\(\int (d+e x) (9+12 x+4 x^2)^{3/2} \, dx\) [1615]

Optimal result

Integrand size = 20, antiderivative size = 50 \[ \int (d+e x) \left (9+12 x+4 x^2\right )^{3/2} \, dx=\frac {1}{16} (2 d-3 e) (3+2 x) \left (9+12 x+4 x^2\right )^{3/2}+\frac {1}{20} e \left (9+12 x+4 x^2\right )^{5/2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 50, 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 = {654, 623} \[ \int (d+e x) \left (9+12 x+4 x^2\right )^{3/2} \, dx=\frac {1}{16} (2 x+3) \left (4 x^2+12 x+9\right )^{3/2} (2 d-3 e)+\frac {1}{20} e \left (4 x^2+12 x+9\right )^{5/2} \]

[In]

Int[(d + e*x)*(9 + 12*x + 4*x^2)^(3/2),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.66 \[ \int (d+e x) \left (9+12 x+4 x^2\right )^{3/2} \, dx=\frac {1}{80} (3+2 x)^3 \sqrt {(3+2 x)^2} (10 d+e (-3+8 x)) \]

[In]

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

[Out]

((3 + 2*x)^3*Sqrt[(3 + 2*x)^2]*(10*d + e*(-3 + 8*x)))/80

Maple [A] (verified)

Time = 2.17 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.24

method	result	size
gosper	\(\frac {x \left (16 e \,x^{4}+20 d \,x^{3}+90 e \,x^{3}+120 d \,x^{2}+180 e \,x^{2}+270 d x +135 e x +270 d \right ) \left (\left (2 x +3\right )^{2}\right )^{\frac {3}{2}}}{10 \left (2 x +3\right )^{3}}\)	\(62\)
default	\(\frac {x \left (16 e \,x^{4}+20 d \,x^{3}+90 e \,x^{3}+120 d \,x^{2}+180 e \,x^{2}+270 d x +135 e x +270 d \right ) \left (\left (2 x +3\right )^{2}\right )^{\frac {3}{2}}}{10 \left (2 x +3\right )^{3}}\)	\(62\)
risch	\(\frac {8 \sqrt {\left (2 x +3\right )^{2}}\, e \,x^{5}}{5 \left (2 x +3\right )}+\frac {\sqrt {\left (2 x +3\right )^{2}}\, \left (8 d +36 e \right ) x^{4}}{8 x +12}+\frac {\sqrt {\left (2 x +3\right )^{2}}\, \left (36 d +54 e \right ) x^{3}}{9+6 x}+\frac {\sqrt {\left (2 x +3\right )^{2}}\, \left (54 d +27 e \right ) x^{2}}{4 x +6}+\frac {27 \sqrt {\left (2 x +3\right )^{2}}\, d x}{2 x +3}\)	\(128\)

[In]

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

[Out]

1/10*x*(16*e*x^4+20*d*x^3+90*e*x^3+120*d*x^2+180*e*x^2+270*d*x+135*e*x+270*d)*((2*x+3)^2)^(3/2)/(2*x+3)^3

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.88 \[ \int (d+e x) \left (9+12 x+4 x^2\right )^{3/2} \, dx=\frac {8}{5} \, e x^{5} + {\left (2 \, d + 9 \, e\right )} x^{4} + 6 \, {\left (2 \, d + 3 \, e\right )} x^{3} + \frac {27}{2} \, {\left (2 \, d + e\right )} x^{2} + 27 \, d x \]

[In]

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

[Out]

8/5*e*x^5 + (2*d + 9*e)*x^4 + 6*(2*d + 3*e)*x^3 + 27/2*(2*d + e)*x^2 + 27*d*x

Sympy [A] (verification not implemented)

Time = 0.61 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.32 \[ \int (d+e x) \left (9+12 x+4 x^2\right )^{3/2} \, dx=\sqrt {4 x^{2} + 12 x + 9} \cdot \left (\frac {27 d}{8} + \frac {4 e x^{4}}{5} - \frac {81 e}{80} + x^{3} \left (d + \frac {33 e}{10}\right ) + x^{2} \cdot \left (\frac {9 d}{2} + \frac {81 e}{20}\right ) + x \left (\frac {27 d}{4} + \frac {27 e}{40}\right )\right ) \]

[In]

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

[Out]

sqrt(4*x**2 + 12*x + 9)*(27*d/8 + 4*e*x**4/5 - 81*e/80 + x**3*(d + 33*e/10) + x**2*(9*d/2 + 81*e/20) + x*(27*d
/4 + 27*e/40))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.56 \[ \int (d+e x) \left (9+12 x+4 x^2\right )^{3/2} \, dx=\frac {1}{20} \, {\left (4 \, x^{2} + 12 \, x + 9\right )}^{\frac {5}{2}} e + \frac {1}{4} \, {\left (4 \, x^{2} + 12 \, x + 9\right )}^{\frac {3}{2}} d x - \frac {3}{8} \, {\left (4 \, x^{2} + 12 \, x + 9\right )}^{\frac {3}{2}} e x + \frac {3}{8} \, {\left (4 \, x^{2} + 12 \, x + 9\right )}^{\frac {3}{2}} d - \frac {9}{16} \, {\left (4 \, x^{2} + 12 \, x + 9\right )}^{\frac {3}{2}} e \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 110 vs. \(2 (42) = 84\).

Time = 0.27 (sec) , antiderivative size = 110, normalized size of antiderivative = 2.20 \[ \int (d+e x) \left (9+12 x+4 x^2\right )^{3/2} \, dx=\frac {8}{5} \, e x^{5} \mathrm {sgn}\left (2 \, x + 3\right ) + 2 \, d x^{4} \mathrm {sgn}\left (2 \, x + 3\right ) + 9 \, e x^{4} \mathrm {sgn}\left (2 \, x + 3\right ) + 12 \, d x^{3} \mathrm {sgn}\left (2 \, x + 3\right ) + 18 \, e x^{3} \mathrm {sgn}\left (2 \, x + 3\right ) + 27 \, d x^{2} \mathrm {sgn}\left (2 \, x + 3\right ) + \frac {27}{2} \, e x^{2} \mathrm {sgn}\left (2 \, x + 3\right ) + 27 \, d x \mathrm {sgn}\left (2 \, x + 3\right ) + \frac {81}{80} \, {\left (10 \, d - 3 \, e\right )} \mathrm {sgn}\left (2 \, x + 3\right ) \]

[In]

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

[Out]

8/5*e*x^5*sgn(2*x + 3) + 2*d*x^4*sgn(2*x + 3) + 9*e*x^4*sgn(2*x + 3) + 12*d*x^3*sgn(2*x + 3) + 18*e*x^3*sgn(2*
x + 3) + 27*d*x^2*sgn(2*x + 3) + 27/2*e*x^2*sgn(2*x + 3) + 27*d*x*sgn(2*x + 3) + 81/80*(10*d - 3*e)*sgn(2*x +
3)

Mupad [B] (verification not implemented)

Time = 9.60 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.34 \[ \int (d+e x) \left (9+12 x+4 x^2\right )^{3/2} \, dx=\frac {e\,{\left (4\,x^2+12\,x+9\right )}^{5/2}}{20}-\frac {9\,e\,{\left (4\,x^2+12\,x+9\right )}^{3/2}}{16}-\frac {3\,e\,x\,{\left (4\,x^2+12\,x+9\right )}^{3/2}}{8}+\frac {d\,\left (2\,x+3\right )\,{\left (4\,x^2+12\,x+9\right )}^{3/2}}{8} \]

[In]

int((d + e*x)*(12*x + 4*x^2 + 9)^(3/2),x)

[Out]

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