3.2397 \(\int \frac{(5-x) (3+2 x)^4}{(2+5 x+3 x^2)^3} \, dx\)

Optimal. Leaf size=59 \[ -\frac{12083 x+11597}{162 \left (3 x^2+5 x+2\right )^2}+\frac{7 (20298 x+16651)}{162 \left (3 x^2+5 x+2\right )}-883 \log (x+1)+\frac{23825}{27} \log (3 x+2) \]

[Out]

-(11597 + 12083*x)/(162*(2 + 5*x + 3*x^2)^2) + (7*(16651 + 20298*x))/(162*(2 + 5*x + 3*x^2)) - 883*Log[1 + x]
+ (23825*Log[2 + 3*x])/27

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Rubi [A]  time = 0.049438, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {816, 1660, 632, 31} \[ -\frac{12083 x+11597}{162 \left (3 x^2+5 x+2\right )^2}+\frac{7 (20298 x+16651)}{162 \left (3 x^2+5 x+2\right )}-883 \log (x+1)+\frac{23825}{27} \log (3 x+2) \]

Antiderivative was successfully verified.

[In]

Int[((5 - x)*(3 + 2*x)^4)/(2 + 5*x + 3*x^2)^3,x]

[Out]

-(11597 + 12083*x)/(162*(2 + 5*x + 3*x^2)^2) + (7*(16651 + 20298*x))/(162*(2 + 5*x + 3*x^2)) - 883*Log[1 + x]
+ (23825*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 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)^4}{\left (2+5 x+3 x^2\right )^3} \, dx &=\int \frac{\frac{13}{2} (3+2 x)^4-\frac{1}{2} (3+2 x)^5}{\left (2+5 x+3 x^2\right )^3} \, dx\\ &=-\frac{11597+12083 x}{162 \left (2+5 x+3 x^2\right )^2}-\frac{1}{2} \int \frac{\frac{13097}{81}-\frac{4624 x}{27}-\frac{64 x^2}{9}+\frac{32 x^3}{3}}{\left (2+5 x+3 x^2\right )^2} \, dx\\ &=-\frac{11597+12083 x}{162 \left (2+5 x+3 x^2\right )^2}+\frac{7 (16651+20298 x)}{162 \left (2+5 x+3 x^2\right )}+\frac{1}{2} \int \frac{\frac{15862}{9}-\frac{32 x}{9}}{2+5 x+3 x^2} \, dx\\ &=-\frac{11597+12083 x}{162 \left (2+5 x+3 x^2\right )^2}+\frac{7 (16651+20298 x)}{162 \left (2+5 x+3 x^2\right )}+\frac{23825}{9} \int \frac{1}{2+3 x} \, dx-2649 \int \frac{1}{3+3 x} \, dx\\ &=-\frac{11597+12083 x}{162 \left (2+5 x+3 x^2\right )^2}+\frac{7 (16651+20298 x)}{162 \left (2+5 x+3 x^2\right )}-883 \log (1+x)+\frac{23825}{27} \log (2+3 x)\\ \end{align*}

Mathematica [A]  time = 0.0383018, size = 62, normalized size = 1.05 \[ \frac{1}{54} \left (47650 \log (-6 x-4)-\frac{3 \left (-47362 x^3-117789 x^2+15894 \left (3 x^2+5 x+2\right )^2 \log (-2 (x+1))-94986 x-24613\right )}{\left (3 x^2+5 x+2\right )^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((5 - x)*(3 + 2*x)^4)/(2 + 5*x + 3*x^2)^3,x]

[Out]

(47650*Log[-4 - 6*x] - (3*(-24613 - 94986*x - 117789*x^2 - 47362*x^3 + 15894*(2 + 5*x + 3*x^2)^2*Log[-2*(1 + x
)]))/(2 + 5*x + 3*x^2)^2)/54

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Maple [A]  time = 0.01, size = 48, normalized size = 0.8 \begin{align*} 3\, \left ( 1+x \right ) ^{-2}+101\, \left ( 1+x \right ) ^{-1}-883\,\ln \left ( 1+x \right ) -{\frac{10625}{54\, \left ( 2+3\,x \right ) ^{2}}}+{\frac{15500}{54+81\,x}}+{\frac{23825\,\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)^4/(3*x^2+5*x+2)^3,x)

[Out]

3/(1+x)^2+101/(1+x)-883*ln(1+x)-10625/54/(2+3*x)^2+15500/27/(2+3*x)+23825/27*ln(2+3*x)

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Maxima [A]  time = 0.99308, size = 73, normalized size = 1.24 \begin{align*} \frac{47362 \, x^{3} + 117789 \, x^{2} + 94986 \, x + 24613}{18 \,{\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )}} + \frac{23825}{27} \, \log \left (3 \, x + 2\right ) - 883 \, \log \left (x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)*(3+2*x)^4/(3*x^2+5*x+2)^3,x, algorithm="maxima")

[Out]

1/18*(47362*x^3 + 117789*x^2 + 94986*x + 24613)/(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4) + 23825/27*log(3*x + 2) -
883*log(x + 1)

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Fricas [A]  time = 1.26285, size = 273, normalized size = 4.63 \begin{align*} \frac{142086 \, x^{3} + 353367 \, x^{2} + 47650 \,{\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (3 \, x + 2\right ) - 47682 \,{\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (x + 1\right ) + 284958 \, x + 73839}{54 \,{\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)^4/(3*x^2+5*x+2)^3,x, algorithm="fricas")

[Out]

1/54*(142086*x^3 + 353367*x^2 + 47650*(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)*log(3*x + 2) - 47682*(9*x^4 + 30*x^
3 + 37*x^2 + 20*x + 4)*log(x + 1) + 284958*x + 73839)/(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)

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Sympy [A]  time = 0.208549, size = 51, normalized size = 0.86 \begin{align*} \frac{47362 x^{3} + 117789 x^{2} + 94986 x + 24613}{162 x^{4} + 540 x^{3} + 666 x^{2} + 360 x + 72} + \frac{23825 \log{\left (x + \frac{2}{3} \right )}}{27} - 883 \log{\left (x + 1 \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)*(3+2*x)**4/(3*x**2+5*x+2)**3,x)

[Out]

(47362*x**3 + 117789*x**2 + 94986*x + 24613)/(162*x**4 + 540*x**3 + 666*x**2 + 360*x + 72) + 23825*log(x + 2/3
)/27 - 883*log(x + 1)

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Giac [A]  time = 1.14848, size = 62, normalized size = 1.05 \begin{align*} \frac{47362 \, x^{3} + 117789 \, x^{2} + 94986 \, x + 24613}{18 \,{\left (3 \, x + 2\right )}^{2}{\left (x + 1\right )}^{2}} + \frac{23825}{27} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) - 883 \, \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)^4/(3*x^2+5*x+2)^3,x, algorithm="giac")

[Out]

1/18*(47362*x^3 + 117789*x^2 + 94986*x + 24613)/((3*x + 2)^2*(x + 1)^2) + 23825/27*log(abs(3*x + 2)) - 883*log
(abs(x + 1))