3.11.35 \(\int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^8} \, dx\)

Optimal. Leaf size=45 \[ \frac {25}{162 (3 x+2)^4}-\frac {13}{27 (3 x+2)^5}+\frac {4}{27 (3 x+2)^6}-\frac {1}{81 (3 x+2)^7} \]

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Rubi [A]  time = 0.02, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {77} \begin {gather*} \frac {25}{162 (3 x+2)^4}-\frac {13}{27 (3 x+2)^5}+\frac {4}{27 (3 x+2)^6}-\frac {1}{81 (3 x+2)^7} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((1 - 2*x)*(3 + 5*x)^2)/(2 + 3*x)^8,x]

[Out]

-1/(81*(2 + 3*x)^7) + 4/(27*(2 + 3*x)^6) - 13/(27*(2 + 3*x)^5) + 25/(162*(2 + 3*x)^4)

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin {align*} \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^8} \, dx &=\int \left (\frac {7}{27 (2+3 x)^8}-\frac {8}{3 (2+3 x)^7}+\frac {65}{9 (2+3 x)^6}-\frac {50}{27 (2+3 x)^5}\right ) \, dx\\ &=-\frac {1}{81 (2+3 x)^7}+\frac {4}{27 (2+3 x)^6}-\frac {13}{27 (2+3 x)^5}+\frac {25}{162 (2+3 x)^4}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 26, normalized size = 0.58 \begin {gather*} \frac {225 x^3+216 x^2+12 x-22}{54 (3 x+2)^7} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((1 - 2*x)*(3 + 5*x)^2)/(2 + 3*x)^8,x]

[Out]

(-22 + 12*x + 216*x^2 + 225*x^3)/(54*(2 + 3*x)^7)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^8} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((1 - 2*x)*(3 + 5*x)^2)/(2 + 3*x)^8,x]

[Out]

IntegrateAlgebraic[((1 - 2*x)*(3 + 5*x)^2)/(2 + 3*x)^8, x]

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fricas [A]  time = 1.21, size = 54, normalized size = 1.20 \begin {gather*} \frac {225 \, x^{3} + 216 \, x^{2} + 12 \, x - 22}{54 \, {\left (2187 \, x^{7} + 10206 \, x^{6} + 20412 \, x^{5} + 22680 \, x^{4} + 15120 \, x^{3} + 6048 \, x^{2} + 1344 \, x + 128\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)*(3+5*x)^2/(2+3*x)^8,x, algorithm="fricas")

[Out]

1/54*(225*x^3 + 216*x^2 + 12*x - 22)/(2187*x^7 + 10206*x^6 + 20412*x^5 + 22680*x^4 + 15120*x^3 + 6048*x^2 + 13
44*x + 128)

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giac [A]  time = 1.21, size = 24, normalized size = 0.53 \begin {gather*} \frac {225 \, x^{3} + 216 \, x^{2} + 12 \, x - 22}{54 \, {\left (3 \, x + 2\right )}^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)*(3+5*x)^2/(2+3*x)^8,x, algorithm="giac")

[Out]

1/54*(225*x^3 + 216*x^2 + 12*x - 22)/(3*x + 2)^7

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maple [A]  time = 0.00, size = 38, normalized size = 0.84 \begin {gather*} -\frac {1}{81 \left (3 x +2\right )^{7}}+\frac {4}{27 \left (3 x +2\right )^{6}}-\frac {13}{27 \left (3 x +2\right )^{5}}+\frac {25}{162 \left (3 x +2\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1-2*x)*(5*x+3)^2/(3*x+2)^8,x)

[Out]

-1/81/(3*x+2)^7+4/27/(3*x+2)^6-13/27/(3*x+2)^5+25/162/(3*x+2)^4

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maxima [A]  time = 0.53, size = 54, normalized size = 1.20 \begin {gather*} \frac {225 \, x^{3} + 216 \, x^{2} + 12 \, x - 22}{54 \, {\left (2187 \, x^{7} + 10206 \, x^{6} + 20412 \, x^{5} + 22680 \, x^{4} + 15120 \, x^{3} + 6048 \, x^{2} + 1344 \, x + 128\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)*(3+5*x)^2/(2+3*x)^8,x, algorithm="maxima")

[Out]

1/54*(225*x^3 + 216*x^2 + 12*x - 22)/(2187*x^7 + 10206*x^6 + 20412*x^5 + 22680*x^4 + 15120*x^3 + 6048*x^2 + 13
44*x + 128)

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mupad [B]  time = 1.11, size = 37, normalized size = 0.82 \begin {gather*} \frac {25}{162\,{\left (3\,x+2\right )}^4}-\frac {13}{27\,{\left (3\,x+2\right )}^5}+\frac {4}{27\,{\left (3\,x+2\right )}^6}-\frac {1}{81\,{\left (3\,x+2\right )}^7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((2*x - 1)*(5*x + 3)^2)/(3*x + 2)^8,x)

[Out]

25/(162*(3*x + 2)^4) - 13/(27*(3*x + 2)^5) + 4/(27*(3*x + 2)^6) - 1/(81*(3*x + 2)^7)

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sympy [A]  time = 0.17, size = 51, normalized size = 1.13 \begin {gather*} - \frac {- 225 x^{3} - 216 x^{2} - 12 x + 22}{118098 x^{7} + 551124 x^{6} + 1102248 x^{5} + 1224720 x^{4} + 816480 x^{3} + 326592 x^{2} + 72576 x + 6912} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)*(3+5*x)**2/(2+3*x)**8,x)

[Out]

-(-225*x**3 - 216*x**2 - 12*x + 22)/(118098*x**7 + 551124*x**6 + 1102248*x**5 + 1224720*x**4 + 816480*x**3 + 3
26592*x**2 + 72576*x + 6912)

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