3.1594 \(\int \frac{(2+3 x)^2}{(1-2 x)^2 (3+5 x)} \, dx\)

Optimal. Leaf size=32 \[ \frac{49}{44 (1-2 x)}+\frac{217}{484} \log (1-2 x)+\frac{1}{605} \log (5 x+3) \]

[Out]

49/(44*(1 - 2*x)) + (217*Log[1 - 2*x])/484 + Log[3 + 5*x]/605

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Rubi [A]  time = 0.0135686, 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} \[ \frac{49}{44 (1-2 x)}+\frac{217}{484} \log (1-2 x)+\frac{1}{605} \log (5 x+3) \]

Antiderivative was successfully verified.

[In]

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

[Out]

49/(44*(1 - 2*x)) + (217*Log[1 - 2*x])/484 + Log[3 + 5*x]/605

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{(2+3 x)^2}{(1-2 x)^2 (3+5 x)} \, dx &=\int \left (\frac{49}{22 (-1+2 x)^2}+\frac{217}{242 (-1+2 x)}+\frac{1}{121 (3+5 x)}\right ) \, dx\\ &=\frac{49}{44 (1-2 x)}+\frac{217}{484} \log (1-2 x)+\frac{1}{605} \log (3+5 x)\\ \end{align*}

Mathematica [A]  time = 0.0134924, size = 40, normalized size = 1.25 \[ -\frac{245}{44 (2 (5 x+3)-11)}+\frac{1}{605} \log (5 x+3)+\frac{217}{484} \log (11-2 (5 x+3)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-245/(44*(-11 + 2*(3 + 5*x))) + Log[3 + 5*x]/605 + (217*Log[11 - 2*(3 + 5*x)])/484

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Maple [A]  time = 0.007, size = 27, normalized size = 0.8 \begin{align*} -{\frac{49}{88\,x-44}}+{\frac{217\,\ln \left ( 2\,x-1 \right ) }{484}}+{\frac{\ln \left ( 3+5\,x \right ) }{605}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-49/44/(2*x-1)+217/484*ln(2*x-1)+1/605*ln(3+5*x)

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Maxima [A]  time = 3.03536, size = 35, normalized size = 1.09 \begin{align*} -\frac{49}{44 \,{\left (2 \, x - 1\right )}} + \frac{1}{605} \, \log \left (5 \, x + 3\right ) + \frac{217}{484} \, \log \left (2 \, x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-49/44/(2*x - 1) + 1/605*log(5*x + 3) + 217/484*log(2*x - 1)

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Fricas [A]  time = 1.30369, size = 111, normalized size = 3.47 \begin{align*} \frac{4 \,{\left (2 \, x - 1\right )} \log \left (5 \, x + 3\right ) + 1085 \,{\left (2 \, x - 1\right )} \log \left (2 \, x - 1\right ) - 2695}{2420 \,{\left (2 \, x - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2420*(4*(2*x - 1)*log(5*x + 3) + 1085*(2*x - 1)*log(2*x - 1) - 2695)/(2*x - 1)

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Sympy [A]  time = 0.125175, size = 24, normalized size = 0.75 \begin{align*} \frac{217 \log{\left (x - \frac{1}{2} \right )}}{484} + \frac{\log{\left (x + \frac{3}{5} \right )}}{605} - \frac{49}{88 x - 44} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

217*log(x - 1/2)/484 + log(x + 3/5)/605 - 49/(88*x - 44)

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Giac [A]  time = 2.46496, size = 58, normalized size = 1.81 \begin{align*} -\frac{49}{44 \,{\left (2 \, x - 1\right )}} - \frac{9}{20} \, \log \left (\frac{{\left | 2 \, x - 1 \right |}}{2 \,{\left (2 \, x - 1\right )}^{2}}\right ) + \frac{1}{605} \, \log \left ({\left | -\frac{11}{2 \, x - 1} - 5 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-49/44/(2*x - 1) - 9/20*log(1/2*abs(2*x - 1)/(2*x - 1)^2) + 1/605*log(abs(-11/(2*x - 1) - 5))