3.65.12 \(\int \frac {-9-27 x+23 x^2+13 x^3+2 x^4}{9 x+6 x^2+x^3} \, dx\)

Optimal. Leaf size=25 \[ -4-x+x^2-\frac {2 (2-x) x}{3+x}-\log (x) \]

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Rubi [A]  time = 0.05, antiderivative size = 16, normalized size of antiderivative = 0.64, number of steps used = 4, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {1594, 27, 1620} \begin {gather*} x^2+x+\frac {30}{x+3}-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-9 - 27*x + 23*x^2 + 13*x^3 + 2*x^4)/(9*x + 6*x^2 + x^3),x]

[Out]

x + x^2 + 30/(3 + x) - 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 {-9-27 x+23 x^2+13 x^3+2 x^4}{x \left (9+6 x+x^2\right )} \, dx\\ &=\int \frac {-9-27 x+23 x^2+13 x^3+2 x^4}{x (3+x)^2} \, dx\\ &=\int \left (1-\frac {1}{x}+2 x-\frac {30}{(3+x)^2}\right ) \, dx\\ &=x+x^2+\frac {30}{3+x}-\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 16, normalized size = 0.64 \begin {gather*} x+x^2+\frac {30}{3+x}-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-9 - 27*x + 23*x^2 + 13*x^3 + 2*x^4)/(9*x + 6*x^2 + x^3),x]

[Out]

x + x^2 + 30/(3 + x) - Log[x]

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fricas [A]  time = 0.64, size = 26, normalized size = 1.04 \begin {gather*} \frac {x^{3} + 4 \, x^{2} - {\left (x + 3\right )} \log \relax (x) + 3 \, x + 30}{x + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^4+13*x^3+23*x^2-27*x-9)/(x^3+6*x^2+9*x),x, algorithm="fricas")

[Out]

(x^3 + 4*x^2 - (x + 3)*log(x) + 3*x + 30)/(x + 3)

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giac [A]  time = 0.12, size = 17, normalized size = 0.68 \begin {gather*} x^{2} + x + \frac {30}{x + 3} - \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^4+13*x^3+23*x^2-27*x-9)/(x^3+6*x^2+9*x),x, algorithm="giac")

[Out]

x^2 + x + 30/(x + 3) - log(abs(x))

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maple [A]  time = 0.03, size = 17, normalized size = 0.68

method	result	size
default	\(x^{2}+x -\ln \relax (x )+\frac {30}{3+x}\)	\(17\)
risch	\(x^{2}+x -\ln \relax (x )+\frac {30}{3+x}\)	\(17\)
norman	\(\frac {x^{3}+4 x^{2}+21}{3+x}-\ln \relax (x )\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^4+13*x^3+23*x^2-27*x-9)/(x^3+6*x^2+9*x),x,method=_RETURNVERBOSE)

[Out]

x^2+x-ln(x)+30/(3+x)

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maxima [A]  time = 0.35, size = 16, normalized size = 0.64 \begin {gather*} x^{2} + x + \frac {30}{x + 3} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^4+13*x^3+23*x^2-27*x-9)/(x^3+6*x^2+9*x),x, algorithm="maxima")

[Out]

x^2 + x + 30/(x + 3) - log(x)

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mupad [B]  time = 4.12, size = 16, normalized size = 0.64 \begin {gather*} x-\ln \relax (x)+\frac {30}{x+3}+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((23*x^2 - 27*x + 13*x^3 + 2*x^4 - 9)/(9*x + 6*x^2 + x^3),x)

[Out]

x - log(x) + 30/(x + 3) + x^2

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sympy [A]  time = 0.09, size = 12, normalized size = 0.48 \begin {gather*} x^{2} + x - \log {\relax (x )} + \frac {30}{x + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**4+13*x**3+23*x**2-27*x-9)/(x**3+6*x**2+9*x),x)

[Out]

x**2 + x - log(x) + 30/(x + 3)

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