\(\int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^6} \, dx\) [1250]

Optimal result

Integrand size = 20, antiderivative size = 45 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^6} \, dx=\frac {49}{405 (2+3 x)^5}-\frac {91}{108 (2+3 x)^4}+\frac {16}{27 (2+3 x)^3}-\frac {10}{81 (2+3 x)^2} \]

[Out]

49/405/(2+3*x)^5-91/108/(2+3*x)^4+16/27/(2+3*x)^3-10/81/(2+3*x)^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 45, 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} \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^6} \, dx=-\frac {10}{81 (3 x+2)^2}+\frac {16}{27 (3 x+2)^3}-\frac {91}{108 (3 x+2)^4}+\frac {49}{405 (3 x+2)^5} \]

[In]

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

[Out]

49/(405*(2 + 3*x)^5) - 91/(108*(2 + 3*x)^4) + 16/(27*(2 + 3*x)^3) - 10/(81*(2 + 3*x)^2)

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 {49}{27 (2+3 x)^6}+\frac {91}{9 (2+3 x)^5}-\frac {16}{3 (2+3 x)^4}+\frac {20}{27 (2+3 x)^3}\right ) \, dx \\ & = \frac {49}{405 (2+3 x)^5}-\frac {91}{108 (2+3 x)^4}+\frac {16}{27 (2+3 x)^3}-\frac {10}{81 (2+3 x)^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.58 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^6} \, dx=-\frac {98-75 x+720 x^2+1800 x^3}{540 (2+3 x)^5} \]

[In]

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

[Out]

-1/540*(98 - 75*x + 720*x^2 + 1800*x^3)/(2 + 3*x)^5

Maple [A] (verified)

Time = 2.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.53

method	result	size
norman	\(\frac {-\frac {10}{3} x^{3}-\frac {4}{3} x^{2}+\frac {5}{36} x -\frac {49}{270}}{\left (2+3 x \right )^{5}}\)	\(24\)
gosper	\(-\frac {1800 x^{3}+720 x^{2}-75 x +98}{540 \left (2+3 x \right )^{5}}\)	\(25\)
risch	\(\frac {-\frac {10}{3} x^{3}-\frac {4}{3} x^{2}+\frac {5}{36} x -\frac {49}{270}}{\left (2+3 x \right )^{5}}\)	\(25\)
parallelrisch	\(\frac {1323 x^{5}+4410 x^{4}+2680 x^{3}+2640 x^{2}+1440 x}{960 \left (2+3 x \right )^{5}}\)	\(34\)
default	\(\frac {49}{405 \left (2+3 x \right )^{5}}-\frac {91}{108 \left (2+3 x \right )^{4}}+\frac {16}{27 \left (2+3 x \right )^{3}}-\frac {10}{81 \left (2+3 x \right )^{2}}\)	\(38\)
meijerg	\(\frac {3 x \left (\frac {81}{16} x^{4}+\frac {135}{8} x^{3}+\frac {45}{2} x^{2}+15 x +5\right )}{320 \left (1+\frac {3 x}{2}\right )^{5}}-\frac {7 x^{2} \left (\frac {27}{8} x^{3}+\frac {45}{4} x^{2}+15 x +10\right )}{1280 \left (1+\frac {3 x}{2}\right )^{5}}-\frac {x^{3} \left (\frac {9}{4} x^{2}+\frac {15}{2} x +10\right )}{240 \left (1+\frac {3 x}{2}\right )^{5}}+\frac {x^{4} \left (\frac {3 x}{2}+5\right )}{64 \left (1+\frac {3 x}{2}\right )^{5}}\)	\(98\)

[In]

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

[Out]

(-10/3*x^3-4/3*x^2+5/36*x-49/270)/(2+3*x)^5

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.98 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^6} \, dx=-\frac {1800 \, x^{3} + 720 \, x^{2} - 75 \, x + 98}{540 \, {\left (243 \, x^{5} + 810 \, x^{4} + 1080 \, x^{3} + 720 \, x^{2} + 240 \, x + 32\right )}} \]

[In]

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

[Out]

-1/540*(1800*x^3 + 720*x^2 - 75*x + 98)/(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 = 39, normalized size of antiderivative = 0.87 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^6} \, dx=\frac {- 1800 x^{3} - 720 x^{2} + 75 x - 98}{131220 x^{5} + 437400 x^{4} + 583200 x^{3} + 388800 x^{2} + 129600 x + 17280} \]

[In]

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

[Out]

(-1800*x**3 - 720*x**2 + 75*x - 98)/(131220*x**5 + 437400*x**4 + 583200*x**3 + 388800*x**2 + 129600*x + 17280)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.98 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^6} \, dx=-\frac {1800 \, x^{3} + 720 \, x^{2} - 75 \, x + 98}{540 \, {\left (243 \, x^{5} + 810 \, x^{4} + 1080 \, x^{3} + 720 \, x^{2} + 240 \, x + 32\right )}} \]

[In]

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

[Out]

-1/540*(1800*x^3 + 720*x^2 - 75*x + 98)/(243*x^5 + 810*x^4 + 1080*x^3 + 720*x^2 + 240*x + 32)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.53 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^6} \, dx=-\frac {1800 \, x^{3} + 720 \, x^{2} - 75 \, x + 98}{540 \, {\left (3 \, x + 2\right )}^{5}} \]

[In]

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

[Out]

-1/540*(1800*x^3 + 720*x^2 - 75*x + 98)/(3*x + 2)^5

Mupad [B] (verification not implemented)

Time = 1.31 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^6} \, dx=\frac {16}{27\,{\left (3\,x+2\right )}^3}-\frac {10}{81\,{\left (3\,x+2\right )}^2}-\frac {91}{108\,{\left (3\,x+2\right )}^4}+\frac {49}{405\,{\left (3\,x+2\right )}^5} \]

[In]

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

[Out]

16/(27*(3*x + 2)^3) - 10/(81*(3*x + 2)^2) - 91/(108*(3*x + 2)^4) + 49/(405*(3*x + 2)^5)