3.12.7 \(\int \frac {6-12 x^3}{(x+3 x^2+x^4) \log (\frac {5 x}{1+3 x+x^3})} \, dx\)

Optimal. Leaf size=22 \[ 6 \log \left (\log \left (\frac {5}{4+\frac {1-x}{x}+x^2}\right )\right ) \]

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Rubi [A]  time = 0.15, antiderivative size = 17, normalized size of antiderivative = 0.77, number of steps used = 2, number of rules used = 2, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {1594, 6684} \begin {gather*} 6 \log \left (\log \left (\frac {5 x}{x^3+3 x+1}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(6 - 12*x^3)/((x + 3*x^2 + x^4)*Log[(5*x)/(1 + 3*x + x^3)]),x]

[Out]

6*Log[Log[(5*x)/(1 + 3*x + x^3)]]

Rule 1594

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

Rule 6684

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {6-12 x^3}{x \left (1+3 x+x^3\right ) \log \left (\frac {5 x}{1+3 x+x^3}\right )} \, dx\\ &=6 \log \left (\log \left (\frac {5 x}{1+3 x+x^3}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 17, normalized size = 0.77 \begin {gather*} 6 \log \left (\log \left (\frac {5 x}{1+3 x+x^3}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6 - 12*x^3)/((x + 3*x^2 + x^4)*Log[(5*x)/(1 + 3*x + x^3)]),x]

[Out]

6*Log[Log[(5*x)/(1 + 3*x + x^3)]]

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fricas [A]  time = 0.59, size = 17, normalized size = 0.77 \begin {gather*} 6 \, \log \left (\log \left (\frac {5 \, x}{x^{3} + 3 \, x + 1}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*x^3+6)/(x^4+3*x^2+x)/log(5*x/(x^3+3*x+1)),x, algorithm="fricas")

[Out]

6*log(log(5*x/(x^3 + 3*x + 1)))

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giac [A]  time = 0.26, size = 17, normalized size = 0.77 \begin {gather*} 6 \, \log \left (\log \left (\frac {5 \, x}{x^{3} + 3 \, x + 1}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*x^3+6)/(x^4+3*x^2+x)/log(5*x/(x^3+3*x+1)),x, algorithm="giac")

[Out]

6*log(log(5*x/(x^3 + 3*x + 1)))

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maple [A]  time = 0.04, size = 18, normalized size = 0.82

method	result	size
norman	\(6 \ln \left (\ln \left (\frac {5 x}{x^{3}+3 x +1}\right )\right )\)	\(18\)
risch	\(6 \ln \left (\ln \left (\frac {5 x}{x^{3}+3 x +1}\right )\right )\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-12*x^3+6)/(x^4+3*x^2+x)/ln(5*x/(x^3+3*x+1)),x,method=_RETURNVERBOSE)

[Out]

6*ln(ln(5*x/(x^3+3*x+1)))

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maxima [A]  time = 0.59, size = 21, normalized size = 0.95 \begin {gather*} 6 \, \log \left (-\log \relax (5) + \log \left (x^{3} + 3 \, x + 1\right ) - \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*x^3+6)/(x^4+3*x^2+x)/log(5*x/(x^3+3*x+1)),x, algorithm="maxima")

[Out]

6*log(-log(5) + log(x^3 + 3*x + 1) - log(x))

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mupad [B]  time = 1.16, size = 17, normalized size = 0.77 \begin {gather*} 6\,\ln \left (\ln \left (\frac {5\,x}{x^3+3\,x+1}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(12*x^3 - 6)/(log((5*x)/(3*x + x^3 + 1))*(x + 3*x^2 + x^4)),x)

[Out]

6*log(log((5*x)/(3*x + x^3 + 1)))

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sympy [A]  time = 0.21, size = 15, normalized size = 0.68 \begin {gather*} 6 \log {\left (\log {\left (\frac {5 x}{x^{3} + 3 x + 1} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*x**3+6)/(x**4+3*x**2+x)/ln(5*x/(x**3+3*x+1)),x)

[Out]

6*log(log(5*x/(x**3 + 3*x + 1)))

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