3.1155 \(\int \frac{(1-2 x) (3+5 x)}{(2+3 x)^4} \, dx\)

Optimal. Leaf size=34 \[ \frac{10}{27 (3 x+2)}-\frac{37}{54 (3 x+2)^2}+\frac{7}{81 (3 x+2)^3} \]

[Out]

7/(81*(2 + 3*x)^3) - 37/(54*(2 + 3*x)^2) + 10/(27*(2 + 3*x))

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Rubi [A]  time = 0.0131873, 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} \[ \frac{10}{27 (3 x+2)}-\frac{37}{54 (3 x+2)^2}+\frac{7}{81 (3 x+2)^3} \]

Antiderivative was successfully verified.

[In]

Int[((1 - 2*x)*(3 + 5*x))/(2 + 3*x)^4,x]

[Out]

7/(81*(2 + 3*x)^3) - 37/(54*(2 + 3*x)^2) + 10/(27*(2 + 3*x))

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)^4} \, dx &=\int \left (-\frac{7}{9 (2+3 x)^4}+\frac{37}{9 (2+3 x)^3}-\frac{10}{9 (2+3 x)^2}\right ) \, dx\\ &=\frac{7}{81 (2+3 x)^3}-\frac{37}{54 (2+3 x)^2}+\frac{10}{27 (2+3 x)}\\ \end{align*}

Mathematica [A]  time = 0.0057269, size = 21, normalized size = 0.62 \[ \frac{540 x^2+387 x+32}{162 (3 x+2)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((1 - 2*x)*(3 + 5*x))/(2 + 3*x)^4,x]

[Out]

(32 + 387*x + 540*x^2)/(162*(2 + 3*x)^3)

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Maple [A]  time = 0.004, size = 29, normalized size = 0.9 \begin{align*}{\frac{7}{81\, \left ( 2+3\,x \right ) ^{3}}}-{\frac{37}{54\, \left ( 2+3\,x \right ) ^{2}}}+{\frac{10}{54+81\,x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1-2*x)*(3+5*x)/(2+3*x)^4,x)

[Out]

7/81/(2+3*x)^3-37/54/(2+3*x)^2+10/27/(2+3*x)

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Maxima [A]  time = 1.12632, size = 39, normalized size = 1.15 \begin{align*} \frac{540 \, x^{2} + 387 \, x + 32}{162 \,{\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)*(3+5*x)/(2+3*x)^4,x, algorithm="maxima")

[Out]

1/162*(540*x^2 + 387*x + 32)/(27*x^3 + 54*x^2 + 36*x + 8)

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Fricas [A]  time = 1.4935, size = 80, normalized size = 2.35 \begin{align*} \frac{540 \, x^{2} + 387 \, x + 32}{162 \,{\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)*(3+5*x)/(2+3*x)^4,x, algorithm="fricas")

[Out]

1/162*(540*x^2 + 387*x + 32)/(27*x^3 + 54*x^2 + 36*x + 8)

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Sympy [A]  time = 0.114882, size = 24, normalized size = 0.71 \begin{align*} \frac{540 x^{2} + 387 x + 32}{4374 x^{3} + 8748 x^{2} + 5832 x + 1296} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)*(3+5*x)/(2+3*x)**4,x)

[Out]

(540*x**2 + 387*x + 32)/(4374*x**3 + 8748*x**2 + 5832*x + 1296)

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Giac [A]  time = 3.07519, size = 26, normalized size = 0.76 \begin{align*} \frac{540 \, x^{2} + 387 \, x + 32}{162 \,{\left (3 \, x + 2\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)*(3+5*x)/(2+3*x)^4,x, algorithm="giac")

[Out]

1/162*(540*x^2 + 387*x + 32)/(3*x + 2)^3