\(\int \frac {(1-2 x) (3+5 x)}{(2+3 x)^2} \, dx\) [1153]

Optimal result

Integrand size = 18, antiderivative size = 27 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^2} \, dx=-\frac {10 x}{9}+\frac {7}{27 (2+3 x)}+\frac {37}{27} \log (2+3 x) \]

[Out]

-10/9*x+7/27/(2+3*x)+37/27*ln(2+3*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, 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} \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^2} \, dx=-\frac {10 x}{9}+\frac {7}{27 (3 x+2)}+\frac {37}{27} \log (3 x+2) \]

[In]

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

[Out]

(-10*x)/9 + 7/(27*(2 + 3*x)) + (37*Log[2 + 3*x])/27

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 {10}{9}-\frac {7}{9 (2+3 x)^2}+\frac {37}{9 (2+3 x)}\right ) \, dx \\ & = -\frac {10 x}{9}+\frac {7}{27 (2+3 x)}+\frac {37}{27} \log (2+3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^2} \, dx=\frac {1}{27} \left (-20-30 x+\frac {7}{2+3 x}+37 \log (2+3 x)\right ) \]

[In]

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

[Out]

(-20 - 30*x + 7/(2 + 3*x) + 37*Log[2 + 3*x])/27

Maple [A] (verified)

Time = 2.15 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74

method	result	size
risch	\(-\frac {10 x}{9}+\frac {7}{81 \left (\frac {2}{3}+x \right )}+\frac {37 \ln \left (2+3 x \right )}{27}\)	\(20\)
default	\(-\frac {10 x}{9}+\frac {7}{27 \left (2+3 x \right )}+\frac {37 \ln \left (2+3 x \right )}{27}\)	\(22\)
norman	\(\frac {-\frac {47}{18} x -\frac {10}{3} x^{2}}{2+3 x}+\frac {37 \ln \left (2+3 x \right )}{27}\)	\(27\)
parallelrisch	\(\frac {222 \ln \left (\frac {2}{3}+x \right ) x -180 x^{2}+148 \ln \left (\frac {2}{3}+x \right )-141 x}{108+162 x}\)	\(32\)
meijerg	\(\frac {11 x}{12 \left (1+\frac {3 x}{2}\right )}+\frac {37 \ln \left (1+\frac {3 x}{2}\right )}{27}-\frac {10 x \left (\frac {9 x}{2}+6\right )}{27 \left (1+\frac {3 x}{2}\right )}\)	\(35\)

[In]

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

[Out]

-10/9*x+7/81/(2/3+x)+37/27*ln(2+3*x)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^2} \, dx=-\frac {90 \, x^{2} - 37 \, {\left (3 \, x + 2\right )} \log \left (3 \, x + 2\right ) + 60 \, x - 7}{27 \, {\left (3 \, x + 2\right )}} \]

[In]

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

[Out]

-1/27*(90*x^2 - 37*(3*x + 2)*log(3*x + 2) + 60*x - 7)/(3*x + 2)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^2} \, dx=- \frac {10 x}{9} + \frac {37 \log {\left (3 x + 2 \right )}}{27} + \frac {7}{81 x + 54} \]

[In]

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

[Out]

-10*x/9 + 37*log(3*x + 2)/27 + 7/(81*x + 54)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^2} \, dx=-\frac {10}{9} \, x + \frac {7}{27 \, {\left (3 \, x + 2\right )}} + \frac {37}{27} \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

-10/9*x + 7/27/(3*x + 2) + 37/27*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^2} \, dx=-\frac {10}{9} \, x + \frac {7}{27 \, {\left (3 \, x + 2\right )}} - \frac {37}{27} \, \log \left (\frac {{\left | 3 \, x + 2 \right |}}{3 \, {\left (3 \, x + 2\right )}^{2}}\right ) - \frac {20}{27} \]

[In]

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

[Out]

-10/9*x + 7/27/(3*x + 2) - 37/27*log(1/3*abs(3*x + 2)/(3*x + 2)^2) - 20/27

Mupad [B] (verification not implemented)

Time = 1.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^2} \, dx=\frac {37\,\ln \left (x+\frac {2}{3}\right )}{27}-\frac {10\,x}{9}+\frac {7}{81\,\left (x+\frac {2}{3}\right )} \]

[In]

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

[Out]

(37*log(x + 2/3))/27 - (10*x)/9 + 7/(81*(x + 2/3))