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

Optimal. Leaf size=21 \[ -\frac{3 \log (x)}{2}+4 \log (x+1)-\frac{5}{2} \log (x+2) \]

[Out]

(-3*Log[x])/2 + 4*Log[1 + x] - (5*Log[2 + x])/2

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Rubi [A]  time = 0.0213803, antiderivative size = 21, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {1594, 800} \[ -\frac{3 \log (x)}{2}+4 \log (x+1)-\frac{5}{2} \log (x+2) \]

Antiderivative was successfully verified.

[In]

Int[(-3 + x)/(2*x + 3*x^2 + x^3),x]

[Out]

(-3*Log[x])/2 + 4*Log[1 + x] - (5*Log[2 + x])/2

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{-3+x}{2 x+3 x^2+x^3} \, dx &=\int \frac{-3+x}{x \left (2+3 x+x^2\right )} \, dx\\ &=\int \left (-\frac{3}{2 x}+\frac{4}{1+x}-\frac{5}{2 (2+x)}\right ) \, dx\\ &=-\frac{3 \log (x)}{2}+4 \log (1+x)-\frac{5}{2} \log (2+x)\\ \end{align*}

Mathematica [A]  time = 0.0054153, size = 21, normalized size = 1. \[ -\frac{3 \log (x)}{2}+4 \log (x+1)-\frac{5}{2} \log (x+2) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Log[x])/2 + 4*Log[1 + x] - (5*Log[2 + x])/2

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Maple [A]  time = 0.006, size = 18, normalized size = 0.9 \begin{align*} -{\frac{3\,\ln \left ( x \right ) }{2}}+4\,\ln \left ( 1+x \right ) -{\frac{5\,\ln \left ( 2+x \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*ln(x)+4*ln(1+x)-5/2*ln(2+x)

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Maxima [A]  time = 0.975493, size = 23, normalized size = 1.1 \begin{align*} -\frac{5}{2} \, \log \left (x + 2\right ) + 4 \, \log \left (x + 1\right ) - \frac{3}{2} \, \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/2*log(x + 2) + 4*log(x + 1) - 3/2*log(x)

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Fricas [A]  time = 0.962911, size = 61, normalized size = 2.9 \begin{align*} -\frac{5}{2} \, \log \left (x + 2\right ) + 4 \, \log \left (x + 1\right ) - \frac{3}{2} \, \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/2*log(x + 2) + 4*log(x + 1) - 3/2*log(x)

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Sympy [A]  time = 0.11801, size = 20, normalized size = 0.95 \begin{align*} - \frac{3 \log{\left (x \right )}}{2} + 4 \log{\left (x + 1 \right )} - \frac{5 \log{\left (x + 2 \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*log(x)/2 + 4*log(x + 1) - 5*log(x + 2)/2

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Giac [A]  time = 1.10726, size = 27, normalized size = 1.29 \begin{align*} -\frac{5}{2} \, \log \left ({\left | x + 2 \right |}\right ) + 4 \, \log \left ({\left | x + 1 \right |}\right ) - \frac{3}{2} \, \log \left ({\left | x \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/2*log(abs(x + 2)) + 4*log(abs(x + 1)) - 3/2*log(abs(x))