ln(ln(1/9*(16*x**3-4)/x**3))+18)/(4*x**4-x)/ln(1/9*(16*x**3-4)/x**3),x)

[Out]

log(log((16*x**3/9 - 4/9)/x**3))**2 + 6*log(log((16*x**3/9 - 4/9)/x**3))

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