\(\int \frac {6-12 x^3}{(x+3 x^2+x^4) \log (\frac {5 x}{1+3 x+x^3})} \, dx\) [1107]

Optimal result

Integrand size = 36, antiderivative size = 22 \[ \int \frac {6-12 x^3}{\left (x+3 x^2+x^4\right ) \log \left (\frac {5 x}{1+3 x+x^3}\right )} \, dx=6 \log \left (\log \left (\frac {5}{4+\frac {1-x}{x}+x^2}\right )\right ) \]

[Out]

6*ln(ln(5/(4+x^2+(1-x)/x)))

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {1608, 6816} \[ \int \frac {6-12 x^3}{\left (x+3 x^2+x^4\right ) \log \left (\frac {5 x}{1+3 x+x^3}\right )} \, dx=6 \log \left (\log \left (\frac {5 x}{x^3+3 x+1}\right )\right ) \]

[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 1608

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 6816

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

Rubi steps \begin{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {6-12 x^3}{\left (x+3 x^2+x^4\right ) \log \left (\frac {5 x}{1+3 x+x^3}\right )} \, dx=6 \log \left (\log \left (\frac {5 x}{1+3 x+x^3}\right )\right ) \]

[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)]]

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82

method	result	size
norman	\(6 \ln \left (\ln \left (\frac {5 x}{x^{3}+3 x +1}\right )\right )\)	\(18\)
risch	\(6 \ln \left (\ln \left (\frac {5 x}{x^{3}+3 x +1}\right )\right )\)	\(18\)
parallelrisch	\(6 \ln \left (\ln \left (\frac {5 x}{x^{3}+3 x +1}\right )\right )\)	\(18\)
default	\(6 \ln \left (\ln \left (5\right )+\ln \left (\frac {x}{x^{3}+3 x +1}\right )\right )\)	\(20\)

[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)))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {6-12 x^3}{\left (x+3 x^2+x^4\right ) \log \left (\frac {5 x}{1+3 x+x^3}\right )} \, dx=6 \, \log \left (\log \left (\frac {5 \, x}{x^{3} + 3 \, x + 1}\right )\right ) \]

[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)))

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {6-12 x^3}{\left (x+3 x^2+x^4\right ) \log \left (\frac {5 x}{1+3 x+x^3}\right )} \, dx=6 \log {\left (\log {\left (\frac {5 x}{x^{3} + 3 x + 1} \right )} \right )} \]

[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)))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {6-12 x^3}{\left (x+3 x^2+x^4\right ) \log \left (\frac {5 x}{1+3 x+x^3}\right )} \, dx=6 \, \log \left (-\log \left (5\right ) + \log \left (x^{3} + 3 \, x + 1\right ) - \log \left (x\right )\right ) \]

[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))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {6-12 x^3}{\left (x+3 x^2+x^4\right ) \log \left (\frac {5 x}{1+3 x+x^3}\right )} \, dx=6 \, \log \left (\log \left (\frac {5 \, x}{x^{3} + 3 \, x + 1}\right )\right ) \]

[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)))

Mupad [B] (verification not implemented)

Time = 9.28 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {6-12 x^3}{\left (x+3 x^2+x^4\right ) \log \left (\frac {5 x}{1+3 x+x^3}\right )} \, dx=6\,\ln \left (\ln \left (\frac {5\,x}{x^3+3\,x+1}\right )\right ) \]

[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)))