3.12.56 \(\int \frac {(1-2 x) (3+5 x)}{(2+3 x)^5} \, dx\) [1156]

Optimal. Leaf size=34 \[ \frac {7}{108 (2+3 x)^4}-\frac {37}{81 (2+3 x)^3}+\frac {5}{27 (2+3 x)^2} \]

[Out]

7/108/(2+3*x)^4-37/81/(2+3*x)^3+5/27/(2+3*x)^2

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Rubi [A]
time = 0.01, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {78} \begin {gather*} \frac {5}{27 (3 x+2)^2}-\frac {37}{81 (3 x+2)^3}+\frac {7}{108 (3 x+2)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

7/(108*(2 + 3*x)^4) - 37/(81*(2 + 3*x)^3) + 5/(27*(2 + 3*x)^2)

Rule 78

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+3 x)^5} \, dx &=\int \left (-\frac {7}{9 (2+3 x)^5}+\frac {37}{9 (2+3 x)^4}-\frac {10}{9 (2+3 x)^3}\right ) \, dx\\ &=\frac {7}{108 (2+3 x)^4}-\frac {37}{81 (2+3 x)^3}+\frac {5}{27 (2+3 x)^2}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 21, normalized size = 0.62 \begin {gather*} \frac {-35+276 x+540 x^2}{324 (2+3 x)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-35 + 276*x + 540*x^2)/(324*(2 + 3*x)^4)

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Maple [A]
time = 0.09, size = 29, normalized size = 0.85

method	result	size
gosper	\(\frac {540 x^{2}+276 x -35}{324 \left (2+3 x \right )^{4}}\)	\(20\)
risch	\(\frac {\frac {5}{3} x^{2}+\frac {23}{27} x -\frac {35}{324}}{\left (2+3 x \right )^{4}}\)	\(20\)
norman	\(\frac {\frac {25}{8} x^{2}+\frac {3}{2} x +\frac {35}{24} x^{3}+\frac {35}{64} x^{4}}{\left (2+3 x \right )^{4}}\)	\(28\)
default	\(\frac {7}{108 \left (2+3 x \right )^{4}}-\frac {37}{81 \left (2+3 x \right )^{3}}+\frac {5}{27 \left (2+3 x \right )^{2}}\)	\(29\)
meijerg	\(\frac {3 x \left (\frac {27}{8} x^{3}+9 x^{2}+9 x +4\right )}{128 \left (1+\frac {3 x}{2}\right )^{4}}-\frac {x^{2} \left (\frac {9}{4} x^{2}+6 x +6\right )}{384 \left (1+\frac {3 x}{2}\right )^{4}}-\frac {5 x^{3} \left (\frac {3 x}{2}+4\right )}{192 \left (1+\frac {3 x}{2}\right )^{4}}\)	\(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1-2*x)*(3+5*x)/(2+3*x)^5,x,method=_RETURNVERBOSE)

[Out]

7/108/(2+3*x)^4-37/81/(2+3*x)^3+5/27/(2+3*x)^2

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Maxima [A]
time = 0.30, size = 34, normalized size = 1.00 \begin {gather*} \frac {540 \, x^{2} + 276 \, x - 35}{324 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/324*(540*x^2 + 276*x - 35)/(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16)

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Fricas [A]
time = 0.51, size = 34, normalized size = 1.00 \begin {gather*} \frac {540 \, x^{2} + 276 \, x - 35}{324 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/324*(540*x^2 + 276*x - 35)/(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16)

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Sympy [A]
time = 0.04, size = 31, normalized size = 0.91 \begin {gather*} - \frac {- 540 x^{2} - 276 x + 35}{26244 x^{4} + 69984 x^{3} + 69984 x^{2} + 31104 x + 5184} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-540*x**2 - 276*x + 35)/(26244*x**4 + 69984*x**3 + 69984*x**2 + 31104*x + 5184)

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Giac [A]
time = 2.23, size = 28, normalized size = 0.82 \begin {gather*} \frac {5}{27 \, {\left (3 \, x + 2\right )}^{2}} - \frac {37}{81 \, {\left (3 \, x + 2\right )}^{3}} + \frac {7}{108 \, {\left (3 \, x + 2\right )}^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/27/(3*x + 2)^2 - 37/81/(3*x + 2)^3 + 7/108/(3*x + 2)^4

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Mupad [B]
time = 1.11, size = 28, normalized size = 0.82 \begin {gather*} \frac {5}{27\,{\left (3\,x+2\right )}^2}-\frac {37}{81\,{\left (3\,x+2\right )}^3}+\frac {7}{108\,{\left (3\,x+2\right )}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/(27*(3*x + 2)^2) - 37/(81*(3*x + 2)^3) + 7/(108*(3*x + 2)^4)

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