### 3.214 $$\int \frac{x}{(9+12 x+4 x^2)^{7/2}} \, dx$$

Optimal. Leaf size=44 $\frac{1}{8 (2 x+3) \left (4 x^2+12 x+9\right )^{5/2}}-\frac{1}{20 \left (4 x^2+12 x+9\right )^{5/2}}$

[Out]

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

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Rubi [A]  time = 0.0091762, antiderivative size = 44, normalized size of antiderivative = 1., 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} $\frac{1}{8 (2 x+3) \left (4 x^2+12 x+9\right )^{5/2}}-\frac{1}{20 \left (4 x^2+12 x+9\right )^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[x/(9 + 12*x + 4*x^2)^(7/2),x]

[Out]

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

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]

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]

Rubi steps

\begin{align*} \int \frac{x}{\left (9+12 x+4 x^2\right )^{7/2}} \, dx &=-\frac{1}{20 \left (9+12 x+4 x^2\right )^{5/2}}-\frac{3}{2} \int \frac{1}{\left (9+12 x+4 x^2\right )^{7/2}} \, dx\\ &=-\frac{1}{20 \left (9+12 x+4 x^2\right )^{5/2}}+\frac{1}{8 (3+2 x) \left (9+12 x+4 x^2\right )^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0086804, size = 27, normalized size = 0.61 $\frac{-4 x-1}{40 (2 x+3)^5 \sqrt{(2 x+3)^2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x/(9 + 12*x + 4*x^2)^(7/2),x]

[Out]

(-1 - 4*x)/(40*(3 + 2*x)^5*Sqrt[(3 + 2*x)^2])

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Maple [A]  time = 0.066, size = 22, normalized size = 0.5 \begin{align*} -{\frac{ \left ( 3+2\,x \right ) \left ( 4\,x+1 \right ) }{40} \left ( \left ( 3+2\,x \right ) ^{2} \right ) ^{-{\frac{7}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x/(4*x^2+12*x+9)^(7/2),x)

[Out]

-1/40*(3+2*x)*(4*x+1)/((3+2*x)^2)^(7/2)

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Maxima [A]  time = 1.72885, size = 32, normalized size = 0.73 \begin{align*} -\frac{1}{20 \,{\left (4 \, x^{2} + 12 \, x + 9\right )}^{\frac{5}{2}}} + \frac{1}{8 \,{\left (2 \, x + 3\right )}^{6}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(4*x^2+12*x+9)^(7/2),x, algorithm="maxima")

[Out]

-1/20/(4*x^2 + 12*x + 9)^(5/2) + 1/8/(2*x + 3)^6

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Fricas [A]  time = 1.59023, size = 113, normalized size = 2.57 \begin{align*} -\frac{4 \, x + 1}{40 \,{\left (64 \, x^{6} + 576 \, x^{5} + 2160 \, x^{4} + 4320 \, x^{3} + 4860 \, x^{2} + 2916 \, x + 729\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(4*x^2+12*x+9)^(7/2),x, algorithm="fricas")

[Out]

-1/40*(4*x + 1)/(64*x^6 + 576*x^5 + 2160*x^4 + 4320*x^3 + 4860*x^2 + 2916*x + 729)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\left (\left (2 x + 3\right )^{2}\right )^{\frac{7}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(4*x**2+12*x+9)**(7/2),x)

[Out]

Integral(x/((2*x + 3)**2)**(7/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(4*x^2+12*x+9)^(7/2),x, algorithm="giac")

[Out]

sage0*x