\(\int \frac {(1-2 x) (3+5 x)}{(2+3 x)^6} \, dx\) [1157]

Optimal result

Integrand size = 18, antiderivative size = 34 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^6} \, dx=\frac {7}{135 (2+3 x)^5}-\frac {37}{108 (2+3 x)^4}+\frac {10}{81 (2+3 x)^3} \]

[Out]

7/135/(2+3*x)^5-37/108/(2+3*x)^4+10/81/(2+3*x)^3

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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {78} \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^6} \, dx=\frac {10}{81 (3 x+2)^3}-\frac {37}{108 (3 x+2)^4}+\frac {7}{135 (3 x+2)^5} \]

[In]

Int[((1 - 2*x)*(3 + 5*x))/(2 + 3*x)^6,x]

[Out]

7/(135*(2 + 3*x)^5) - 37/(108*(2 + 3*x)^4) + 10/(81*(2 + 3*x)^3)

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 {7}{9 (2+3 x)^6}+\frac {37}{9 (2+3 x)^5}-\frac {10}{9 (2+3 x)^4}\right ) \, dx \\ & = \frac {7}{135 (2+3 x)^5}-\frac {37}{108 (2+3 x)^4}+\frac {10}{81 (2+3 x)^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.62 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^6} \, dx=\frac {-226+735 x+1800 x^2}{1620 (2+3 x)^5} \]

[In]

Integrate[((1 - 2*x)*(3 + 5*x))/(2 + 3*x)^6,x]

[Out]

(-226 + 735*x + 1800*x^2)/(1620*(2 + 3*x)^5)

Maple [A] (verified)

Time = 2.13 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.56

method	result	size
norman	\(\frac {\frac {10}{9} x^{2}+\frac {49}{108} x -\frac {113}{810}}{\left (2+3 x \right )^{5}}\)	\(19\)
gosper	\(\frac {1800 x^{2}+735 x -226}{1620 \left (2+3 x \right )^{5}}\)	\(20\)
risch	\(\frac {\frac {10}{9} x^{2}+\frac {49}{108} x -\frac {113}{810}}{\left (2+3 x \right )^{5}}\)	\(20\)
default	\(\frac {7}{135 \left (2+3 x \right )^{5}}-\frac {37}{108 \left (2+3 x \right )^{4}}+\frac {10}{81 \left (2+3 x \right )^{3}}\)	\(29\)
parallelrisch	\(\frac {1017 x^{5}+3390 x^{4}+4520 x^{3}+4080 x^{2}+1440 x}{960 \left (2+3 x \right )^{5}}\)	\(34\)
meijerg	\(\frac {3 x \left (\frac {81}{16} x^{4}+\frac {135}{8} x^{3}+\frac {45}{2} x^{2}+15 x +5\right )}{320 \left (1+\frac {3 x}{2}\right )^{5}}-\frac {x^{2} \left (\frac {27}{8} x^{3}+\frac {45}{4} x^{2}+15 x +10\right )}{1280 \left (1+\frac {3 x}{2}\right )^{5}}-\frac {x^{3} \left (\frac {9}{4} x^{2}+\frac {15}{2} x +10\right )}{192 \left (1+\frac {3 x}{2}\right )^{5}}\)	\(81\)

[In]

int((1-2*x)*(3+5*x)/(2+3*x)^6,x,method=_RETURNVERBOSE)

[Out]

(10/9*x^2+49/108*x-113/810)/(2+3*x)^5

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^6} \, dx=\frac {1800 \, x^{2} + 735 \, x - 226}{1620 \, {\left (243 \, x^{5} + 810 \, x^{4} + 1080 \, x^{3} + 720 \, x^{2} + 240 \, x + 32\right )}} \]

[In]

integrate((1-2*x)*(3+5*x)/(2+3*x)^6,x, algorithm="fricas")

[Out]

1/1620*(1800*x^2 + 735*x - 226)/(243*x^5 + 810*x^4 + 1080*x^3 + 720*x^2 + 240*x + 32)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^6} \, dx=- \frac {- 1800 x^{2} - 735 x + 226}{393660 x^{5} + 1312200 x^{4} + 1749600 x^{3} + 1166400 x^{2} + 388800 x + 51840} \]

[In]

integrate((1-2*x)*(3+5*x)/(2+3*x)**6,x)

[Out]

-(-1800*x**2 - 735*x + 226)/(393660*x**5 + 1312200*x**4 + 1749600*x**3 + 1166400*x**2 + 388800*x + 51840)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^6} \, dx=\frac {1800 \, x^{2} + 735 \, x - 226}{1620 \, {\left (243 \, x^{5} + 810 \, x^{4} + 1080 \, x^{3} + 720 \, x^{2} + 240 \, x + 32\right )}} \]

[In]

integrate((1-2*x)*(3+5*x)/(2+3*x)^6,x, algorithm="maxima")

[Out]

1/1620*(1800*x^2 + 735*x - 226)/(243*x^5 + 810*x^4 + 1080*x^3 + 720*x^2 + 240*x + 32)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.56 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^6} \, dx=\frac {1800 \, x^{2} + 735 \, x - 226}{1620 \, {\left (3 \, x + 2\right )}^{5}} \]

[In]

integrate((1-2*x)*(3+5*x)/(2+3*x)^6,x, algorithm="giac")

[Out]

1/1620*(1800*x^2 + 735*x - 226)/(3*x + 2)^5

Mupad [B] (verification not implemented)

Time = 1.21 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x) (3+5 x)}{(2+3 x)^6} \, dx=\frac {10}{81\,{\left (3\,x+2\right )}^3}-\frac {37}{108\,{\left (3\,x+2\right )}^4}+\frac {7}{135\,{\left (3\,x+2\right )}^5} \]

[In]

int(-((2*x - 1)*(5*x + 3))/(3*x + 2)^6,x)

[Out]

10/(81*(3*x + 2)^3) - 37/(108*(3*x + 2)^4) + 7/(135*(3*x + 2)^5)