3.1615 \(\int (d+e x) (9+12 x+4 x^2)^{3/2} \, dx\)

Optimal. Leaf size=50 \[ \frac {1}{16} (2 x+3) \left (4 x^2+12 x+9\right )^{3/2} (2 d-3 e)+\frac {1}{20} e \left (4 x^2+12 x+9\right )^{5/2} \]

[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)

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Rubi [A]  time = 0.01, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {640, 609} \[ \frac {1}{16} (2 x+3) \left (4 x^2+12 x+9\right )^{3/2} (2 d-3 e)+\frac {1}{20} e \left (4 x^2+12 x+9\right )^{5/2} \]

Antiderivative was successfully verified.

[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 609

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 640

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*} \int (d+e x) \left (9+12 x+4 x^2\right )^{3/2} \, dx &=\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*}

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Mathematica [A]  time = 0.02, size = 57, normalized size = 1.14 \[ \frac {x \sqrt {(2 x+3)^2} \left (10 d \left (2 x^3+12 x^2+27 x+27\right )+e x \left (16 x^3+90 x^2+180 x+135\right )\right )}{20 x+30} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[(3 + 2*x)^2]*(10*d*(27 + 27*x + 12*x^2 + 2*x^3) + e*x*(135 + 180*x + 90*x^2 + 16*x^3)))/(30 + 20*x)

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fricas [A]  time = 0.90, size = 44, normalized size = 0.88 \[ \frac {8}{5} \, e x^{5} + {\left (2 \, d + 9 \, e\right )} x^{4} + 6 \, {\left (2 \, d + 3 \, e\right )} x^{3} + \frac {27}{2} \, {\left (2 \, d + e\right )} x^{2} + 27 \, d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [B]  time = 0.16, size = 115, normalized size = 2.30 \[ \frac {8}{5} \, x^{5} e \mathrm {sgn}\left (2 \, x + 3\right ) + 2 \, d x^{4} \mathrm {sgn}\left (2 \, x + 3\right ) + 9 \, x^{4} e \mathrm {sgn}\left (2 \, x + 3\right ) + 12 \, d x^{3} \mathrm {sgn}\left (2 \, x + 3\right ) + 18 \, x^{3} e \mathrm {sgn}\left (2 \, x + 3\right ) + 27 \, d x^{2} \mathrm {sgn}\left (2 \, x + 3\right ) + \frac {27}{2} \, x^{2} e \mathrm {sgn}\left (2 \, x + 3\right ) + 27 \, d x \mathrm {sgn}\left (2 \, x + 3\right ) + \frac {81}{80} \, {\left (10 \, d - 3 \, e\right )} \mathrm {sgn}\left (2 \, x + 3\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 62, normalized size = 1.24 \[ \frac {\left (16 x^{4} e +20 d \,x^{3}+90 x^{3} e +120 d \,x^{2}+180 e \,x^{2}+270 d x +135 e x +270 d \right ) \left (\left (2 x +3\right )^{2}\right )^{\frac {3}{2}} x}{10 \left (2 x +3\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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

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maxima [A]  time = 2.47, size = 78, normalized size = 1.56 \[ \frac {1}{20} \, {\left (4 \, x^{2} + 12 \, x + 9\right )}^{\frac {5}{2}} e + \frac {1}{4} \, {\left (4 \, x^{2} + 12 \, x + 9\right )}^{\frac {3}{2}} d x - \frac {3}{8} \, {\left (4 \, x^{2} + 12 \, x + 9\right )}^{\frac {3}{2}} e x + \frac {3}{8} \, {\left (4 \, x^{2} + 12 \, x + 9\right )}^{\frac {3}{2}} d - \frac {9}{16} \, {\left (4 \, x^{2} + 12 \, x + 9\right )}^{\frac {3}{2}} e \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 0.62, size = 67, normalized size = 1.34 \[ \frac {e\,{\left (4\,x^2+12\,x+9\right )}^{5/2}}{20}-\frac {9\,e\,{\left (4\,x^2+12\,x+9\right )}^{3/2}}{16}-\frac {3\,e\,x\,{\left (4\,x^2+12\,x+9\right )}^{3/2}}{8}+\frac {d\,\left (2\,x+3\right )\,{\left (4\,x^2+12\,x+9\right )}^{3/2}}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d + e x\right ) \left (\left (2 x + 3\right )^{2}\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((d + e*x)*((2*x + 3)**2)**(3/2), x)

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