3.441 \(\int \frac{1+x}{-6 x+x^2+x^3} \, dx\)

Optimal. Leaf size=25 \[ \frac{3}{10} \log (2-x)-\frac{\log (x)}{6}-\frac{2}{15} \log (x+3) \]

[Out]

(3*Log[2 - x])/10 - Log[x]/6 - (2*Log[3 + x])/15

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Rubi [A]  time = 0.0208231, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {1594, 800} \[ \frac{3}{10} \log (2-x)-\frac{\log (x)}{6}-\frac{2}{15} \log (x+3) \]

Antiderivative was successfully verified.

[In]

Int[(1 + x)/(-6*x + x^2 + x^3),x]

[Out]

(3*Log[2 - x])/10 - Log[x]/6 - (2*Log[3 + x])/15

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 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rubi steps

\begin{align*} \int \frac{1+x}{-6 x+x^2+x^3} \, dx &=\int \frac{1+x}{x \left (-6+x+x^2\right )} \, dx\\ &=\int \left (\frac{3}{10 (-2+x)}-\frac{1}{6 x}-\frac{2}{15 (3+x)}\right ) \, dx\\ &=\frac{3}{10} \log (2-x)-\frac{\log (x)}{6}-\frac{2}{15} \log (3+x)\\ \end{align*}

Mathematica [A]  time = 0.0051272, size = 25, normalized size = 1. \[ \frac{3}{10} \log (2-x)-\frac{\log (x)}{6}-\frac{2}{15} \log (x+3) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x)/(-6*x + x^2 + x^3),x]

[Out]

(3*Log[2 - x])/10 - Log[x]/6 - (2*Log[3 + x])/15

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Maple [A]  time = 0.005, size = 18, normalized size = 0.7 \begin{align*} -{\frac{\ln \left ( x \right ) }{6}}-{\frac{2\,\ln \left ( 3+x \right ) }{15}}+{\frac{3\,\ln \left ( -2+x \right ) }{10}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+x)/(x^3+x^2-6*x),x)

[Out]

-1/6*ln(x)-2/15*ln(3+x)+3/10*ln(-2+x)

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Maxima [A]  time = 0.962201, size = 23, normalized size = 0.92 \begin{align*} -\frac{2}{15} \, \log \left (x + 3\right ) + \frac{3}{10} \, \log \left (x - 2\right ) - \frac{1}{6} \, \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)/(x^3+x^2-6*x),x, algorithm="maxima")

[Out]

-2/15*log(x + 3) + 3/10*log(x - 2) - 1/6*log(x)

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Fricas [A]  time = 0.976519, size = 66, normalized size = 2.64 \begin{align*} -\frac{2}{15} \, \log \left (x + 3\right ) + \frac{3}{10} \, \log \left (x - 2\right ) - \frac{1}{6} \, \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)/(x^3+x^2-6*x),x, algorithm="fricas")

[Out]

-2/15*log(x + 3) + 3/10*log(x - 2) - 1/6*log(x)

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Sympy [A]  time = 0.119831, size = 20, normalized size = 0.8 \begin{align*} - \frac{\log{\left (x \right )}}{6} + \frac{3 \log{\left (x - 2 \right )}}{10} - \frac{2 \log{\left (x + 3 \right )}}{15} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)/(x**3+x**2-6*x),x)

[Out]

-log(x)/6 + 3*log(x - 2)/10 - 2*log(x + 3)/15

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Giac [A]  time = 1.16003, size = 27, normalized size = 1.08 \begin{align*} -\frac{2}{15} \, \log \left ({\left | x + 3 \right |}\right ) + \frac{3}{10} \, \log \left ({\left | x - 2 \right |}\right ) - \frac{1}{6} \, \log \left ({\left | x \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)/(x^3+x^2-6*x),x, algorithm="giac")

[Out]

-2/15*log(abs(x + 3)) + 3/10*log(abs(x - 2)) - 1/6*log(abs(x))