\(\int \frac {(3+5 x)^3}{(1-2 x)^2} \, dx\) [1579]

Optimal result

Integrand size = 15, antiderivative size = 34 \[ \int \frac {(3+5 x)^3}{(1-2 x)^2} \, dx=\frac {1331}{16 (1-2 x)}+\frac {175 x}{2}+\frac {125 x^2}{8}+\frac {1815}{16} \log (1-2 x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {45} \[ \int \frac {(3+5 x)^3}{(1-2 x)^2} \, dx=\frac {125 x^2}{8}+\frac {175 x}{2}+\frac {1331}{16 (1-2 x)}+\frac {1815}{16} \log (1-2 x) \]

[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 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*} \text {integral}& = \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] (verified)

Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {(3+5 x)^3}{(1-2 x)^2} \, dx=\frac {-1137-5850 x+5100 x^2+1000 x^3+3630 (-1+2 x) \log (1-2 x)}{-32+64 x} \]

[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)

Maple [A] (verified)

Time = 2.66 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74

method	result	size
risch	\(\frac {125 x^{2}}{8}+\frac {175 x}{2}-\frac {1331}{32 \left (x -\frac {1}{2}\right )}+\frac {1815 \ln \left (-1+2 x \right )}{16}\)	\(25\)
default	\(\frac {125 x^{2}}{8}+\frac {175 x}{2}+\frac {1815 \ln \left (-1+2 x \right )}{16}-\frac {1331}{16 \left (-1+2 x \right )}\)	\(27\)
norman	\(\frac {-\frac {2031}{8} x +\frac {1275}{8} x^{2}+\frac {125}{4} x^{3}}{-1+2 x}+\frac {1815 \ln \left (-1+2 x \right )}{16}\)	\(32\)
parallelrisch	\(\frac {500 x^{3}+3630 \ln \left (x -\frac {1}{2}\right ) x +2550 x^{2}-1815 \ln \left (x -\frac {1}{2}\right )-4062 x}{-16+32 x}\)	\(37\)
meijerg	\(\frac {189 x}{2 \left (1-2 x \right )}+\frac {1815 \ln \left (1-2 x \right )}{16}+\frac {75 x \left (-6 x +6\right )}{4 \left (1-2 x \right )}+\frac {125 x \left (-8 x^{2}-12 x +12\right )}{32 \left (1-2 x \right )}\)	\(55\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int \frac {(3+5 x)^3}{(1-2 x)^2} \, dx=\frac {500 \, x^{3} + 2550 \, x^{2} + 1815 \, {\left (2 \, x - 1\right )} \log \left (2 \, x - 1\right ) - 1400 \, x - 1331}{16 \, {\left (2 \, x - 1\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79 \[ \int \frac {(3+5 x)^3}{(1-2 x)^2} \, dx=\frac {125 x^{2}}{8} + \frac {175 x}{2} + \frac {1815 \log {\left (2 x - 1 \right )}}{16} - \frac {1331}{32 x - 16} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \frac {(3+5 x)^3}{(1-2 x)^2} \, dx=\frac {125}{8} \, x^{2} + \frac {175}{2} \, x - \frac {1331}{16 \, {\left (2 \, x - 1\right )}} + \frac {1815}{16} \, \log \left (2 \, x - 1\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.41 \[ \int \frac {(3+5 x)^3}{(1-2 x)^2} \, dx=\frac {25}{32} \, {\left (2 \, x - 1\right )}^{2} {\left (\frac {66}{2 \, x - 1} + 5\right )} - \frac {1331}{16 \, {\left (2 \, x - 1\right )}} - \frac {1815}{16} \, \log \left (\frac {{\left | 2 \, x - 1 \right |}}{2 \, {\left (2 \, x - 1\right )}^{2}}\right ) \]

[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)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int \frac {(3+5 x)^3}{(1-2 x)^2} \, dx=\frac {175\,x}{2}+\frac {1815\,\ln \left (x-\frac {1}{2}\right )}{16}-\frac {1331}{32\,\left (x-\frac {1}{2}\right )}+\frac {125\,x^2}{8} \]

[In]

int((5*x + 3)^3/(2*x - 1)^2,x)

[Out]

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