\(\int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^2} \, dx\) [1168]

Optimal result

Integrand size = 20, antiderivative size = 34 \[ \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^2} \, dx=\frac {95 x}{27}-\frac {25 x^2}{9}-\frac {7}{81 (2+3 x)}-\frac {8}{9} \log (2+3 x) \]

[Out]

95/27*x-25/9*x^2-7/81/(2+3*x)-8/9*ln(2+3*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, 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} \[ \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^2} \, dx=-\frac {25 x^2}{9}+\frac {95 x}{27}-\frac {7}{81 (3 x+2)}-\frac {8}{9} \log (3 x+2) \]

[In]

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

[Out]

(95*x)/27 - (25*x^2)/9 - 7/(81*(2 + 3*x)) - (8*Log[2 + 3*x])/9

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 {95}{27}-\frac {50 x}{9}+\frac {7}{27 (2+3 x)^2}-\frac {8}{3 (2+3 x)}\right ) \, dx \\ & = \frac {95 x}{27}-\frac {25 x^2}{9}-\frac {7}{81 (2+3 x)}-\frac {8}{9} \log (2+3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^2} \, dx=\frac {191+480 x+135 x^2-225 x^3-24 (2+3 x) \log (2+3 x)}{54+81 x} \]

[In]

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

[Out]

(191 + 480*x + 135*x^2 - 225*x^3 - 24*(2 + 3*x)*Log[2 + 3*x])/(54 + 81*x)

Maple [A] (verified)

Time = 1.79 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74

method	result	size
risch	\(-\frac {25 x^{2}}{9}+\frac {95 x}{27}-\frac {7}{243 \left (\frac {2}{3}+x \right )}-\frac {8 \ln \left (2+3 x \right )}{9}\)	\(25\)
default	\(\frac {95 x}{27}-\frac {25 x^{2}}{9}-\frac {7}{81 \left (2+3 x \right )}-\frac {8 \ln \left (2+3 x \right )}{9}\)	\(27\)
norman	\(\frac {\frac {43}{6} x +5 x^{2}-\frac {25}{3} x^{3}}{2+3 x}-\frac {8 \ln \left (2+3 x \right )}{9}\)	\(32\)
parallelrisch	\(-\frac {150 x^{3}+48 \ln \left (\frac {2}{3}+x \right ) x -90 x^{2}+32 \ln \left (\frac {2}{3}+x \right )-129 x}{18 \left (2+3 x \right )}\)	\(37\)
meijerg	\(\frac {x}{4+6 x}-\frac {8 \ln \left (1+\frac {3 x}{2}\right )}{9}-\frac {35 x \left (\frac {9 x}{2}+6\right )}{27 \left (1+\frac {3 x}{2}\right )}+\frac {25 x \left (-\frac {9}{2} x^{2}+9 x +12\right )}{27 \left (1+\frac {3 x}{2}\right )}\)	\(55\)

[In]

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

[Out]

-25/9*x^2+95/27*x-7/243/(2/3+x)-8/9*ln(2+3*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^2} \, dx=-\frac {675 \, x^{3} - 405 \, x^{2} + 72 \, {\left (3 \, x + 2\right )} \log \left (3 \, x + 2\right ) - 570 \, x + 7}{81 \, {\left (3 \, x + 2\right )}} \]

[In]

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

[Out]

-1/81*(675*x^3 - 405*x^2 + 72*(3*x + 2)*log(3*x + 2) - 570*x + 7)/(3*x + 2)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79 \[ \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^2} \, dx=- \frac {25 x^{2}}{9} + \frac {95 x}{27} - \frac {8 \log {\left (3 x + 2 \right )}}{9} - \frac {7}{243 x + 162} \]

[In]

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

[Out]

-25*x**2/9 + 95*x/27 - 8*log(3*x + 2)/9 - 7/(243*x + 162)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^2} \, dx=-\frac {25}{9} \, x^{2} + \frac {95}{27} \, x - \frac {7}{81 \, {\left (3 \, x + 2\right )}} - \frac {8}{9} \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

-25/9*x^2 + 95/27*x - 7/81/(3*x + 2) - 8/9*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.41 \[ \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^2} \, dx=\frac {5}{81} \, {\left (3 \, x + 2\right )}^{2} {\left (\frac {39}{3 \, x + 2} - 5\right )} - \frac {7}{81 \, {\left (3 \, x + 2\right )}} + \frac {8}{9} \, \log \left (\frac {{\left | 3 \, x + 2 \right |}}{3 \, {\left (3 \, x + 2\right )}^{2}}\right ) \]

[In]

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

[Out]

5/81*(3*x + 2)^2*(39/(3*x + 2) - 5) - 7/81/(3*x + 2) + 8/9*log(1/3*abs(3*x + 2)/(3*x + 2)^2)

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^2} \, dx=\frac {95\,x}{27}-\frac {8\,\ln \left (x+\frac {2}{3}\right )}{9}-\frac {7}{243\,\left (x+\frac {2}{3}\right )}-\frac {25\,x^2}{9} \]

[In]

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

[Out]

(95*x)/27 - (8*log(x + 2/3))/9 - 7/(243*(x + 2/3)) - (25*x^2)/9