3.213 \(\int \frac {x}{(9+12 x+4 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=44 \[ \frac {3}{16 (2 x+3) \left (4 x^2+12 x+9\right )^{3/2}}-\frac {1}{12 \left (4 x^2+12 x+9\right )^{3/2}} \]

[Out]

-1/12/(4*x^2+12*x+9)^(3/2)+3/16/(3+2*x)/(4*x^2+12*x+9)^(3/2)

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Rubi [A]  time = 0.01, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {640, 607} \[ \frac {3}{16 (2 x+3) \left (4 x^2+12 x+9\right )^{3/2}}-\frac {1}{12 \left (4 x^2+12 x+9\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x/(9 + 12*x + 4*x^2)^(5/2),x]

[Out]

-1/(12*(9 + 12*x + 4*x^2)^(3/2)) + 3/(16*(3 + 2*x)*(9 + 12*x + 4*x^2)^(3/2))

Rule 607

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(2*(a + b*x + c*x^2)^(p + 1))/((2*p + 1)*(b + 2
*c*x)), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && LtQ[p, -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 \frac {x}{\left (9+12 x+4 x^2\right )^{5/2}} \, dx &=-\frac {1}{12 \left (9+12 x+4 x^2\right )^{3/2}}-\frac {3}{2} \int \frac {1}{\left (9+12 x+4 x^2\right )^{5/2}} \, dx\\ &=-\frac {1}{12 \left (9+12 x+4 x^2\right )^{3/2}}+\frac {3}{16 (3+2 x) \left (9+12 x+4 x^2\right )^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 27, normalized size = 0.61 \[ \frac {-8 x-3}{48 (2 x+3)^3 \sqrt {(2 x+3)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3 - 8*x)/(48*(3 + 2*x)^3*Sqrt[(3 + 2*x)^2])

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fricas [A]  time = 1.00, size = 29, normalized size = 0.66 \[ -\frac {8 \, x + 3}{48 \, {\left (16 \, x^{4} + 96 \, x^{3} + 216 \, x^{2} + 216 \, x + 81\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(8*x + 3)/(16*x^4 + 96*x^3 + 216*x^2 + 216*x + 81)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.05, size = 22, normalized size = 0.50 \[ -\frac {\left (2 x +3\right ) \left (8 x +3\right )}{48 \left (\left (2 x +3\right )^{2}\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(2*x+3)*(8*x+3)/((2*x+3)^2)^(5/2)

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maxima [A]  time = 2.95, size = 24, normalized size = 0.55 \[ -\frac {1}{12 \, {\left (4 \, x^{2} + 12 \, x + 9\right )}^{\frac {3}{2}}} + \frac {3}{16 \, {\left (2 \, x + 3\right )}^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12/(4*x^2 + 12*x + 9)^(3/2) + 3/16/(2*x + 3)^4

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mupad [B]  time = 0.18, size = 26, normalized size = 0.59 \[ -\frac {\left (8\,x+3\right )\,\sqrt {4\,x^2+12\,x+9}}{48\,{\left (2\,x+3\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(12*x + 4*x^2 + 9)^(5/2),x)

[Out]

-((8*x + 3)*(12*x + 4*x^2 + 9)^(1/2))/(48*(2*x + 3)^5)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\left (\left (2 x + 3\right )^{2}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/((2*x + 3)**2)**(5/2), x)

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