\(\int (d+e x) (9+12 x+4 x^2)^{5/2} \, dx\) [1614]

Optimal result

Integrand size = 20, antiderivative size = 50 \[ \int (d+e x) \left (9+12 x+4 x^2\right )^{5/2} \, dx=\frac {1}{24} (2 d-3 e) (3+2 x) \left (9+12 x+4 x^2\right )^{5/2}+\frac {1}{28} e \left (9+12 x+4 x^2\right )^{7/2} \]

[Out]

1/24*(2*d-3*e)*(3+2*x)*(4*x^2+12*x+9)^(5/2)+1/28*e*(4*x^2+12*x+9)^(7/2)

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} \[ \int (d+e x) \left (9+12 x+4 x^2\right )^{5/2} \, dx=\frac {1}{24} (2 x+3) \left (4 x^2+12 x+9\right )^{5/2} (2 d-3 e)+\frac {1}{28} e \left (4 x^2+12 x+9\right )^{7/2} \]

[In]

Int[(d + e*x)*(9 + 12*x + 4*x^2)^(5/2),x]

[Out]

((2*d - 3*e)*(3 + 2*x)*(9 + 12*x + 4*x^2)^(5/2))/24 + (e*(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} e \left (9+12 x+4 x^2\right )^{7/2}+\frac {1}{2} (2 d-3 e) \int \left (9+12 x+4 x^2\right )^{5/2} \, dx \\ & = \frac {1}{24} (2 d-3 e) (3+2 x) \left (9+12 x+4 x^2\right )^{5/2}+\frac {1}{28} e \left (9+12 x+4 x^2\right )^{7/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.68 \[ \int (d+e x) \left (9+12 x+4 x^2\right )^{5/2} \, dx=\frac {1}{168} (3+2 x)^5 \sqrt {(3+2 x)^2} (14 d+3 e (-1+4 x)) \]

[In]

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

[Out]

((3 + 2*x)^5*Sqrt[(3 + 2*x)^2]*(14*d + 3*e*(-1 + 4*x)))/168

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(85\) vs. \(2(42)=84\).

Time = 2.18 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.72

method	result	size
gosper	\(\frac {x \left (192 e \,x^{6}+224 d \,x^{5}+1680 e \,x^{5}+2016 d \,x^{4}+6048 e \,x^{4}+7560 d \,x^{3}+11340 e \,x^{3}+15120 d \,x^{2}+11340 e \,x^{2}+17010 d x +5103 e x +10206 d \right ) \left (\left (2 x +3\right )^{2}\right )^{\frac {5}{2}}}{42 \left (2 x +3\right )^{5}}\)	\(86\)
default	\(\frac {x \left (192 e \,x^{6}+224 d \,x^{5}+1680 e \,x^{5}+2016 d \,x^{4}+6048 e \,x^{4}+7560 d \,x^{3}+11340 e \,x^{3}+15120 d \,x^{2}+11340 e \,x^{2}+17010 d x +5103 e x +10206 d \right ) \left (\left (2 x +3\right )^{2}\right )^{\frac {5}{2}}}{42 \left (2 x +3\right )^{5}}\)	\(86\)
risch	\(\frac {32 \sqrt {\left (2 x +3\right )^{2}}\, e \,x^{7}}{7 \left (2 x +3\right )}+\frac {\sqrt {\left (2 x +3\right )^{2}}\, \left (32 d +240 e \right ) x^{6}}{12 x +18}+\frac {\sqrt {\left (2 x +3\right )^{2}}\, \left (240 d +720 e \right ) x^{5}}{10 x +15}+\frac {\sqrt {\left (2 x +3\right )^{2}}\, \left (720 d +1080 e \right ) x^{4}}{8 x +12}+\frac {\sqrt {\left (2 x +3\right )^{2}}\, \left (1080 d +810 e \right ) x^{3}}{9+6 x}+\frac {\sqrt {\left (2 x +3\right )^{2}}\, \left (810 d +243 e \right ) x^{2}}{4 x +6}+\frac {243 \sqrt {\left (2 x +3\right )^{2}}\, d x}{2 x +3}\)	\(184\)

[In]

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

[Out]

1/42*x*(192*e*x^6+224*d*x^5+1680*e*x^5+2016*d*x^4+6048*e*x^4+7560*d*x^3+11340*e*x^3+15120*d*x^2+11340*e*x^2+17
010*d*x+5103*e*x+10206*d)*((2*x+3)^2)^(5/2)/(2*x+3)^5

Fricas [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.38 \[ \int (d+e x) \left (9+12 x+4 x^2\right )^{5/2} \, dx=\frac {32}{7} \, e x^{7} + \frac {8}{3} \, {\left (2 \, d + 15 \, e\right )} x^{6} + 48 \, {\left (d + 3 \, e\right )} x^{5} + 90 \, {\left (2 \, d + 3 \, e\right )} x^{4} + 90 \, {\left (4 \, d + 3 \, e\right )} x^{3} + \frac {81}{2} \, {\left (10 \, d + 3 \, e\right )} x^{2} + 243 \, d x \]

[In]

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

[Out]

32/7*e*x^7 + 8/3*(2*d + 15*e)*x^6 + 48*(d + 3*e)*x^5 + 90*(2*d + 3*e)*x^4 + 90*(4*d + 3*e)*x^3 + 81/2*(10*d +
3*e)*x^2 + 243*d*x

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (41) = 82\).

Time = 0.69 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.84 \[ \int (d+e x) \left (9+12 x+4 x^2\right )^{5/2} \, dx=\sqrt {4 x^{2} + 12 x + 9} \cdot \left (\frac {81 d}{4} + \frac {16 e x^{6}}{7} - \frac {243 e}{56} + x^{5} \cdot \left (\frac {8 d}{3} + \frac {116 e}{7}\right ) + x^{4} \cdot \left (20 d + \frac {330 e}{7}\right ) + x^{3} \cdot \left (60 d + \frac {450 e}{7}\right ) + x^{2} \cdot \left (90 d + \frac {270 e}{7}\right ) + x \left (\frac {135 d}{2} + \frac {81 e}{28}\right )\right ) \]

[In]

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

[Out]

sqrt(4*x**2 + 12*x + 9)*(81*d/4 + 16*e*x**6/7 - 243*e/56 + x**5*(8*d/3 + 116*e/7) + x**4*(20*d + 330*e/7) + x*
*3*(60*d + 450*e/7) + x**2*(90*d + 270*e/7) + x*(135*d/2 + 81*e/28))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.56 \[ \int (d+e x) \left (9+12 x+4 x^2\right )^{5/2} \, dx=\frac {1}{28} \, {\left (4 \, x^{2} + 12 \, x + 9\right )}^{\frac {7}{2}} e + \frac {1}{6} \, {\left (4 \, x^{2} + 12 \, x + 9\right )}^{\frac {5}{2}} d x - \frac {1}{4} \, {\left (4 \, x^{2} + 12 \, x + 9\right )}^{\frac {5}{2}} e x + \frac {1}{4} \, {\left (4 \, x^{2} + 12 \, x + 9\right )}^{\frac {5}{2}} d - \frac {3}{8} \, {\left (4 \, x^{2} + 12 \, x + 9\right )}^{\frac {5}{2}} e \]

[In]

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

[Out]

1/28*(4*x^2 + 12*x + 9)^(7/2)*e + 1/6*(4*x^2 + 12*x + 9)^(5/2)*d*x - 1/4*(4*x^2 + 12*x + 9)^(5/2)*e*x + 1/4*(4
*x^2 + 12*x + 9)^(5/2)*d - 3/8*(4*x^2 + 12*x + 9)^(5/2)*e

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (42) = 84\).

Time = 0.27 (sec) , antiderivative size = 158, normalized size of antiderivative = 3.16 \[ \int (d+e x) \left (9+12 x+4 x^2\right )^{5/2} \, dx=\frac {32}{7} \, e x^{7} \mathrm {sgn}\left (2 \, x + 3\right ) + \frac {16}{3} \, d x^{6} \mathrm {sgn}\left (2 \, x + 3\right ) + 40 \, e x^{6} \mathrm {sgn}\left (2 \, x + 3\right ) + 48 \, d x^{5} \mathrm {sgn}\left (2 \, x + 3\right ) + 144 \, e x^{5} \mathrm {sgn}\left (2 \, x + 3\right ) + 180 \, d x^{4} \mathrm {sgn}\left (2 \, x + 3\right ) + 270 \, e x^{4} \mathrm {sgn}\left (2 \, x + 3\right ) + 360 \, d x^{3} \mathrm {sgn}\left (2 \, x + 3\right ) + 270 \, e x^{3} \mathrm {sgn}\left (2 \, x + 3\right ) + 405 \, d x^{2} \mathrm {sgn}\left (2 \, x + 3\right ) + \frac {243}{2} \, e x^{2} \mathrm {sgn}\left (2 \, x + 3\right ) + 243 \, d x \mathrm {sgn}\left (2 \, x + 3\right ) + \frac {243}{56} \, {\left (14 \, d - 3 \, e\right )} \mathrm {sgn}\left (2 \, x + 3\right ) \]

[In]

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

[Out]

32/7*e*x^7*sgn(2*x + 3) + 16/3*d*x^6*sgn(2*x + 3) + 40*e*x^6*sgn(2*x + 3) + 48*d*x^5*sgn(2*x + 3) + 144*e*x^5*
sgn(2*x + 3) + 180*d*x^4*sgn(2*x + 3) + 270*e*x^4*sgn(2*x + 3) + 360*d*x^3*sgn(2*x + 3) + 270*e*x^3*sgn(2*x +
3) + 405*d*x^2*sgn(2*x + 3) + 243/2*e*x^2*sgn(2*x + 3) + 243*d*x*sgn(2*x + 3) + 243/56*(14*d - 3*e)*sgn(2*x +
3)

Mupad [F(-1)]

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

[In]

int((d + e*x)*(12*x + 4*x^2 + 9)^(5/2),x)

[Out]

int((d + e*x)*(12*x + 4*x^2 + 9)^(5/2), x)