\(\int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^4} \, dx\) [1350]

Optimal result

Integrand size = 20, antiderivative size = 49 \[ \int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^4} \, dx=-\frac {40 x}{81}+\frac {343}{729 (2+3 x)^3}-\frac {2009}{486 (2+3 x)^2}+\frac {518}{81 (2+3 x)}+\frac {428}{243} \log (2+3 x) \]

[Out]

-40/81*x+343/729/(2+3*x)^3-2009/486/(2+3*x)^2+518/81/(2+3*x)+428/243*ln(2+3*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 49, 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} \[ \int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^4} \, dx=-\frac {40 x}{81}+\frac {518}{81 (3 x+2)}-\frac {2009}{486 (3 x+2)^2}+\frac {343}{729 (3 x+2)^3}+\frac {428}{243} \log (3 x+2) \]

[In]

Int[((1 - 2*x)^3*(3 + 5*x))/(2 + 3*x)^4,x]

[Out]

(-40*x)/81 + 343/(729*(2 + 3*x)^3) - 2009/(486*(2 + 3*x)^2) + 518/(81*(2 + 3*x)) + (428*Log[2 + 3*x])/243

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 {40}{81}-\frac {343}{81 (2+3 x)^4}+\frac {2009}{81 (2+3 x)^3}-\frac {518}{27 (2+3 x)^2}+\frac {428}{81 (2+3 x)}\right ) \, dx \\ & = -\frac {40 x}{81}+\frac {343}{729 (2+3 x)^3}-\frac {2009}{486 (2+3 x)^2}+\frac {518}{81 (2+3 x)}+\frac {428}{243} \log (2+3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.94 \[ \int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^4} \, dx=\frac {22088+70767 x+32076 x^2-51840 x^3-19440 x^4+2568 (2+3 x)^3 \log (2+3 x)}{1458 (2+3 x)^3} \]

[In]

Integrate[((1 - 2*x)^3*(3 + 5*x))/(2 + 3*x)^4,x]

[Out]

(22088 + 70767*x + 32076*x^2 - 51840*x^3 - 19440*x^4 + 2568*(2 + 3*x)^3*Log[2 + 3*x])/(1458*(2 + 3*x)^3)

Maple [A] (verified)

Time = 2.38 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.65

method	result	size
risch	\(-\frac {40 x}{81}+\frac {\frac {518}{9} x^{2}+\frac {10423}{162} x +\frac {12964}{729}}{\left (2+3 x \right )^{3}}+\frac {428 \ln \left (2+3 x \right )}{243}\)	\(32\)
norman	\(\frac {-\frac {3181}{162} x -\frac {2167}{27} x^{2}-\frac {4681}{54} x^{3}-\frac {40}{3} x^{4}}{\left (2+3 x \right )^{3}}+\frac {428 \ln \left (2+3 x \right )}{243}\)	\(37\)
default	\(-\frac {40 x}{81}+\frac {343}{729 \left (2+3 x \right )^{3}}-\frac {2009}{486 \left (2+3 x \right )^{2}}+\frac {518}{81 \left (2+3 x \right )}+\frac {428 \ln \left (2+3 x \right )}{243}\)	\(40\)
parallelrisch	\(\frac {92448 \ln \left (\frac {2}{3}+x \right ) x^{3}-25920 x^{4}+184896 \ln \left (\frac {2}{3}+x \right ) x^{2}-168516 x^{3}+123264 \ln \left (\frac {2}{3}+x \right ) x -156024 x^{2}+27392 \ln \left (\frac {2}{3}+x \right )-38172 x}{1944 \left (2+3 x \right )^{3}}\)	\(60\)
meijerg	\(\frac {x \left (\frac {9}{4} x^{2}+\frac {9}{2} x +3\right )}{16 \left (1+\frac {3 x}{2}\right )^{3}}-\frac {13 x^{2} \left (3+\frac {3 x}{2}\right )}{96 \left (1+\frac {3 x}{2}\right )^{3}}+\frac {x^{3}}{8 \left (1+\frac {3 x}{2}\right )^{3}}-\frac {x \left (\frac {99}{2} x^{2}+45 x +12\right )}{18 \left (1+\frac {3 x}{2}\right )^{3}}+\frac {428 \ln \left (1+\frac {3 x}{2}\right )}{243}-\frac {8 x \left (\frac {405}{8} x^{3}+\frac {495}{2} x^{2}+225 x +60\right )}{243 \left (1+\frac {3 x}{2}\right )^{3}}\)	\(104\)

[In]

int((1-2*x)^3*(3+5*x)/(2+3*x)^4,x,method=_RETURNVERBOSE)

[Out]

-40/81*x+27*(518/243*x^2+10423/4374*x+12964/19683)/(2+3*x)^3+428/243*ln(2+3*x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.27 \[ \int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^4} \, dx=-\frac {19440 \, x^{4} + 38880 \, x^{3} - 57996 \, x^{2} - 2568 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )} \log \left (3 \, x + 2\right ) - 88047 \, x - 25928}{1458 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} \]

[In]

integrate((1-2*x)^3*(3+5*x)/(2+3*x)^4,x, algorithm="fricas")

[Out]

-1/1458*(19440*x^4 + 38880*x^3 - 57996*x^2 - 2568*(27*x^3 + 54*x^2 + 36*x + 8)*log(3*x + 2) - 88047*x - 25928)
/(27*x^3 + 54*x^2 + 36*x + 8)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.84 \[ \int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^4} \, dx=- \frac {40 x}{81} - \frac {- 83916 x^{2} - 93807 x - 25928}{39366 x^{3} + 78732 x^{2} + 52488 x + 11664} + \frac {428 \log {\left (3 x + 2 \right )}}{243} \]

[In]

integrate((1-2*x)**3*(3+5*x)/(2+3*x)**4,x)

[Out]

-40*x/81 - (-83916*x**2 - 93807*x - 25928)/(39366*x**3 + 78732*x**2 + 52488*x + 11664) + 428*log(3*x + 2)/243

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.84 \[ \int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^4} \, dx=-\frac {40}{81} \, x + \frac {7 \, {\left (11988 \, x^{2} + 13401 \, x + 3704\right )}}{1458 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} + \frac {428}{243} \, \log \left (3 \, x + 2\right ) \]

[In]

integrate((1-2*x)^3*(3+5*x)/(2+3*x)^4,x, algorithm="maxima")

[Out]

-40/81*x + 7/1458*(11988*x^2 + 13401*x + 3704)/(27*x^3 + 54*x^2 + 36*x + 8) + 428/243*log(3*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.65 \[ \int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^4} \, dx=-\frac {40}{81} \, x + \frac {7 \, {\left (11988 \, x^{2} + 13401 \, x + 3704\right )}}{1458 \, {\left (3 \, x + 2\right )}^{3}} + \frac {428}{243} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \]

[In]

integrate((1-2*x)^3*(3+5*x)/(2+3*x)^4,x, algorithm="giac")

[Out]

-40/81*x + 7/1458*(11988*x^2 + 13401*x + 3704)/(3*x + 2)^3 + 428/243*log(abs(3*x + 2))

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.73 \[ \int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^4} \, dx=\frac {428\,\ln \left (x+\frac {2}{3}\right )}{243}-\frac {40\,x}{81}+\frac {\frac {518\,x^2}{243}+\frac {10423\,x}{4374}+\frac {12964}{19683}}{x^3+2\,x^2+\frac {4\,x}{3}+\frac {8}{27}} \]

[In]

int(-((2*x - 1)^3*(5*x + 3))/(3*x + 2)^4,x)

[Out]

(428*log(x + 2/3))/243 - (40*x)/81 + ((10423*x)/4374 + (518*x^2)/243 + 12964/19683)/((4*x)/3 + 2*x^2 + x^3 + 8
/27)