3.17.65 \(\int \frac {(3+5 x)^3}{(1-2 x)^3} \, dx\) [1665]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 38, normalized size of antiderivative = 1.00, 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 = {45} \begin {gather*} -\frac {125 x}{8}-\frac {1815}{16 (1-2 x)}+\frac {1331}{32 (1-2 x)^2}-\frac {825}{16} \log (1-2 x) \end {gather*}

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 45

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*}

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Mathematica [A]
time = 0.02, size = 34, normalized size = 0.89 \begin {gather*} \frac {1}{32} \left (-500 x+\frac {-2049+6260 x+1000 x^2}{(1-2 x)^2}-1650 \log (1-2 x)\right ) \end {gather*}

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 = 0.10, size = 31, normalized size = 0.82

method	result	size
risch	\(-\frac {125 x}{8}+\frac {\frac {1815 x}{8}-\frac {2299}{32}}{\left (-1+2 x \right )^{2}}-\frac {825 \ln \left (-1+2 x \right )}{16}\)	\(27\)
default	\(-\frac {125 x}{8}+\frac {1815}{16 \left (-1+2 x \right )}+\frac {1331}{32 \left (-1+2 x \right )^{2}}-\frac {825 \ln \left (-1+2 x \right )}{16}\)	\(31\)
norman	\(\frac {-\frac {609}{8} x +\frac {2799}{8} x^{2}-\frac {125}{2} x^{3}}{\left (-1+2 x \right )^{2}}-\frac {825 \ln \left (-1+2 x \right )}{16}\)	\(32\)
meijerg	\(\frac {27 x \left (2-2 x \right )}{2 \left (1-2 x \right )^{2}}+\frac {135 x^{2}}{2 \left (1-2 x \right )^{2}}-\frac {75 x \left (-18 x +6\right )}{8 \left (1-2 x \right )^{2}}-\frac {825 \ln \left (1-2 x \right )}{16}-\frac {125 x \left (16 x^{2}-36 x +12\right )}{32 \left (1-2 x \right )^{2}}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 31, normalized size = 0.82 \begin {gather*} -\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 {gather*}

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 = 0.40, size = 47, normalized size = 1.24 \begin {gather*} -\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 {gather*}

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 = 0.05, size = 31, normalized size = 0.82 \begin {gather*} - \frac {125 x}{8} - \frac {2299 - 7260 x}{128 x^{2} - 128 x + 32} - \frac {825 \log {\left (2 x - 1 \right )}}{16} \end {gather*}

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

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Giac [A]
time = 1.19, size = 27, normalized size = 0.71 \begin {gather*} -\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 {gather*}

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))

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Mupad [B]
time = 1.05, size = 26, normalized size = 0.68 \begin {gather*} \frac {\frac {1815\,x}{32}-\frac {2299}{128}}{x^2-x+\frac {1}{4}}-\frac {825\,\ln \left (x-\frac {1}{2}\right )}{16}-\frac {125\,x}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*x + 3)^3/(2*x - 1)^3,x)

[Out]

((1815*x)/32 - 2299/128)/(x^2 - x + 1/4) - (825*log(x - 1/2))/16 - (125*x)/8

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