3.2399 \(\int \frac{(5-x) (3+2 x)^2}{(2+5 x+3 x^2)^3} \, dx\)

Optimal. Leaf size=57 \[ -\frac{587 x+533}{18 \left (3 x^2+5 x+2\right )^2}+\frac{9918 x+8269}{18 \left (3 x^2+5 x+2\right )}-551 \log (x+1)+551 \log (3 x+2) \]

[Out]

-(533 + 587*x)/(18*(2 + 5*x + 3*x^2)^2) + (8269 + 9918*x)/(18*(2 + 5*x + 3*x^2)) - 551*Log[1 + x] + 551*Log[2
+ 3*x]

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Rubi [A]  time = 0.0428569, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {816, 1660, 638, 616, 31} \[ -\frac{587 x+533}{18 \left (3 x^2+5 x+2\right )^2}+\frac{9918 x+8269}{18 \left (3 x^2+5 x+2\right )}-551 \log (x+1)+551 \log (3 x+2) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(533 + 587*x)/(18*(2 + 5*x + 3*x^2)^2) + (8269 + 9918*x)/(18*(2 + 5*x + 3*x^2)) - 551*Log[1 + x] + 551*Log[2
+ 3*x]

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 638

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*p + 3)*(2*c*d - b*e))/((p + 1)*(b^2
- 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 616

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[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)^2}{\left (2+5 x+3 x^2\right )^3} \, dx &=\int \frac{45+51 x+8 x^2-4 x^3}{\left (2+5 x+3 x^2\right )^3} \, dx\\ &=-\frac{533+587 x}{18 \left (2+5 x+3 x^2\right )^2}-\frac{1}{2} \int \frac{\frac{1673}{9}+\frac{8 x}{3}}{\left (2+5 x+3 x^2\right )^2} \, dx\\ &=-\frac{533+587 x}{18 \left (2+5 x+3 x^2\right )^2}+\frac{8269+9918 x}{18 \left (2+5 x+3 x^2\right )}+551 \int \frac{1}{2+5 x+3 x^2} \, dx\\ &=-\frac{533+587 x}{18 \left (2+5 x+3 x^2\right )^2}+\frac{8269+9918 x}{18 \left (2+5 x+3 x^2\right )}+1653 \int \frac{1}{2+3 x} \, dx-1653 \int \frac{1}{3+3 x} \, dx\\ &=-\frac{533+587 x}{18 \left (2+5 x+3 x^2\right )^2}+\frac{8269+9918 x}{18 \left (2+5 x+3 x^2\right )}-551 \log (1+x)+551 \log (2+3 x)\\ \end{align*}

Mathematica [A]  time = 0.0308496, size = 56, normalized size = 0.98 \[ -\frac{587 x+533}{18 \left (3 x^2+5 x+2\right )^2}+\frac{9918 x+8269}{54 x^2+90 x+36}+551 \log (-6 x-4)-551 \log (-2 (x+1)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(533 + 587*x)/(18*(2 + 5*x + 3*x^2)^2) + (8269 + 9918*x)/(36 + 90*x + 54*x^2) + 551*Log[-4 - 6*x] - 551*Log[-
2*(1 + x)]

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Maple [A]  time = 0.01, size = 48, normalized size = 0.8 \begin{align*} 3\, \left ( 1+x \right ) ^{-2}+77\, \left ( 1+x \right ) ^{-1}-551\,\ln \left ( 1+x \right ) -{\frac{425}{6\, \left ( 2+3\,x \right ) ^{2}}}+320\, \left ( 2+3\,x \right ) ^{-1}+551\,\ln \left ( 2+3\,x \right ) \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)^3,x)

[Out]

3/(1+x)^2+77/(1+x)-551*ln(1+x)-425/6/(2+3*x)^2+320/(2+3*x)+551*ln(2+3*x)

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Maxima [A]  time = 1.02848, size = 73, normalized size = 1.28 \begin{align*} \frac{9918 \, x^{3} + 24799 \, x^{2} + 20198 \, x + 5335}{6 \,{\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )}} + 551 \, \log \left (3 \, x + 2\right ) - 551 \, \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)^3,x, algorithm="maxima")

[Out]

1/6*(9918*x^3 + 24799*x^2 + 20198*x + 5335)/(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4) + 551*log(3*x + 2) - 551*log(
x + 1)

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Fricas [A]  time = 1.28388, size = 262, normalized size = 4.6 \begin{align*} \frac{9918 \, x^{3} + 24799 \, x^{2} + 3306 \,{\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (3 \, x + 2\right ) - 3306 \,{\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (x + 1\right ) + 20198 \, x + 5335}{6 \,{\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\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)^3,x, algorithm="fricas")

[Out]

1/6*(9918*x^3 + 24799*x^2 + 3306*(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)*log(3*x + 2) - 3306*(9*x^4 + 30*x^3 + 37
*x^2 + 20*x + 4)*log(x + 1) + 20198*x + 5335)/(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)

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Sympy [A]  time = 0.184523, size = 49, normalized size = 0.86 \begin{align*} \frac{9918 x^{3} + 24799 x^{2} + 20198 x + 5335}{54 x^{4} + 180 x^{3} + 222 x^{2} + 120 x + 24} + 551 \log{\left (x + \frac{2}{3} \right )} - 551 \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)**3,x)

[Out]

(9918*x**3 + 24799*x**2 + 20198*x + 5335)/(54*x**4 + 180*x**3 + 222*x**2 + 120*x + 24) + 551*log(x + 2/3) - 55
1*log(x + 1)

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Giac [A]  time = 1.156, size = 62, normalized size = 1.09 \begin{align*} \frac{9918 \, x^{3} + 24799 \, x^{2} + 20198 \, x + 5335}{6 \,{\left (3 \, x^{2} + 5 \, x + 2\right )}^{2}} + 551 \, \log \left ({\left | 3 \, x + 2 \right |}\right ) - 551 \, \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)^2/(3*x^2+5*x+2)^3,x, algorithm="giac")

[Out]

1/6*(9918*x^3 + 24799*x^2 + 20198*x + 5335)/(3*x^2 + 5*x + 2)^2 + 551*log(abs(3*x + 2)) - 551*log(abs(x + 1))