\(\int \frac {(1-2 x)^3}{(3+5 x)^2} \, dx\) [1409]

Optimal result

Integrand size = 15, antiderivative size = 34 \[ \int \frac {(1-2 x)^3}{(3+5 x)^2} \, dx=\frac {108 x}{125}-\frac {4 x^2}{25}-\frac {1331}{625 (3+5 x)}-\frac {726}{625} \log (3+5 x) \]

[Out]

108/125*x-4/25*x^2-1331/625/(3+5*x)-726/625*ln(3+5*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} \[ \int \frac {(1-2 x)^3}{(3+5 x)^2} \, dx=-\frac {4 x^2}{25}+\frac {108 x}{125}-\frac {1331}{625 (5 x+3)}-\frac {726}{625} \log (5 x+3) \]

[In]

Int[(1 - 2*x)^3/(3 + 5*x)^2,x]

[Out]

(108*x)/125 - (4*x^2)/25 - 1331/(625*(3 + 5*x)) - (726*Log[3 + 5*x])/625

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 {108}{125}-\frac {8 x}{25}+\frac {1331}{125 (3+5 x)^2}-\frac {726}{125 (3+5 x)}\right ) \, dx \\ & = \frac {108 x}{125}-\frac {4 x^2}{25}-\frac {1331}{625 (3+5 x)}-\frac {726}{625} \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15 \[ \int \frac {(1-2 x)^3}{(3+5 x)^2} \, dx=\frac {-2066+395 x+2400 x^2-500 x^3-726 (3+5 x) \log (6+10 x)}{625 (3+5 x)} \]

[In]

Integrate[(1 - 2*x)^3/(3 + 5*x)^2,x]

[Out]

(-2066 + 395*x + 2400*x^2 - 500*x^3 - 726*(3 + 5*x)*Log[6 + 10*x])/(625*(3 + 5*x))

Maple [A] (verified)

Time = 2.45 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74

method	result	size
risch	\(-\frac {4 x^{2}}{25}+\frac {108 x}{125}-\frac {1331}{3125 \left (x +\frac {3}{5}\right )}-\frac {726 \ln \left (3+5 x \right )}{625}\)	\(25\)
default	\(\frac {108 x}{125}-\frac {4 x^{2}}{25}-\frac {1331}{625 \left (3+5 x \right )}-\frac {726 \ln \left (3+5 x \right )}{625}\)	\(27\)
norman	\(\frac {\frac {2303}{375} x +\frac {96}{25} x^{2}-\frac {4}{5} x^{3}}{3+5 x}-\frac {726 \ln \left (3+5 x \right )}{625}\)	\(32\)
parallelrisch	\(-\frac {1500 x^{3}+10890 \ln \left (x +\frac {3}{5}\right ) x -7200 x^{2}+6534 \ln \left (x +\frac {3}{5}\right )-11515 x}{1875 \left (3+5 x \right )}\)	\(37\)
meijerg	\(\frac {23 x}{45 \left (1+\frac {5 x}{3}\right )}-\frac {726 \ln \left (1+\frac {5 x}{3}\right )}{625}+\frac {4 x \left (5 x +6\right )}{25 \left (1+\frac {5 x}{3}\right )}+\frac {6 x \left (-\frac {50}{9} x^{2}+10 x +12\right )}{125 \left (1+\frac {5 x}{3}\right )}\)	\(55\)

[In]

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

[Out]

-4/25*x^2+108/125*x-1331/3125/(x+3/5)-726/625*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int \frac {(1-2 x)^3}{(3+5 x)^2} \, dx=-\frac {500 \, x^{3} - 2400 \, x^{2} + 726 \, {\left (5 \, x + 3\right )} \log \left (5 \, x + 3\right ) - 1620 \, x + 1331}{625 \, {\left (5 \, x + 3\right )}} \]

[In]

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

[Out]

-1/625*(500*x^3 - 2400*x^2 + 726*(5*x + 3)*log(5*x + 3) - 1620*x + 1331)/(5*x + 3)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79 \[ \int \frac {(1-2 x)^3}{(3+5 x)^2} \, dx=- \frac {4 x^{2}}{25} + \frac {108 x}{125} - \frac {726 \log {\left (5 x + 3 \right )}}{625} - \frac {1331}{3125 x + 1875} \]

[In]

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

[Out]

-4*x**2/25 + 108*x/125 - 726*log(5*x + 3)/625 - 1331/(3125*x + 1875)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \frac {(1-2 x)^3}{(3+5 x)^2} \, dx=-\frac {4}{25} \, x^{2} + \frac {108}{125} \, x - \frac {1331}{625 \, {\left (5 \, x + 3\right )}} - \frac {726}{625} \, \log \left (5 \, x + 3\right ) \]

[In]

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

[Out]

-4/25*x^2 + 108/125*x - 1331/625/(5*x + 3) - 726/625*log(5*x + 3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.41 \[ \int \frac {(1-2 x)^3}{(3+5 x)^2} \, dx=\frac {4}{625} \, {\left (5 \, x + 3\right )}^{2} {\left (\frac {33}{5 \, x + 3} - 1\right )} - \frac {1331}{625 \, {\left (5 \, x + 3\right )}} + \frac {726}{625} \, \log \left (\frac {{\left | 5 \, x + 3 \right |}}{5 \, {\left (5 \, x + 3\right )}^{2}}\right ) \]

[In]

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

[Out]

4/625*(5*x + 3)^2*(33/(5*x + 3) - 1) - 1331/625/(5*x + 3) + 726/625*log(1/5*abs(5*x + 3)/(5*x + 3)^2)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int \frac {(1-2 x)^3}{(3+5 x)^2} \, dx=\frac {108\,x}{125}-\frac {726\,\ln \left (x+\frac {3}{5}\right )}{625}-\frac {1331}{3125\,\left (x+\frac {3}{5}\right )}-\frac {4\,x^2}{25} \]

[In]

int(-(2*x - 1)^3/(5*x + 3)^2,x)

[Out]

(108*x)/125 - (726*log(x + 3/5))/625 - 1331/(3125*(x + 3/5)) - (4*x^2)/25