3.1156 \(\int \frac{(1-2 x) (3+5 x)}{(2+3 x)^5} \, dx\)

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.01388, antiderivative size = 34, normalized size of antiderivative = 1., 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 = {77} \[ \frac{5}{27 (3 x+2)^2}-\frac{37}{81 (3 x+2)^3}+\frac{7}{108 (3 x+2)^4} \]

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 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+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*}

Mathematica [A]  time = 0.0050599, size = 21, normalized size = 0.62 \[ \frac{540 x^2+276 x-35}{324 (3 x+2)^4} \]

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)

________________________________________________________________________________________

Maple [A]  time = 0.004, size = 29, normalized size = 0.9 \begin{align*}{\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}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

Maxima [A]  time = 1.33924, size = 46, normalized size = 1.35 \begin{align*} \frac{540 \, x^{2} + 276 \, x - 35}{324 \,{\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )}} \end{align*}

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)

________________________________________________________________________________________

Fricas [A]  time = 1.4504, size = 96, normalized size = 2.82 \begin{align*} \frac{540 \, x^{2} + 276 \, x - 35}{324 \,{\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )}} \end{align*}

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)

________________________________________________________________________________________

Sympy [A]  time = 0.126262, size = 29, normalized size = 0.85 \begin{align*} \frac{540 x^{2} + 276 x - 35}{26244 x^{4} + 69984 x^{3} + 69984 x^{2} + 31104 x + 5184} \end{align*}

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)

________________________________________________________________________________________

Giac [A]  time = 3.30621, size = 38, normalized size = 1.12 \begin{align*} \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{align*}

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