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

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

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 21, normalized size of antiderivative = 1.00, 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 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 {-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*}

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Mathematica [A]  time = 0.01, size = 21, normalized size = 1.00 \[ -\frac {3 \log (x)}{2}+4 \log (x+1)-\frac {5}{2} \log (x+2) \]

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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fricas [A]  time = 0.40, size = 17, normalized size = 0.81 \[ -\frac {5}{2} \, \log \left (x + 2\right ) + 4 \, \log \left (x + 1\right ) - \frac {3}{2} \, \log \relax (x) \]

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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giac [A]  time = 0.01, size = 20, normalized size = 0.95 \[ -\frac {5}{2} \, \log \left ({\left | x + 2 \right |}\right ) + 4 \, \log \left ({\left | x + 1 \right |}\right ) - \frac {3}{2} \, \log \left ({\left | x \right |}\right ) \]

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))

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maple [A]  time = 0.01, size = 18, normalized size = 0.86 \[ -\frac {3 \ln \relax (x )}{2}+4 \ln \left (x +1\right )-\frac {5 \ln \left (x +2\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.63, size = 17, normalized size = 0.81 \[ -\frac {5}{2} \, \log \left (x + 2\right ) + 4 \, \log \left (x + 1\right ) - \frac {3}{2} \, \log \relax (x) \]

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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mupad [B]  time = 0.06, size = 17, normalized size = 0.81 \[ 4\,\ln \left (x+1\right )-\frac {5\,\ln \left (x+2\right )}{2}-\frac {3\,\ln \relax (x)}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - 3)/(2*x + 3*x^2 + x^3),x)

[Out]

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

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sympy [A]  time = 0.13, size = 20, normalized size = 0.95 \[ - \frac {3 \log {\relax (x )}}{2} + 4 \log {\left (x + 1 \right )} - \frac {5 \log {\left (x + 2 \right )}}{2} \]

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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