3.427 \(\int \frac{-x+x^3}{6+2 x} \, dx\)

Optimal. Leaf size=24 \[ \frac{x^3}{6}-\frac{3 x^2}{4}+4 x-12 \log (x+3) \]

[Out]

4*x - (3*x^2)/4 + x^3/6 - 12*Log[3 + x]

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Rubi [A]  time = 0.0188889, antiderivative size = 24, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {1593, 772} \[ \frac{x^3}{6}-\frac{3 x^2}{4}+4 x-12 \log (x+3) \]

Antiderivative was successfully verified.

[In]

Int[(-x + x^3)/(6 + 2*x),x]

[Out]

4*x - (3*x^2)/4 + x^3/6 - 12*Log[3 + x]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 772

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \frac{-x+x^3}{6+2 x} \, dx &=\int \frac{x \left (-1+x^2\right )}{6+2 x} \, dx\\ &=\int \left (4-\frac{3 x}{2}+\frac{x^2}{2}-\frac{12}{3+x}\right ) \, dx\\ &=4 x-\frac{3 x^2}{4}+\frac{x^3}{6}-12 \log (3+x)\\ \end{align*}

Mathematica [A]  time = 0.0058096, size = 31, normalized size = 1.29 \[ \frac{1}{2} \left (\frac{x^3}{3}-\frac{3 x^2}{2}+8 x-24 \log (x+3)+\frac{93}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(-x + x^3)/(6 + 2*x),x]

[Out]

(93/2 + 8*x - (3*x^2)/2 + x^3/3 - 24*Log[3 + x])/2

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Maple [A]  time = 0.003, size = 21, normalized size = 0.9 \begin{align*} 4\,x-{\frac{3\,{x}^{2}}{4}}+{\frac{{x}^{3}}{6}}-12\,\ln \left ( 3+x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*x-3/4*x^2+1/6*x^3-12*ln(3+x)

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Maxima [A]  time = 1.04315, size = 27, normalized size = 1.12 \begin{align*} \frac{1}{6} \, x^{3} - \frac{3}{4} \, x^{2} + 4 \, x - 12 \, \log \left (x + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^3 - 3/4*x^2 + 4*x - 12*log(x + 3)

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Fricas [A]  time = 0.965709, size = 55, normalized size = 2.29 \begin{align*} \frac{1}{6} \, x^{3} - \frac{3}{4} \, x^{2} + 4 \, x - 12 \, \log \left (x + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^3 - 3/4*x^2 + 4*x - 12*log(x + 3)

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Sympy [A]  time = 0.072741, size = 20, normalized size = 0.83 \begin{align*} \frac{x^{3}}{6} - \frac{3 x^{2}}{4} + 4 x - 12 \log{\left (x + 3 \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**3/6 - 3*x**2/4 + 4*x - 12*log(x + 3)

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Giac [A]  time = 1.15991, size = 28, normalized size = 1.17 \begin{align*} \frac{1}{6} \, x^{3} - \frac{3}{4} \, x^{2} + 4 \, x - 12 \, \log \left ({\left | x + 3 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^3 - 3/4*x^2 + 4*x - 12*log(abs(x + 3))