\(\int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^5} \, dx\) [1351]

Optimal result

Integrand size = 20, antiderivative size = 55 \[ \int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^5} \, dx=\frac {343}{972 (2+3 x)^4}-\frac {2009}{729 (2+3 x)^3}+\frac {259}{81 (2+3 x)^2}-\frac {428}{243 (2+3 x)}-\frac {40}{243} \log (2+3 x) \]

[Out]

343/972/(2+3*x)^4-2009/729/(2+3*x)^3+259/81/(2+3*x)^2-428/243/(2+3*x)-40/243*ln(2+3*x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 55, 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} \[ \int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^5} \, dx=-\frac {428}{243 (3 x+2)}+\frac {259}{81 (3 x+2)^2}-\frac {2009}{729 (3 x+2)^3}+\frac {343}{972 (3 x+2)^4}-\frac {40}{243} \log (3 x+2) \]

[In]

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

[Out]

343/(972*(2 + 3*x)^4) - 2009/(729*(2 + 3*x)^3) + 259/(81*(2 + 3*x)^2) - 428/(243*(2 + 3*x)) - (40*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 {343}{81 (2+3 x)^5}+\frac {2009}{81 (2+3 x)^4}-\frac {518}{27 (2+3 x)^3}+\frac {428}{81 (2+3 x)^2}-\frac {40}{81 (2+3 x)}\right ) \, dx \\ & = \frac {343}{972 (2+3 x)^4}-\frac {2009}{729 (2+3 x)^3}+\frac {259}{81 (2+3 x)^2}-\frac {428}{243 (2+3 x)}-\frac {40}{243} \log (2+3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.75 \[ \int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^5} \, dx=-\frac {18835+97116 x+193428 x^2+138672 x^3+480 (2+3 x)^4 \log (4+6 x)}{2916 (2+3 x)^4} \]

[In]

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

[Out]

-1/2916*(18835 + 97116*x + 193428*x^2 + 138672*x^3 + 480*(2 + 3*x)^4*Log[4 + 6*x])/(2 + 3*x)^4

Maple [A] (verified)

Time = 2.39 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.62

method	result	size
risch	\(\frac {-\frac {428}{9} x^{3}-\frac {199}{3} x^{2}-\frac {8093}{243} x -\frac {18835}{2916}}{\left (2+3 x \right )^{4}}-\frac {40 \ln \left (2+3 x \right )}{243}\)	\(34\)
norman	\(\frac {\frac {4507}{216} x^{2}+\frac {8563}{216} x^{3}+\frac {883}{162} x +\frac {18835}{576} x^{4}}{\left (2+3 x \right )^{4}}-\frac {40 \ln \left (2+3 x \right )}{243}\)	\(37\)
default	\(\frac {343}{972 \left (2+3 x \right )^{4}}-\frac {2009}{729 \left (2+3 x \right )^{3}}+\frac {259}{81 \left (2+3 x \right )^{2}}-\frac {428}{243 \left (2+3 x \right )}-\frac {40 \ln \left (2+3 x \right )}{243}\)	\(46\)
parallelrisch	\(-\frac {207360 \ln \left (\frac {2}{3}+x \right ) x^{4}+552960 \ln \left (\frac {2}{3}+x \right ) x^{3}-508545 x^{4}+552960 \ln \left (\frac {2}{3}+x \right ) x^{2}-616536 x^{3}+245760 \ln \left (\frac {2}{3}+x \right ) x -324504 x^{2}+40960 \ln \left (\frac {2}{3}+x \right )-84768 x}{15552 \left (2+3 x \right )^{4}}\)	\(69\)
meijerg	\(\frac {3 x \left (\frac {27}{8} x^{3}+9 x^{2}+9 x +4\right )}{128 \left (1+\frac {3 x}{2}\right )^{4}}-\frac {13 x^{2} \left (\frac {9}{4} x^{2}+6 x +6\right )}{384 \left (1+\frac {3 x}{2}\right )^{4}}+\frac {x^{3} \left (\frac {3 x}{2}+4\right )}{64 \left (1+\frac {3 x}{2}\right )^{4}}+\frac {9 x^{4}}{32 \left (1+\frac {3 x}{2}\right )^{4}}+\frac {x \left (\frac {3375}{8} x^{3}+585 x^{2}+315 x +60\right )}{243 \left (1+\frac {3 x}{2}\right )^{4}}-\frac {40 \ln \left (1+\frac {3 x}{2}\right )}{243}\)	\(111\)

[In]

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

[Out]

81*(-428/729*x^3-199/243*x^2-8093/19683*x-18835/236196)/(2+3*x)^4-40/243*ln(2+3*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.22 \[ \int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^5} \, dx=-\frac {138672 \, x^{3} + 193428 \, x^{2} + 480 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )} \log \left (3 \, x + 2\right ) + 97116 \, x + 18835}{2916 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )}} \]

[In]

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

[Out]

-1/2916*(138672*x^3 + 193428*x^2 + 480*(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16)*log(3*x + 2) + 97116*x + 18835
)/(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.84 \[ \int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^5} \, dx=- \frac {138672 x^{3} + 193428 x^{2} + 97116 x + 18835}{236196 x^{4} + 629856 x^{3} + 629856 x^{2} + 279936 x + 46656} - \frac {40 \log {\left (3 x + 2 \right )}}{243} \]

[In]

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

[Out]

-(138672*x**3 + 193428*x**2 + 97116*x + 18835)/(236196*x**4 + 629856*x**3 + 629856*x**2 + 279936*x + 46656) -
40*log(3*x + 2)/243

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.87 \[ \int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^5} \, dx=-\frac {138672 \, x^{3} + 193428 \, x^{2} + 97116 \, x + 18835}{2916 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )}} - \frac {40}{243} \, \log \left (3 \, x + 2\right ) \]

[In]

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

[Out]

-1/2916*(138672*x^3 + 193428*x^2 + 97116*x + 18835)/(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16) - 40/243*log(3*x
+ 2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00 \[ \int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^5} \, dx=-\frac {428}{243 \, {\left (3 \, x + 2\right )}} + \frac {259}{81 \, {\left (3 \, x + 2\right )}^{2}} - \frac {2009}{729 \, {\left (3 \, x + 2\right )}^{3}} + \frac {343}{972 \, {\left (3 \, x + 2\right )}^{4}} + \frac {40}{243} \, \log \left (\frac {{\left | 3 \, x + 2 \right |}}{3 \, {\left (3 \, x + 2\right )}^{2}}\right ) \]

[In]

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

[Out]

-428/243/(3*x + 2) + 259/81/(3*x + 2)^2 - 2009/729/(3*x + 2)^3 + 343/972/(3*x + 2)^4 + 40/243*log(1/3*abs(3*x
+ 2)/(3*x + 2)^2)

Mupad [B] (verification not implemented)

Time = 1.35 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.80 \[ \int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^5} \, dx=-\frac {40\,\ln \left (x+\frac {2}{3}\right )}{243}-\frac {\frac {428\,x^3}{729}+\frac {199\,x^2}{243}+\frac {8093\,x}{19683}+\frac {18835}{236196}}{x^4+\frac {8\,x^3}{3}+\frac {8\,x^2}{3}+\frac {32\,x}{27}+\frac {16}{81}} \]

[In]

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

[Out]

- (40*log(x + 2/3))/243 - ((8093*x)/19683 + (199*x^2)/243 + (428*x^3)/729 + 18835/236196)/((32*x)/27 + (8*x^2)
/3 + (8*x^3)/3 + x^4 + 16/81)