### 3.921 $$\int \frac{(2+e x)^{5/2}}{(12-3 e^2 x^2)^{3/2}} \, dx$$

Optimal. Leaf size=45 $\frac{2 \sqrt{2-e x}}{3 \sqrt{3} e}+\frac{8}{3 \sqrt{3} e \sqrt{2-e x}}$

[Out]

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

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Rubi [A]  time = 0.0171074, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.083, Rules used = {627, 43} $\frac{2 \sqrt{2-e x}}{3 \sqrt{3} e}+\frac{8}{3 \sqrt{3} e \sqrt{2-e x}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(2 + e*x)^(5/2)/(12 - 3*e^2*x^2)^(3/2),x]

[Out]

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

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{(2+e x)^{5/2}}{\left (12-3 e^2 x^2\right )^{3/2}} \, dx &=\int \frac{2+e x}{(6-3 e x)^{3/2}} \, dx\\ &=\int \left (\frac{4}{(6-3 e x)^{3/2}}-\frac{1}{3 \sqrt{6-3 e x}}\right ) \, dx\\ &=\frac{8}{3 \sqrt{3} e \sqrt{2-e x}}+\frac{2 \sqrt{2-e x}}{3 \sqrt{3} e}\\ \end{align*}

Mathematica [A]  time = 0.0566121, size = 35, normalized size = 0.78 $-\frac{2 (e x-6) \sqrt{e x+2}}{3 e \sqrt{12-3 e^2 x^2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(2 + e*x)^(5/2)/(12 - 3*e^2*x^2)^(3/2),x]

[Out]

(-2*(-6 + e*x)*Sqrt[2 + e*x])/(3*e*Sqrt[12 - 3*e^2*x^2])

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Maple [A]  time = 0.04, size = 35, normalized size = 0.8 \begin{align*} 2\,{\frac{ \left ( ex-2 \right ) \left ( ex-6 \right ) \left ( ex+2 \right ) ^{3/2}}{e \left ( -3\,{e}^{2}{x}^{2}+12 \right ) ^{3/2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+2)^(5/2)/(-3*e^2*x^2+12)^(3/2),x)

[Out]

2*(e*x-2)*(e*x-6)*(e*x+2)^(3/2)/e/(-3*e^2*x^2+12)^(3/2)

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Maxima [C]  time = 1.55537, size = 27, normalized size = 0.6 \begin{align*} \frac{2 i \, \sqrt{3}{\left (e x - 6\right )}}{9 \, \sqrt{e x - 2} e} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+2)^(5/2)/(-3*e^2*x^2+12)^(3/2),x, algorithm="maxima")

[Out]

2/9*I*sqrt(3)*(e*x - 6)/(sqrt(e*x - 2)*e)

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Fricas [A]  time = 1.82831, size = 90, normalized size = 2. \begin{align*} \frac{2 \, \sqrt{-3 \, e^{2} x^{2} + 12} \sqrt{e x + 2}{\left (e x - 6\right )}}{9 \,{\left (e^{3} x^{2} - 4 \, e\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+2)^(5/2)/(-3*e^2*x^2+12)^(3/2),x, algorithm="fricas")

[Out]

2/9*sqrt(-3*e^2*x^2 + 12)*sqrt(e*x + 2)*(e*x - 6)/(e^3*x^2 - 4*e)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+2)**(5/2)/(-3*e**2*x**2+12)**(3/2),x)

[Out]

Timed out

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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+2)^(5/2)/(-3*e^2*x^2+12)^(3/2),x, algorithm="giac")

[Out]

sage0*x