\(\int \frac {(1-2 x) (2+3 x)^3}{3+5 x} \, dx\) [1196]

Optimal result

Integrand size = 20, antiderivative size = 37 \[ \int \frac {(1-2 x) (2+3 x)^3}{3+5 x} \, dx=\frac {1663 x}{625}+\frac {279 x^2}{250}-\frac {81 x^3}{25}-\frac {27 x^4}{10}+\frac {11 \log (3+5 x)}{3125} \]

[Out]

1663/625*x+279/250*x^2-81/25*x^3-27/10*x^4+11/3125*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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {78} \[ \int \frac {(1-2 x) (2+3 x)^3}{3+5 x} \, dx=-\frac {27 x^4}{10}-\frac {81 x^3}{25}+\frac {279 x^2}{250}+\frac {1663 x}{625}+\frac {11 \log (5 x+3)}{3125} \]

[In]

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

[Out]

(1663*x)/625 + (279*x^2)/250 - (81*x^3)/25 - (27*x^4)/10 + (11*Log[3 + 5*x])/3125

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 {1663}{625}+\frac {279 x}{125}-\frac {243 x^2}{25}-\frac {54 x^3}{5}+\frac {11}{625 (3+5 x)}\right ) \, dx \\ & = \frac {1663 x}{625}+\frac {279 x^2}{250}-\frac {81 x^3}{25}-\frac {27 x^4}{10}+\frac {11 \log (3+5 x)}{3125} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.95 \[ \int \frac {(1-2 x) (2+3 x)^3}{3+5 x} \, dx=\frac {5 \left (1056+3326 x+1395 x^2-4050 x^3-3375 x^4\right )+22 \log (3+5 x)}{6250} \]

[In]

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

[Out]

(5*(1056 + 3326*x + 1395*x^2 - 4050*x^3 - 3375*x^4) + 22*Log[3 + 5*x])/6250

Maple [A] (verified)

Time = 2.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.70

method	result	size
parallelrisch	\(-\frac {27 x^{4}}{10}-\frac {81 x^{3}}{25}+\frac {279 x^{2}}{250}+\frac {1663 x}{625}+\frac {11 \ln \left (x +\frac {3}{5}\right )}{3125}\)	\(26\)
default	\(\frac {1663 x}{625}+\frac {279 x^{2}}{250}-\frac {81 x^{3}}{25}-\frac {27 x^{4}}{10}+\frac {11 \ln \left (3+5 x \right )}{3125}\)	\(28\)
norman	\(\frac {1663 x}{625}+\frac {279 x^{2}}{250}-\frac {81 x^{3}}{25}-\frac {27 x^{4}}{10}+\frac {11 \ln \left (3+5 x \right )}{3125}\)	\(28\)
risch	\(\frac {1663 x}{625}+\frac {279 x^{2}}{250}-\frac {81 x^{3}}{25}-\frac {27 x^{4}}{10}+\frac {11 \ln \left (3+5 x \right )}{3125}\)	\(28\)
meijerg	\(\frac {11 \ln \left (1+\frac {5 x}{3}\right )}{3125}+4 x +\frac {9 x \left (-5 x +6\right )}{25}-\frac {243 x \left (\frac {100}{9} x^{2}-10 x +12\right )}{500}+\frac {243 x \left (-\frac {625}{9} x^{3}+\frac {500}{9} x^{2}-50 x +60\right )}{6250}\)	\(52\)

[In]

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

[Out]

-27/10*x^4-81/25*x^3+279/250*x^2+1663/625*x+11/3125*ln(x+3/5)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.73 \[ \int \frac {(1-2 x) (2+3 x)^3}{3+5 x} \, dx=-\frac {27}{10} \, x^{4} - \frac {81}{25} \, x^{3} + \frac {279}{250} \, x^{2} + \frac {1663}{625} \, x + \frac {11}{3125} \, \log \left (5 \, x + 3\right ) \]

[In]

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

[Out]

-27/10*x^4 - 81/25*x^3 + 279/250*x^2 + 1663/625*x + 11/3125*log(5*x + 3)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.92 \[ \int \frac {(1-2 x) (2+3 x)^3}{3+5 x} \, dx=- \frac {27 x^{4}}{10} - \frac {81 x^{3}}{25} + \frac {279 x^{2}}{250} + \frac {1663 x}{625} + \frac {11 \log {\left (5 x + 3 \right )}}{3125} \]

[In]

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

[Out]

-27*x**4/10 - 81*x**3/25 + 279*x**2/250 + 1663*x/625 + 11*log(5*x + 3)/3125

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.73 \[ \int \frac {(1-2 x) (2+3 x)^3}{3+5 x} \, dx=-\frac {27}{10} \, x^{4} - \frac {81}{25} \, x^{3} + \frac {279}{250} \, x^{2} + \frac {1663}{625} \, x + \frac {11}{3125} \, \log \left (5 \, x + 3\right ) \]

[In]

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

[Out]

-27/10*x^4 - 81/25*x^3 + 279/250*x^2 + 1663/625*x + 11/3125*log(5*x + 3)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.76 \[ \int \frac {(1-2 x) (2+3 x)^3}{3+5 x} \, dx=-\frac {27}{10} \, x^{4} - \frac {81}{25} \, x^{3} + \frac {279}{250} \, x^{2} + \frac {1663}{625} \, x + \frac {11}{3125} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) \]

[In]

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

[Out]

-27/10*x^4 - 81/25*x^3 + 279/250*x^2 + 1663/625*x + 11/3125*log(abs(5*x + 3))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.68 \[ \int \frac {(1-2 x) (2+3 x)^3}{3+5 x} \, dx=\frac {1663\,x}{625}+\frac {11\,\ln \left (x+\frac {3}{5}\right )}{3125}+\frac {279\,x^2}{250}-\frac {81\,x^3}{25}-\frac {27\,x^4}{10} \]

[In]

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

[Out]

(1663*x)/625 + (11*log(x + 3/5))/3125 + (279*x^2)/250 - (81*x^3)/25 - (27*x^4)/10