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

Optimal. Leaf size=36 \[ -\frac {8 x^3}{9}+\frac {26 x^2}{9}+\frac {922 x}{27}-6 \log (x+1)+\frac {2125}{81} \log (3 x+2) \]

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Rubi [A]  time = 0.02, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {800, 632, 31} \begin {gather*} -\frac {8 x^3}{9}+\frac {26 x^2}{9}+\frac {922 x}{27}-6 \log (x+1)+\frac {2125}{81} \log (3 x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(922*x)/27 + (26*x^2)/9 - (8*x^3)/9 - 6*Log[1 + x] + (2125*Log[2 + 3*x])/81

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 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), 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] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {(5-x) (3+2 x)^3}{2+5 x+3 x^2} \, dx &=\int \left (\frac {922}{27}+\frac {52 x}{9}-\frac {8 x^2}{3}+\frac {1801+1639 x}{27 \left (2+5 x+3 x^2\right )}\right ) \, dx\\ &=\frac {922 x}{27}+\frac {26 x^2}{9}-\frac {8 x^3}{9}+\frac {1}{27} \int \frac {1801+1639 x}{2+5 x+3 x^2} \, dx\\ &=\frac {922 x}{27}+\frac {26 x^2}{9}-\frac {8 x^3}{9}-18 \int \frac {1}{3+3 x} \, dx+\frac {2125}{27} \int \frac {1}{2+3 x} \, dx\\ &=\frac {922 x}{27}+\frac {26 x^2}{9}-\frac {8 x^3}{9}-6 \log (1+x)+\frac {2125}{81} \log (2+3 x)\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 35, normalized size = 0.97 \begin {gather*} \frac {1}{162} \left (-144 x^3+468 x^2+5532 x+4250 \log (-6 x-4)-972 \log (-2 (x+1))+6759\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(6759 + 5532*x + 468*x^2 - 144*x^3 + 4250*Log[-4 - 6*x] - 972*Log[-2*(1 + x)])/162

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.50, size = 28, normalized size = 0.78 \begin {gather*} -\frac {8}{9} \, x^{3} + \frac {26}{9} \, x^{2} + \frac {922}{27} \, x + \frac {2125}{81} \, \log \left (3 \, x + 2\right ) - 6 \, \log \left (x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/9*x^3 + 26/9*x^2 + 922/27*x + 2125/81*log(3*x + 2) - 6*log(x + 1)

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giac [A]  time = 0.15, size = 30, normalized size = 0.83 \begin {gather*} -\frac {8}{9} \, x^{3} + \frac {26}{9} \, x^{2} + \frac {922}{27} \, x + \frac {2125}{81} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) - 6 \, \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)^3/(3*x^2+5*x+2),x, algorithm="giac")

[Out]

-8/9*x^3 + 26/9*x^2 + 922/27*x + 2125/81*log(abs(3*x + 2)) - 6*log(abs(x + 1))

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maple [A]  time = 0.05, size = 29, normalized size = 0.81 \begin {gather*} -\frac {8 x^{3}}{9}+\frac {26 x^{2}}{9}+\frac {922 x}{27}+\frac {2125 \ln \left (3 x +2\right )}{81}-6 \ln \left (x +1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

922/27*x+26/9*x^2-8/9*x^3-6*ln(x+1)+2125/81*ln(3*x+2)

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maxima [A]  time = 0.50, size = 28, normalized size = 0.78 \begin {gather*} -\frac {8}{9} \, x^{3} + \frac {26}{9} \, x^{2} + \frac {922}{27} \, x + \frac {2125}{81} \, \log \left (3 \, x + 2\right ) - 6 \, \log \left (x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/9*x^3 + 26/9*x^2 + 922/27*x + 2125/81*log(3*x + 2) - 6*log(x + 1)

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mupad [B]  time = 2.31, size = 26, normalized size = 0.72 \begin {gather*} \frac {922\,x}{27}-6\,\ln \left (x+1\right )+\frac {2125\,\ln \left (x+\frac {2}{3}\right )}{81}+\frac {26\,x^2}{9}-\frac {8\,x^3}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(922*x)/27 - 6*log(x + 1) + (2125*log(x + 2/3))/81 + (26*x^2)/9 - (8*x^3)/9

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sympy [A]  time = 0.13, size = 34, normalized size = 0.94 \begin {gather*} - \frac {8 x^{3}}{9} + \frac {26 x^{2}}{9} + \frac {922 x}{27} + \frac {2125 \log {\left (x + \frac {2}{3} \right )}}{81} - 6 \log {\left (x + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8*x**3/9 + 26*x**2/9 + 922*x/27 + 2125*log(x + 2/3)/81 - 6*log(x + 1)

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