\(\int \frac {(1-2 x) (2+3 x)^2}{(3+5 x)^2} \, dx\) [1212]

Optimal result

Integrand size = 20, antiderivative size = 34 \[ \int \frac {(1-2 x) (2+3 x)^2}{(3+5 x)^2} \, dx=\frac {33 x}{125}-\frac {9 x^2}{25}-\frac {11}{625 (3+5 x)}+\frac {64}{625} \log (3+5 x) \]

[Out]

33/125*x-9/25*x^2-11/625/(3+5*x)+64/625*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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {78} \[ \int \frac {(1-2 x) (2+3 x)^2}{(3+5 x)^2} \, dx=-\frac {9 x^2}{25}+\frac {33 x}{125}-\frac {11}{625 (5 x+3)}+\frac {64}{625} \log (5 x+3) \]

[In]

Int[((1 - 2*x)*(2 + 3*x)^2)/(3 + 5*x)^2,x]

[Out]

(33*x)/125 - (9*x^2)/25 - 11/(625*(3 + 5*x)) + (64*Log[3 + 5*x])/625

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {33}{125}-\frac {18 x}{25}+\frac {11}{125 (3+5 x)^2}+\frac {64}{125 (3+5 x)}\right ) \, dx \\ & = \frac {33 x}{125}-\frac {9 x^2}{25}-\frac {11}{625 (3+5 x)}+\frac {64}{625} \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21 \[ \int \frac {(1-2 x) (2+3 x)^2}{(3+5 x)^2} \, dx=\frac {619+1545 x+150 x^2-1125 x^3+64 (3+5 x) \log (-3 (3+5 x))}{625 (3+5 x)} \]

[In]

Integrate[((1 - 2*x)*(2 + 3*x)^2)/(3 + 5*x)^2,x]

[Out]

(619 + 1545*x + 150*x^2 - 1125*x^3 + 64*(3 + 5*x)*Log[-3*(3 + 5*x)])/(625*(3 + 5*x))

Maple [A] (verified)

Time = 2.22 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74

method	result	size
risch	\(-\frac {9 x^{2}}{25}+\frac {33 x}{125}-\frac {11}{3125 \left (x +\frac {3}{5}\right )}+\frac {64 \ln \left (3+5 x \right )}{625}\)	\(25\)
default	\(\frac {33 x}{125}-\frac {9 x^{2}}{25}-\frac {11}{625 \left (3+5 x \right )}+\frac {64 \ln \left (3+5 x \right )}{625}\)	\(27\)
norman	\(\frac {\frac {308}{375} x +\frac {6}{25} x^{2}-\frac {9}{5} x^{3}}{3+5 x}+\frac {64 \ln \left (3+5 x \right )}{625}\)	\(32\)
parallelrisch	\(\frac {-3375 x^{3}+960 \ln \left (x +\frac {3}{5}\right ) x +450 x^{2}+576 \ln \left (x +\frac {3}{5}\right )+1540 x}{5625+9375 x}\)	\(37\)
meijerg	\(\frac {8 x}{45 \left (1+\frac {5 x}{3}\right )}+\frac {64 \ln \left (1+\frac {5 x}{3}\right )}{625}-\frac {x \left (5 x +6\right )}{5 \left (1+\frac {5 x}{3}\right )}+\frac {27 x \left (-\frac {50}{9} x^{2}+10 x +12\right )}{250 \left (1+\frac {5 x}{3}\right )}\)	\(55\)

[In]

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

[Out]

-9/25*x^2+33/125*x-11/3125/(x+3/5)+64/625*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int \frac {(1-2 x) (2+3 x)^2}{(3+5 x)^2} \, dx=-\frac {1125 \, x^{3} - 150 \, x^{2} - 64 \, {\left (5 \, x + 3\right )} \log \left (5 \, x + 3\right ) - 495 \, x + 11}{625 \, {\left (5 \, x + 3\right )}} \]

[In]

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

[Out]

-1/625*(1125*x^3 - 150*x^2 - 64*(5*x + 3)*log(5*x + 3) - 495*x + 11)/(5*x + 3)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79 \[ \int \frac {(1-2 x) (2+3 x)^2}{(3+5 x)^2} \, dx=- \frac {9 x^{2}}{25} + \frac {33 x}{125} + \frac {64 \log {\left (5 x + 3 \right )}}{625} - \frac {11}{3125 x + 1875} \]

[In]

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

[Out]

-9*x**2/25 + 33*x/125 + 64*log(5*x + 3)/625 - 11/(3125*x + 1875)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \frac {(1-2 x) (2+3 x)^2}{(3+5 x)^2} \, dx=-\frac {9}{25} \, x^{2} + \frac {33}{125} \, x - \frac {11}{625 \, {\left (5 \, x + 3\right )}} + \frac {64}{625} \, \log \left (5 \, x + 3\right ) \]

[In]

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

[Out]

-9/25*x^2 + 33/125*x - 11/625/(5*x + 3) + 64/625*log(5*x + 3)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.41 \[ \int \frac {(1-2 x) (2+3 x)^2}{(3+5 x)^2} \, dx=\frac {3}{625} \, {\left (5 \, x + 3\right )}^{2} {\left (\frac {29}{5 \, x + 3} - 3\right )} - \frac {11}{625 \, {\left (5 \, x + 3\right )}} - \frac {64}{625} \, \log \left (\frac {{\left | 5 \, x + 3 \right |}}{5 \, {\left (5 \, x + 3\right )}^{2}}\right ) \]

[In]

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

[Out]

3/625*(5*x + 3)^2*(29/(5*x + 3) - 3) - 11/625/(5*x + 3) - 64/625*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 \[ \int \frac {(1-2 x) (2+3 x)^2}{(3+5 x)^2} \, dx=\frac {33\,x}{125}+\frac {64\,\ln \left (x+\frac {3}{5}\right )}{625}-\frac {11}{3125\,\left (x+\frac {3}{5}\right )}-\frac {9\,x^2}{25} \]

[In]

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

[Out]

(33*x)/125 + (64*log(x + 3/5))/625 - 11/(3125*(x + 3/5)) - (9*x^2)/25