\(\int \frac {(1-2 x)^2 (2+3 x)}{3+5 x} \, dx\) [1295]

Optimal result

Integrand size = 20, antiderivative size = 30 \[ \int \frac {(1-2 x)^2 (2+3 x)}{3+5 x} \, dx=\frac {43 x}{125}-\frac {28 x^2}{25}+\frac {4 x^3}{5}+\frac {121}{625} \log (3+5 x) \]

[Out]

43/125*x-28/25*x^2+4/5*x^3+121/625*ln(3+5*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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {78} \[ \int \frac {(1-2 x)^2 (2+3 x)}{3+5 x} \, dx=\frac {4 x^3}{5}-\frac {28 x^2}{25}+\frac {43 x}{125}+\frac {121}{625} \log (5 x+3) \]

[In]

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

[Out]

(43*x)/125 - (28*x^2)/25 + (4*x^3)/5 + (121*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 {43}{125}-\frac {56 x}{25}+\frac {12 x^2}{5}+\frac {121}{125 (3+5 x)}\right ) \, dx \\ & = \frac {43 x}{125}-\frac {28 x^2}{25}+\frac {4 x^3}{5}+\frac {121}{625} \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {(1-2 x)^2 (2+3 x)}{3+5 x} \, dx=\frac {1}{625} \left (489+215 x-700 x^2+500 x^3+121 \log (3+5 x)\right ) \]

[In]

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

[Out]

(489 + 215*x - 700*x^2 + 500*x^3 + 121*Log[3 + 5*x])/625

Maple [A] (verified)

Time = 2.30 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.70

method	result	size
parallelrisch	\(\frac {4 x^{3}}{5}-\frac {28 x^{2}}{25}+\frac {43 x}{125}+\frac {121 \ln \left (x +\frac {3}{5}\right )}{625}\)	\(21\)
default	\(\frac {43 x}{125}-\frac {28 x^{2}}{25}+\frac {4 x^{3}}{5}+\frac {121 \ln \left (3+5 x \right )}{625}\)	\(23\)
norman	\(\frac {43 x}{125}-\frac {28 x^{2}}{25}+\frac {4 x^{3}}{5}+\frac {121 \ln \left (3+5 x \right )}{625}\)	\(23\)
risch	\(\frac {43 x}{125}-\frac {28 x^{2}}{25}+\frac {4 x^{3}}{5}+\frac {121 \ln \left (3+5 x \right )}{625}\)	\(23\)
meijerg	\(\frac {121 \ln \left (1+\frac {5 x}{3}\right )}{625}-x +\frac {2 x \left (-5 x +6\right )}{25}+\frac {9 x \left (\frac {100}{9} x^{2}-10 x +12\right )}{125}\)	\(34\)

[In]

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

[Out]

4/5*x^3-28/25*x^2+43/125*x+121/625*ln(x+3/5)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {(1-2 x)^2 (2+3 x)}{3+5 x} \, dx=\frac {4}{5} \, x^{3} - \frac {28}{25} \, x^{2} + \frac {43}{125} \, x + \frac {121}{625} \, \log \left (5 \, x + 3\right ) \]

[In]

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

[Out]

4/5*x^3 - 28/25*x^2 + 43/125*x + 121/625*log(5*x + 3)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {(1-2 x)^2 (2+3 x)}{3+5 x} \, dx=\frac {4 x^{3}}{5} - \frac {28 x^{2}}{25} + \frac {43 x}{125} + \frac {121 \log {\left (5 x + 3 \right )}}{625} \]

[In]

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

[Out]

4*x**3/5 - 28*x**2/25 + 43*x/125 + 121*log(5*x + 3)/625

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {(1-2 x)^2 (2+3 x)}{3+5 x} \, dx=\frac {4}{5} \, x^{3} - \frac {28}{25} \, x^{2} + \frac {43}{125} \, x + \frac {121}{625} \, \log \left (5 \, x + 3\right ) \]

[In]

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

[Out]

4/5*x^3 - 28/25*x^2 + 43/125*x + 121/625*log(5*x + 3)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {(1-2 x)^2 (2+3 x)}{3+5 x} \, dx=\frac {4}{5} \, x^{3} - \frac {28}{25} \, x^{2} + \frac {43}{125} \, x + \frac {121}{625} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) \]

[In]

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

[Out]

4/5*x^3 - 28/25*x^2 + 43/125*x + 121/625*log(abs(5*x + 3))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67 \[ \int \frac {(1-2 x)^2 (2+3 x)}{3+5 x} \, dx=\frac {43\,x}{125}+\frac {121\,\ln \left (x+\frac {3}{5}\right )}{625}-\frac {28\,x^2}{25}+\frac {4\,x^3}{5} \]

[In]

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

[Out]

(43*x)/125 + (121*log(x + 3/5))/625 - (28*x^2)/25 + (4*x^3)/5