\(\int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^5} \, dx\) [1249]

Optimal result

Integrand size = 20, antiderivative size = 37 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^5} \, dx=\frac {(1-2 x)^3}{84 (2+3 x)^4}-\frac {23 (1-2 x)^3}{294 (2+3 x)^3} \]

[Out]

1/84*(1-2*x)^3/(2+3*x)^4-23/294*(1-2*x)^3/(2+3*x)^3

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {79, 37} \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^5} \, dx=\frac {(1-2 x)^3}{84 (3 x+2)^4}-\frac {23 (1-2 x)^3}{294 (3 x+2)^3} \]

[In]

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

[Out]

(1 - 2*x)^3/(84*(2 + 3*x)^4) - (23*(1 - 2*x)^3)/(294*(2 + 3*x)^3)

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rubi steps \begin{align*} \text {integral}& = \frac {(1-2 x)^3}{84 (2+3 x)^4}+\frac {23}{14} \int \frac {(1-2 x)^2}{(2+3 x)^4} \, dx \\ & = \frac {(1-2 x)^3}{84 (2+3 x)^4}-\frac {23 (1-2 x)^3}{294 (2+3 x)^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.70 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^5} \, dx=-\frac {167+516 x+1728 x^2+2160 x^3}{324 (2+3 x)^4} \]

[In]

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

[Out]

-1/324*(167 + 516*x + 1728*x^2 + 2160*x^3)/(2 + 3*x)^4

Maple [A] (verified)

Time = 2.22 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.68

method	result	size
gosper	\(-\frac {2160 x^{3}+1728 x^{2}+516 x +167}{324 \left (2+3 x \right )^{4}}\)	\(25\)
risch	\(\frac {-\frac {20}{3} x^{3}-\frac {16}{3} x^{2}-\frac {43}{27} x -\frac {167}{324}}{\left (2+3 x \right )^{4}}\)	\(25\)
norman	\(\frac {\frac {7}{24} x^{3}+\frac {13}{8} x^{2}+\frac {3}{2} x +\frac {167}{64} x^{4}}{\left (2+3 x \right )^{4}}\)	\(28\)
parallelrisch	\(\frac {501 x^{4}+56 x^{3}+312 x^{2}+288 x}{192 \left (2+3 x \right )^{4}}\)	\(29\)
default	\(-\frac {20}{81 \left (2+3 x \right )}+\frac {8}{9 \left (2+3 x \right )^{2}}+\frac {49}{324 \left (2+3 x \right )^{4}}-\frac {91}{81 \left (2+3 x \right )^{3}}\)	\(38\)
meijerg	\(\frac {3 x \left (\frac {27}{8} x^{3}+9 x^{2}+9 x +4\right )}{128 \left (1+\frac {3 x}{2}\right )^{4}}-\frac {7 x^{2} \left (\frac {9}{4} x^{2}+6 x +6\right )}{384 \left (1+\frac {3 x}{2}\right )^{4}}-\frac {x^{3} \left (\frac {3 x}{2}+4\right )}{48 \left (1+\frac {3 x}{2}\right )^{4}}+\frac {5 x^{4}}{32 \left (1+\frac {3 x}{2}\right )^{4}}\)	\(78\)

[In]

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

[Out]

-1/324*(2160*x^3+1728*x^2+516*x+167)/(2+3*x)^4

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.05 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^5} \, dx=-\frac {2160 \, x^{3} + 1728 \, x^{2} + 516 \, x + 167}{324 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )}} \]

[In]

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

[Out]

-1/324*(2160*x^3 + 1728*x^2 + 516*x + 167)/(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.97 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^5} \, dx=\frac {- 2160 x^{3} - 1728 x^{2} - 516 x - 167}{26244 x^{4} + 69984 x^{3} + 69984 x^{2} + 31104 x + 5184} \]

[In]

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

[Out]

(-2160*x**3 - 1728*x**2 - 516*x - 167)/(26244*x**4 + 69984*x**3 + 69984*x**2 + 31104*x + 5184)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.05 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^5} \, dx=-\frac {2160 \, x^{3} + 1728 \, x^{2} + 516 \, x + 167}{324 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )}} \]

[In]

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

[Out]

-1/324*(2160*x^3 + 1728*x^2 + 516*x + 167)/(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^5} \, dx=-\frac {20}{81 \, {\left (3 \, x + 2\right )}} + \frac {8}{9 \, {\left (3 \, x + 2\right )}^{2}} - \frac {91}{81 \, {\left (3 \, x + 2\right )}^{3}} + \frac {49}{324 \, {\left (3 \, x + 2\right )}^{4}} \]

[In]

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

[Out]

-20/81/(3*x + 2) + 8/9/(3*x + 2)^2 - 91/81/(3*x + 2)^3 + 49/324/(3*x + 2)^4

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int \frac {(1-2 x)^2 (3+5 x)}{(2+3 x)^5} \, dx=\frac {8}{9\,{\left (3\,x+2\right )}^2}-\frac {20}{81\,\left (3\,x+2\right )}-\frac {91}{81\,{\left (3\,x+2\right )}^3}+\frac {49}{324\,{\left (3\,x+2\right )}^4} \]

[In]

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

[Out]

8/(9*(3*x + 2)^2) - 20/(81*(3*x + 2)) - 91/(81*(3*x + 2)^3) + 49/(324*(3*x + 2)^4)