\(\int \frac {(1-2 x)^2 (3+5 x)^2}{(2+3 x)^6} \, dx\) [1267]

   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 = 22, antiderivative size = 56 \[ \int \frac {(1-2 x)^2 (3+5 x)^2}{(2+3 x)^6} \, dx=-\frac {49}{1215 (2+3 x)^5}+\frac {259}{486 (2+3 x)^4}-\frac {503}{243 (2+3 x)^3}+\frac {370}{243 (2+3 x)^2}-\frac {100}{243 (2+3 x)} \]

[Out]

-49/1215/(2+3*x)^5+259/486/(2+3*x)^4-503/243/(2+3*x)^3+370/243/(2+3*x)^2-100/243/(2+3*x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {90} \[ \int \frac {(1-2 x)^2 (3+5 x)^2}{(2+3 x)^6} \, dx=-\frac {100}{243 (3 x+2)}+\frac {370}{243 (3 x+2)^2}-\frac {503}{243 (3 x+2)^3}+\frac {259}{486 (3 x+2)^4}-\frac {49}{1215 (3 x+2)^5} \]

[In]

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

[Out]

-49/(1215*(2 + 3*x)^5) + 259/(486*(2 + 3*x)^4) - 503/(243*(2 + 3*x)^3) + 370/(243*(2 + 3*x)^2) - 100/(243*(2 +
 3*x))

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {49}{81 (2+3 x)^6}-\frac {518}{81 (2+3 x)^5}+\frac {503}{27 (2+3 x)^4}-\frac {740}{81 (2+3 x)^3}+\frac {100}{81 (2+3 x)^2}\right ) \, dx \\ & = -\frac {49}{1215 (2+3 x)^5}+\frac {259}{486 (2+3 x)^4}-\frac {503}{243 (2+3 x)^3}+\frac {370}{243 (2+3 x)^2}-\frac {100}{243 (2+3 x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.55 \[ \int \frac {(1-2 x)^2 (3+5 x)^2}{(2+3 x)^6} \, dx=-\frac {4028+19275 x+61470 x^2+116100 x^3+81000 x^4}{2430 (2+3 x)^5} \]

[In]

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

[Out]

-1/2430*(4028 + 19275*x + 61470*x^2 + 116100*x^3 + 81000*x^4)/(2 + 3*x)^5

Maple [A] (verified)

Time = 2.29 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.52

method result size
norman \(\frac {-\frac {100}{3} x^{4}-\frac {430}{9} x^{3}-\frac {683}{27} x^{2}-\frac {1285}{162} x -\frac {2014}{1215}}{\left (2+3 x \right )^{5}}\) \(29\)
gosper \(-\frac {81000 x^{4}+116100 x^{3}+61470 x^{2}+19275 x +4028}{2430 \left (2+3 x \right )^{5}}\) \(30\)
risch \(\frac {-\frac {100}{3} x^{4}-\frac {430}{9} x^{3}-\frac {683}{27} x^{2}-\frac {1285}{162} x -\frac {2014}{1215}}{\left (2+3 x \right )^{5}}\) \(30\)
parallelrisch \(\frac {6042 x^{5}+4140 x^{4}+3920 x^{3}+5760 x^{2}+2160 x}{480 \left (2+3 x \right )^{5}}\) \(34\)
default \(-\frac {49}{1215 \left (2+3 x \right )^{5}}+\frac {259}{486 \left (2+3 x \right )^{4}}-\frac {503}{243 \left (2+3 x \right )^{3}}+\frac {370}{243 \left (2+3 x \right )^{2}}-\frac {100}{243 \left (2+3 x \right )}\) \(47\)
meijerg \(\frac {9 x \left (\frac {81}{16} x^{4}+\frac {135}{8} x^{3}+\frac {45}{2} x^{2}+15 x +5\right )}{320 \left (1+\frac {3 x}{2}\right )^{5}}-\frac {3 x^{2} \left (\frac {27}{8} x^{3}+\frac {45}{4} x^{2}+15 x +10\right )}{640 \left (1+\frac {3 x}{2}\right )^{5}}-\frac {59 x^{3} \left (\frac {9}{4} x^{2}+\frac {15}{2} x +10\right )}{1920 \left (1+\frac {3 x}{2}\right )^{5}}+\frac {x^{4} \left (\frac {3 x}{2}+5\right )}{64 \left (1+\frac {3 x}{2}\right )^{5}}+\frac {5 x^{5}}{16 \left (1+\frac {3 x}{2}\right )^{5}}\) \(110\)

[In]

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

[Out]

(-100/3*x^4-430/9*x^3-683/27*x^2-1285/162*x-2014/1215)/(2+3*x)^5

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.88 \[ \int \frac {(1-2 x)^2 (3+5 x)^2}{(2+3 x)^6} \, dx=-\frac {81000 \, x^{4} + 116100 \, x^{3} + 61470 \, x^{2} + 19275 \, x + 4028}{2430 \, {\left (243 \, x^{5} + 810 \, x^{4} + 1080 \, x^{3} + 720 \, x^{2} + 240 \, x + 32\right )}} \]

[In]

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

[Out]

-1/2430*(81000*x^4 + 116100*x^3 + 61470*x^2 + 19275*x + 4028)/(243*x^5 + 810*x^4 + 1080*x^3 + 720*x^2 + 240*x
+ 32)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x)^2 (3+5 x)^2}{(2+3 x)^6} \, dx=\frac {- 81000 x^{4} - 116100 x^{3} - 61470 x^{2} - 19275 x - 4028}{590490 x^{5} + 1968300 x^{4} + 2624400 x^{3} + 1749600 x^{2} + 583200 x + 77760} \]

[In]

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

[Out]

(-81000*x**4 - 116100*x**3 - 61470*x**2 - 19275*x - 4028)/(590490*x**5 + 1968300*x**4 + 2624400*x**3 + 1749600
*x**2 + 583200*x + 77760)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.88 \[ \int \frac {(1-2 x)^2 (3+5 x)^2}{(2+3 x)^6} \, dx=-\frac {81000 \, x^{4} + 116100 \, x^{3} + 61470 \, x^{2} + 19275 \, x + 4028}{2430 \, {\left (243 \, x^{5} + 810 \, x^{4} + 1080 \, x^{3} + 720 \, x^{2} + 240 \, x + 32\right )}} \]

[In]

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

[Out]

-1/2430*(81000*x^4 + 116100*x^3 + 61470*x^2 + 19275*x + 4028)/(243*x^5 + 810*x^4 + 1080*x^3 + 720*x^2 + 240*x
+ 32)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.52 \[ \int \frac {(1-2 x)^2 (3+5 x)^2}{(2+3 x)^6} \, dx=-\frac {81000 \, x^{4} + 116100 \, x^{3} + 61470 \, x^{2} + 19275 \, x + 4028}{2430 \, {\left (3 \, x + 2\right )}^{5}} \]

[In]

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

[Out]

-1/2430*(81000*x^4 + 116100*x^3 + 61470*x^2 + 19275*x + 4028)/(3*x + 2)^5

Mupad [B] (verification not implemented)

Time = 1.17 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.82 \[ \int \frac {(1-2 x)^2 (3+5 x)^2}{(2+3 x)^6} \, dx=\frac {370}{243\,{\left (3\,x+2\right )}^2}-\frac {100}{243\,\left (3\,x+2\right )}-\frac {503}{243\,{\left (3\,x+2\right )}^3}+\frac {259}{486\,{\left (3\,x+2\right )}^4}-\frac {49}{1215\,{\left (3\,x+2\right )}^5} \]

[In]

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

[Out]

370/(243*(3*x + 2)^2) - 100/(243*(3*x + 2)) - 503/(243*(3*x + 2)^3) + 259/(486*(3*x + 2)^4) - 49/(1215*(3*x +
2)^5)