\(\int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^8} \, dx\) [1354]

Optimal result

Integrand size = 20, antiderivative size = 56 \[ \int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^8} \, dx=\frac {49}{243 (2+3 x)^7}-\frac {2009}{1458 (2+3 x)^6}+\frac {518}{405 (2+3 x)^5}-\frac {107}{243 (2+3 x)^4}+\frac {40}{729 (2+3 x)^3} \]

[Out]

49/243/(2+3*x)^7-2009/1458/(2+3*x)^6+518/405/(2+3*x)^5-107/243/(2+3*x)^4+40/729/(2+3*x)^3

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 56, 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} \[ \int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^8} \, dx=\frac {40}{729 (3 x+2)^3}-\frac {107}{243 (3 x+2)^4}+\frac {518}{405 (3 x+2)^5}-\frac {2009}{1458 (3 x+2)^6}+\frac {49}{243 (3 x+2)^7} \]

[In]

Int[((1 - 2*x)^3*(3 + 5*x))/(2 + 3*x)^8,x]

[Out]

49/(243*(2 + 3*x)^7) - 2009/(1458*(2 + 3*x)^6) + 518/(405*(2 + 3*x)^5) - 107/(243*(2 + 3*x)^4) + 40/(729*(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 {343}{81 (2+3 x)^8}+\frac {2009}{81 (2+3 x)^7}-\frac {518}{27 (2+3 x)^6}+\frac {428}{81 (2+3 x)^5}-\frac {40}{81 (2+3 x)^4}\right ) \, dx \\ & = \frac {49}{243 (2+3 x)^7}-\frac {2009}{1458 (2+3 x)^6}+\frac {518}{405 (2+3 x)^5}-\frac {107}{243 (2+3 x)^4}+\frac {40}{729 (2+3 x)^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.55 \[ \int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^8} \, dx=\frac {-604+4593 x-3024 x^2-270 x^3+32400 x^4}{7290 (2+3 x)^7} \]

[In]

Integrate[((1 - 2*x)^3*(3 + 5*x))/(2 + 3*x)^8,x]

[Out]

(-604 + 4593*x - 3024*x^2 - 270*x^3 + 32400*x^4)/(7290*(2 + 3*x)^7)

Maple [A] (verified)

Time = 2.39 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.52

method	result	size
norman	\(\frac {-\frac {56}{135} x^{2}-\frac {1}{27} x^{3}+\frac {40}{9} x^{4}+\frac {1531}{2430} x -\frac {302}{3645}}{\left (2+3 x \right )^{7}}\)	\(29\)
gosper	\(\frac {32400 x^{4}-270 x^{3}-3024 x^{2}+4593 x -604}{7290 \left (2+3 x \right )^{7}}\)	\(30\)
risch	\(\frac {-\frac {56}{135} x^{2}-\frac {1}{27} x^{3}+\frac {40}{9} x^{4}+\frac {1531}{2430} x -\frac {302}{3645}}{\left (2+3 x \right )^{7}}\)	\(30\)
parallelrisch	\(\frac {906 x^{7}+4228 x^{6}+8456 x^{5}+12240 x^{4}+6240 x^{3}+2240 x^{2}+960 x}{640 \left (2+3 x \right )^{7}}\)	\(44\)
default	\(\frac {49}{243 \left (2+3 x \right )^{7}}-\frac {2009}{1458 \left (2+3 x \right )^{6}}+\frac {518}{405 \left (2+3 x \right )^{5}}-\frac {107}{243 \left (2+3 x \right )^{4}}+\frac {40}{729 \left (2+3 x \right )^{3}}\)	\(47\)
meijerg	\(\frac {3 x \left (\frac {729}{64} x^{6}+\frac {1701}{32} x^{5}+\frac {1701}{16} x^{4}+\frac {945}{8} x^{3}+\frac {315}{4} x^{2}+\frac {63}{2} x +7\right )}{1792 \left (1+\frac {3 x}{2}\right )^{7}}-\frac {13 x^{2} \left (\frac {243}{32} x^{5}+\frac {567}{16} x^{4}+\frac {567}{8} x^{3}+\frac {315}{4} x^{2}+\frac {105}{2} x +21\right )}{10752 \left (1+\frac {3 x}{2}\right )^{7}}+\frac {x^{3} \left (\frac {81}{16} x^{4}+\frac {189}{8} x^{3}+\frac {189}{4} x^{2}+\frac {105}{2} x +35\right )}{4480 \left (1+\frac {3 x}{2}\right )^{7}}+\frac {9 x^{4} \left (\frac {27}{8} x^{3}+\frac {63}{4} x^{2}+\frac {63}{2} x +35\right )}{8960 \left (1+\frac {3 x}{2}\right )^{7}}-\frac {x^{5} \left (\frac {9}{4} x^{2}+\frac {21}{2} x +21\right )}{672 \left (1+\frac {3 x}{2}\right )^{7}}\)	\(160\)

[In]

int((1-2*x)^3*(3+5*x)/(2+3*x)^8,x,method=_RETURNVERBOSE)

[Out]

(-56/135*x^2-1/27*x^3+40/9*x^4+1531/2430*x-302/3645)/(2+3*x)^7

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.05 \[ \int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^8} \, dx=\frac {32400 \, x^{4} - 270 \, x^{3} - 3024 \, x^{2} + 4593 \, x - 604}{7290 \, {\left (2187 \, x^{7} + 10206 \, x^{6} + 20412 \, x^{5} + 22680 \, x^{4} + 15120 \, x^{3} + 6048 \, x^{2} + 1344 \, x + 128\right )}} \]

[In]

integrate((1-2*x)^3*(3+5*x)/(2+3*x)^8,x, algorithm="fricas")

[Out]

1/7290*(32400*x^4 - 270*x^3 - 3024*x^2 + 4593*x - 604)/(2187*x^7 + 10206*x^6 + 20412*x^5 + 22680*x^4 + 15120*x
^3 + 6048*x^2 + 1344*x + 128)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00 \[ \int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^8} \, dx=- \frac {- 32400 x^{4} + 270 x^{3} + 3024 x^{2} - 4593 x + 604}{15943230 x^{7} + 74401740 x^{6} + 148803480 x^{5} + 165337200 x^{4} + 110224800 x^{3} + 44089920 x^{2} + 9797760 x + 933120} \]

[In]

integrate((1-2*x)**3*(3+5*x)/(2+3*x)**8,x)

[Out]

-(-32400*x**4 + 270*x**3 + 3024*x**2 - 4593*x + 604)/(15943230*x**7 + 74401740*x**6 + 148803480*x**5 + 1653372
00*x**4 + 110224800*x**3 + 44089920*x**2 + 9797760*x + 933120)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.05 \[ \int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^8} \, dx=\frac {32400 \, x^{4} - 270 \, x^{3} - 3024 \, x^{2} + 4593 \, x - 604}{7290 \, {\left (2187 \, x^{7} + 10206 \, x^{6} + 20412 \, x^{5} + 22680 \, x^{4} + 15120 \, x^{3} + 6048 \, x^{2} + 1344 \, x + 128\right )}} \]

[In]

integrate((1-2*x)^3*(3+5*x)/(2+3*x)^8,x, algorithm="maxima")

[Out]

1/7290*(32400*x^4 - 270*x^3 - 3024*x^2 + 4593*x - 604)/(2187*x^7 + 10206*x^6 + 20412*x^5 + 22680*x^4 + 15120*x
^3 + 6048*x^2 + 1344*x + 128)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.52 \[ \int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^8} \, dx=\frac {32400 \, x^{4} - 270 \, x^{3} - 3024 \, x^{2} + 4593 \, x - 604}{7290 \, {\left (3 \, x + 2\right )}^{7}} \]

[In]

integrate((1-2*x)^3*(3+5*x)/(2+3*x)^8,x, algorithm="giac")

[Out]

1/7290*(32400*x^4 - 270*x^3 - 3024*x^2 + 4593*x - 604)/(3*x + 2)^7

Mupad [B] (verification not implemented)

Time = 1.20 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x)^3 (3+5 x)}{(2+3 x)^8} \, dx=\frac {40}{729\,{\left (3\,x+2\right )}^3}-\frac {107}{243\,{\left (3\,x+2\right )}^4}+\frac {518}{405\,{\left (3\,x+2\right )}^5}-\frac {2009}{1458\,{\left (3\,x+2\right )}^6}+\frac {49}{243\,{\left (3\,x+2\right )}^7} \]

[In]

int(-((2*x - 1)^3*(5*x + 3))/(3*x + 2)^8,x)

[Out]

40/(729*(3*x + 2)^3) - 107/(243*(3*x + 2)^4) + 518/(405*(3*x + 2)^5) - 2009/(1458*(3*x + 2)^6) + 49/(243*(3*x
+ 2)^7)