3.2384 \(\int \frac{5-x}{(3+2 x)^2 (2+5 x+3 x^2)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0289858, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.04, Rules used = {800} \[ -\frac{13}{5 (2 x+3)}-6 \log (x+1)+\frac{99}{25} \log (2 x+3)+\frac{51}{25} \log (3 x+2) \]

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*}

Mathematica [A]  time = 0.0272577, size = 38, normalized size = 1. \[ \frac{1}{25} \left (-\frac{65}{2 x+3}+51 \log (-6 x-4)-150 \log (-2 (x+1))+99 \log (2 x+3)\right ) \]

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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Maple [A]  time = 0.007, size = 33, normalized size = 0.9 \begin{align*} -{\frac{13}{15+10\,x}}-6\,\ln \left ( 1+x \right ) +{\frac{99\,\ln \left ( 3+2\,x \right ) }{25}}+{\frac{51\,\ln \left ( 2+3\,x \right ) }{25}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.22218, size = 43, normalized size = 1.13 \begin{align*} -\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{align*}

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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Fricas [A]  time = 1.28028, size = 140, normalized size = 3.68 \begin{align*} \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{align*}

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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Sympy [A]  time = 0.169717, size = 32, normalized size = 0.84 \begin{align*} \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{align*}

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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Giac [A]  time = 1.1113, size = 54, normalized size = 1.42 \begin{align*} -\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{align*}

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))