3.1563 \(\int \frac{(3+5 x)^2}{(1-2 x)^2 (2+3 x)} \, dx\)

Optimal. Leaf size=32 \[ \frac{121}{28 (1-2 x)}+\frac{407}{196} \log (1-2 x)+\frac{1}{147} \log (3 x+2) \]

[Out]

121/(28*(1 - 2*x)) + (407*Log[1 - 2*x])/196 + Log[2 + 3*x]/147

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Rubi [A]  time = 0.0141535, 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} \[ \frac{121}{28 (1-2 x)}+\frac{407}{196} \log (1-2 x)+\frac{1}{147} \log (3 x+2) \]

Antiderivative was successfully verified.

[In]

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

[Out]

121/(28*(1 - 2*x)) + (407*Log[1 - 2*x])/196 + Log[2 + 3*x]/147

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{(3+5 x)^2}{(1-2 x)^2 (2+3 x)} \, dx &=\int \left (\frac{121}{14 (-1+2 x)^2}+\frac{407}{98 (-1+2 x)}+\frac{1}{49 (2+3 x)}\right ) \, dx\\ &=\frac{121}{28 (1-2 x)}+\frac{407}{196} \log (1-2 x)+\frac{1}{147} \log (2+3 x)\\ \end{align*}

Mathematica [A]  time = 0.0141993, size = 40, normalized size = 1.25 \[ -\frac{363}{28 (2 (3 x+2)-7)}+\frac{1}{147} \log (3 x+2)+\frac{407}{196} \log (7-2 (3 x+2)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-363/(28*(-7 + 2*(2 + 3*x))) + Log[2 + 3*x]/147 + (407*Log[7 - 2*(2 + 3*x)])/196

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Maple [A]  time = 0.006, size = 27, normalized size = 0.8 \begin{align*} -{\frac{121}{56\,x-28}}+{\frac{407\,\ln \left ( 2\,x-1 \right ) }{196}}+{\frac{\ln \left ( 2+3\,x \right ) }{147}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-121/28/(2*x-1)+407/196*ln(2*x-1)+1/147*ln(2+3*x)

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Maxima [A]  time = 1.12048, size = 35, normalized size = 1.09 \begin{align*} -\frac{121}{28 \,{\left (2 \, x - 1\right )}} + \frac{1}{147} \, \log \left (3 \, x + 2\right ) + \frac{407}{196} \, \log \left (2 \, x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-121/28/(2*x - 1) + 1/147*log(3*x + 2) + 407/196*log(2*x - 1)

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Fricas [A]  time = 1.22879, size = 109, normalized size = 3.41 \begin{align*} \frac{4 \,{\left (2 \, x - 1\right )} \log \left (3 \, x + 2\right ) + 1221 \,{\left (2 \, x - 1\right )} \log \left (2 \, x - 1\right ) - 2541}{588 \,{\left (2 \, x - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/588*(4*(2*x - 1)*log(3*x + 2) + 1221*(2*x - 1)*log(2*x - 1) - 2541)/(2*x - 1)

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Sympy [A]  time = 0.13524, size = 24, normalized size = 0.75 \begin{align*} \frac{407 \log{\left (x - \frac{1}{2} \right )}}{196} + \frac{\log{\left (x + \frac{2}{3} \right )}}{147} - \frac{121}{56 x - 28} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

407*log(x - 1/2)/196 + log(x + 2/3)/147 - 121/(56*x - 28)

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Giac [A]  time = 2.28602, size = 58, normalized size = 1.81 \begin{align*} -\frac{121}{28 \,{\left (2 \, x - 1\right )}} - \frac{25}{12} \, \log \left (\frac{{\left | 2 \, x - 1 \right |}}{2 \,{\left (2 \, x - 1\right )}^{2}}\right ) + \frac{1}{147} \, \log \left ({\left | -\frac{7}{2 \, x - 1} - 3 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-121/28/(2*x - 1) - 25/12*log(1/2*abs(2*x - 1)/(2*x - 1)^2) + 1/147*log(abs(-7/(2*x - 1) - 3))