\(\int \frac {-6+2 x+x^4}{-2 x+x^2+x^3} \, dx\) [118]

Optimal result

Integrand size = 21, antiderivative size = 27 \[ \int \frac {-6+2 x+x^4}{-2 x+x^2+x^3} \, dx=-x+\frac {x^2}{2}-\log (1-x)+3 \log (x)+\log (2+x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {1608, 1642} \[ \int \frac {-6+2 x+x^4}{-2 x+x^2+x^3} \, dx=\frac {x^2}{2}-x-\log (1-x)+3 \log (x)+\log (x+2) \]

[In]

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

[Out]

-x + x^2/2 - Log[1 - x] + 3*Log[x] + Log[2 + 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 1642

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {-6+2 x+x^4}{-2 x+x^2+x^3} \, dx=-x+\frac {x^2}{2}-\log (1-x)+3 \log (x)+\log (2+x) \]

[In]

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

[Out]

-x + x^2/2 - Log[1 - x] + 3*Log[x] + Log[2 + x]

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89

method	result	size
default	\(-x +\frac {x^{2}}{2}-\ln \left (-1+x \right )+\ln \left (2+x \right )+3 \ln \left (x \right )\)	\(24\)
norman	\(-x +\frac {x^{2}}{2}-\ln \left (-1+x \right )+\ln \left (2+x \right )+3 \ln \left (x \right )\)	\(24\)
risch	\(-x +\frac {x^{2}}{2}-\ln \left (-1+x \right )+\ln \left (2+x \right )+3 \ln \left (x \right )\)	\(24\)
parallelrisch	\(-x +\frac {x^{2}}{2}-\ln \left (-1+x \right )+\ln \left (2+x \right )+3 \ln \left (x \right )\)	\(24\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {-6+2 x+x^4}{-2 x+x^2+x^3} \, dx=\frac {1}{2} \, x^{2} - x + \log \left (x + 2\right ) - \log \left (x - 1\right ) + 3 \, \log \left (x\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {-6+2 x+x^4}{-2 x+x^2+x^3} \, dx=\frac {x^{2}}{2} - x + 3 \log {\left (x \right )} - \log {\left (x - 1 \right )} + \log {\left (x + 2 \right )} \]

[In]

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

[Out]

x**2/2 - x + 3*log(x) - log(x - 1) + log(x + 2)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {-6+2 x+x^4}{-2 x+x^2+x^3} \, dx=\frac {1}{2} \, x^{2} - x + \log \left (x + 2\right ) - \log \left (x - 1\right ) + 3 \, \log \left (x\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {-6+2 x+x^4}{-2 x+x^2+x^3} \, dx=\frac {1}{2} \, x^{2} - x + \log \left ({\left | x + 2 \right |}\right ) - \log \left ({\left | x - 1 \right |}\right ) + 3 \, \log \left ({\left | x \right |}\right ) \]

[In]

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

[Out]

1/2*x^2 - x + log(abs(x + 2)) - log(abs(x - 1)) + 3*log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {-6+2 x+x^4}{-2 x+x^2+x^3} \, dx=3\,\ln \left (x\right )-x+\frac {x^2}{2}+\mathrm {atan}\left (\frac {192{}\mathrm {i}}{7\,\left (28\,x-40\right )}+\frac {9}{7}{}\mathrm {i}\right )\,2{}\mathrm {i} \]

[In]

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

[Out]

atan(192i/(7*(28*x - 40)) + 9i/7)*2i - x + 3*log(x) + x^2/2