\(\int x (9+12 x+4 x^2)^{3/2} \, dx\) [209]

Optimal result

Integrand size = 16, antiderivative size = 42 \[ \int x \left (9+12 x+4 x^2\right )^{3/2} \, dx=-\frac {3}{16} (3+2 x) \left (9+12 x+4 x^2\right )^{3/2}+\frac {1}{20} \left (9+12 x+4 x^2\right )^{5/2} \]

[Out]

-3/16*(3+2*x)*(4*x^2+12*x+9)^(3/2)+1/20*(4*x^2+12*x+9)^(5/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {654, 623} \[ \int x \left (9+12 x+4 x^2\right )^{3/2} \, dx=\frac {1}{20} \left (4 x^2+12 x+9\right )^{5/2}-\frac {3}{16} (2 x+3) \left (4 x^2+12 x+9\right )^{3/2} \]

[In]

Int[x*(9 + 12*x + 4*x^2)^(3/2),x]

[Out]

(-3*(3 + 2*x)*(9 + 12*x + 4*x^2)^(3/2))/16 + (9 + 12*x + 4*x^2)^(5/2)/20

Rule 623

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x)*((a + b*x + c*x^2)^p/(2*c*(2*p + 1)
)), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && NeQ[p, -2^(-1)]

Rule 654

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*((a + b*x + c*x^2)^(p +
1)/(2*c*(p + 1))), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{20} \left (9+12 x+4 x^2\right )^{5/2}-\frac {3}{2} \int \left (9+12 x+4 x^2\right )^{3/2} \, dx \\ & = -\frac {3}{16} (3+2 x) \left (9+12 x+4 x^2\right )^{3/2}+\frac {1}{20} \left (9+12 x+4 x^2\right )^{5/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.88 \[ \int x \left (9+12 x+4 x^2\right )^{3/2} \, dx=\frac {x^2 \sqrt {(3+2 x)^2} \left (135+180 x+90 x^2+16 x^3\right )}{30+20 x} \]

[In]

Integrate[x*(9 + 12*x + 4*x^2)^(3/2),x]

[Out]

(x^2*Sqrt[(3 + 2*x)^2]*(135 + 180*x + 90*x^2 + 16*x^3))/(30 + 20*x)

Maple [A] (verified)

Time = 2.08 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.88

method	result	size
gosper	\(\frac {x^{2} \left (16 x^{3}+90 x^{2}+180 x +135\right ) \left (\left (2 x +3\right )^{2}\right )^{\frac {3}{2}}}{10 \left (2 x +3\right )^{3}}\)	\(37\)
default	\(\frac {x^{2} \left (16 x^{3}+90 x^{2}+180 x +135\right ) \left (\left (2 x +3\right )^{2}\right )^{\frac {3}{2}}}{10 \left (2 x +3\right )^{3}}\)	\(37\)
risch	\(\frac {8 \sqrt {\left (2 x +3\right )^{2}}\, x^{5}}{5 \left (2 x +3\right )}+\frac {9 \sqrt {\left (2 x +3\right )^{2}}\, x^{4}}{2 x +3}+\frac {18 \sqrt {\left (2 x +3\right )^{2}}\, x^{3}}{2 x +3}+\frac {27 \sqrt {\left (2 x +3\right )^{2}}\, x^{2}}{2 \left (2 x +3\right )}\)	\(86\)

[In]

int(x*(4*x^2+12*x+9)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/10*x^2*(16*x^3+90*x^2+180*x+135)*((2*x+3)^2)^(3/2)/(2*x+3)^3

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.50 \[ \int x \left (9+12 x+4 x^2\right )^{3/2} \, dx=\frac {8}{5} \, x^{5} + 9 \, x^{4} + 18 \, x^{3} + \frac {27}{2} \, x^{2} \]

[In]

integrate(x*(4*x^2+12*x+9)^(3/2),x, algorithm="fricas")

[Out]

8/5*x^5 + 9*x^4 + 18*x^3 + 27/2*x^2

Sympy [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.98 \[ \int x \left (9+12 x+4 x^2\right )^{3/2} \, dx=\sqrt {4 x^{2} + 12 x + 9} \cdot \left (\frac {4 x^{4}}{5} + \frac {33 x^{3}}{10} + \frac {81 x^{2}}{20} + \frac {27 x}{40} - \frac {81}{80}\right ) \]

[In]

integrate(x*(4*x**2+12*x+9)**(3/2),x)

[Out]

sqrt(4*x**2 + 12*x + 9)*(4*x**4/5 + 33*x**3/10 + 81*x**2/20 + 27*x/40 - 81/80)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.05 \[ \int x \left (9+12 x+4 x^2\right )^{3/2} \, dx=\frac {1}{20} \, {\left (4 \, x^{2} + 12 \, x + 9\right )}^{\frac {5}{2}} - \frac {3}{8} \, {\left (4 \, x^{2} + 12 \, x + 9\right )}^{\frac {3}{2}} x - \frac {9}{16} \, {\left (4 \, x^{2} + 12 \, x + 9\right )}^{\frac {3}{2}} \]

[In]

integrate(x*(4*x^2+12*x+9)^(3/2),x, algorithm="maxima")

[Out]

1/20*(4*x^2 + 12*x + 9)^(5/2) - 3/8*(4*x^2 + 12*x + 9)^(3/2)*x - 9/16*(4*x^2 + 12*x + 9)^(3/2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.26 \[ \int x \left (9+12 x+4 x^2\right )^{3/2} \, dx=\frac {8}{5} \, x^{5} \mathrm {sgn}\left (2 \, x + 3\right ) + 9 \, x^{4} \mathrm {sgn}\left (2 \, x + 3\right ) + 18 \, x^{3} \mathrm {sgn}\left (2 \, x + 3\right ) + \frac {27}{2} \, x^{2} \mathrm {sgn}\left (2 \, x + 3\right ) - \frac {243}{80} \, \mathrm {sgn}\left (2 \, x + 3\right ) \]

[In]

integrate(x*(4*x^2+12*x+9)^(3/2),x, algorithm="giac")

[Out]

8/5*x^5*sgn(2*x + 3) + 9*x^4*sgn(2*x + 3) + 18*x^3*sgn(2*x + 3) + 27/2*x^2*sgn(2*x + 3) - 243/80*sgn(2*x + 3)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.57 \[ \int x \left (9+12 x+4 x^2\right )^{3/2} \, dx=\frac {{\left (4\,x^2+12\,x+9\right )}^{3/2}\,\left (16\,x^2+18\,x-9\right )}{80} \]

[In]

int(x*(12*x + 4*x^2 + 9)^(3/2),x)

[Out]

((12*x + 4*x^2 + 9)^(3/2)*(18*x + 16*x^2 - 9))/80