\(\int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^3} \, dx\) [1169]

Optimal result

Integrand size = 20, antiderivative size = 38 \[ \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^3} \, dx=-\frac {50 x}{27}-\frac {7}{162 (2+3 x)^2}+\frac {8}{9 (2+3 x)}+\frac {65}{27} \log (2+3 x) \]

[Out]

-50/27*x-7/162/(2+3*x)^2+8/9/(2+3*x)+65/27*ln(2+3*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, 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} \[ \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^3} \, dx=-\frac {50 x}{27}+\frac {8}{9 (3 x+2)}-\frac {7}{162 (3 x+2)^2}+\frac {65}{27} \log (3 x+2) \]

[In]

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

[Out]

(-50*x)/27 - 7/(162*(2 + 3*x)^2) + 8/(9*(2 + 3*x)) + (65*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 {50}{27}+\frac {7}{27 (2+3 x)^3}-\frac {8}{3 (2+3 x)^2}+\frac {65}{9 (2+3 x)}\right ) \, dx \\ & = -\frac {50 x}{27}-\frac {7}{162 (2+3 x)^2}+\frac {8}{9 (2+3 x)}+\frac {65}{27} \log (2+3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.92 \[ \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^3} \, dx=\frac {1}{162} \left (-300 x-\frac {3 \left (173+656 x+600 x^2\right )}{(2+3 x)^2}+390 \log (2+3 x)\right ) \]

[In]

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

[Out]

(-300*x - (3*(173 + 656*x + 600*x^2))/(2 + 3*x)^2 + 390*Log[2 + 3*x])/162

Maple [A] (verified)

Time = 3.56 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.71

method	result	size
risch	\(-\frac {50 x}{27}+\frac {\frac {8 x}{3}+\frac {281}{162}}{\left (2+3 x \right )^{2}}+\frac {65 \ln \left (2+3 x \right )}{27}\)	\(27\)
default	\(-\frac {50 x}{27}-\frac {7}{162 \left (2+3 x \right )^{2}}+\frac {8}{9 \left (2+3 x \right )}+\frac {65 \ln \left (2+3 x \right )}{27}\)	\(31\)
norman	\(\frac {-\frac {179}{18} x -\frac {209}{8} x^{2}-\frac {50}{3} x^{3}}{\left (2+3 x \right )^{2}}+\frac {65 \ln \left (2+3 x \right )}{27}\)	\(32\)
parallelrisch	\(\frac {4680 \ln \left (\frac {2}{3}+x \right ) x^{2}-3600 x^{3}+6240 \ln \left (\frac {2}{3}+x \right ) x -5643 x^{2}+2080 \ln \left (\frac {2}{3}+x \right )-2148 x}{216 \left (2+3 x \right )^{2}}\)	\(46\)
meijerg	\(\frac {9 x \left (\frac {3 x}{2}+2\right )}{16 \left (1+\frac {3 x}{2}\right )^{2}}+\frac {3 x^{2}}{4 \left (1+\frac {3 x}{2}\right )^{2}}+\frac {35 x \left (\frac {27 x}{2}+6\right )}{108 \left (1+\frac {3 x}{2}\right )^{2}}+\frac {65 \ln \left (1+\frac {3 x}{2}\right )}{27}-\frac {25 x \left (9 x^{2}+27 x +12\right )}{54 \left (1+\frac {3 x}{2}\right )^{2}}\)	\(72\)

[In]

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

[Out]

-50/27*x+9*(8/27*x+281/1458)/(2+3*x)^2+65/27*ln(2+3*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.24 \[ \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^3} \, dx=-\frac {2700 \, x^{3} + 3600 \, x^{2} - 390 \, {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (3 \, x + 2\right ) + 768 \, x - 281}{162 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \]

[In]

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

[Out]

-1/162*(2700*x^3 + 3600*x^2 - 390*(9*x^2 + 12*x + 4)*log(3*x + 2) + 768*x - 281)/(9*x^2 + 12*x + 4)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^3} \, dx=- \frac {50 x}{27} - \frac {- 432 x - 281}{1458 x^{2} + 1944 x + 648} + \frac {65 \log {\left (3 x + 2 \right )}}{27} \]

[In]

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

[Out]

-50*x/27 - (-432*x - 281)/(1458*x**2 + 1944*x + 648) + 65*log(3*x + 2)/27

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^3} \, dx=-\frac {50}{27} \, x + \frac {432 \, x + 281}{162 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} + \frac {65}{27} \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

-50/27*x + 1/162*(432*x + 281)/(9*x^2 + 12*x + 4) + 65/27*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.71 \[ \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^3} \, dx=-\frac {50}{27} \, x + \frac {432 \, x + 281}{162 \, {\left (3 \, x + 2\right )}^{2}} + \frac {65}{27} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \]

[In]

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

[Out]

-50/27*x + 1/162*(432*x + 281)/(3*x + 2)^2 + 65/27*log(abs(3*x + 2))

Mupad [B] (verification not implemented)

Time = 1.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.68 \[ \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^3} \, dx=\frac {65\,\ln \left (x+\frac {2}{3}\right )}{27}-\frac {50\,x}{27}+\frac {\frac {8\,x}{27}+\frac {281}{1458}}{x^2+\frac {4\,x}{3}+\frac {4}{9}} \]

[In]

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

[Out]

(65*log(x + 2/3))/27 - (50*x)/27 + ((8*x)/27 + 281/1458)/((4*x)/3 + x^2 + 4/9)