\(\int x (9+12 x+4 x^2)^{5/2} \, dx\) [208]

Optimal result

Integrand size = 16, antiderivative size = 42 \[ \int x \left (9+12 x+4 x^2\right )^{5/2} \, dx=-\frac {1}{8} (3+2 x) \left (9+12 x+4 x^2\right )^{5/2}+\frac {1}{28} \left (9+12 x+4 x^2\right )^{7/2} \]

[Out]

-1/8*(3+2*x)*(4*x^2+12*x+9)^(5/2)+1/28*(4*x^2+12*x+9)^(7/2)

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} \[ \int x \left (9+12 x+4 x^2\right )^{5/2} \, dx=\frac {1}{28} \left (4 x^2+12 x+9\right )^{7/2}-\frac {1}{8} (2 x+3) \left (4 x^2+12 x+9\right )^{5/2} \]

[In]

Int[x*(9 + 12*x + 4*x^2)^(5/2),x]

[Out]

-1/8*((3 + 2*x)*(9 + 12*x + 4*x^2)^(5/2)) + (9 + 12*x + 4*x^2)^(7/2)/28

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}{28} \left (9+12 x+4 x^2\right )^{7/2}-\frac {3}{2} \int \left (9+12 x+4 x^2\right )^{5/2} \, dx \\ & = -\frac {1}{8} (3+2 x) \left (9+12 x+4 x^2\right )^{5/2}+\frac {1}{28} \left (9+12 x+4 x^2\right )^{7/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.64 \[ \int x \left (9+12 x+4 x^2\right )^{5/2} \, dx=\frac {1}{56} (3+2 x)^5 \sqrt {(3+2 x)^2} (-1+4 x) \]

[In]

Integrate[x*(9 + 12*x + 4*x^2)^(5/2),x]

[Out]

((3 + 2*x)^5*Sqrt[(3 + 2*x)^2]*(-1 + 4*x))/56

Maple [A] (verified)

Time = 2.06 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.12

method	result	size
gosper	\(\frac {x^{2} \left (64 x^{5}+560 x^{4}+2016 x^{3}+3780 x^{2}+3780 x +1701\right ) \left (\left (2 x +3\right )^{2}\right )^{\frac {5}{2}}}{14 \left (2 x +3\right )^{5}}\)	\(47\)
default	\(\frac {x^{2} \left (64 x^{5}+560 x^{4}+2016 x^{3}+3780 x^{2}+3780 x +1701\right ) \left (\left (2 x +3\right )^{2}\right )^{\frac {5}{2}}}{14 \left (2 x +3\right )^{5}}\)	\(47\)
risch	\(\frac {32 \sqrt {\left (2 x +3\right )^{2}}\, x^{7}}{7 \left (2 x +3\right )}+\frac {40 \sqrt {\left (2 x +3\right )^{2}}\, x^{6}}{2 x +3}+\frac {144 \sqrt {\left (2 x +3\right )^{2}}\, x^{5}}{2 x +3}+\frac {270 \sqrt {\left (2 x +3\right )^{2}}\, x^{4}}{2 x +3}+\frac {270 \sqrt {\left (2 x +3\right )^{2}}\, x^{3}}{2 x +3}+\frac {243 \sqrt {\left (2 x +3\right )^{2}}\, x^{2}}{2 \left (2 x +3\right )}\)	\(128\)

[In]

int(x*(4*x^2+12*x+9)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/14*x^2*(64*x^5+560*x^4+2016*x^3+3780*x^2+3780*x+1701)*((2*x+3)^2)^(5/2)/(2*x+3)^5

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.74 \[ \int x \left (9+12 x+4 x^2\right )^{5/2} \, dx=\frac {32}{7} \, x^{7} + 40 \, x^{6} + 144 \, x^{5} + 270 \, x^{4} + 270 \, x^{3} + \frac {243}{2} \, x^{2} \]

[In]

integrate(x*(4*x^2+12*x+9)^(5/2),x, algorithm="fricas")

[Out]

32/7*x^7 + 40*x^6 + 144*x^5 + 270*x^4 + 270*x^3 + 243/2*x^2

Sympy [A] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.29 \[ \int x \left (9+12 x+4 x^2\right )^{5/2} \, dx=\sqrt {4 x^{2} + 12 x + 9} \cdot \left (\frac {16 x^{6}}{7} + \frac {116 x^{5}}{7} + \frac {330 x^{4}}{7} + \frac {450 x^{3}}{7} + \frac {270 x^{2}}{7} + \frac {81 x}{28} - \frac {243}{56}\right ) \]

[In]

integrate(x*(4*x**2+12*x+9)**(5/2),x)

[Out]

sqrt(4*x**2 + 12*x + 9)*(16*x**6/7 + 116*x**5/7 + 330*x**4/7 + 450*x**3/7 + 270*x**2/7 + 81*x/28 - 243/56)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.05 \[ \int x \left (9+12 x+4 x^2\right )^{5/2} \, dx=\frac {1}{28} \, {\left (4 \, x^{2} + 12 \, x + 9\right )}^{\frac {7}{2}} - \frac {1}{4} \, {\left (4 \, x^{2} + 12 \, x + 9\right )}^{\frac {5}{2}} x - \frac {3}{8} \, {\left (4 \, x^{2} + 12 \, x + 9\right )}^{\frac {5}{2}} \]

[In]

integrate(x*(4*x^2+12*x+9)^(5/2),x, algorithm="maxima")

[Out]

1/28*(4*x^2 + 12*x + 9)^(7/2) - 1/4*(4*x^2 + 12*x + 9)^(5/2)*x - 3/8*(4*x^2 + 12*x + 9)^(5/2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (34) = 68\).

Time = 0.28 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.79 \[ \int x \left (9+12 x+4 x^2\right )^{5/2} \, dx=\frac {32}{7} \, x^{7} \mathrm {sgn}\left (2 \, x + 3\right ) + 40 \, x^{6} \mathrm {sgn}\left (2 \, x + 3\right ) + 144 \, x^{5} \mathrm {sgn}\left (2 \, x + 3\right ) + 270 \, x^{4} \mathrm {sgn}\left (2 \, x + 3\right ) + 270 \, x^{3} \mathrm {sgn}\left (2 \, x + 3\right ) + \frac {243}{2} \, x^{2} \mathrm {sgn}\left (2 \, x + 3\right ) - \frac {729}{56} \, \mathrm {sgn}\left (2 \, x + 3\right ) \]

[In]

integrate(x*(4*x^2+12*x+9)^(5/2),x, algorithm="giac")

[Out]

32/7*x^7*sgn(2*x + 3) + 40*x^6*sgn(2*x + 3) + 144*x^5*sgn(2*x + 3) + 270*x^4*sgn(2*x + 3) + 270*x^3*sgn(2*x +
3) + 243/2*x^2*sgn(2*x + 3) - 729/56*sgn(2*x + 3)

Mupad [F(-1)]

Timed out. \[ \int x \left (9+12 x+4 x^2\right )^{5/2} \, dx=\int x\,{\left (4\,x^2+12\,x+9\right )}^{5/2} \,d x \]

[In]

int(x*(12*x + 4*x^2 + 9)^(5/2),x)

[Out]

int(x*(12*x + 4*x^2 + 9)^(5/2), x)