\(\int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^2} \, dx\) [1246]

Optimal result

Integrand size = 20, antiderivative size = 34 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^2} \, dx=-\frac {104 x}{27}+\frac {10 x^2}{9}+\frac {49}{81 (2+3 x)}+\frac {91}{27} \log (2+3 x) \]

[Out]

-104/27*x+10/9*x^2+49/81/(2+3*x)+91/27*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} \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^2} \, dx=\frac {10 x^2}{9}-\frac {104 x}{27}+\frac {49}{81 (3 x+2)}+\frac {91}{27} \log (3 x+2) \]

[In]

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

[Out]

(-104*x)/27 + (10*x^2)/9 + 49/(81*(2 + 3*x)) + (91*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 {104}{27}+\frac {20 x}{9}-\frac {49}{27 (2+3 x)^2}+\frac {91}{9 (2+3 x)}\right ) \, dx \\ & = -\frac {104 x}{27}+\frac {10 x^2}{9}+\frac {49}{81 (2+3 x)}+\frac {91}{27} \log (2+3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^2} \, dx=\frac {632-447 x-1512 x^2+540 x^3+546 (2+3 x) \log (4+6 x)}{162 (2+3 x)} \]

[In]

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

[Out]

(632 - 447*x - 1512*x^2 + 540*x^3 + 546*(2 + 3*x)*Log[4 + 6*x])/(162*(2 + 3*x))

Maple [A] (verified)

Time = 2.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74

method	result	size
risch	\(\frac {10 x^{2}}{9}-\frac {104 x}{27}+\frac {49}{243 \left (\frac {2}{3}+x \right )}+\frac {91 \ln \left (2+3 x \right )}{27}\)	\(25\)
default	\(-\frac {104 x}{27}+\frac {10 x^{2}}{9}+\frac {49}{81 \left (2+3 x \right )}+\frac {91 \ln \left (2+3 x \right )}{27}\)	\(27\)
norman	\(\frac {-\frac {155}{18} x -\frac {28}{3} x^{2}+\frac {10}{3} x^{3}}{2+3 x}+\frac {91 \ln \left (2+3 x \right )}{27}\)	\(32\)
parallelrisch	\(\frac {180 x^{3}+546 \ln \left (\frac {2}{3}+x \right ) x -504 x^{2}+364 \ln \left (\frac {2}{3}+x \right )-465 x}{108+162 x}\)	\(37\)
meijerg	\(\frac {23 x}{12 \left (1+\frac {3 x}{2}\right )}+\frac {91 \ln \left (1+\frac {3 x}{2}\right )}{27}-\frac {8 x \left (\frac {9 x}{2}+6\right )}{27 \left (1+\frac {3 x}{2}\right )}-\frac {10 x \left (-\frac {9}{2} x^{2}+9 x +12\right )}{27 \left (1+\frac {3 x}{2}\right )}\)	\(55\)

[In]

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

[Out]

10/9*x^2-104/27*x+49/243/(2/3+x)+91/27*ln(2+3*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^2} \, dx=\frac {270 \, x^{3} - 756 \, x^{2} + 273 \, {\left (3 \, x + 2\right )} \log \left (3 \, x + 2\right ) - 624 \, x + 49}{81 \, {\left (3 \, x + 2\right )}} \]

[In]

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

[Out]

1/81*(270*x^3 - 756*x^2 + 273*(3*x + 2)*log(3*x + 2) - 624*x + 49)/(3*x + 2)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^2} \, dx=\frac {10 x^{2}}{9} - \frac {104 x}{27} + \frac {91 \log {\left (3 x + 2 \right )}}{27} + \frac {49}{243 x + 162} \]

[In]

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

[Out]

10*x**2/9 - 104*x/27 + 91*log(3*x + 2)/27 + 49/(243*x + 162)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^2} \, dx=\frac {10}{9} \, x^{2} - \frac {104}{27} \, x + \frac {49}{81 \, {\left (3 \, x + 2\right )}} + \frac {91}{27} \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

10/9*x^2 - 104/27*x + 49/81/(3*x + 2) + 91/27*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.41 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^2} \, dx=-\frac {2}{81} \, {\left (3 \, x + 2\right )}^{2} {\left (\frac {72}{3 \, x + 2} - 5\right )} + \frac {49}{81 \, {\left (3 \, x + 2\right )}} - \frac {91}{27} \, \log \left (\frac {{\left | 3 \, x + 2 \right |}}{3 \, {\left (3 \, x + 2\right )}^{2}}\right ) \]

[In]

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

[Out]

-2/81*(3*x + 2)^2*(72/(3*x + 2) - 5) + 49/81/(3*x + 2) - 91/27*log(1/3*abs(3*x + 2)/(3*x + 2)^2)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^2} \, dx=\frac {91\,\ln \left (x+\frac {2}{3}\right )}{27}-\frac {104\,x}{27}+\frac {49}{243\,\left (x+\frac {2}{3}\right )}+\frac {10\,x^2}{9} \]

[In]

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

[Out]

(91*log(x + 2/3))/27 - (104*x)/27 + 49/(243*(x + 2/3)) + (10*x^2)/9