3.29.34 \(\int \frac {3-15 x+5 x^2-x^3}{x-2 x^2+x^3} \, dx\)

Optimal. Leaf size=18 \[ 2+\frac {8}{-1+x}-x+\log (x)+\log \left (x^2\right ) \]

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Rubi [A]  time = 0.04, antiderivative size = 17, normalized size of antiderivative = 0.94, 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 {8}{1-x}+3 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 - 15*x + 5*x^2 - x^3)/(x - 2*x^2 + x^3),x]

[Out]

-8/(1 - x) - x + 3*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 {3-15 x+5 x^2-x^3}{x \left (1-2 x+x^2\right )} \, dx\\ &=\int \frac {3-15 x+5 x^2-x^3}{(-1+x)^2 x} \, dx\\ &=\int \left (-1-\frac {8}{(-1+x)^2}+\frac {3}{x}\right ) \, dx\\ &=-\frac {8}{1-x}-x+3 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 15, normalized size = 0.83 \begin {gather*} \frac {8}{-1+x}-x+3 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 - 15*x + 5*x^2 - x^3)/(x - 2*x^2 + x^3),x]

[Out]

8/(-1 + x) - x + 3*Log[x]

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fricas [A]  time = 1.11, size = 22, normalized size = 1.22 \begin {gather*} -\frac {x^{2} - 3 \, {\left (x - 1\right )} \log \relax (x) - x - 8}{x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^3+5*x^2-15*x+3)/(x^3-2*x^2+x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.22, size = 16, normalized size = 0.89 \begin {gather*} -x + \frac {8}{x - 1} + 3 \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^3+5*x^2-15*x+3)/(x^3-2*x^2+x),x, algorithm="giac")

[Out]

-x + 8/(x - 1) + 3*log(abs(x))

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maple [A]  time = 0.02, size = 16, normalized size = 0.89

method	result	size
default	\(-x +3 \ln \relax (x )+\frac {8}{x -1}\)	\(16\)
risch	\(-x +3 \ln \relax (x )+\frac {8}{x -1}\)	\(16\)
norman	\(\frac {-x^{2}+9}{x -1}+3 \ln \relax (x )\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^3+5*x^2-15*x+3)/(x^3-2*x^2+x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.64, size = 15, normalized size = 0.83 \begin {gather*} -x + \frac {8}{x - 1} + 3 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^3+5*x^2-15*x+3)/(x^3-2*x^2+x),x, algorithm="maxima")

[Out]

-x + 8/(x - 1) + 3*log(x)

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mupad [B]  time = 0.05, size = 15, normalized size = 0.83 \begin {gather*} 3\,\ln \relax (x)-x+\frac {8}{x-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(15*x - 5*x^2 + x^3 - 3)/(x - 2*x^2 + x^3),x)

[Out]

3*log(x) - x + 8/(x - 1)

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sympy [A]  time = 0.08, size = 10, normalized size = 0.56 \begin {gather*} - x + 3 \log {\relax (x )} + \frac {8}{x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**3+5*x**2-15*x+3)/(x**3-2*x**2+x),x)

[Out]

-x + 3*log(x) + 8/(x - 1)

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