\(\int \frac {(1-2 x)^2 (2+3 x)^2}{3+5 x} \, dx\) [1294]

Optimal result

Integrand size = 22, antiderivative size = 37 \[ \int \frac {(1-2 x)^2 (2+3 x)^2}{3+5 x} \, dx=\frac {793 x}{625}-\frac {431 x^2}{250}-\frac {16 x^3}{25}+\frac {9 x^4}{5}+\frac {121 \log (3+5 x)}{3125} \]

[Out]

793/625*x-431/250*x^2-16/25*x^3+9/5*x^4+121/3125*ln(3+5*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {90} \[ \int \frac {(1-2 x)^2 (2+3 x)^2}{3+5 x} \, dx=\frac {9 x^4}{5}-\frac {16 x^3}{25}-\frac {431 x^2}{250}+\frac {793 x}{625}+\frac {121 \log (5 x+3)}{3125} \]

[In]

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

[Out]

(793*x)/625 - (431*x^2)/250 - (16*x^3)/25 + (9*x^4)/5 + (121*Log[3 + 5*x])/3125

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {793}{625}-\frac {431 x}{125}-\frac {48 x^2}{25}+\frac {36 x^3}{5}+\frac {121}{625 (3+5 x)}\right ) \, dx \\ & = \frac {793 x}{625}-\frac {431 x^2}{250}-\frac {16 x^3}{25}+\frac {9 x^4}{5}+\frac {121 \log (3+5 x)}{3125} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.95 \[ \int \frac {(1-2 x)^2 (2+3 x)^2}{3+5 x} \, dx=\frac {5 \left (1263+1586 x-2155 x^2-800 x^3+2250 x^4\right )+242 \log (3+5 x)}{6250} \]

[In]

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

[Out]

(5*(1263 + 1586*x - 2155*x^2 - 800*x^3 + 2250*x^4) + 242*Log[3 + 5*x])/6250

Maple [A] (verified)

Time = 2.34 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.70

method	result	size
parallelrisch	\(\frac {9 x^{4}}{5}-\frac {16 x^{3}}{25}-\frac {431 x^{2}}{250}+\frac {793 x}{625}+\frac {121 \ln \left (x +\frac {3}{5}\right )}{3125}\)	\(26\)
default	\(\frac {793 x}{625}-\frac {431 x^{2}}{250}-\frac {16 x^{3}}{25}+\frac {9 x^{4}}{5}+\frac {121 \ln \left (3+5 x \right )}{3125}\)	\(28\)
norman	\(\frac {793 x}{625}-\frac {431 x^{2}}{250}-\frac {16 x^{3}}{25}+\frac {9 x^{4}}{5}+\frac {121 \ln \left (3+5 x \right )}{3125}\)	\(28\)
risch	\(\frac {793 x}{625}-\frac {431 x^{2}}{250}-\frac {16 x^{3}}{25}+\frac {9 x^{4}}{5}+\frac {121 \ln \left (3+5 x \right )}{3125}\)	\(28\)
meijerg	\(\frac {121 \ln \left (1+\frac {5 x}{3}\right )}{3125}-\frac {4 x}{5}+\frac {23 x \left (-5 x +6\right )}{50}+\frac {9 x \left (\frac {100}{9} x^{2}-10 x +12\right )}{125}-\frac {81 x \left (-\frac {625}{9} x^{3}+\frac {500}{9} x^{2}-50 x +60\right )}{3125}\)	\(52\)

[In]

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

[Out]

9/5*x^4-16/25*x^3-431/250*x^2+793/625*x+121/3125*ln(x+3/5)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.73 \[ \int \frac {(1-2 x)^2 (2+3 x)^2}{3+5 x} \, dx=\frac {9}{5} \, x^{4} - \frac {16}{25} \, x^{3} - \frac {431}{250} \, x^{2} + \frac {793}{625} \, x + \frac {121}{3125} \, \log \left (5 \, x + 3\right ) \]

[In]

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

[Out]

9/5*x^4 - 16/25*x^3 - 431/250*x^2 + 793/625*x + 121/3125*log(5*x + 3)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.92 \[ \int \frac {(1-2 x)^2 (2+3 x)^2}{3+5 x} \, dx=\frac {9 x^{4}}{5} - \frac {16 x^{3}}{25} - \frac {431 x^{2}}{250} + \frac {793 x}{625} + \frac {121 \log {\left (5 x + 3 \right )}}{3125} \]

[In]

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

[Out]

9*x**4/5 - 16*x**3/25 - 431*x**2/250 + 793*x/625 + 121*log(5*x + 3)/3125

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.73 \[ \int \frac {(1-2 x)^2 (2+3 x)^2}{3+5 x} \, dx=\frac {9}{5} \, x^{4} - \frac {16}{25} \, x^{3} - \frac {431}{250} \, x^{2} + \frac {793}{625} \, x + \frac {121}{3125} \, \log \left (5 \, x + 3\right ) \]

[In]

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

[Out]

9/5*x^4 - 16/25*x^3 - 431/250*x^2 + 793/625*x + 121/3125*log(5*x + 3)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.76 \[ \int \frac {(1-2 x)^2 (2+3 x)^2}{3+5 x} \, dx=\frac {9}{5} \, x^{4} - \frac {16}{25} \, x^{3} - \frac {431}{250} \, x^{2} + \frac {793}{625} \, x + \frac {121}{3125} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) \]

[In]

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

[Out]

9/5*x^4 - 16/25*x^3 - 431/250*x^2 + 793/625*x + 121/3125*log(abs(5*x + 3))

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.68 \[ \int \frac {(1-2 x)^2 (2+3 x)^2}{3+5 x} \, dx=\frac {793\,x}{625}+\frac {121\,\ln \left (x+\frac {3}{5}\right )}{3125}-\frac {431\,x^2}{250}-\frac {16\,x^3}{25}+\frac {9\,x^4}{5} \]

[In]

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

[Out]

(793*x)/625 + (121*log(x + 3/5))/3125 - (431*x^2)/250 - (16*x^3)/25 + (9*x^4)/5