3.1213 \(\int \frac{(1-2 x) (2+3 x)}{(3+5 x)^2} \, dx\)

Optimal. Leaf size=27 \[ -\frac{6 x}{25}-\frac{11}{125 (5 x+3)}+\frac{31}{125} \log (5 x+3) \]

[Out]

(-6*x)/25 - 11/(125*(3 + 5*x)) + (31*Log[3 + 5*x])/125

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Rubi [A]  time = 0.0116215, antiderivative size = 27, 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} \[ -\frac{6 x}{25}-\frac{11}{125 (5 x+3)}+\frac{31}{125} \log (5 x+3) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0091958, size = 26, normalized size = 0.96 \[ \frac{1}{125} \left (-30 x-\frac{11}{5 x+3}+31 \log (5 x+3)-18\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-18 - 30*x - 11/(3 + 5*x) + 31*Log[3 + 5*x])/125

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Maple [A]  time = 0.005, size = 22, normalized size = 0.8 \begin{align*} -{\frac{6\,x}{25}}-{\frac{11}{375+625\,x}}+{\frac{31\,\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)^2,x)

[Out]

-6/25*x-11/125/(3+5*x)+31/125*ln(3+5*x)

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Maxima [A]  time = 2.18226, size = 28, normalized size = 1.04 \begin{align*} -\frac{6}{25} \, x - \frac{11}{125 \,{\left (5 \, x + 3\right )}} + \frac{31}{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)^2,x, algorithm="maxima")

[Out]

-6/25*x - 11/125/(5*x + 3) + 31/125*log(5*x + 3)

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Fricas [A]  time = 1.47497, size = 92, normalized size = 3.41 \begin{align*} -\frac{150 \, x^{2} - 31 \,{\left (5 \, x + 3\right )} \log \left (5 \, x + 3\right ) + 90 \, x + 11}{125 \,{\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)^2,x, algorithm="fricas")

[Out]

-1/125*(150*x^2 - 31*(5*x + 3)*log(5*x + 3) + 90*x + 11)/(5*x + 3)

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Sympy [A]  time = 0.091536, size = 20, normalized size = 0.74 \begin{align*} - \frac{6 x}{25} + \frac{31 \log{\left (5 x + 3 \right )}}{125} - \frac{11}{625 x + 375} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6*x/25 + 31*log(5*x + 3)/125 - 11/(625*x + 375)

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Giac [A]  time = 2.95158, size = 43, normalized size = 1.59 \begin{align*} -\frac{6}{25} \, x - \frac{11}{125 \,{\left (5 \, x + 3\right )}} - \frac{31}{125} \, \log \left (\frac{{\left | 5 \, x + 3 \right |}}{5 \,{\left (5 \, x + 3\right )}^{2}}\right ) - \frac{18}{125} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6/25*x - 11/125/(5*x + 3) - 31/125*log(1/5*abs(5*x + 3)/(5*x + 3)^2) - 18/125