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]

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

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Rubi [A]  time = 0.05, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, 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 31

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

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 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
]

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*}

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Mathematica [A]  time = 0.04, 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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fricas [A]  time = 0.64, size = 93, normalized size = 1.58 \[ \frac {142086 \, x^{3} + 353367 \, x^{2} + 47650 \, {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (3 \, x + 2\right ) - 47682 \, {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (x + 1\right ) + 284958 \, x + 73839}{54 \, {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )}} \]

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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giac [A]  time = 0.17, size = 46, normalized size = 0.78 \[ \frac {47362 \, x^{3} + 117789 \, x^{2} + 94986 \, x + 24613}{18 \, {\left (3 \, x + 2\right )}^{2} {\left (x + 1\right )}^{2}} + \frac {23825}{27} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) - 883 \, \log \left ({\left | x + 1 \right |}\right ) \]

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))

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maple [A]  time = 0.05, size = 48, normalized size = 0.81 \[ \frac {23825 \ln \left (3 x +2\right )}{27}-883 \ln \left (x +1\right )-\frac {10625}{54 \left (3 x +2\right )^{2}}+\frac {15500}{27 \left (3 x +2\right )}+\frac {3}{\left (x +1\right )^{2}}+\frac {101}{x +1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.49, size = 54, normalized size = 0.92 \[ \frac {47362 \, x^{3} + 117789 \, x^{2} + 94986 \, x + 24613}{18 \, {\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )}} + \frac {23825}{27} \, \log \left (3 \, x + 2\right ) - 883 \, \log \left (x + 1\right ) \]

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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mupad [B]  time = 2.45, size = 49, normalized size = 0.83 \[ \frac {23825\,\ln \left (x+\frac {2}{3}\right )}{27}-883\,\ln \left (x+1\right )+\frac {\frac {23681\,x^3}{81}+\frac {39263\,x^2}{54}+\frac {1759\,x}{3}+\frac {24613}{162}}{x^4+\frac {10\,x^3}{3}+\frac {37\,x^2}{9}+\frac {20\,x}{9}+\frac {4}{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(23825*log(x + 2/3))/27 - 883*log(x + 1) + ((1759*x)/3 + (39263*x^2)/54 + (23681*x^3)/81 + 24613/162)/((20*x)/
9 + (37*x^2)/9 + (10*x^3)/3 + x^4 + 4/9)

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sympy [A]  time = 0.19, size = 53, normalized size = 0.90 \[ - \frac {- 47362 x^{3} - 117789 x^{2} - 94986 x - 24613}{162 x^{4} + 540 x^{3} + 666 x^{2} + 360 x + 72} + \frac {23825 \log {\left (x + \frac {2}{3} \right )}}{27} - 883 \log {\left (x + 1 \right )} \]

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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