\(\int \frac {-11+6 x+28 x^3}{-11 x+2 x^2+4 x^4} \, dx\) [2598]

Optimal result

Integrand size = 27, antiderivative size = 18 \[ \int \frac {-11+6 x+28 x^3}{-11 x+2 x^2+4 x^4} \, dx=\log \left (\frac {3}{25} x \left (-11+2 x+4 x^3\right )^2\right ) \]

[Out]

ln(3/25*x*(4*x^3+2*x-11)^2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1608, 6874, 1601} \[ \int \frac {-11+6 x+28 x^3}{-11 x+2 x^2+4 x^4} \, dx=2 \log \left (-4 x^3-2 x+11\right )+\log (x) \]

[In]

Int[(-11 + 6*x + 28*x^3)/(-11*x + 2*x^2 + 4*x^4),x]

[Out]

Log[x] + 2*Log[11 - 2*x - 4*x^3]

Rule 1601

Int[(Pp_)/(Qq_), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*(Log[RemoveConte
nt[Qq, x]]/(q*Coeff[Qq, x, q])), x] /; EqQ[p, q - 1] && EqQ[Pp, Simplify[(Coeff[Pp, x, p]/(q*Coeff[Qq, x, q]))
*D[Qq, x]]]] /; PolyQ[Pp, x] && PolyQ[Qq, x]

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 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-11+6 x+28 x^3}{x \left (-11+2 x+4 x^3\right )} \, dx \\ & = \int \left (\frac {1}{x}+\frac {4 \left (1+6 x^2\right )}{-11+2 x+4 x^3}\right ) \, dx \\ & = \log (x)+4 \int \frac {1+6 x^2}{-11+2 x+4 x^3} \, dx \\ & = \log (x)+2 \log \left (11-2 x-4 x^3\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {-11+6 x+28 x^3}{-11 x+2 x^2+4 x^4} \, dx=\log (x)+2 \log \left (11-2 x-4 x^3\right ) \]

[In]

Integrate[(-11 + 6*x + 28*x^3)/(-11*x + 2*x^2 + 4*x^4),x]

[Out]

Log[x] + 2*Log[11 - 2*x - 4*x^3]

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83

method	result	size
parallelrisch	\(\ln \left (x \right )+2 \ln \left (x^{3}+\frac {1}{2} x -\frac {11}{4}\right )\)	\(15\)
default	\(\ln \left (x \right )+2 \ln \left (4 x^{3}+2 x -11\right )\)	\(17\)
norman	\(\ln \left (x \right )+2 \ln \left (4 x^{3}+2 x -11\right )\)	\(17\)
risch	\(\ln \left (x \right )+2 \ln \left (4 x^{3}+2 x -11\right )\)	\(17\)

[In]

int((28*x^3+6*x-11)/(4*x^4+2*x^2-11*x),x,method=_RETURNVERBOSE)

[Out]

ln(x)+2*ln(x^3+1/2*x-11/4)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {-11+6 x+28 x^3}{-11 x+2 x^2+4 x^4} \, dx=2 \, \log \left (4 \, x^{3} + 2 \, x - 11\right ) + \log \left (x\right ) \]

[In]

integrate((28*x^3+6*x-11)/(4*x^4+2*x^2-11*x),x, algorithm="fricas")

[Out]

2*log(4*x^3 + 2*x - 11) + log(x)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {-11+6 x+28 x^3}{-11 x+2 x^2+4 x^4} \, dx=\log {\left (x \right )} + 2 \log {\left (4 x^{3} + 2 x - 11 \right )} \]

[In]

integrate((28*x**3+6*x-11)/(4*x**4+2*x**2-11*x),x)

[Out]

log(x) + 2*log(4*x**3 + 2*x - 11)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {-11+6 x+28 x^3}{-11 x+2 x^2+4 x^4} \, dx=2 \, \log \left (4 \, x^{3} + 2 \, x - 11\right ) + \log \left (x\right ) \]

[In]

integrate((28*x^3+6*x-11)/(4*x^4+2*x^2-11*x),x, algorithm="maxima")

[Out]

2*log(4*x^3 + 2*x - 11) + log(x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {-11+6 x+28 x^3}{-11 x+2 x^2+4 x^4} \, dx=2 \, \log \left ({\left | 4 \, x^{3} + 2 \, x - 11 \right |}\right ) + \log \left ({\left | x \right |}\right ) \]

[In]

integrate((28*x^3+6*x-11)/(4*x^4+2*x^2-11*x),x, algorithm="giac")

[Out]

2*log(abs(4*x^3 + 2*x - 11)) + log(abs(x))

Mupad [B] (verification not implemented)

Time = 7.89 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {-11+6 x+28 x^3}{-11 x+2 x^2+4 x^4} \, dx=2\,\ln \left (x^3+\frac {x}{2}-\frac {11}{4}\right )+\ln \left (x\right ) \]

[In]

int((6*x + 28*x^3 - 11)/(2*x^2 - 11*x + 4*x^4),x)

[Out]

2*log(x/2 + x^3 - 11/4) + log(x)