3.37.45 \(\int \frac {8+2 x-2 x^2+x^3}{4 x-4 x^2+x^3} \, dx\)

Optimal. Leaf size=18 \[ -4+\frac {6}{2-x}+x+\log \left (-x^2\right ) \]

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Rubi [A]  time = 0.04, antiderivative size = 15, normalized size of antiderivative = 0.83, number of steps used = 4, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {1594, 27, 1620} \begin {gather*} x+\frac {6}{2-x}+2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(8 + 2*x - 2*x^2 + x^3)/(4*x - 4*x^2 + x^3),x]

[Out]

6/(2 - x) + x + 2*Log[x]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {8+2 x-2 x^2+x^3}{x \left (4-4 x+x^2\right )} \, dx\\ &=\int \frac {8+2 x-2 x^2+x^3}{(-2+x)^2 x} \, dx\\ &=\int \left (1+\frac {6}{(-2+x)^2}+\frac {2}{x}\right ) \, dx\\ &=\frac {6}{2-x}+x+2 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 13, normalized size = 0.72 \begin {gather*} -\frac {6}{-2+x}+x+2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8 + 2*x - 2*x^2 + x^3)/(4*x - 4*x^2 + x^3),x]

[Out]

-6/(-2 + x) + x + 2*Log[x]

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fricas [A]  time = 0.61, size = 21, normalized size = 1.17 \begin {gather*} \frac {x^{2} + 2 \, {\left (x - 2\right )} \log \relax (x) - 2 \, x - 6}{x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3-2*x^2+2*x+8)/(x^3-4*x^2+4*x),x, algorithm="fricas")

[Out]

(x^2 + 2*(x - 2)*log(x) - 2*x - 6)/(x - 2)

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giac [A]  time = 0.12, size = 14, normalized size = 0.78 \begin {gather*} x - \frac {6}{x - 2} + 2 \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3-2*x^2+2*x+8)/(x^3-4*x^2+4*x),x, algorithm="giac")

[Out]

x - 6/(x - 2) + 2*log(abs(x))

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maple [A]  time = 0.02, size = 14, normalized size = 0.78

method	result	size
default	\(x +2 \ln \relax (x )-\frac {6}{x -2}\)	\(14\)
risch	\(x +2 \ln \relax (x )-\frac {6}{x -2}\)	\(14\)
norman	\(\frac {x^{2}-10}{x -2}+2 \ln \relax (x )\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3-2*x^2+2*x+8)/(x^3-4*x^2+4*x),x,method=_RETURNVERBOSE)

[Out]

x+2*ln(x)-6/(x-2)

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maxima [A]  time = 0.38, size = 13, normalized size = 0.72 \begin {gather*} x - \frac {6}{x - 2} + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3-2*x^2+2*x+8)/(x^3-4*x^2+4*x),x, algorithm="maxima")

[Out]

x - 6/(x - 2) + 2*log(x)

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mupad [B]  time = 0.04, size = 13, normalized size = 0.72 \begin {gather*} x+2\,\ln \relax (x)-\frac {6}{x-2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x - 2*x^2 + x^3 + 8)/(4*x - 4*x^2 + x^3),x)

[Out]

x + 2*log(x) - 6/(x - 2)

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sympy [A]  time = 0.09, size = 10, normalized size = 0.56 \begin {gather*} x + 2 \log {\relax (x )} - \frac {6}{x - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**3-2*x**2+2*x+8)/(x**3-4*x**2+4*x),x)

[Out]

x + 2*log(x) - 6/(x - 2)

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