3.22.41 \(\int \frac {5-x}{2+5 x+3 x^2} \, dx\)

Optimal. Leaf size=17 \[ \frac {17}{3} \log (3 x+2)-6 \log (x+1) \]

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Rubi [A]  time = 0.01, antiderivative size = 17, 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 = {632, 31} \begin {gather*} \frac {17}{3} \log (3 x+2)-6 \log (x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 - x)/(2 + 5*x + 3*x^2),x]

[Out]

-6*Log[1 + x] + (17*Log[2 + 3*x])/3

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 632

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && NiceSqrtQ[b^2 - 4*a*
c]

Rubi steps

\begin {align*} \int \frac {5-x}{2+5 x+3 x^2} \, dx &=17 \int \frac {1}{2+3 x} \, dx-18 \int \frac {1}{3+3 x} \, dx\\ &=-6 \log (1+x)+\frac {17}{3} \log (2+3 x)\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 17, normalized size = 1.00 \begin {gather*} \frac {17}{3} \log (3 x+2)-6 \log (x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 - x)/(2 + 5*x + 3*x^2),x]

[Out]

-6*Log[1 + x] + (17*Log[2 + 3*x])/3

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {5-x}{2+5 x+3 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(5 - x)/(2 + 5*x + 3*x^2),x]

[Out]

IntegrateAlgebraic[(5 - x)/(2 + 5*x + 3*x^2), x]

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fricas [A]  time = 0.38, size = 15, normalized size = 0.88 \begin {gather*} \frac {17}{3} \, \log \left (3 \, x + 2\right ) - 6 \, \log \left (x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)/(3*x^2+5*x+2),x, algorithm="fricas")

[Out]

17/3*log(3*x + 2) - 6*log(x + 1)

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giac [A]  time = 0.15, size = 17, normalized size = 1.00 \begin {gather*} \frac {17}{3} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) - 6 \, \log \left ({\left | x + 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)/(3*x^2+5*x+2),x, algorithm="giac")

[Out]

17/3*log(abs(3*x + 2)) - 6*log(abs(x + 1))

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maple [A]  time = 0.05, size = 16, normalized size = 0.94 \begin {gather*} \frac {17 \ln \left (3 x +2\right )}{3}-6 \ln \left (x +1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5-x)/(3*x^2+5*x+2),x)

[Out]

-6*ln(x+1)+17/3*ln(3*x+2)

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maxima [A]  time = 0.73, size = 15, normalized size = 0.88 \begin {gather*} \frac {17}{3} \, \log \left (3 \, x + 2\right ) - 6 \, \log \left (x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)/(3*x^2+5*x+2),x, algorithm="maxima")

[Out]

17/3*log(3*x + 2) - 6*log(x + 1)

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mupad [B]  time = 2.32, size = 13, normalized size = 0.76 \begin {gather*} \frac {17\,\ln \left (x+\frac {2}{3}\right )}{3}-6\,\ln \left (x+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x - 5)/(5*x + 3*x^2 + 2),x)

[Out]

(17*log(x + 2/3))/3 - 6*log(x + 1)

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sympy [A]  time = 0.10, size = 15, normalized size = 0.88 \begin {gather*} \frac {17 \log {\left (x + \frac {2}{3} \right )}}{3} - 6 \log {\left (x + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)/(3*x**2+5*x+2),x)

[Out]

17*log(x + 2/3)/3 - 6*log(x + 1)

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