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

Optimal. Leaf size=27 \[ -\frac {10 x}{9}+\frac {7}{27 (3 x+2)}+\frac {37}{27} \log (3 x+2) \]

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Rubi [A]  time = 0.01, antiderivative size = 27, 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 = {77} \begin {gather*} -\frac {10 x}{9}+\frac {7}{27 (3 x+2)}+\frac {37}{27} \log (3 x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-10*x)/9 + 7/(27*(2 + 3*x)) + (37*Log[2 + 3*x])/27

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

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Mathematica [A]  time = 0.02, size = 26, normalized size = 0.96 \begin {gather*} \frac {1}{27} \left (-30 x+\frac {7}{3 x+2}+37 \log (3 x+2)-20\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-20 - 30*x + 7/(2 + 3*x) + 37*Log[2 + 3*x])/27

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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+3 x)^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.47, size = 32, normalized size = 1.19 \begin {gather*} -\frac {90 \, x^{2} - 37 \, {\left (3 \, x + 2\right )} \log \left (3 \, x + 2\right ) + 60 \, x - 7}{27 \, {\left (3 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/27*(90*x^2 - 37*(3*x + 2)*log(3*x + 2) + 60*x - 7)/(3*x + 2)

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giac [A]  time = 1.19, size = 32, normalized size = 1.19 \begin {gather*} -\frac {10}{9} \, x + \frac {7}{27 \, {\left (3 \, x + 2\right )}} - \frac {37}{27} \, \log \left (\frac {{\left | 3 \, x + 2 \right |}}{3 \, {\left (3 \, x + 2\right )}^{2}}\right ) - \frac {20}{27} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-10/9*x + 7/27/(3*x + 2) - 37/27*log(1/3*abs(3*x + 2)/(3*x + 2)^2) - 20/27

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maple [A]  time = 0.01, size = 22, normalized size = 0.81 \begin {gather*} -\frac {10 x}{9}+\frac {37 \ln \left (3 x +2\right )}{27}+\frac {7}{27 \left (3 x +2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-10/9*x+7/27/(3*x+2)+37/27*ln(3*x+2)

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maxima [A]  time = 0.52, size = 21, normalized size = 0.78 \begin {gather*} -\frac {10}{9} \, x + \frac {7}{27 \, {\left (3 \, x + 2\right )}} + \frac {37}{27} \, \log \left (3 \, x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-10/9*x + 7/27/(3*x + 2) + 37/27*log(3*x + 2)

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mupad [B]  time = 1.10, size = 19, normalized size = 0.70 \begin {gather*} \frac {37\,\ln \left (x+\frac {2}{3}\right )}{27}-\frac {10\,x}{9}+\frac {7}{81\,\left (x+\frac {2}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(37*log(x + 2/3))/27 - (10*x)/9 + 7/(81*(x + 2/3))

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sympy [A]  time = 0.10, size = 20, normalized size = 0.74 \begin {gather*} - \frac {10 x}{9} + \frac {37 \log {\left (3 x + 2 \right )}}{27} + \frac {7}{81 x + 54} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-10*x/9 + 37*log(3*x + 2)/27 + 7/(81*x + 54)

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