\(\int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^8} \, dx\) [1174]

Optimal result

Integrand size = 20, antiderivative size = 45 \[ \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^8} \, dx=-\frac {1}{81 (2+3 x)^7}+\frac {4}{27 (2+3 x)^6}-\frac {13}{27 (2+3 x)^5}+\frac {25}{162 (2+3 x)^4} \]

[Out]

-1/81/(2+3*x)^7+4/27/(2+3*x)^6-13/27/(2+3*x)^5+25/162/(2+3*x)^4

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} \[ \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^8} \, dx=\frac {25}{162 (3 x+2)^4}-\frac {13}{27 (3 x+2)^5}+\frac {4}{27 (3 x+2)^6}-\frac {1}{81 (3 x+2)^7} \]

[In]

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

[Out]

-1/81*1/(2 + 3*x)^7 + 4/(27*(2 + 3*x)^6) - 13/(27*(2 + 3*x)^5) + 25/(162*(2 + 3*x)^4)

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}{27 (2+3 x)^8}-\frac {8}{3 (2+3 x)^7}+\frac {65}{9 (2+3 x)^6}-\frac {50}{27 (2+3 x)^5}\right ) \, dx \\ & = -\frac {1}{81 (2+3 x)^7}+\frac {4}{27 (2+3 x)^6}-\frac {13}{27 (2+3 x)^5}+\frac {25}{162 (2+3 x)^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.58 \[ \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^8} \, dx=\frac {-22+12 x+216 x^2+225 x^3}{54 (2+3 x)^7} \]

[In]

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

[Out]

(-22 + 12*x + 216*x^2 + 225*x^3)/(54*(2 + 3*x)^7)

Maple [A] (verified)

Time = 2.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.53

method	result	size
norman	\(\frac {4 x^{2}+\frac {2}{9} x +\frac {25}{6} x^{3}-\frac {11}{27}}{\left (2+3 x \right )^{7}}\)	\(24\)
gosper	\(\frac {225 x^{3}+216 x^{2}+12 x -22}{54 \left (2+3 x \right )^{7}}\)	\(25\)
risch	\(\frac {4 x^{2}+\frac {2}{9} x +\frac {25}{6} x^{3}-\frac {11}{27}}{\left (2+3 x \right )^{7}}\)	\(25\)
default	\(-\frac {1}{81 \left (2+3 x \right )^{7}}+\frac {4}{27 \left (2+3 x \right )^{6}}-\frac {13}{27 \left (2+3 x \right )^{5}}+\frac {25}{162 \left (2+3 x \right )^{4}}\)	\(38\)
parallelrisch	\(\frac {2673 x^{7}+12474 x^{6}+24948 x^{5}+27720 x^{4}+20080 x^{3}+8928 x^{2}+1728 x}{384 \left (2+3 x \right )^{7}}\)	\(44\)
meijerg	\(\frac {9 x \left (\frac {729}{64} x^{6}+\frac {1701}{32} x^{5}+\frac {1701}{16} x^{4}+\frac {945}{8} x^{3}+\frac {315}{4} x^{2}+\frac {63}{2} x +7\right )}{1792 \left (1+\frac {3 x}{2}\right )^{7}}+\frac {x^{2} \left (\frac {243}{32} x^{5}+\frac {567}{16} x^{4}+\frac {567}{8} x^{3}+\frac {315}{4} x^{2}+\frac {105}{2} x +21\right )}{896 \left (1+\frac {3 x}{2}\right )^{7}}-\frac {x^{3} \left (\frac {81}{16} x^{4}+\frac {189}{8} x^{3}+\frac {189}{4} x^{2}+\frac {105}{2} x +35\right )}{768 \left (1+\frac {3 x}{2}\right )^{7}}-\frac {5 x^{4} \left (\frac {27}{8} x^{3}+\frac {63}{4} x^{2}+\frac {63}{2} x +35\right )}{3584 \left (1+\frac {3 x}{2}\right )^{7}}\)	\(138\)

[In]

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

[Out]

(4*x^2+2/9*x+25/6*x^3-11/27)/(2+3*x)^7

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.20 \[ \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^8} \, dx=\frac {225 \, x^{3} + 216 \, x^{2} + 12 \, x - 22}{54 \, {\left (2187 \, x^{7} + 10206 \, x^{6} + 20412 \, x^{5} + 22680 \, x^{4} + 15120 \, x^{3} + 6048 \, x^{2} + 1344 \, x + 128\right )}} \]

[In]

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

[Out]

1/54*(225*x^3 + 216*x^2 + 12*x - 22)/(2187*x^7 + 10206*x^6 + 20412*x^5 + 22680*x^4 + 15120*x^3 + 6048*x^2 + 13
44*x + 128)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.13 \[ \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^8} \, dx=- \frac {- 225 x^{3} - 216 x^{2} - 12 x + 22}{118098 x^{7} + 551124 x^{6} + 1102248 x^{5} + 1224720 x^{4} + 816480 x^{3} + 326592 x^{2} + 72576 x + 6912} \]

[In]

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

[Out]

-(-225*x**3 - 216*x**2 - 12*x + 22)/(118098*x**7 + 551124*x**6 + 1102248*x**5 + 1224720*x**4 + 816480*x**3 + 3
26592*x**2 + 72576*x + 6912)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.20 \[ \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^8} \, dx=\frac {225 \, x^{3} + 216 \, x^{2} + 12 \, x - 22}{54 \, {\left (2187 \, x^{7} + 10206 \, x^{6} + 20412 \, x^{5} + 22680 \, x^{4} + 15120 \, x^{3} + 6048 \, x^{2} + 1344 \, x + 128\right )}} \]

[In]

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

[Out]

1/54*(225*x^3 + 216*x^2 + 12*x - 22)/(2187*x^7 + 10206*x^6 + 20412*x^5 + 22680*x^4 + 15120*x^3 + 6048*x^2 + 13
44*x + 128)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.53 \[ \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^8} \, dx=\frac {225 \, x^{3} + 216 \, x^{2} + 12 \, x - 22}{54 \, {\left (3 \, x + 2\right )}^{7}} \]

[In]

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

[Out]

1/54*(225*x^3 + 216*x^2 + 12*x - 22)/(3*x + 2)^7

Mupad [B] (verification not implemented)

Time = 1.31 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x) (3+5 x)^2}{(2+3 x)^8} \, dx=\frac {25}{162\,{\left (3\,x+2\right )}^4}-\frac {13}{27\,{\left (3\,x+2\right )}^5}+\frac {4}{27\,{\left (3\,x+2\right )}^6}-\frac {1}{81\,{\left (3\,x+2\right )}^7} \]

[In]

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

[Out]

25/(162*(3*x + 2)^4) - 13/(27*(3*x + 2)^5) + 4/(27*(3*x + 2)^6) - 1/(81*(3*x + 2)^7)