3.1298 \(\int \frac{(1-2 x)^2}{(2+3 x)^2 (3+5 x)} \, dx\)

Optimal. Leaf size=32 \[ \frac{49}{9 (3 x+2)}-\frac{217}{9} \log (3 x+2)+\frac{121}{5} \log (5 x+3) \]

[Out]

49/(9*(2 + 3*x)) - (217*Log[2 + 3*x])/9 + (121*Log[3 + 5*x])/5

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Rubi [A]  time = 0.0131435, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.045, Rules used = {88} \[ \frac{49}{9 (3 x+2)}-\frac{217}{9} \log (3 x+2)+\frac{121}{5} \log (5 x+3) \]

Antiderivative was successfully verified.

[In]

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

[Out]

49/(9*(2 + 3*x)) - (217*Log[2 + 3*x])/9 + (121*Log[3 + 5*x])/5

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int \frac{(1-2 x)^2}{(2+3 x)^2 (3+5 x)} \, dx &=\int \left (-\frac{49}{3 (2+3 x)^2}-\frac{217}{3 (2+3 x)}+\frac{121}{3+5 x}\right ) \, dx\\ &=\frac{49}{9 (2+3 x)}-\frac{217}{9} \log (2+3 x)+\frac{121}{5} \log (3+5 x)\\ \end{align*}

Mathematica [A]  time = 0.017176, size = 32, normalized size = 1. \[ \frac{49}{27 x+18}-\frac{217}{9} \log (5 (3 x+2))+\frac{121}{5} \log (5 x+3) \]

Antiderivative was successfully verified.

[In]

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

[Out]

49/(18 + 27*x) - (217*Log[5*(2 + 3*x)])/9 + (121*Log[3 + 5*x])/5

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Maple [A]  time = 0.006, size = 27, normalized size = 0.8 \begin{align*}{\frac{49}{18+27\,x}}-{\frac{217\,\ln \left ( 2+3\,x \right ) }{9}}+{\frac{121\,\ln \left ( 3+5\,x \right ) }{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

49/9/(2+3*x)-217/9*ln(2+3*x)+121/5*ln(3+5*x)

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Maxima [A]  time = 1.05437, size = 35, normalized size = 1.09 \begin{align*} \frac{49}{9 \,{\left (3 \, x + 2\right )}} + \frac{121}{5} \, \log \left (5 \, x + 3\right ) - \frac{217}{9} \, \log \left (3 \, x + 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

49/9/(3*x + 2) + 121/5*log(5*x + 3) - 217/9*log(3*x + 2)

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Fricas [A]  time = 1.55478, size = 111, normalized size = 3.47 \begin{align*} \frac{1089 \,{\left (3 \, x + 2\right )} \log \left (5 \, x + 3\right ) - 1085 \,{\left (3 \, x + 2\right )} \log \left (3 \, x + 2\right ) + 245}{45 \,{\left (3 \, x + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/45*(1089*(3*x + 2)*log(5*x + 3) - 1085*(3*x + 2)*log(3*x + 2) + 245)/(3*x + 2)

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Sympy [A]  time = 0.128503, size = 26, normalized size = 0.81 \begin{align*} \frac{121 \log{\left (x + \frac{3}{5} \right )}}{5} - \frac{217 \log{\left (x + \frac{2}{3} \right )}}{9} + \frac{49}{27 x + 18} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

121*log(x + 3/5)/5 - 217*log(x + 2/3)/9 + 49/(27*x + 18)

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Giac [A]  time = 1.96185, size = 58, normalized size = 1.81 \begin{align*} \frac{49}{9 \,{\left (3 \, x + 2\right )}} - \frac{4}{45} \, \log \left (\frac{{\left | 3 \, x + 2 \right |}}{3 \,{\left (3 \, x + 2\right )}^{2}}\right ) + \frac{121}{5} \, \log \left ({\left | -\frac{1}{3 \, x + 2} + 5 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

49/9/(3*x + 2) - 4/45*log(1/3*abs(3*x + 2)/(3*x + 2)^2) + 121/5*log(abs(-1/(3*x + 2) + 5))