\(\int \frac {(1-2 x) (2+3 x)}{3+5 x} \, dx\) [1198]

Optimal result

Integrand size = 18, antiderivative size = 23 \[ \int \frac {(1-2 x) (2+3 x)}{3+5 x} \, dx=\frac {13 x}{25}-\frac {3 x^2}{5}+\frac {11}{125} \log (3+5 x) \]

[Out]

13/25*x-3/5*x^2+11/125*ln(3+5*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, 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} \[ \int \frac {(1-2 x) (2+3 x)}{3+5 x} \, dx=-\frac {3 x^2}{5}+\frac {13 x}{25}+\frac {11}{125} \log (5 x+3) \]

[In]

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

[Out]

(13*x)/25 - (3*x^2)/5 + (11*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 {13}{25}-\frac {6 x}{5}+\frac {11}{25 (3+5 x)}\right ) \, dx \\ & = \frac {13 x}{25}-\frac {3 x^2}{5}+\frac {11}{125} \log (3+5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {(1-2 x) (2+3 x)}{3+5 x} \, dx=\frac {1}{125} \left (66+65 x-75 x^2+11 \log (3+5 x)\right ) \]

[In]

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

[Out]

(66 + 65*x - 75*x^2 + 11*Log[3 + 5*x])/125

Maple [A] (verified)

Time = 1.80 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.70

method	result	size
parallelrisch	\(-\frac {3 x^{2}}{5}+\frac {13 x}{25}+\frac {11 \ln \left (x +\frac {3}{5}\right )}{125}\)	\(16\)
default	\(\frac {13 x}{25}-\frac {3 x^{2}}{5}+\frac {11 \ln \left (3+5 x \right )}{125}\)	\(18\)
norman	\(\frac {13 x}{25}-\frac {3 x^{2}}{5}+\frac {11 \ln \left (3+5 x \right )}{125}\)	\(18\)
risch	\(\frac {13 x}{25}-\frac {3 x^{2}}{5}+\frac {11 \ln \left (3+5 x \right )}{125}\)	\(18\)
meijerg	\(\frac {11 \ln \left (1+\frac {5 x}{3}\right )}{125}-\frac {x}{5}+\frac {3 x \left (-5 x +6\right )}{25}\)	\(21\)

[In]

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

[Out]

-3/5*x^2+13/25*x+11/125*ln(x+3/5)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {(1-2 x) (2+3 x)}{3+5 x} \, dx=-\frac {3}{5} \, x^{2} + \frac {13}{25} \, x + \frac {11}{125} \, \log \left (5 \, x + 3\right ) \]

[In]

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

[Out]

-3/5*x^2 + 13/25*x + 11/125*log(5*x + 3)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {(1-2 x) (2+3 x)}{3+5 x} \, dx=- \frac {3 x^{2}}{5} + \frac {13 x}{25} + \frac {11 \log {\left (5 x + 3 \right )}}{125} \]

[In]

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

[Out]

-3*x**2/5 + 13*x/25 + 11*log(5*x + 3)/125

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {(1-2 x) (2+3 x)}{3+5 x} \, dx=-\frac {3}{5} \, x^{2} + \frac {13}{25} \, x + \frac {11}{125} \, \log \left (5 \, x + 3\right ) \]

[In]

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

[Out]

-3/5*x^2 + 13/25*x + 11/125*log(5*x + 3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {(1-2 x) (2+3 x)}{3+5 x} \, dx=-\frac {3}{5} \, x^{2} + \frac {13}{25} \, x + \frac {11}{125} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) \]

[In]

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

[Out]

-3/5*x^2 + 13/25*x + 11/125*log(abs(5*x + 3))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65 \[ \int \frac {(1-2 x) (2+3 x)}{3+5 x} \, dx=\frac {13\,x}{25}+\frac {11\,\ln \left (x+\frac {3}{5}\right )}{125}-\frac {3\,x^2}{5} \]

[In]

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

[Out]

(13*x)/25 + (11*log(x + 3/5))/125 - (3*x^2)/5