3.27.77 \(\int \frac {5-6 x^2+2 x^3}{-5 x+9 x^2-6 x^3+x^4} \, dx\)

Optimal. Leaf size=12 \[ \log \left ((-3+x)^2-\frac {5}{x}\right ) \]

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Rubi [A]  time = 0.05, antiderivative size = 21, normalized size of antiderivative = 1.75, number of steps used = 3, number of rules used = 2, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {2074, 1587} \begin {gather*} \log \left (-x^3+6 x^2-9 x+5\right )-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 - 6*x^2 + 2*x^3)/(-5*x + 9*x^2 - 6*x^3 + x^4),x]

[Out]

-Log[x] + Log[5 - 9*x + 6*x^2 - x^3]

Rule 1587

Int[(Pp_)/(Qq_), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[(Coeff[Pp, x, p]*Log[RemoveConte
nt[Qq, x]])/(q*Coeff[Qq, x, q]), x] /; EqQ[p, q - 1] && EqQ[Pp, Simplify[(Coeff[Pp, x, p]*D[Qq, x])/(q*Coeff[Q
q, x, q])]]] /; PolyQ[Pp, x] && PolyQ[Qq, x]

Rule 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {1}{x}+\frac {3 \left (3-4 x+x^2\right )}{-5+9 x-6 x^2+x^3}\right ) \, dx\\ &=-\log (x)+3 \int \frac {3-4 x+x^2}{-5+9 x-6 x^2+x^3} \, dx\\ &=-\log (x)+\log \left (5-9 x+6 x^2-x^3\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 21, normalized size = 1.75 \begin {gather*} -\log (x)+\log \left (5-9 x+6 x^2-x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 - 6*x^2 + 2*x^3)/(-5*x + 9*x^2 - 6*x^3 + x^4),x]

[Out]

-Log[x] + Log[5 - 9*x + 6*x^2 - x^3]

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fricas [A]  time = 1.18, size = 19, normalized size = 1.58 \begin {gather*} \log \left (x^{3} - 6 \, x^{2} + 9 \, x - 5\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3-6*x^2+5)/(x^4-6*x^3+9*x^2-5*x),x, algorithm="fricas")

[Out]

log(x^3 - 6*x^2 + 9*x - 5) - log(x)

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giac [A]  time = 0.19, size = 21, normalized size = 1.75 \begin {gather*} \log \left ({\left | x^{3} - 6 \, x^{2} + 9 \, x - 5 \right |}\right ) - \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3-6*x^2+5)/(x^4-6*x^3+9*x^2-5*x),x, algorithm="giac")

[Out]

log(abs(x^3 - 6*x^2 + 9*x - 5)) - log(abs(x))

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maple [A]  time = 0.02, size = 20, normalized size = 1.67

method	result	size
default	\(-\ln \relax (x )+\ln \left (x^{3}-6 x^{2}+9 x -5\right )\)	\(20\)
norman	\(-\ln \relax (x )+\ln \left (x^{3}-6 x^{2}+9 x -5\right )\)	\(20\)
risch	\(-\ln \relax (x )+\ln \left (x^{3}-6 x^{2}+9 x -5\right )\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^3-6*x^2+5)/(x^4-6*x^3+9*x^2-5*x),x,method=_RETURNVERBOSE)

[Out]

-ln(x)+ln(x^3-6*x^2+9*x-5)

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maxima [A]  time = 0.43, size = 19, normalized size = 1.58 \begin {gather*} \log \left (x^{3} - 6 \, x^{2} + 9 \, x - 5\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^3-6*x^2+5)/(x^4-6*x^3+9*x^2-5*x),x, algorithm="maxima")

[Out]

log(x^3 - 6*x^2 + 9*x - 5) - log(x)

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mupad [B]  time = 0.08, size = 19, normalized size = 1.58 \begin {gather*} \ln \left (x^3-6\,x^2+9\,x-5\right )-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x^3 - 6*x^2 + 5)/(5*x - 9*x^2 + 6*x^3 - x^4),x)

[Out]

log(9*x - 6*x^2 + x^3 - 5) - log(x)

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sympy [B]  time = 0.10, size = 17, normalized size = 1.42 \begin {gather*} - \log {\relax (x )} + \log {\left (x^{3} - 6 x^{2} + 9 x - 5 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**3-6*x**2+5)/(x**4-6*x**3+9*x**2-5*x),x)

[Out]

-log(x) + log(x**3 - 6*x**2 + 9*x - 5)

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