3.1228 \(\int \frac{(1-2 x) (2+3 x)}{(3+5 x)^3} \, dx\)

Optimal. Leaf size=33 \[ -\frac{31}{125 (5 x+3)}-\frac{11}{250 (5 x+3)^2}-\frac{6}{125} \log (5 x+3) \]

[Out]

-11/(250*(3 + 5*x)^2) - 31/(125*(3 + 5*x)) - (6*Log[3 + 5*x])/125

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Rubi [A]  time = 0.0125276, antiderivative size = 33, 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} \[ -\frac{31}{125 (5 x+3)}-\frac{11}{250 (5 x+3)^2}-\frac{6}{125} \log (5 x+3) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-11/(250*(3 + 5*x)^2) - 31/(125*(3 + 5*x)) - (6*Log[3 + 5*x])/125

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) (2+3 x)}{(3+5 x)^3} \, dx &=\int \left (\frac{11}{25 (3+5 x)^3}+\frac{31}{25 (3+5 x)^2}-\frac{6}{25 (3+5 x)}\right ) \, dx\\ &=-\frac{11}{250 (3+5 x)^2}-\frac{31}{125 (3+5 x)}-\frac{6}{125} \log (3+5 x)\\ \end{align*}

Mathematica [A]  time = 0.0079225, size = 33, normalized size = 1. \[ -\frac{31}{125 (5 x+3)}-\frac{11}{250 (5 x+3)^2}-\frac{6}{125} \log (5 x+3) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-11/(250*(3 + 5*x)^2) - 31/(125*(3 + 5*x)) - (6*Log[3 + 5*x])/125

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Maple [A]  time = 0.004, size = 28, normalized size = 0.9 \begin{align*} -{\frac{11}{250\, \left ( 3+5\,x \right ) ^{2}}}-{\frac{31}{375+625\,x}}-{\frac{6\,\ln \left ( 3+5\,x \right ) }{125}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-11/250/(3+5*x)^2-31/125/(3+5*x)-6/125*ln(3+5*x)

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Maxima [A]  time = 2.07221, size = 38, normalized size = 1.15 \begin{align*} -\frac{310 \, x + 197}{250 \,{\left (25 \, x^{2} + 30 \, x + 9\right )}} - \frac{6}{125} \, \log \left (5 \, x + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/250*(310*x + 197)/(25*x^2 + 30*x + 9) - 6/125*log(5*x + 3)

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Fricas [A]  time = 1.49881, size = 108, normalized size = 3.27 \begin{align*} -\frac{12 \,{\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (5 \, x + 3\right ) + 310 \, x + 197}{250 \,{\left (25 \, x^{2} + 30 \, x + 9\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/250*(12*(25*x^2 + 30*x + 9)*log(5*x + 3) + 310*x + 197)/(25*x^2 + 30*x + 9)

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Sympy [A]  time = 0.106527, size = 26, normalized size = 0.79 \begin{align*} - \frac{310 x + 197}{6250 x^{2} + 7500 x + 2250} - \frac{6 \log{\left (5 x + 3 \right )}}{125} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(310*x + 197)/(6250*x**2 + 7500*x + 2250) - 6*log(5*x + 3)/125

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Giac [A]  time = 2.60364, size = 32, normalized size = 0.97 \begin{align*} -\frac{310 \, x + 197}{250 \,{\left (5 \, x + 3\right )}^{2}} - \frac{6}{125} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/250*(310*x + 197)/(5*x + 3)^2 - 6/125*log(abs(5*x + 3))