\(\int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^7} \, dx\) [1251]

Optimal result

Integrand size = 20, antiderivative size = 45 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^7} \, dx=\frac {49}{486 (2+3 x)^6}-\frac {91}{135 (2+3 x)^5}+\frac {4}{9 (2+3 x)^4}-\frac {20}{243 (2+3 x)^3} \]

[Out]

49/486/(2+3*x)^6-91/135/(2+3*x)^5+4/9/(2+3*x)^4-20/243/(2+3*x)^3

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} \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^7} \, dx=-\frac {20}{243 (3 x+2)^3}+\frac {4}{9 (3 x+2)^4}-\frac {91}{135 (3 x+2)^5}+\frac {49}{486 (3 x+2)^6} \]

[In]

Int[((1 - 2*x)^2*(3 + 5*x))/(2 + 3*x)^7,x]

[Out]

49/(486*(2 + 3*x)^6) - 91/(135*(2 + 3*x)^5) + 4/(9*(2 + 3*x)^4) - 20/(243*(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 {49}{27 (2+3 x)^7}+\frac {91}{9 (2+3 x)^6}-\frac {16}{3 (2+3 x)^5}+\frac {20}{27 (2+3 x)^4}\right ) \, dx \\ & = \frac {49}{486 (2+3 x)^6}-\frac {91}{135 (2+3 x)^5}+\frac {4}{9 (2+3 x)^4}-\frac {20}{243 (2+3 x)^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.58 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^7} \, dx=-\frac {311-846 x+1080 x^2+5400 x^3}{2430 (2+3 x)^6} \]

[In]

Integrate[((1 - 2*x)^2*(3 + 5*x))/(2 + 3*x)^7,x]

[Out]

-1/2430*(311 - 846*x + 1080*x^2 + 5400*x^3)/(2 + 3*x)^6

Maple [A] (verified)

Time = 0.74 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.53

method	result	size
norman	\(\frac {-\frac {20}{9} x^{3}-\frac {4}{9} x^{2}+\frac {47}{135} x -\frac {311}{2430}}{\left (2+3 x \right )^{6}}\)	\(24\)
gosper	\(-\frac {5400 x^{3}+1080 x^{2}-846 x +311}{2430 \left (2+3 x \right )^{6}}\)	\(25\)
risch	\(\frac {-\frac {20}{9} x^{3}-\frac {4}{9} x^{2}+\frac {47}{135} x -\frac {311}{2430}}{\left (2+3 x \right )^{6}}\)	\(25\)
default	\(\frac {49}{486 \left (2+3 x \right )^{6}}-\frac {91}{135 \left (2+3 x \right )^{5}}+\frac {4}{9 \left (2+3 x \right )^{4}}-\frac {20}{243 \left (2+3 x \right )^{3}}\)	\(38\)
parallelrisch	\(\frac {2799 x^{6}+11196 x^{5}+18660 x^{4}+12320 x^{3}+7440 x^{2}+2880 x}{1920 \left (2+3 x \right )^{6}}\)	\(39\)
meijerg	\(\frac {x \left (\frac {243}{32} x^{5}+\frac {243}{8} x^{4}+\frac {405}{8} x^{3}+45 x^{2}+\frac {45}{2} x +6\right )}{256 \left (1+\frac {3 x}{2}\right )^{6}}-\frac {7 x^{2} \left (\frac {81}{16} x^{4}+\frac {81}{4} x^{3}+\frac {135}{4} x^{2}+30 x +15\right )}{3840 \left (1+\frac {3 x}{2}\right )^{6}}-\frac {x^{3} \left (\frac {27}{8} x^{3}+\frac {27}{2} x^{2}+\frac {45}{2} x +20\right )}{960 \left (1+\frac {3 x}{2}\right )^{6}}+\frac {x^{4} \left (\frac {9}{4} x^{2}+9 x +15\right )}{384 \left (1+\frac {3 x}{2}\right )^{6}}\)	\(118\)

[In]

int((1-2*x)^2*(3+5*x)/(2+3*x)^7,x,method=_RETURNVERBOSE)

[Out]

(-20/9*x^3-4/9*x^2+47/135*x-311/2430)/(2+3*x)^6

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.09 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^7} \, dx=-\frac {5400 \, x^{3} + 1080 \, x^{2} - 846 \, x + 311}{2430 \, {\left (729 \, x^{6} + 2916 \, x^{5} + 4860 \, x^{4} + 4320 \, x^{3} + 2160 \, x^{2} + 576 \, x + 64\right )}} \]

[In]

integrate((1-2*x)^2*(3+5*x)/(2+3*x)^7,x, algorithm="fricas")

[Out]

-1/2430*(5400*x^3 + 1080*x^2 - 846*x + 311)/(729*x^6 + 2916*x^5 + 4860*x^4 + 4320*x^3 + 2160*x^2 + 576*x + 64)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.98 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^7} \, dx=\frac {- 5400 x^{3} - 1080 x^{2} + 846 x - 311}{1771470 x^{6} + 7085880 x^{5} + 11809800 x^{4} + 10497600 x^{3} + 5248800 x^{2} + 1399680 x + 155520} \]

[In]

integrate((1-2*x)**2*(3+5*x)/(2+3*x)**7,x)

[Out]

(-5400*x**3 - 1080*x**2 + 846*x - 311)/(1771470*x**6 + 7085880*x**5 + 11809800*x**4 + 10497600*x**3 + 5248800*
x**2 + 1399680*x + 155520)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.09 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^7} \, dx=-\frac {5400 \, x^{3} + 1080 \, x^{2} - 846 \, x + 311}{2430 \, {\left (729 \, x^{6} + 2916 \, x^{5} + 4860 \, x^{4} + 4320 \, x^{3} + 2160 \, x^{2} + 576 \, x + 64\right )}} \]

[In]

integrate((1-2*x)^2*(3+5*x)/(2+3*x)^7,x, algorithm="maxima")

[Out]

-1/2430*(5400*x^3 + 1080*x^2 - 846*x + 311)/(729*x^6 + 2916*x^5 + 4860*x^4 + 4320*x^3 + 2160*x^2 + 576*x + 64)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.53 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^7} \, dx=-\frac {5400 \, x^{3} + 1080 \, x^{2} - 846 \, x + 311}{2430 \, {\left (3 \, x + 2\right )}^{6}} \]

[In]

integrate((1-2*x)^2*(3+5*x)/(2+3*x)^7,x, algorithm="giac")

[Out]

-1/2430*(5400*x^3 + 1080*x^2 - 846*x + 311)/(3*x + 2)^6

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^7} \, dx=\frac {4}{9\,{\left (3\,x+2\right )}^4}-\frac {20}{243\,{\left (3\,x+2\right )}^3}-\frac {91}{135\,{\left (3\,x+2\right )}^5}+\frac {49}{486\,{\left (3\,x+2\right )}^6} \]

[In]

int(((2*x - 1)^2*(5*x + 3))/(3*x + 2)^7,x)

[Out]

4/(9*(3*x + 2)^4) - 20/(243*(3*x + 2)^3) - 91/(135*(3*x + 2)^5) + 49/(486*(3*x + 2)^6)