3.1665 \(\int \frac{(3+5 x)^3}{(1-2 x)^3} \, dx\)

Optimal. Leaf size=38 \[ -\frac{125 x}{8}-\frac{1815}{16 (1-2 x)}+\frac{1331}{32 (1-2 x)^2}-\frac{825}{16} \log (1-2 x) \]

[Out]

1331/(32*(1 - 2*x)^2) - 1815/(16*(1 - 2*x)) - (125*x)/8 - (825*Log[1 - 2*x])/16

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Rubi [A]  time = 0.0157222, antiderivative size = 38, 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} \[ -\frac{125 x}{8}-\frac{1815}{16 (1-2 x)}+\frac{1331}{32 (1-2 x)^2}-\frac{825}{16} \log (1-2 x) \]

Antiderivative was successfully verified.

[In]

Int[(3 + 5*x)^3/(1 - 2*x)^3,x]

[Out]

1331/(32*(1 - 2*x)^2) - 1815/(16*(1 - 2*x)) - (125*x)/8 - (825*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)^3} \, dx &=\int \left (-\frac{125}{8}-\frac{1331}{8 (-1+2 x)^3}-\frac{1815}{8 (-1+2 x)^2}-\frac{825}{8 (-1+2 x)}\right ) \, dx\\ &=\frac{1331}{32 (1-2 x)^2}-\frac{1815}{16 (1-2 x)}-\frac{125 x}{8}-\frac{825}{16} \log (1-2 x)\\ \end{align*}

Mathematica [A]  time = 0.0187076, size = 34, normalized size = 0.89 \[ \frac{1}{32} \left (\frac{1000 x^2+6260 x-2049}{(1-2 x)^2}-500 x-1650 \log (1-2 x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(3 + 5*x)^3/(1 - 2*x)^3,x]

[Out]

(-500*x + (-2049 + 6260*x + 1000*x^2)/(1 - 2*x)^2 - 1650*Log[1 - 2*x])/32

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Maple [A]  time = 0.005, size = 31, normalized size = 0.8 \begin{align*} -{\frac{125\,x}{8}}-{\frac{825\,\ln \left ( 2\,x-1 \right ) }{16}}+{\frac{1331}{32\, \left ( 2\,x-1 \right ) ^{2}}}+{\frac{1815}{32\,x-16}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-125/8*x-825/16*ln(2*x-1)+1331/32/(2*x-1)^2+1815/16/(2*x-1)

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Maxima [A]  time = 1.06819, size = 42, normalized size = 1.11 \begin{align*} -\frac{125}{8} \, x + \frac{121 \,{\left (60 \, x - 19\right )}}{32 \,{\left (4 \, x^{2} - 4 \, x + 1\right )}} - \frac{825}{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)^3,x, algorithm="maxima")

[Out]

-125/8*x + 121/32*(60*x - 19)/(4*x^2 - 4*x + 1) - 825/16*log(2*x - 1)

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Fricas [A]  time = 1.54561, size = 136, normalized size = 3.58 \begin{align*} -\frac{2000 \, x^{3} - 2000 \, x^{2} + 1650 \,{\left (4 \, x^{2} - 4 \, x + 1\right )} \log \left (2 \, x - 1\right ) - 6760 \, x + 2299}{32 \,{\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)^3/(1-2*x)^3,x, algorithm="fricas")

[Out]

-1/32*(2000*x^3 - 2000*x^2 + 1650*(4*x^2 - 4*x + 1)*log(2*x - 1) - 6760*x + 2299)/(4*x^2 - 4*x + 1)

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Sympy [A]  time = 0.114712, size = 29, normalized size = 0.76 \begin{align*} - \frac{125 x}{8} + \frac{7260 x - 2299}{128 x^{2} - 128 x + 32} - \frac{825 \log{\left (2 x - 1 \right )}}{16} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-125*x/8 + (7260*x - 2299)/(128*x**2 - 128*x + 32) - 825*log(2*x - 1)/16

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Giac [A]  time = 2.24147, size = 36, normalized size = 0.95 \begin{align*} -\frac{125}{8} \, x + \frac{121 \,{\left (60 \, x - 19\right )}}{32 \,{\left (2 \, x - 1\right )}^{2}} - \frac{825}{16} \, \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)^3/(1-2*x)^3,x, algorithm="giac")

[Out]

-125/8*x + 121/32*(60*x - 19)/(2*x - 1)^2 - 825/16*log(abs(2*x - 1))