\(\int \frac {3-x+3 x^2-2 x^3+x^4}{3 x-2 x^2+x^3} \, dx\) [273]

Optimal result

Integrand size = 33, antiderivative size = 23 \[ \int \frac {3-x+3 x^2-2 x^3+x^4}{3 x-2 x^2+x^3} \, dx=\frac {x^2}{2}+\log (x)-\frac {1}{2} \log \left (3-2 x+x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1608, 1642, 642} \[ \int \frac {3-x+3 x^2-2 x^3+x^4}{3 x-2 x^2+x^3} \, dx=\frac {x^2}{2}-\frac {1}{2} \log \left (x^2-2 x+3\right )+\log (x) \]

[In]

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

[Out]

x^2/2 + Log[x] - Log[3 - 2*x + x^2]/2

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {3-x+3 x^2-2 x^3+x^4}{3 x-2 x^2+x^3} \, dx=\frac {x^2}{2}+\log (x)-\frac {1}{2} \log \left (3-2 x+x^2\right ) \]

[In]

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

[Out]

x^2/2 + Log[x] - Log[3 - 2*x + x^2]/2

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87

method	result	size
default	\(\frac {x^{2}}{2}+\ln \left (x \right )-\frac {\ln \left (x^{2}-2 x +3\right )}{2}\)	\(20\)
norman	\(\frac {x^{2}}{2}+\ln \left (x \right )-\frac {\ln \left (x^{2}-2 x +3\right )}{2}\)	\(20\)
risch	\(\frac {x^{2}}{2}+\ln \left (x \right )-\frac {\ln \left (x^{2}-2 x +3\right )}{2}\)	\(20\)
parallelrisch	\(\frac {x^{2}}{2}+\ln \left (x \right )-\frac {\ln \left (x^{2}-2 x +3\right )}{2}\)	\(20\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {3-x+3 x^2-2 x^3+x^4}{3 x-2 x^2+x^3} \, dx=\frac {1}{2} \, x^{2} - \frac {1}{2} \, \log \left (x^{2} - 2 \, x + 3\right ) + \log \left (x\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {3-x+3 x^2-2 x^3+x^4}{3 x-2 x^2+x^3} \, dx=\frac {x^{2}}{2} + \log {\left (x \right )} - \frac {\log {\left (x^{2} - 2 x + 3 \right )}}{2} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {3-x+3 x^2-2 x^3+x^4}{3 x-2 x^2+x^3} \, dx=\frac {1}{2} \, x^{2} - \frac {1}{2} \, \log \left (x^{2} - 2 \, x + 3\right ) + \log \left (x\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {3-x+3 x^2-2 x^3+x^4}{3 x-2 x^2+x^3} \, dx=\frac {1}{2} \, x^{2} - \frac {1}{2} \, \log \left (x^{2} - 2 \, x + 3\right ) + \log \left ({\left | x \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {3-x+3 x^2-2 x^3+x^4}{3 x-2 x^2+x^3} \, dx=\ln \left (x\right )-\frac {\ln \left (x^2-2\,x+3\right )}{2}+\frac {x^2}{2} \]

[In]

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

[Out]

log(x) - log(x^2 - 2*x + 3)/2 + x^2/2