3.1579 \(\int \frac{(3+5 x)^3}{(1-2 x)^2} \, dx\)

Optimal. Leaf size=34 \[ \frac{125 x^2}{8}+\frac{175 x}{2}+\frac{1331}{16 (1-2 x)}+\frac{1815}{16} \log (1-2 x) \]

[Out]

1331/(16*(1 - 2*x)) + (175*x)/2 + (125*x^2)/8 + (1815*Log[1 - 2*x])/16

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Rubi [A]  time = 0.0146051, antiderivative size = 34, 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} \[ \frac{125 x^2}{8}+\frac{175 x}{2}+\frac{1331}{16 (1-2 x)}+\frac{1815}{16} \log (1-2 x) \]

Antiderivative was successfully verified.

[In]

Int[(3 + 5*x)^3/(1 - 2*x)^2,x]

[Out]

1331/(16*(1 - 2*x)) + (175*x)/2 + (125*x^2)/8 + (1815*Log[1 - 2*x])/16

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)^3}{(1-2 x)^2} \, dx &=\int \left (\frac{175}{2}+\frac{125 x}{4}+\frac{1331}{8 (-1+2 x)^2}+\frac{1815}{8 (-1+2 x)}\right ) \, dx\\ &=\frac{1331}{16 (1-2 x)}+\frac{175 x}{2}+\frac{125 x^2}{8}+\frac{1815}{16} \log (1-2 x)\\ \end{align*}

Mathematica [A]  time = 0.00736, size = 36, normalized size = 1.06 \[ \frac{1000 x^3+5100 x^2-5850 x+3630 (2 x-1) \log (1-2 x)-1137}{64 x-32} \]

Antiderivative was successfully verified.

[In]

Integrate[(3 + 5*x)^3/(1 - 2*x)^2,x]

[Out]

(-1137 - 5850*x + 5100*x^2 + 1000*x^3 + 3630*(-1 + 2*x)*Log[1 - 2*x])/(-32 + 64*x)

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Maple [A]  time = 0.005, size = 27, normalized size = 0.8 \begin{align*}{\frac{125\,{x}^{2}}{8}}+{\frac{175\,x}{2}}+{\frac{1815\,\ln \left ( 2\,x-1 \right ) }{16}}-{\frac{1331}{32\,x-16}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

125/8*x^2+175/2*x+1815/16*ln(2*x-1)-1331/16/(2*x-1)

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Maxima [A]  time = 1.1324, size = 35, normalized size = 1.03 \begin{align*} \frac{125}{8} \, x^{2} + \frac{175}{2} \, x - \frac{1331}{16 \,{\left (2 \, x - 1\right )}} + \frac{1815}{16} \, \log \left (2 \, x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

125/8*x^2 + 175/2*x - 1331/16/(2*x - 1) + 1815/16*log(2*x - 1)

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Fricas [A]  time = 1.24352, size = 112, normalized size = 3.29 \begin{align*} \frac{500 \, x^{3} + 2550 \, x^{2} + 1815 \,{\left (2 \, x - 1\right )} \log \left (2 \, x - 1\right ) - 1400 \, x - 1331}{16 \,{\left (2 \, x - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(500*x^3 + 2550*x^2 + 1815*(2*x - 1)*log(2*x - 1) - 1400*x - 1331)/(2*x - 1)

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Sympy [A]  time = 0.092911, size = 27, normalized size = 0.79 \begin{align*} \frac{125 x^{2}}{8} + \frac{175 x}{2} + \frac{1815 \log{\left (2 x - 1 \right )}}{16} - \frac{1331}{32 x - 16} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

125*x**2/8 + 175*x/2 + 1815*log(2*x - 1)/16 - 1331/(32*x - 16)

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Giac [A]  time = 1.49988, size = 65, normalized size = 1.91 \begin{align*} \frac{25}{32} \,{\left (2 \, x - 1\right )}^{2}{\left (\frac{66}{2 \, x - 1} + 5\right )} - \frac{1331}{16 \,{\left (2 \, x - 1\right )}} - \frac{1815}{16} \, \log \left (\frac{{\left | 2 \, x - 1 \right |}}{2 \,{\left (2 \, x - 1\right )}^{2}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25/32*(2*x - 1)^2*(66/(2*x - 1) + 5) - 1331/16/(2*x - 1) - 1815/16*log(1/2*abs(2*x - 1)/(2*x - 1)^2)