\(\int (4+12 x+9 x^2)^{3/2} \, dx\) [64]

Optimal result

Integrand size = 14, antiderivative size = 23 \[ \int \left (4+12 x+9 x^2\right )^{3/2} \, dx=\frac {1}{12} (2+3 x) \left (4+12 x+9 x^2\right )^{3/2} \]

[Out]

1/12*(2+3*x)*(9*x^2+12*x+4)^(3/2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {623} \[ \int \left (4+12 x+9 x^2\right )^{3/2} \, dx=\frac {1}{12} (3 x+2) \left (9 x^2+12 x+4\right )^{3/2} \]

[In]

Int[(4 + 12*x + 9*x^2)^(3/2),x]

[Out]

((2 + 3*x)*(4 + 12*x + 9*x^2)^(3/2))/12

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)]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \left (4+12 x+9 x^2\right )^{3/2} \, dx=\frac {1}{12} (2+3 x) \left ((2+3 x)^2\right )^{3/2} \]

[In]

Integrate[(4 + 12*x + 9*x^2)^(3/2),x]

[Out]

((2 + 3*x)*((2 + 3*x)^2)^(3/2))/12

Maple [A] (verified)

Time = 2.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74

method	result	size
default	\(\frac {\left (2+3 x \right ) \left (\left (2+3 x \right )^{2}\right )^{\frac {3}{2}}}{12}\)	\(17\)
risch	\(\frac {\sqrt {\left (2+3 x \right )^{2}}\, \left (2+3 x \right )^{3}}{12}\)	\(19\)
gosper	\(\frac {x \left (27 x^{3}+72 x^{2}+72 x +32\right ) \left (\left (2+3 x \right )^{2}\right )^{\frac {3}{2}}}{4 \left (2+3 x \right )^{3}}\)	\(35\)

[In]

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

[Out]

1/12*(2+3*x)*((2+3*x)^2)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \left (4+12 x+9 x^2\right )^{3/2} \, dx=\frac {27}{4} \, x^{4} + 18 \, x^{3} + 18 \, x^{2} + 8 \, x \]

[In]

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

[Out]

27/4*x^4 + 18*x^3 + 18*x^2 + 8*x

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (19) = 38\).

Time = 0.46 (sec) , antiderivative size = 80, normalized size of antiderivative = 3.48 \[ \int \left (4+12 x+9 x^2\right )^{3/2} \, dx=4 \left (\frac {x}{2} + \frac {1}{3}\right ) \sqrt {9 x^{2} + 12 x + 4} + 12 \left (\frac {x^{2}}{3} + \frac {x}{9} - \frac {2}{27}\right ) \sqrt {9 x^{2} + 12 x + 4} + 9 \sqrt {9 x^{2} + 12 x + 4} \left (\frac {x^{3}}{4} + \frac {x^{2}}{18} - \frac {x}{27} + \frac {2}{81}\right ) \]

[In]

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

[Out]

4*(x/2 + 1/3)*sqrt(9*x**2 + 12*x + 4) + 12*(x**2/3 + x/9 - 2/27)*sqrt(9*x**2 + 12*x + 4) + 9*sqrt(9*x**2 + 12*
x + 4)*(x**3/4 + x**2/18 - x/27 + 2/81)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \left (4+12 x+9 x^2\right )^{3/2} \, dx=\frac {1}{4} \, {\left (9 \, x^{2} + 12 \, x + 4\right )}^{\frac {3}{2}} x + \frac {1}{6} \, {\left (9 \, x^{2} + 12 \, x + 4\right )}^{\frac {3}{2}} \]

[In]

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

[Out]

1/4*(9*x^2 + 12*x + 4)^(3/2)*x + 1/6*(9*x^2 + 12*x + 4)^(3/2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (19) = 38\).

Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.96 \[ \int \left (4+12 x+9 x^2\right )^{3/2} \, dx=\frac {3}{4} \, {\left (3 \, x^{2} + 4 \, x\right )}^{2} \mathrm {sgn}\left (3 \, x + 2\right ) + 2 \, {\left (3 \, x^{2} + 4 \, x\right )} \mathrm {sgn}\left (3 \, x + 2\right ) + \frac {4}{3} \, \mathrm {sgn}\left (3 \, x + 2\right ) \]

[In]

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

[Out]

3/4*(3*x^2 + 4*x)^2*sgn(3*x + 2) + 2*(3*x^2 + 4*x)*sgn(3*x + 2) + 4/3*sgn(3*x + 2)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \left (4+12 x+9 x^2\right )^{3/2} \, dx=\frac {\left (9\,x+6\right )\,{\left (9\,x^2+12\,x+4\right )}^{3/2}}{36} \]

[In]

int((12*x + 9*x^2 + 4)^(3/2),x)

[Out]

((9*x + 6)*(12*x + 9*x^2 + 4)^(3/2))/36