\(\int \frac {(1-2 x) (2+3 x)^3}{(3+5 x)^3} \, dx\) [1226]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 20, antiderivative size = 45 \[ \int \frac {(1-2 x) (2+3 x)^3}{(3+5 x)^3} \, dx=\frac {81 x}{625}-\frac {27 x^2}{125}-\frac {11}{6250 (3+5 x)^2}-\frac {97}{3125 (3+5 x)}+\frac {279 \log (3+5 x)}{3125} \]

[Out]

81/625*x-27/125*x^2-11/6250/(3+5*x)^2-97/3125/(3+5*x)+279/3125*ln(3+5*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 45, 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} \[ \int \frac {(1-2 x) (2+3 x)^3}{(3+5 x)^3} \, dx=-\frac {27 x^2}{125}+\frac {81 x}{625}-\frac {97}{3125 (5 x+3)}-\frac {11}{6250 (5 x+3)^2}+\frac {279 \log (5 x+3)}{3125} \]

[In]

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

[Out]

(81*x)/625 - (27*x^2)/125 - 11/(6250*(3 + 5*x)^2) - 97/(3125*(3 + 5*x)) + (279*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 {81}{625}-\frac {54 x}{125}+\frac {11}{625 (3+5 x)^3}+\frac {97}{625 (3+5 x)^2}+\frac {279}{625 (3+5 x)}\right ) \, dx \\ & = \frac {81 x}{625}-\frac {27 x^2}{125}-\frac {11}{6250 (3+5 x)^2}-\frac {97}{3125 (3+5 x)}+\frac {279 \log (3+5 x)}{3125} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.07 \[ \int \frac {(1-2 x) (2+3 x)^3}{(3+5 x)^3} \, dx=\frac {9667+40520 x+40650 x^2-20250 x^3-33750 x^4+558 (3+5 x)^2 \log (-3 (3+5 x))}{6250 (3+5 x)^2} \]

[In]

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

[Out]

(9667 + 40520*x + 40650*x^2 - 20250*x^3 - 33750*x^4 + 558*(3 + 5*x)^2*Log[-3*(3 + 5*x)])/(6250*(3 + 5*x)^2)

Maple [A] (verified)

Time = 0.73 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.71

method result size
risch \(-\frac {27 x^{2}}{125}+\frac {81 x}{625}+\frac {-\frac {97 x}{625}-\frac {593}{6250}}{\left (3+5 x \right )^{2}}+\frac {279 \ln \left (3+5 x \right )}{3125}\) \(32\)
default \(\frac {81 x}{625}-\frac {27 x^{2}}{125}-\frac {11}{6250 \left (3+5 x \right )^{2}}-\frac {97}{3125 \left (3+5 x \right )}+\frac {279 \ln \left (3+5 x \right )}{3125}\) \(36\)
norman \(\frac {\frac {2489}{1875} x +\frac {4967}{2250} x^{2}-\frac {81}{25} x^{3}-\frac {27}{5} x^{4}}{\left (3+5 x \right )^{2}}+\frac {279 \ln \left (3+5 x \right )}{3125}\) \(37\)
parallelrisch \(\frac {-303750 x^{4}+125550 \ln \left (x +\frac {3}{5}\right ) x^{2}-182250 x^{3}+150660 \ln \left (x +\frac {3}{5}\right ) x +124175 x^{2}+45198 \ln \left (x +\frac {3}{5}\right )+74670 x}{56250 \left (3+5 x \right )^{2}}\) \(51\)
meijerg \(\frac {4 x \left (\frac {5 x}{3}+2\right )}{27 \left (1+\frac {5 x}{3}\right )^{2}}+\frac {10 x^{2}}{27 \left (1+\frac {5 x}{3}\right )^{2}}+\frac {x \left (15 x +6\right )}{25 \left (1+\frac {5 x}{3}\right )^{2}}+\frac {279 \ln \left (1+\frac {5 x}{3}\right )}{3125}-\frac {81 x \left (\frac {100}{9} x^{2}+30 x +12\right )}{500 \left (1+\frac {5 x}{3}\right )^{2}}+\frac {81 x \left (-\frac {625}{27} x^{3}+\frac {500}{9} x^{2}+150 x +60\right )}{3125 \left (1+\frac {5 x}{3}\right )^{2}}\) \(97\)

[In]

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

[Out]

-27/125*x^2+81/625*x+25*(-97/15625*x-593/156250)/(3+5*x)^2+279/3125*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.16 \[ \int \frac {(1-2 x) (2+3 x)^3}{(3+5 x)^3} \, dx=-\frac {33750 \, x^{4} + 20250 \, x^{3} - 12150 \, x^{2} - 558 \, {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (5 \, x + 3\right ) - 6320 \, x + 593}{6250 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]

[In]

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

[Out]

-1/6250*(33750*x^4 + 20250*x^3 - 12150*x^2 - 558*(25*x^2 + 30*x + 9)*log(5*x + 3) - 6320*x + 593)/(25*x^2 + 30
*x + 9)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.80 \[ \int \frac {(1-2 x) (2+3 x)^3}{(3+5 x)^3} \, dx=- \frac {27 x^{2}}{125} + \frac {81 x}{625} - \frac {970 x + 593}{156250 x^{2} + 187500 x + 56250} + \frac {279 \log {\left (5 x + 3 \right )}}{3125} \]

[In]

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

[Out]

-27*x**2/125 + 81*x/625 - (970*x + 593)/(156250*x**2 + 187500*x + 56250) + 279*log(5*x + 3)/3125

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.80 \[ \int \frac {(1-2 x) (2+3 x)^3}{(3+5 x)^3} \, dx=-\frac {27}{125} \, x^{2} + \frac {81}{625} \, x - \frac {970 \, x + 593}{6250 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} + \frac {279}{3125} \, \log \left (5 \, x + 3\right ) \]

[In]

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

[Out]

-27/125*x^2 + 81/625*x - 1/6250*(970*x + 593)/(25*x^2 + 30*x + 9) + 279/3125*log(5*x + 3)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.71 \[ \int \frac {(1-2 x) (2+3 x)^3}{(3+5 x)^3} \, dx=-\frac {27}{125} \, x^{2} + \frac {81}{625} \, x - \frac {970 \, x + 593}{6250 \, {\left (5 \, x + 3\right )}^{2}} + \frac {279}{3125} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) \]

[In]

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

[Out]

-27/125*x^2 + 81/625*x - 1/6250*(970*x + 593)/(5*x + 3)^2 + 279/3125*log(abs(5*x + 3))

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.71 \[ \int \frac {(1-2 x) (2+3 x)^3}{(3+5 x)^3} \, dx=\frac {81\,x}{625}+\frac {279\,\ln \left (x+\frac {3}{5}\right )}{3125}-\frac {\frac {97\,x}{15625}+\frac {593}{156250}}{x^2+\frac {6\,x}{5}+\frac {9}{25}}-\frac {27\,x^2}{125} \]

[In]

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

[Out]

(81*x)/625 + (279*log(x + 3/5))/3125 - ((97*x)/15625 + 593/156250)/((6*x)/5 + x^2 + 9/25) - (27*x^2)/125