\(\int \frac {(3+5 x)^3}{1-2 x} \, dx\) [1475]

Optimal result

Integrand size = 15, antiderivative size = 30 \[ \int \frac {(3+5 x)^3}{1-2 x} \, dx=-\frac {1115 x}{8}-\frac {575 x^2}{8}-\frac {125 x^3}{6}-\frac {1331}{16} \log (1-2 x) \]

[Out]

-1115/8*x-575/8*x^2-125/6*x^3-1331/16*ln(1-2*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, 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} \[ \int \frac {(3+5 x)^3}{1-2 x} \, dx=-\frac {125 x^3}{6}-\frac {575 x^2}{8}-\frac {1115 x}{8}-\frac {1331}{16} \log (1-2 x) \]

[In]

Int[(3 + 5*x)^3/(1 - 2*x),x]

[Out]

(-1115*x)/8 - (575*x^2)/8 - (125*x^3)/6 - (1331*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 {1115}{8}-\frac {575 x}{4}-\frac {125 x^2}{2}-\frac {1331}{8 (-1+2 x)}\right ) \, dx \\ & = -\frac {1115 x}{8}-\frac {575 x^2}{8}-\frac {125 x^3}{6}-\frac {1331}{16} \log (1-2 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {(3+5 x)^3}{1-2 x} \, dx=\frac {1}{96} \left (-5 \left (-1733+2676 x+1380 x^2+400 x^3\right )-7986 \log (1-2 x)\right ) \]

[In]

Integrate[(3 + 5*x)^3/(1 - 2*x),x]

[Out]

(-5*(-1733 + 2676*x + 1380*x^2 + 400*x^3) - 7986*Log[1 - 2*x])/96

Maple [A] (verified)

Time = 2.51 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.70

method	result	size
parallelrisch	\(-\frac {125 x^{3}}{6}-\frac {575 x^{2}}{8}-\frac {1115 x}{8}-\frac {1331 \ln \left (x -\frac {1}{2}\right )}{16}\)	\(21\)
default	\(-\frac {125 x^{3}}{6}-\frac {575 x^{2}}{8}-\frac {1115 x}{8}-\frac {1331 \ln \left (-1+2 x \right )}{16}\)	\(23\)
norman	\(-\frac {125 x^{3}}{6}-\frac {575 x^{2}}{8}-\frac {1115 x}{8}-\frac {1331 \ln \left (-1+2 x \right )}{16}\)	\(23\)
risch	\(-\frac {125 x^{3}}{6}-\frac {575 x^{2}}{8}-\frac {1115 x}{8}-\frac {1331 \ln \left (-1+2 x \right )}{16}\)	\(23\)
meijerg	\(-\frac {1331 \ln \left (1-2 x \right )}{16}-\frac {135 x}{2}-\frac {75 x \left (6 x +6\right )}{8}-\frac {125 x \left (16 x^{2}+12 x +12\right )}{96}\)	\(34\)

[In]

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

[Out]

-125/6*x^3-575/8*x^2-1115/8*x-1331/16*ln(x-1/2)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {(3+5 x)^3}{1-2 x} \, dx=-\frac {125}{6} \, x^{3} - \frac {575}{8} \, x^{2} - \frac {1115}{8} \, x - \frac {1331}{16} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-125/6*x^3 - 575/8*x^2 - 1115/8*x - 1331/16*log(2*x - 1)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {(3+5 x)^3}{1-2 x} \, dx=- \frac {125 x^{3}}{6} - \frac {575 x^{2}}{8} - \frac {1115 x}{8} - \frac {1331 \log {\left (2 x - 1 \right )}}{16} \]

[In]

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

[Out]

-125*x**3/6 - 575*x**2/8 - 1115*x/8 - 1331*log(2*x - 1)/16

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {(3+5 x)^3}{1-2 x} \, dx=-\frac {125}{6} \, x^{3} - \frac {575}{8} \, x^{2} - \frac {1115}{8} \, x - \frac {1331}{16} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

-125/6*x^3 - 575/8*x^2 - 1115/8*x - 1331/16*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {(3+5 x)^3}{1-2 x} \, dx=-\frac {125}{6} \, x^{3} - \frac {575}{8} \, x^{2} - \frac {1115}{8} \, x - \frac {1331}{16} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \]

[In]

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

[Out]

-125/6*x^3 - 575/8*x^2 - 1115/8*x - 1331/16*log(abs(2*x - 1))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67 \[ \int \frac {(3+5 x)^3}{1-2 x} \, dx=-\frac {1115\,x}{8}-\frac {1331\,\ln \left (x-\frac {1}{2}\right )}{16}-\frac {575\,x^2}{8}-\frac {125\,x^3}{6} \]

[In]

int(-(5*x + 3)^3/(2*x - 1),x)

[Out]

- (1115*x)/8 - (1331*log(x - 1/2))/16 - (575*x^2)/8 - (125*x^3)/6