\(\int \frac {(1-2 x) (2+3 x)}{(3+5 x)^3} \, dx\) [1228]

Optimal result

Integrand size = 18, antiderivative size = 33 \[ \int \frac {(1-2 x) (2+3 x)}{(3+5 x)^3} \, dx=-\frac {11}{250 (3+5 x)^2}-\frac {31}{125 (3+5 x)}-\frac {6}{125} \log (3+5 x) \]

[Out]

-11/250/(3+5*x)^2-31/125/(3+5*x)-6/125*ln(3+5*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {78} \[ \int \frac {(1-2 x) (2+3 x)}{(3+5 x)^3} \, dx=-\frac {31}{125 (5 x+3)}-\frac {11}{250 (5 x+3)^2}-\frac {6}{125} \log (5 x+3) \]

[In]

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

[Out]

-11/(250*(3 + 5*x)^2) - 31/(125*(3 + 5*x)) - (6*Log[3 + 5*x])/125

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 {11}{25 (3+5 x)^3}+\frac {31}{25 (3+5 x)^2}-\frac {6}{25 (3+5 x)}\right ) \, dx \\ & = -\frac {11}{250 (3+5 x)^2}-\frac {31}{125 (3+5 x)}-\frac {6}{125} \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00 \[ \int \frac {(1-2 x) (2+3 x)}{(3+5 x)^3} \, dx=-\frac {11}{250 (3+5 x)^2}-\frac {31}{125 (3+5 x)}-\frac {6}{125} \log (3+5 x) \]

[In]

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

[Out]

-11/(250*(3 + 5*x)^2) - 31/(125*(3 + 5*x)) - (6*Log[3 + 5*x])/125

Maple [A] (verified)

Time = 2.23 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73

method	result	size
risch	\(\frac {-\frac {31 x}{25}-\frac {197}{250}}{\left (3+5 x \right )^{2}}-\frac {6 \ln \left (3+5 x \right )}{125}\)	\(24\)
norman	\(\frac {\frac {104}{75} x +\frac {197}{90} x^{2}}{\left (3+5 x \right )^{2}}-\frac {6 \ln \left (3+5 x \right )}{125}\)	\(27\)
default	\(-\frac {11}{250 \left (3+5 x \right )^{2}}-\frac {31}{125 \left (3+5 x \right )}-\frac {6 \ln \left (3+5 x \right )}{125}\)	\(28\)
parallelrisch	\(-\frac {2700 \ln \left (x +\frac {3}{5}\right ) x^{2}+3240 \ln \left (x +\frac {3}{5}\right ) x -4925 x^{2}+972 \ln \left (x +\frac {3}{5}\right )-3120 x}{2250 \left (3+5 x \right )^{2}}\)	\(41\)
meijerg	\(\frac {x \left (\frac {5 x}{3}+2\right )}{27 \left (1+\frac {5 x}{3}\right )^{2}}-\frac {x^{2}}{54 \left (1+\frac {5 x}{3}\right )^{2}}+\frac {x \left (15 x +6\right )}{75 \left (1+\frac {5 x}{3}\right )^{2}}-\frac {6 \ln \left (1+\frac {5 x}{3}\right )}{125}\)	\(52\)

[In]

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

[Out]

25*(-31/625*x-197/6250)/(3+5*x)^2-6/125*ln(3+5*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.12 \[ \int \frac {(1-2 x) (2+3 x)}{(3+5 x)^3} \, dx=-\frac {12 \, {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (5 \, x + 3\right ) + 310 \, x + 197}{250 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]

[In]

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

[Out]

-1/250*(12*(25*x^2 + 30*x + 9)*log(5*x + 3) + 310*x + 197)/(25*x^2 + 30*x + 9)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \frac {(1-2 x) (2+3 x)}{(3+5 x)^3} \, dx=- \frac {310 x + 197}{6250 x^{2} + 7500 x + 2250} - \frac {6 \log {\left (5 x + 3 \right )}}{125} \]

[In]

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

[Out]

-(310*x + 197)/(6250*x**2 + 7500*x + 2250) - 6*log(5*x + 3)/125

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {(1-2 x) (2+3 x)}{(3+5 x)^3} \, dx=-\frac {310 \, x + 197}{250 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} - \frac {6}{125} \, \log \left (5 \, x + 3\right ) \]

[In]

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

[Out]

-1/250*(310*x + 197)/(25*x^2 + 30*x + 9) - 6/125*log(5*x + 3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73 \[ \int \frac {(1-2 x) (2+3 x)}{(3+5 x)^3} \, dx=-\frac {310 \, x + 197}{250 \, {\left (5 \, x + 3\right )}^{2}} - \frac {6}{125} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) \]

[In]

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

[Out]

-1/250*(310*x + 197)/(5*x + 3)^2 - 6/125*log(abs(5*x + 3))

Mupad [B] (verification not implemented)

Time = 1.17 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73 \[ \int \frac {(1-2 x) (2+3 x)}{(3+5 x)^3} \, dx=-\frac {6\,\ln \left (x+\frac {3}{5}\right )}{125}-\frac {\frac {31\,x}{625}+\frac {197}{6250}}{x^2+\frac {6\,x}{5}+\frac {9}{25}} \]

[In]

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

[Out]

- (6*log(x + 3/5))/125 - ((31*x)/625 + 197/6250)/((6*x)/5 + x^2 + 9/25)