\(\int \frac {(1-2 x)^2 (3+5 x)}{2+3 x} \, dx\) [1245]

Optimal result

Integrand size = 20, antiderivative size = 30 \[ \int \frac {(1-2 x)^2 (3+5 x)}{2+3 x} \, dx=\frac {65 x}{27}-\frac {32 x^2}{9}+\frac {20 x^3}{9}-\frac {49}{81} \log (2+3 x) \]

[Out]

65/27*x-32/9*x^2+20/9*x^3-49/81*ln(2+3*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, 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} \[ \int \frac {(1-2 x)^2 (3+5 x)}{2+3 x} \, dx=\frac {20 x^3}{9}-\frac {32 x^2}{9}+\frac {65 x}{27}-\frac {49}{81} \log (3 x+2) \]

[In]

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

[Out]

(65*x)/27 - (32*x^2)/9 + (20*x^3)/9 - (49*Log[2 + 3*x])/81

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 {65}{27}-\frac {64 x}{9}+\frac {20 x^2}{3}-\frac {49}{27 (2+3 x)}\right ) \, dx \\ & = \frac {65 x}{27}-\frac {32 x^2}{9}+\frac {20 x^3}{9}-\frac {49}{81} \log (2+3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {(1-2 x)^2 (3+5 x)}{2+3 x} \, dx=\frac {1}{243} \left (934+585 x-864 x^2+540 x^3-147 \log (2+3 x)\right ) \]

[In]

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

[Out]

(934 + 585*x - 864*x^2 + 540*x^3 - 147*Log[2 + 3*x])/243

Maple [A] (verified)

Time = 0.74 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.70

method	result	size
parallelrisch	\(\frac {20 x^{3}}{9}-\frac {32 x^{2}}{9}+\frac {65 x}{27}-\frac {49 \ln \left (\frac {2}{3}+x \right )}{81}\)	\(21\)
default	\(\frac {65 x}{27}-\frac {32 x^{2}}{9}+\frac {20 x^{3}}{9}-\frac {49 \ln \left (2+3 x \right )}{81}\)	\(23\)
norman	\(\frac {65 x}{27}-\frac {32 x^{2}}{9}+\frac {20 x^{3}}{9}-\frac {49 \ln \left (2+3 x \right )}{81}\)	\(23\)
risch	\(\frac {65 x}{27}-\frac {32 x^{2}}{9}+\frac {20 x^{3}}{9}-\frac {49 \ln \left (2+3 x \right )}{81}\)	\(23\)
meijerg	\(-\frac {49 \ln \left (1+\frac {3 x}{2}\right )}{81}-\frac {7 x}{3}+\frac {8 x \left (-\frac {9 x}{2}+6\right )}{27}+\frac {20 x \left (9 x^{2}-9 x +12\right )}{81}\)	\(34\)

[In]

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

[Out]

20/9*x^3-32/9*x^2+65/27*x-49/81*ln(2/3+x)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {(1-2 x)^2 (3+5 x)}{2+3 x} \, dx=\frac {20}{9} \, x^{3} - \frac {32}{9} \, x^{2} + \frac {65}{27} \, x - \frac {49}{81} \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

20/9*x^3 - 32/9*x^2 + 65/27*x - 49/81*log(3*x + 2)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {(1-2 x)^2 (3+5 x)}{2+3 x} \, dx=\frac {20 x^{3}}{9} - \frac {32 x^{2}}{9} + \frac {65 x}{27} - \frac {49 \log {\left (3 x + 2 \right )}}{81} \]

[In]

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

[Out]

20*x**3/9 - 32*x**2/9 + 65*x/27 - 49*log(3*x + 2)/81

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {(1-2 x)^2 (3+5 x)}{2+3 x} \, dx=\frac {20}{9} \, x^{3} - \frac {32}{9} \, x^{2} + \frac {65}{27} \, x - \frac {49}{81} \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

20/9*x^3 - 32/9*x^2 + 65/27*x - 49/81*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {(1-2 x)^2 (3+5 x)}{2+3 x} \, dx=\frac {20}{9} \, x^{3} - \frac {32}{9} \, x^{2} + \frac {65}{27} \, x - \frac {49}{81} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \]

[In]

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

[Out]

20/9*x^3 - 32/9*x^2 + 65/27*x - 49/81*log(abs(3*x + 2))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67 \[ \int \frac {(1-2 x)^2 (3+5 x)}{2+3 x} \, dx=\frac {65\,x}{27}-\frac {49\,\ln \left (x+\frac {2}{3}\right )}{81}-\frac {32\,x^2}{9}+\frac {20\,x^3}{9} \]

[In]

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

[Out]

(65*x)/27 - (49*log(x + 2/3))/81 - (32*x^2)/9 + (20*x^3)/9