\(\int \frac {(3+5 x)^2}{1-2 x} \, dx\) [1459]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 15, antiderivative size = 23 \[ \int \frac {(3+5 x)^2}{1-2 x} \, dx=-\frac {85 x}{4}-\frac {25 x^2}{4}-\frac {121}{8} \log (1-2 x) \]

[Out]

-85/4*x-25/4*x^2-121/8*ln(1-2*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, 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} \[ \int \frac {(3+5 x)^2}{1-2 x} \, dx=-\frac {25 x^2}{4}-\frac {85 x}{4}-\frac {121}{8} \log (1-2 x) \]

[In]

Int[(3 + 5*x)^2/(1 - 2*x),x]

[Out]

(-85*x)/4 - (25*x^2)/4 - (121*Log[1 - 2*x])/8

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 {85}{4}-\frac {25 x}{2}-\frac {121}{4 (-1+2 x)}\right ) \, dx \\ & = -\frac {85 x}{4}-\frac {25 x^2}{4}-\frac {121}{8} \log (1-2 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {(3+5 x)^2}{1-2 x} \, dx=\frac {1}{16} \left (-5 \left (-39+68 x+20 x^2\right )-242 \log (1-2 x)\right ) \]

[In]

Integrate[(3 + 5*x)^2/(1 - 2*x),x]

[Out]

(-5*(-39 + 68*x + 20*x^2) - 242*Log[1 - 2*x])/16

Maple [A] (verified)

Time = 2.50 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.70

method result size
parallelrisch \(-\frac {25 x^{2}}{4}-\frac {85 x}{4}-\frac {121 \ln \left (x -\frac {1}{2}\right )}{8}\) \(16\)
default \(-\frac {25 x^{2}}{4}-\frac {85 x}{4}-\frac {121 \ln \left (-1+2 x \right )}{8}\) \(18\)
norman \(-\frac {25 x^{2}}{4}-\frac {85 x}{4}-\frac {121 \ln \left (-1+2 x \right )}{8}\) \(18\)
risch \(-\frac {25 x^{2}}{4}-\frac {85 x}{4}-\frac {121 \ln \left (-1+2 x \right )}{8}\) \(18\)
meijerg \(-\frac {121 \ln \left (1-2 x \right )}{8}-15 x -\frac {25 x \left (6 x +6\right )}{24}\) \(21\)

[In]

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

[Out]

-25/4*x^2-85/4*x-121/8*ln(x-1/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {(3+5 x)^2}{1-2 x} \, dx=-\frac {25}{4} \, x^{2} - \frac {85}{4} \, x - \frac {121}{8} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-25/4*x^2 - 85/4*x - 121/8*log(2*x - 1)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {(3+5 x)^2}{1-2 x} \, dx=- \frac {25 x^{2}}{4} - \frac {85 x}{4} - \frac {121 \log {\left (2 x - 1 \right )}}{8} \]

[In]

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

[Out]

-25*x**2/4 - 85*x/4 - 121*log(2*x - 1)/8

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {(3+5 x)^2}{1-2 x} \, dx=-\frac {25}{4} \, x^{2} - \frac {85}{4} \, x - \frac {121}{8} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-25/4*x^2 - 85/4*x - 121/8*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {(3+5 x)^2}{1-2 x} \, dx=-\frac {25}{4} \, x^{2} - \frac {85}{4} \, x - \frac {121}{8} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \]

[In]

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

[Out]

-25/4*x^2 - 85/4*x - 121/8*log(abs(2*x - 1))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int \frac {(3+5 x)^2}{1-2 x} \, dx=-\frac {85\,x}{4}-\frac {121\,\ln \left (x-\frac {1}{2}\right )}{8}-\frac {25\,x^2}{4} \]

[In]

int(-(5*x + 3)^2/(2*x - 1),x)

[Out]

- (85*x)/4 - (121*log(x - 1/2))/8 - (25*x^2)/4