3.2388 \(\int \frac{(5-x) (3+2 x)^3}{(2+5 x+3 x^2)^2} \, dx\)

Optimal. Leaf size=43 \[ -\frac{2611 x+2449}{27 \left (3 x^2+5 x+2\right )}-\frac{8 x}{9}+71 \log (x+1)-\frac{1825}{27} \log (3 x+2) \]

[Out]

(-8*x)/9 - (2449 + 2611*x)/(27*(2 + 5*x + 3*x^2)) + 71*Log[1 + x] - (1825*Log[2 + 3*x])/27

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Rubi [A]  time = 0.0470334, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {816, 1660, 1657, 632, 31} \[ -\frac{2611 x+2449}{27 \left (3 x^2+5 x+2\right )}-\frac{8 x}{9}+71 \log (x+1)-\frac{1825}{27} \log (3 x+2) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*x)/9 - (2449 + 2611*x)/(27*(2 + 5*x + 3*x^2)) + 71*Log[1 + x] - (1825*Log[2 + 3*x])/27

Rule 816

Int[((d_) + (e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Int[(a
+ b*x + c*x^2)^p*ExpandIntegrand[(d + e*x)^m*(f + g*x), 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] && ILtQ[p, -1] && IGtQ[m, 0] && RationalQ[a, b, c, d, e, f, g]

Rule 1660

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x + c*
x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b
*x + c*x^2, x], x, 1]}, Simp[((b*f - 2*a*g + (2*c*f - b*g)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c
)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (
2*p + 3)*(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

Rule 1657

Int[(Pq_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x + c*x^2)^p, x
], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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]

Rule 31

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

Rubi steps

\begin{align*} \int \frac{(5-x) (3+2 x)^3}{\left (2+5 x+3 x^2\right )^2} \, dx &=\int \frac{\frac{13}{2} (3+2 x)^3-\frac{1}{2} (3+2 x)^4}{\left (2+5 x+3 x^2\right )^2} \, dx\\ &=-\frac{2449+2611 x}{27 \left (2+5 x+3 x^2\right )}-\int \frac{\frac{563}{9}-\frac{52 x}{9}+\frac{8 x^2}{3}}{2+5 x+3 x^2} \, dx\\ &=-\frac{2449+2611 x}{27 \left (2+5 x+3 x^2\right )}-\int \left (\frac{8}{9}+\frac{547-92 x}{9 \left (2+5 x+3 x^2\right )}\right ) \, dx\\ &=-\frac{8 x}{9}-\frac{2449+2611 x}{27 \left (2+5 x+3 x^2\right )}-\frac{1}{9} \int \frac{547-92 x}{2+5 x+3 x^2} \, dx\\ &=-\frac{8 x}{9}-\frac{2449+2611 x}{27 \left (2+5 x+3 x^2\right )}-\frac{1825}{9} \int \frac{1}{2+3 x} \, dx+213 \int \frac{1}{3+3 x} \, dx\\ &=-\frac{8 x}{9}-\frac{2449+2611 x}{27 \left (2+5 x+3 x^2\right )}+71 \log (1+x)-\frac{1825}{27} \log (2+3 x)\\ \end{align*}

Mathematica [A]  time = 0.0351499, size = 47, normalized size = 1.09 \[ -\frac{2611 x+2449}{81 x^2+135 x+54}-\frac{4}{9} (2 x+3)-\frac{1825}{27} \log (-6 x-4)+71 \log (-2 (x+1)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(3 + 2*x))/9 - (2449 + 2611*x)/(54 + 135*x + 81*x^2) - (1825*Log[-4 - 6*x])/27 + 71*Log[-2*(1 + x)]

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Maple [A]  time = 0.007, size = 35, normalized size = 0.8 \begin{align*} -{\frac{8\,x}{9}}-6\, \left ( 1+x \right ) ^{-1}+71\,\ln \left ( 1+x \right ) -{\frac{2125}{54+81\,x}}-{\frac{1825\,\ln \left ( 2+3\,x \right ) }{27}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/9*x-6/(1+x)+71*ln(1+x)-2125/27/(2+3*x)-1825/27*ln(2+3*x)

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Maxima [A]  time = 1.26974, size = 50, normalized size = 1.16 \begin{align*} -\frac{8}{9} \, x - \frac{2611 \, x + 2449}{27 \,{\left (3 \, x^{2} + 5 \, x + 2\right )}} - \frac{1825}{27} \, \log \left (3 \, x + 2\right ) + 71 \, \log \left (x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/9*x - 1/27*(2611*x + 2449)/(3*x^2 + 5*x + 2) - 1825/27*log(3*x + 2) + 71*log(x + 1)

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Fricas [A]  time = 1.29478, size = 181, normalized size = 4.21 \begin{align*} -\frac{72 \, x^{3} + 120 \, x^{2} + 1825 \,{\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (3 \, x + 2\right ) - 1917 \,{\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (x + 1\right ) + 2659 \, x + 2449}{27 \,{\left (3 \, x^{2} + 5 \, x + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/27*(72*x^3 + 120*x^2 + 1825*(3*x^2 + 5*x + 2)*log(3*x + 2) - 1917*(3*x^2 + 5*x + 2)*log(x + 1) + 2659*x + 2
449)/(3*x^2 + 5*x + 2)

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Sympy [A]  time = 0.172657, size = 36, normalized size = 0.84 \begin{align*} - \frac{8 x}{9} - \frac{2611 x + 2449}{81 x^{2} + 135 x + 54} - \frac{1825 \log{\left (x + \frac{2}{3} \right )}}{27} + 71 \log{\left (x + 1 \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8*x/9 - (2611*x + 2449)/(81*x**2 + 135*x + 54) - 1825*log(x + 2/3)/27 + 71*log(x + 1)

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Giac [A]  time = 1.15975, size = 53, normalized size = 1.23 \begin{align*} -\frac{8}{9} \, x - \frac{2611 \, x + 2449}{27 \,{\left (3 \, x + 2\right )}{\left (x + 1\right )}} - \frac{1825}{27} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) + 71 \, \log \left ({\left | x + 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/9*x - 1/27*(2611*x + 2449)/((3*x + 2)*(x + 1)) - 1825/27*log(abs(3*x + 2)) + 71*log(abs(x + 1))