3.2263 \(\int \frac {x^3}{2-3 x+x^2} \, dx\)

Optimal. Leaf size=27 \[ \frac {x^2}{2}+3 x-\log (1-x)+8 \log (2-x) \]

[Out]

3*x+1/2*x^2-ln(1-x)+8*ln(2-x)

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Rubi [A]  time = 0.01, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {701, 632, 31} \[ \frac {x^2}{2}+3 x-\log (1-x)+8 \log (2-x) \]

Antiderivative was successfully verified.

[In]

Int[x^3/(2 - 3*x + x^2),x]

[Out]

3*x + x^2/2 - Log[1 - x] + 8*Log[2 - x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 632

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && NiceSqrtQ[b^2 - 4*a*
c]

Rule 701

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[(d + e*x)
^m, a + b*x + c*x^2, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2,
0] && NeQ[2*c*d - b*e, 0] && IGtQ[m, 1] && (NeQ[d, 0] || GtQ[m, 2])

Rubi steps

\begin {align*} \int \frac {x^3}{2-3 x+x^2} \, dx &=\int \left (3+x-\frac {6-7 x}{2-3 x+x^2}\right ) \, dx\\ &=3 x+\frac {x^2}{2}-\int \frac {6-7 x}{2-3 x+x^2} \, dx\\ &=3 x+\frac {x^2}{2}+8 \int \frac {1}{-2+x} \, dx-\int \frac {1}{-1+x} \, dx\\ &=3 x+\frac {x^2}{2}-\log (1-x)+8 \log (2-x)\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 27, normalized size = 1.00 \[ \frac {x^2}{2}+3 x-\log (1-x)+8 \log (2-x) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/(2 - 3*x + x^2),x]

[Out]

3*x + x^2/2 - Log[1 - x] + 8*Log[2 - x]

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fricas [A]  time = 0.84, size = 21, normalized size = 0.78 \[ \frac {1}{2} \, x^{2} + 3 \, x - \log \left (x - 1\right ) + 8 \, \log \left (x - 2\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2 + 3*x - log(x - 1) + 8*log(x - 2)

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giac [A]  time = 0.15, size = 23, normalized size = 0.85 \[ \frac {1}{2} \, x^{2} + 3 \, x - \log \left ({\left | x - 1 \right |}\right ) + 8 \, \log \left ({\left | x - 2 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2 + 3*x - log(abs(x - 1)) + 8*log(abs(x - 2))

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maple [A]  time = 0.06, size = 22, normalized size = 0.81 \[ \frac {x^{2}}{2}+3 x +8 \ln \left (x -2\right )-\ln \left (x -1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2+3*x-ln(x-1)+8*ln(x-2)

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maxima [A]  time = 0.87, size = 21, normalized size = 0.78 \[ \frac {1}{2} \, x^{2} + 3 \, x - \log \left (x - 1\right ) + 8 \, \log \left (x - 2\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2 + 3*x - log(x - 1) + 8*log(x - 2)

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mupad [B]  time = 0.03, size = 21, normalized size = 0.78 \[ 3\,x-\ln \left (x-1\right )+8\,\ln \left (x-2\right )+\frac {x^2}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(x^2 - 3*x + 2),x)

[Out]

3*x - log(x - 1) + 8*log(x - 2) + x^2/2

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sympy [A]  time = 0.11, size = 19, normalized size = 0.70 \[ \frac {x^{2}}{2} + 3 x + 8 \log {\left (x - 2 \right )} - \log {\left (x - 1 \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2/2 + 3*x + 8*log(x - 2) - log(x - 1)

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