### 3.1619 $$\int \frac{d+e x}{(9+12 x+4 x^2)^{5/2}} \, dx$$

Optimal. Leaf size=52 $-\frac{2 d-3 e}{16 (2 x+3) \left (4 x^2+12 x+9\right )^{3/2}}-\frac{e}{12 \left (4 x^2+12 x+9\right )^{3/2}}$

[Out]

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

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Rubi [A]  time = 0.0126159, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.1, Rules used = {640, 607} $-\frac{2 d-3 e}{16 (2 x+3) \left (4 x^2+12 x+9\right )^{3/2}}-\frac{e}{12 \left (4 x^2+12 x+9\right )^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0140349, size = 34, normalized size = 0.65 $\frac{-6 d-e (8 x+3)}{48 (2 x+3)^3 \sqrt{(2 x+3)^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-6*d - e*(3 + 8*x))/(48*(3 + 2*x)^3*Sqrt[(3 + 2*x)^2])

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Maple [A]  time = 0.078, size = 28, normalized size = 0.5 \begin{align*} -{\frac{ \left ( 3+2\,x \right ) \left ( 8\,ex+6\,d+3\,e \right ) }{48} \left ( \left ( 3+2\,x \right ) ^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(3+2*x)*(8*e*x+6*d+3*e)/((3+2*x)^2)^(5/2)

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Maxima [A]  time = 1.60159, size = 49, normalized size = 0.94 \begin{align*} -\frac{e}{12 \,{\left (4 \, x^{2} + 12 \, x + 9\right )}^{\frac{3}{2}}} - \frac{d}{8 \,{\left (2 \, x + 3\right )}^{4}} + \frac{3 \, e}{16 \,{\left (2 \, x + 3\right )}^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.51083, size = 92, normalized size = 1.77 \begin{align*} -\frac{8 \, e x + 6 \, d + 3 \, e}{48 \,{\left (16 \, x^{4} + 96 \, x^{3} + 216 \, x^{2} + 216 \, x + 81\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(8*e*x + 6*d + 3*e)/(16*x^4 + 96*x^3 + 216*x^2 + 216*x + 81)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{d + e x}{\left (\left (2 x + 3\right )^{2}\right )^{\frac{5}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((d + e*x)/((2*x + 3)**2)**(5/2), 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((e*x+d)/(4*x^2+12*x+9)^(5/2),x, algorithm="giac")

[Out]

sage0*x