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

Optimal. Leaf size=50 \[ -\frac {8 x^2}{9}-\frac {12083 x+11597}{81 \left (3 x^2+5 x+2\right )}+\frac {112 x}{27}+83 \log (x+1)-\frac {1625}{27} \log (3 x+2) \]

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Rubi [A]  time = 0.06, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {816, 1660, 1657, 632, 31} \begin {gather*} -\frac {8 x^2}{9}-\frac {12083 x+11597}{81 \left (3 x^2+5 x+2\right )}+\frac {112 x}{27}+83 \log (x+1)-\frac {1625}{27} \log (3 x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(112*x)/27 - (8*x^2)/9 - (11597 + 12083*x)/(81*(2 + 5*x + 3*x^2)) + 83*Log[1 + x] - (1625*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 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 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 )^2} \, 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 )^2} \, dx\\ &=-\frac {11597+12083 x}{81 \left (2+5 x+3 x^2\right )}-\int \frac {\frac {169}{27}-\frac {2312 x}{27}-\frac {32 x^2}{9}+\frac {16 x^3}{3}}{2+5 x+3 x^2} \, dx\\ &=-\frac {11597+12083 x}{81 \left (2+5 x+3 x^2\right )}-\int \left (-\frac {112}{27}+\frac {16 x}{9}+\frac {131-616 x}{9 \left (2+5 x+3 x^2\right )}\right ) \, dx\\ &=\frac {112 x}{27}-\frac {8 x^2}{9}-\frac {11597+12083 x}{81 \left (2+5 x+3 x^2\right )}-\frac {1}{9} \int \frac {131-616 x}{2+5 x+3 x^2} \, dx\\ &=\frac {112 x}{27}-\frac {8 x^2}{9}-\frac {11597+12083 x}{81 \left (2+5 x+3 x^2\right )}-\frac {1625}{9} \int \frac {1}{2+3 x} \, dx+249 \int \frac {1}{3+3 x} \, dx\\ &=\frac {112 x}{27}-\frac {8 x^2}{9}-\frac {11597+12083 x}{81 \left (2+5 x+3 x^2\right )}+83 \log (1+x)-\frac {1625}{27} \log (2+3 x)\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 56, normalized size = 1.12 \begin {gather*} \frac {1}{81} \left (-\frac {12083 x+11597}{3 x^2+5 x+2}-18 (2 x+3)^2+276 (2 x+3)-4875 \log (-6 x-4)+6723 \log (-2 (x+1))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(276*(3 + 2*x) - 18*(3 + 2*x)^2 - (11597 + 12083*x)/(2 + 5*x + 3*x^2) - 4875*Log[-4 - 6*x] + 6723*Log[-2*(1 +
x)])/81

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(5-x) (3+2 x)^4}{\left (2+5 x+3 x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.39, size = 68, normalized size = 1.36 \begin {gather*} -\frac {216 \, x^{4} - 648 \, x^{3} - 1536 \, x^{2} + 4875 \, {\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (3 \, x + 2\right ) - 6723 \, {\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (x + 1\right ) + 11411 \, x + 11597}{81 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/81*(216*x^4 - 648*x^3 - 1536*x^2 + 4875*(3*x^2 + 5*x + 2)*log(3*x + 2) - 6723*(3*x^2 + 5*x + 2)*log(x + 1)
+ 11411*x + 11597)/(3*x^2 + 5*x + 2)

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giac [A]  time = 0.16, size = 44, normalized size = 0.88 \begin {gather*} -\frac {8}{9} \, x^{2} + \frac {112}{27} \, x - \frac {12083 \, x + 11597}{81 \, {\left (3 \, x + 2\right )} {\left (x + 1\right )}} - \frac {1625}{27} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) + 83 \, \log \left ({\left | x + 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/9*x^2 + 112/27*x - 1/81*(12083*x + 11597)/((3*x + 2)*(x + 1)) - 1625/27*log(abs(3*x + 2)) + 83*log(abs(x +
1))

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maple [A]  time = 0.05, size = 40, normalized size = 0.80 \begin {gather*} -\frac {8 x^{2}}{9}+\frac {112 x}{27}-\frac {1625 \ln \left (3 x +2\right )}{27}+83 \ln \left (x +1\right )-\frac {10625}{81 \left (3 x +2\right )}-\frac {6}{x +1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/9*x^2+112/27*x-10625/81/(3*x+2)-1625/27*ln(3*x+2)-6/(x+1)+83*ln(x+1)

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maxima [A]  time = 0.82, size = 42, normalized size = 0.84 \begin {gather*} -\frac {8}{9} \, x^{2} + \frac {112}{27} \, x - \frac {12083 \, x + 11597}{81 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}} - \frac {1625}{27} \, \log \left (3 \, x + 2\right ) + 83 \, \log \left (x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/9*x^2 + 112/27*x - 1/81*(12083*x + 11597)/(3*x^2 + 5*x + 2) - 1625/27*log(3*x + 2) + 83*log(x + 1)

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mupad [B]  time = 0.04, size = 38, normalized size = 0.76 \begin {gather*} \frac {112\,x}{27}+83\,\ln \left (x+1\right )-\frac {1625\,\ln \left (x+\frac {2}{3}\right )}{27}-\frac {\frac {12083\,x}{243}+\frac {11597}{243}}{x^2+\frac {5\,x}{3}+\frac {2}{3}}-\frac {8\,x^2}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(112*x)/27 + 83*log(x + 1) - (1625*log(x + 2/3))/27 - ((12083*x)/243 + 11597/243)/((5*x)/3 + x^2 + 2/3) - (8*x
^2)/9

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sympy [A]  time = 0.16, size = 42, normalized size = 0.84 \begin {gather*} - \frac {8 x^{2}}{9} + \frac {112 x}{27} - \frac {12083 x + 11597}{243 x^{2} + 405 x + 162} - \frac {1625 \log {\left (x + \frac {2}{3} \right )}}{27} + 83 \log {\left (x + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8*x**2/9 + 112*x/27 - (12083*x + 11597)/(243*x**2 + 405*x + 162) - 1625*log(x + 2/3)/27 + 83*log(x + 1)

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