3.12.54 \(\int \frac {(1-2 x) (3+5 x)}{(2+3 x)^3} \, dx\) [1154]

Optimal. Leaf size=33 \[ \frac {7}{54 (2+3 x)^2}-\frac {37}{27 (2+3 x)}-\frac {10}{27} \log (2+3 x) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 33, 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 {37}{27 (3 x+2)}+\frac {7}{54 (3 x+2)^2}-\frac {10}{27} \log (3 x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

7/(54*(2 + 3*x)^2) - 37/(27*(2 + 3*x)) - (10*Log[2 + 3*x])/27

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

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Mathematica [A]
time = 0.01, size = 27, normalized size = 0.82 \begin {gather*} \frac {1}{54} \left (-\frac {3 (47+74 x)}{(2+3 x)^2}-20 \log (2+3 x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*(47 + 74*x))/(2 + 3*x)^2 - 20*Log[2 + 3*x])/54

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Maple [A]
time = 0.10, size = 28, normalized size = 0.85

method	result	size
risch	\(\frac {-\frac {37 x}{9}-\frac {47}{18}}{\left (2+3 x \right )^{2}}-\frac {10 \ln \left (2+3 x \right )}{27}\)	\(24\)
norman	\(\frac {\frac {67}{18} x +\frac {47}{8} x^{2}}{\left (2+3 x \right )^{2}}-\frac {10 \ln \left (2+3 x \right )}{27}\)	\(27\)
default	\(\frac {7}{54 \left (2+3 x \right )^{2}}-\frac {37}{27 \left (2+3 x \right )}-\frac {10 \ln \left (2+3 x \right )}{27}\)	\(28\)
meijerg	\(\frac {3 x \left (\frac {3 x}{2}+2\right )}{16 \left (1+\frac {3 x}{2}\right )^{2}}-\frac {x^{2}}{16 \left (1+\frac {3 x}{2}\right )^{2}}+\frac {5 x \left (\frac {27 x}{2}+6\right )}{54 \left (1+\frac {3 x}{2}\right )^{2}}-\frac {10 \ln \left (1+\frac {3 x}{2}\right )}{27}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.31, size = 28, normalized size = 0.85 \begin {gather*} -\frac {74 \, x + 47}{18 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} - \frac {10}{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)^3,x, algorithm="maxima")

[Out]

-1/18*(74*x + 47)/(9*x^2 + 12*x + 4) - 10/27*log(3*x + 2)

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Fricas [A]
time = 0.50, size = 37, normalized size = 1.12 \begin {gather*} -\frac {20 \, {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (3 \, x + 2\right ) + 222 \, x + 141}{54 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/54*(20*(9*x^2 + 12*x + 4)*log(3*x + 2) + 222*x + 141)/(9*x^2 + 12*x + 4)

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Sympy [A]
time = 0.05, size = 26, normalized size = 0.79 \begin {gather*} - \frac {74 x + 47}{162 x^{2} + 216 x + 72} - \frac {10 \log {\left (3 x + 2 \right )}}{27} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(74*x + 47)/(162*x**2 + 216*x + 72) - 10*log(3*x + 2)/27

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Giac [A]
time = 1.31, size = 24, normalized size = 0.73 \begin {gather*} -\frac {74 \, x + 47}{18 \, {\left (3 \, x + 2\right )}^{2}} - \frac {10}{27} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/18*(74*x + 47)/(3*x + 2)^2 - 10/27*log(abs(3*x + 2))

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Mupad [B]
time = 1.10, size = 24, normalized size = 0.73 \begin {gather*} -\frac {10\,\ln \left (x+\frac {2}{3}\right )}{27}-\frac {\frac {37\,x}{81}+\frac {47}{162}}{x^2+\frac {4\,x}{3}+\frac {4}{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (10*log(x + 2/3))/27 - ((37*x)/81 + 47/162)/((4*x)/3 + x^2 + 4/9)

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