3.20.34 \(\int \frac {x^3}{1-2 x+x^2} \, dx\)

Optimal. Leaf size=26 \[ \frac {x^2}{2}+2 x+\frac {1}{1-x}+3 \log (1-x) \]

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Rubi [A]  time = 0.01, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {27, 43} \begin {gather*} \frac {x^2}{2}+2 x+\frac {1}{1-x}+3 \log (1-x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/(1 - 2*x + x^2),x]

[Out]

(1 - x)^(-1) + 2*x + x^2/2 + 3*Log[1 - 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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 25, normalized size = 0.96 \begin {gather*} \frac {1}{2} \left (x^2+4 x-\frac {2}{x-1}+6 \log (x-1)-5\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/(1 - 2*x + x^2),x]

[Out]

(-5 - 2/(-1 + x) + 4*x + x^2 + 6*Log[-1 + x])/2

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^3}{1-2 x+x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[x^3/(1 - 2*x + x^2),x]

[Out]

IntegrateAlgebraic[x^3/(1 - 2*x + x^2), x]

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fricas [A]  time = 0.40, size = 29, normalized size = 1.12 \begin {gather*} \frac {x^{3} + 3 \, x^{2} + 6 \, {\left (x - 1\right )} \log \left (x - 1\right ) - 4 \, x - 2}{2 \, {\left (x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(x^2-2*x+1),x, algorithm="fricas")

[Out]

1/2*(x^3 + 3*x^2 + 6*(x - 1)*log(x - 1) - 4*x - 2)/(x - 1)

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giac [A]  time = 0.16, size = 23, normalized size = 0.88 \begin {gather*} \frac {1}{2} \, x^{2} + 2 \, x - \frac {1}{x - 1} + 3 \, \log \left ({\left | x - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(x^2-2*x+1),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.08, size = 23, normalized size = 0.88 \begin {gather*} \frac {x^{2}}{2}+2 x +3 \ln \left (x -1\right )-\frac {1}{x -1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(x^2-2*x+1),x)

[Out]

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

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maxima [A]  time = 0.90, size = 22, normalized size = 0.85 \begin {gather*} \frac {1}{2} \, x^{2} + 2 \, x - \frac {1}{x - 1} + 3 \, \log \left (x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(x^2-2*x+1),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.04, size = 22, normalized size = 0.85 \begin {gather*} 2\,x+3\,\ln \left (x-1\right )-\frac {1}{x-1}+\frac {x^2}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(x^2 - 2*x + 1),x)

[Out]

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

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sympy [A]  time = 0.08, size = 19, normalized size = 0.73 \begin {gather*} \frac {x^{2}}{2} + 2 x + 3 \log {\left (x - 1 \right )} - \frac {1}{x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(x**2-2*x+1),x)

[Out]

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

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