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]

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

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Rubi [A]  time = 0.02, antiderivative size = 25, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.01, size = 25, normalized size = 1.00 \[ \frac {3}{10} \log (2-x)-\frac {\log (x)}{6}-\frac {2}{15} \log (x+3) \]

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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fricas [A]  time = 0.76, size = 17, normalized size = 0.68 \[ -\frac {2}{15} \, \log \left (x + 3\right ) + \frac {3}{10} \, \log \left (x - 2\right ) - \frac {1}{6} \, \log \relax (x) \]

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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giac [A]  time = 0.36, size = 20, normalized size = 0.80 \[ -\frac {2}{15} \, \log \left ({\left | x + 3 \right |}\right ) + \frac {3}{10} \, \log \left ({\left | x - 2 \right |}\right ) - \frac {1}{6} \, \log \left ({\left | x \right |}\right ) \]

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))

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maple [A]  time = 0.01, size = 18, normalized size = 0.72 \[ -\frac {\ln \relax (x )}{6}+\frac {3 \ln \left (x -2\right )}{10}-\frac {2 \ln \left (x +3\right )}{15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.65, size = 17, normalized size = 0.68 \[ -\frac {2}{15} \, \log \left (x + 3\right ) + \frac {3}{10} \, \log \left (x - 2\right ) - \frac {1}{6} \, \log \relax (x) \]

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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mupad [B]  time = 0.11, size = 17, normalized size = 0.68 \[ \frac {3\,\ln \left (x-2\right )}{10}-\frac {2\,\ln \left (x+3\right )}{15}-\frac {\ln \relax (x)}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + 1)/(x^2 - 6*x + x^3),x)

[Out]

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

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sympy [A]  time = 0.13, size = 20, normalized size = 0.80 \[ - \frac {\log {\relax (x )}}{6} + \frac {3 \log {\left (x - 2 \right )}}{10} - \frac {2 \log {\left (x + 3 \right )}}{15} \]

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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