\(\int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^8} \, dx\) [1252]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 20, antiderivative size = 45 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^8} \, dx=\frac {7}{81 (2+3 x)^7}-\frac {91}{162 (2+3 x)^6}+\frac {16}{45 (2+3 x)^5}-\frac {5}{81 (2+3 x)^4} \]

[Out]

7/81/(2+3*x)^7-91/162/(2+3*x)^6+16/45/(2+3*x)^5-5/81/(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} \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^8} \, dx=-\frac {5}{81 (3 x+2)^4}+\frac {16}{45 (3 x+2)^5}-\frac {91}{162 (3 x+2)^6}+\frac {7}{81 (3 x+2)^7} \]

[In]

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

[Out]

7/(81*(2 + 3*x)^7) - 91/(162*(2 + 3*x)^6) + 16/(45*(2 + 3*x)^5) - 5/(81*(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 {49}{27 (2+3 x)^8}+\frac {91}{9 (2+3 x)^7}-\frac {16}{3 (2+3 x)^6}+\frac {20}{27 (2+3 x)^5}\right ) \, dx \\ & = \frac {7}{81 (2+3 x)^7}-\frac {91}{162 (2+3 x)^6}+\frac {16}{45 (2+3 x)^5}-\frac {5}{81 (2+3 x)^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.58 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^8} \, dx=-\frac {88-291 x+108 x^2+1350 x^3}{810 (2+3 x)^7} \]

[In]

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

[Out]

-1/810*(88 - 291*x + 108*x^2 + 1350*x^3)/(2 + 3*x)^7

Maple [A] (verified)

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

method result size
norman \(\frac {-\frac {5}{3} x^{3}-\frac {2}{15} x^{2}+\frac {97}{270} x -\frac {44}{405}}{\left (2+3 x \right )^{7}}\) \(24\)
gosper \(-\frac {1350 x^{3}+108 x^{2}-291 x +88}{810 \left (2+3 x \right )^{7}}\) \(25\)
risch \(\frac {-\frac {5}{3} x^{3}-\frac {2}{15} x^{2}+\frac {97}{270} x -\frac {44}{405}}{\left (2+3 x \right )^{7}}\) \(25\)
default \(\frac {7}{81 \left (2+3 x \right )^{7}}-\frac {91}{162 \left (2+3 x \right )^{6}}+\frac {16}{45 \left (2+3 x \right )^{5}}-\frac {5}{81 \left (2+3 x \right )^{4}}\) \(38\)
parallelrisch \(\frac {3564 x^{7}+16632 x^{6}+33264 x^{5}+36960 x^{4}+21440 x^{3}+9600 x^{2}+2880 x}{1920 \left (2+3 x \right )^{7}}\) \(44\)
meijerg \(\frac {3 x \left (\frac {729}{64} x^{6}+\frac {1701}{32} x^{5}+\frac {1701}{16} x^{4}+\frac {945}{8} x^{3}+\frac {315}{4} x^{2}+\frac {63}{2} x +7\right )}{1792 \left (1+\frac {3 x}{2}\right )^{7}}-\frac {x^{2} \left (\frac {243}{32} x^{5}+\frac {567}{16} x^{4}+\frac {567}{8} x^{3}+\frac {315}{4} x^{2}+\frac {105}{2} x +21\right )}{1536 \left (1+\frac {3 x}{2}\right )^{7}}-\frac {x^{3} \left (\frac {81}{16} x^{4}+\frac {189}{8} x^{3}+\frac {189}{4} x^{2}+\frac {105}{2} x +35\right )}{3360 \left (1+\frac {3 x}{2}\right )^{7}}+\frac {x^{4} \left (\frac {27}{8} x^{3}+\frac {63}{4} x^{2}+\frac {63}{2} x +35\right )}{1792 \left (1+\frac {3 x}{2}\right )^{7}}\) \(138\)

[In]

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

[Out]

(-5/3*x^3-2/15*x^2+97/270*x-44/405)/(2+3*x)^7

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.20 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^8} \, dx=-\frac {1350 \, x^{3} + 108 \, x^{2} - 291 \, x + 88}{810 \, {\left (2187 \, x^{7} + 10206 \, x^{6} + 20412 \, x^{5} + 22680 \, x^{4} + 15120 \, x^{3} + 6048 \, x^{2} + 1344 \, x + 128\right )}} \]

[In]

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

[Out]

-1/810*(1350*x^3 + 108*x^2 - 291*x + 88)/(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 = 0.07 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.09 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^8} \, dx=\frac {- 1350 x^{3} - 108 x^{2} + 291 x - 88}{1771470 x^{7} + 8266860 x^{6} + 16533720 x^{5} + 18370800 x^{4} + 12247200 x^{3} + 4898880 x^{2} + 1088640 x + 103680} \]

[In]

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

[Out]

(-1350*x**3 - 108*x**2 + 291*x - 88)/(1771470*x**7 + 8266860*x**6 + 16533720*x**5 + 18370800*x**4 + 12247200*x
**3 + 4898880*x**2 + 1088640*x + 103680)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.20 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^8} \, dx=-\frac {1350 \, x^{3} + 108 \, x^{2} - 291 \, x + 88}{810 \, {\left (2187 \, x^{7} + 10206 \, x^{6} + 20412 \, x^{5} + 22680 \, x^{4} + 15120 \, x^{3} + 6048 \, x^{2} + 1344 \, x + 128\right )}} \]

[In]

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

[Out]

-1/810*(1350*x^3 + 108*x^2 - 291*x + 88)/(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 = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.53 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^8} \, dx=-\frac {1350 \, x^{3} + 108 \, x^{2} - 291 \, x + 88}{810 \, {\left (3 \, x + 2\right )}^{7}} \]

[In]

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

[Out]

-1/810*(1350*x^3 + 108*x^2 - 291*x + 88)/(3*x + 2)^7

Mupad [B] (verification not implemented)

Time = 1.15 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^8} \, dx=\frac {16}{45\,{\left (3\,x+2\right )}^5}-\frac {5}{81\,{\left (3\,x+2\right )}^4}-\frac {91}{162\,{\left (3\,x+2\right )}^6}+\frac {7}{81\,{\left (3\,x+2\right )}^7} \]

[In]

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

[Out]

16/(45*(3*x + 2)^5) - 5/(81*(3*x + 2)^4) - 91/(162*(3*x + 2)^6) + 7/(81*(3*x + 2)^7)