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

Optimal. Leaf size=38 \[ -\frac {13}{5 (2 x+3)}-6 \log (x+1)+\frac {99}{25} \log (2 x+3)+\frac {51}{25} \log (3 x+2) \]

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Rubi [A]  time = 0.03, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {800} \begin {gather*} -\frac {13}{5 (2 x+3)}-6 \log (x+1)+\frac {99}{25} \log (2 x+3)+\frac {51}{25} \log (3 x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-13/(5*(3 + 2*x)) - 6*Log[1 + x] + (99*Log[3 + 2*x])/25 + (51*Log[2 + 3*x])/25

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 {5-x}{(3+2 x)^2 \left (2+5 x+3 x^2\right )} \, dx &=\int \left (-\frac {6}{1+x}+\frac {26}{5 (3+2 x)^2}+\frac {198}{25 (3+2 x)}+\frac {153}{25 (2+3 x)}\right ) \, dx\\ &=-\frac {13}{5 (3+2 x)}-6 \log (1+x)+\frac {99}{25} \log (3+2 x)+\frac {51}{25} \log (2+3 x)\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 38, normalized size = 1.00 \begin {gather*} \frac {1}{25} \left (-\frac {65}{2 x+3}+51 \log (-6 x-4)-150 \log (-2 (x+1))+99 \log (2 x+3)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-65/(3 + 2*x) + 51*Log[-4 - 6*x] - 150*Log[-2*(1 + x)] + 99*Log[3 + 2*x])/25

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.39, size = 48, normalized size = 1.26 \begin {gather*} \frac {51 \, {\left (2 \, x + 3\right )} \log \left (3 \, x + 2\right ) + 99 \, {\left (2 \, x + 3\right )} \log \left (2 \, x + 3\right ) - 150 \, {\left (2 \, x + 3\right )} \log \left (x + 1\right ) - 65}{25 \, {\left (2 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/25*(51*(2*x + 3)*log(3*x + 2) + 99*(2*x + 3)*log(2*x + 3) - 150*(2*x + 3)*log(x + 1) - 65)/(2*x + 3)

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giac [A]  time = 0.19, size = 40, normalized size = 1.05 \begin {gather*} -\frac {13}{5 \, {\left (2 \, x + 3\right )}} - 6 \, \log \left ({\left | -\frac {1}{2 \, x + 3} + 1 \right |}\right ) + \frac {51}{25} \, \log \left ({\left | -\frac {5}{2 \, x + 3} + 3 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-13/5/(2*x + 3) - 6*log(abs(-1/(2*x + 3) + 1)) + 51/25*log(abs(-5/(2*x + 3) + 3))

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maple [A]  time = 0.06, size = 33, normalized size = 0.87 \begin {gather*} \frac {51 \ln \left (3 x +2\right )}{25}+\frac {99 \ln \left (2 x +3\right )}{25}-6 \ln \left (x +1\right )-\frac {13}{5 \left (2 x +3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-13/5/(2*x+3)-6*ln(x+1)+99/25*ln(2*x+3)+51/25*ln(3*x+2)

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maxima [A]  time = 0.73, size = 32, normalized size = 0.84 \begin {gather*} -\frac {13}{5 \, {\left (2 \, x + 3\right )}} + \frac {51}{25} \, \log \left (3 \, x + 2\right ) + \frac {99}{25} \, \log \left (2 \, x + 3\right ) - 6 \, \log \left (x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-13/5/(2*x + 3) + 51/25*log(3*x + 2) + 99/25*log(2*x + 3) - 6*log(x + 1)

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mupad [B]  time = 2.29, size = 28, normalized size = 0.74 \begin {gather*} \frac {51\,\ln \left (x+\frac {2}{3}\right )}{25}-6\,\ln \left (x+1\right )+\frac {99\,\ln \left (x+\frac {3}{2}\right )}{25}-\frac {13}{10\,\left (x+\frac {3}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(51*log(x + 2/3))/25 - 6*log(x + 1) + (99*log(x + 3/2))/25 - 13/(10*(x + 3/2))

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sympy [A]  time = 0.16, size = 32, normalized size = 0.84 \begin {gather*} \frac {51 \log {\left (x + \frac {2}{3} \right )}}{25} - 6 \log {\left (x + 1 \right )} + \frac {99 \log {\left (x + \frac {3}{2} \right )}}{25} - \frac {13}{10 x + 15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

51*log(x + 2/3)/25 - 6*log(x + 1) + 99*log(x + 3/2)/25 - 13/(10*x + 15)

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