3.1562 \(\int \frac{(3+5 x)^2}{(1-2 x)^2} \, dx\)

Optimal. Leaf size=27 \[ \frac{25 x}{4}+\frac{121}{8 (1-2 x)}+\frac{55}{4} \log (1-2 x) \]

[Out]

121/(8*(1 - 2*x)) + (25*x)/4 + (55*Log[1 - 2*x])/4

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Rubi [A]  time = 0.0112766, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {43} \[ \frac{25 x}{4}+\frac{121}{8 (1-2 x)}+\frac{55}{4} \log (1-2 x) \]

Antiderivative was successfully verified.

[In]

Int[(3 + 5*x)^2/(1 - 2*x)^2,x]

[Out]

121/(8*(1 - 2*x)) + (25*x)/4 + (55*Log[1 - 2*x])/4

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{(3+5 x)^2}{(1-2 x)^2} \, dx &=\int \left (\frac{25}{4}+\frac{121}{4 (-1+2 x)^2}+\frac{55}{2 (-1+2 x)}\right ) \, dx\\ &=\frac{121}{8 (1-2 x)}+\frac{25 x}{4}+\frac{55}{4} \log (1-2 x)\\ \end{align*}

Mathematica [A]  time = 0.0110537, size = 26, normalized size = 0.96 \[ \frac{1}{8} \left (50 x+\frac{121}{1-2 x}+110 \log (1-2 x)-25\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(3 + 5*x)^2/(1 - 2*x)^2,x]

[Out]

(-25 + 121/(1 - 2*x) + 50*x + 110*Log[1 - 2*x])/8

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Maple [A]  time = 0.004, size = 22, normalized size = 0.8 \begin{align*}{\frac{25\,x}{4}}+{\frac{55\,\ln \left ( 2\,x-1 \right ) }{4}}-{\frac{121}{16\,x-8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25/4*x+55/4*ln(2*x-1)-121/8/(2*x-1)

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Maxima [A]  time = 2.29747, size = 28, normalized size = 1.04 \begin{align*} \frac{25}{4} \, x - \frac{121}{8 \,{\left (2 \, x - 1\right )}} + \frac{55}{4} \, \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,x, algorithm="maxima")

[Out]

25/4*x - 121/8/(2*x - 1) + 55/4*log(2*x - 1)

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Fricas [A]  time = 1.23393, size = 90, normalized size = 3.33 \begin{align*} \frac{100 \, x^{2} + 110 \,{\left (2 \, x - 1\right )} \log \left (2 \, x - 1\right ) - 50 \, x - 121}{8 \,{\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,x, algorithm="fricas")

[Out]

1/8*(100*x^2 + 110*(2*x - 1)*log(2*x - 1) - 50*x - 121)/(2*x - 1)

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Sympy [A]  time = 0.090339, size = 20, normalized size = 0.74 \begin{align*} \frac{25 x}{4} + \frac{55 \log{\left (2 x - 1 \right )}}{4} - \frac{121}{16 x - 8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25*x/4 + 55*log(2*x - 1)/4 - 121/(16*x - 8)

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Giac [A]  time = 3.03121, size = 43, normalized size = 1.59 \begin{align*} \frac{25}{4} \, x - \frac{121}{8 \,{\left (2 \, x - 1\right )}} - \frac{55}{4} \, \log \left (\frac{{\left | 2 \, x - 1 \right |}}{2 \,{\left (2 \, x - 1\right )}^{2}}\right ) - \frac{25}{8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25/4*x - 121/8/(2*x - 1) - 55/4*log(1/2*abs(2*x - 1)/(2*x - 1)^2) - 25/8