3.1652 \(\int \frac{(3+5 x)^2}{(1-2 x)^3} \, dx\)

Optimal. Leaf size=33 \[ -\frac{55}{4 (1-2 x)}+\frac{121}{16 (1-2 x)^2}-\frac{25}{8} \log (1-2 x) \]

[Out]

121/(16*(1 - 2*x)^2) - 55/(4*(1 - 2*x)) - (25*Log[1 - 2*x])/8

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Rubi [A]  time = 0.012367, antiderivative size = 33, 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} \[ -\frac{55}{4 (1-2 x)}+\frac{121}{16 (1-2 x)^2}-\frac{25}{8} \log (1-2 x) \]

Antiderivative was successfully verified.

[In]

Int[(3 + 5*x)^2/(1 - 2*x)^3,x]

[Out]

121/(16*(1 - 2*x)^2) - 55/(4*(1 - 2*x)) - (25*Log[1 - 2*x])/8

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)^3} \, dx &=\int \left (-\frac{121}{4 (-1+2 x)^3}-\frac{55}{2 (-1+2 x)^2}-\frac{25}{4 (-1+2 x)}\right ) \, dx\\ &=\frac{121}{16 (1-2 x)^2}-\frac{55}{4 (1-2 x)}-\frac{25}{8} \log (1-2 x)\\ \end{align*}

Mathematica [A]  time = 0.0074873, size = 33, normalized size = 1. \[ -\frac{55}{4 (1-2 x)}+\frac{121}{16 (1-2 x)^2}-\frac{25}{8} \log (1-2 x) \]

Antiderivative was successfully verified.

[In]

Integrate[(3 + 5*x)^2/(1 - 2*x)^3,x]

[Out]

121/(16*(1 - 2*x)^2) - 55/(4*(1 - 2*x)) - (25*Log[1 - 2*x])/8

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Maple [A]  time = 0.006, size = 28, normalized size = 0.9 \begin{align*} -{\frac{25\,\ln \left ( 2\,x-1 \right ) }{8}}+{\frac{121}{16\, \left ( 2\,x-1 \right ) ^{2}}}+{\frac{55}{8\,x-4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.05154, size = 38, normalized size = 1.15 \begin{align*} \frac{11 \,{\left (40 \, x - 9\right )}}{16 \,{\left (4 \, x^{2} - 4 \, x + 1\right )}} - \frac{25}{8} \, \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)^3,x, algorithm="maxima")

[Out]

11/16*(40*x - 9)/(4*x^2 - 4*x + 1) - 25/8*log(2*x - 1)

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Fricas [A]  time = 1.49271, size = 100, normalized size = 3.03 \begin{align*} -\frac{50 \,{\left (4 \, x^{2} - 4 \, x + 1\right )} \log \left (2 \, x - 1\right ) - 440 \, x + 99}{16 \,{\left (4 \, x^{2} - 4 \, x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(50*(4*x^2 - 4*x + 1)*log(2*x - 1) - 440*x + 99)/(4*x^2 - 4*x + 1)

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Sympy [A]  time = 0.109294, size = 24, normalized size = 0.73 \begin{align*} \frac{440 x - 99}{64 x^{2} - 64 x + 16} - \frac{25 \log{\left (2 x - 1 \right )}}{8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(440*x - 99)/(64*x**2 - 64*x + 16) - 25*log(2*x - 1)/8

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Giac [A]  time = 2.6247, size = 32, normalized size = 0.97 \begin{align*} \frac{11 \,{\left (40 \, x - 9\right )}}{16 \,{\left (2 \, x - 1\right )}^{2}} - \frac{25}{8} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

11/16*(40*x - 9)/(2*x - 1)^2 - 25/8*log(abs(2*x - 1))