### 3.916 $$\int \frac{1}{(2+e x)^{3/2} \sqrt{12-3 e^2 x^2}} \, dx$$

Optimal. Leaf size=57 $-\frac{\sqrt{2-e x}}{4 \sqrt{3} e (e x+2)}-\frac{\tanh ^{-1}\left (\frac{1}{2} \sqrt{2-e x}\right )}{8 \sqrt{3} e}$

[Out]

-Sqrt[2 - e*x]/(4*Sqrt[3]*e*(2 + e*x)) - ArcTanh[Sqrt[2 - e*x]/2]/(8*Sqrt[3]*e)

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Rubi [A]  time = 0.0207186, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.167, Rules used = {627, 51, 63, 206} $-\frac{\sqrt{2-e x}}{4 \sqrt{3} e (e x+2)}-\frac{\tanh ^{-1}\left (\frac{1}{2} \sqrt{2-e x}\right )}{8 \sqrt{3} e}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/((2 + e*x)^(3/2)*Sqrt[12 - 3*e^2*x^2]),x]

[Out]

-Sqrt[2 - e*x]/(4*Sqrt[3]*e*(2 + e*x)) - ArcTanh[Sqrt[2 - e*x]/2]/(8*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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0614101, size = 54, normalized size = 0.95 $\frac{-2 \sqrt{2-e x}-(e x+2) \tanh ^{-1}\left (\frac{1}{2} \sqrt{2-e x}\right )}{8 \sqrt{3} e (e x+2)}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((2 + e*x)^(3/2)*Sqrt[12 - 3*e^2*x^2]),x]

[Out]

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

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Maple [B]  time = 0.127, size = 88, normalized size = 1.5 \begin{align*} -{\frac{\sqrt{3}}{24\,e}\sqrt{-{e}^{2}{x}^{2}+4} \left ( \sqrt{3}{\it Artanh} \left ({\frac{\sqrt{3}}{6}\sqrt{-3\,ex+6}} \right ) xe+2\,\sqrt{3}{\it Artanh} \left ( 1/6\,\sqrt{3}\sqrt{-3\,ex+6} \right ) +2\,\sqrt{-3\,ex+6} \right ){\frac{1}{\sqrt{ \left ( ex+2 \right ) ^{3}}}}{\frac{1}{\sqrt{-3\,ex+6}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+2)^(3/2)/(-3*e^2*x^2+12)^(1/2),x)

[Out]

-1/24*(-e^2*x^2+4)^(1/2)*(3^(1/2)*arctanh(1/6*3^(1/2)*(-3*e*x+6)^(1/2))*x*e+2*3^(1/2)*arctanh(1/6*3^(1/2)*(-3*
e*x+6)^(1/2))+2*(-3*e*x+6)^(1/2))/((e*x+2)^3)^(1/2)*3^(1/2)/(-3*e*x+6)^(1/2)/e

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-3 \, e^{2} x^{2} + 12}{\left (e x + 2\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+2)^(3/2)/(-3*e^2*x^2+12)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(-3*e^2*x^2 + 12)*(e*x + 2)^(3/2)), x)

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Fricas [B]  time = 1.78991, size = 278, normalized size = 4.88 \begin{align*} \frac{\sqrt{3}{\left (e^{2} x^{2} + 4 \, e x + 4\right )} \log \left (-\frac{3 \, e^{2} x^{2} - 12 \, e x + 4 \, \sqrt{3} \sqrt{-3 \, e^{2} x^{2} + 12} \sqrt{e x + 2} - 36}{e^{2} x^{2} + 4 \, e x + 4}\right ) - 4 \, \sqrt{-3 \, e^{2} x^{2} + 12} \sqrt{e x + 2}}{48 \,{\left (e^{3} x^{2} + 4 \, e^{2} x + 4 \, e\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+2)^(3/2)/(-3*e^2*x^2+12)^(1/2),x, algorithm="fricas")

[Out]

1/48*(sqrt(3)*(e^2*x^2 + 4*e*x + 4)*log(-(3*e^2*x^2 - 12*e*x + 4*sqrt(3)*sqrt(-3*e^2*x^2 + 12)*sqrt(e*x + 2) -
36)/(e^2*x^2 + 4*e*x + 4)) - 4*sqrt(-3*e^2*x^2 + 12)*sqrt(e*x + 2))/(e^3*x^2 + 4*e^2*x + 4*e)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\sqrt{3} \int \frac{1}{e x \sqrt{e x + 2} \sqrt{- e^{2} x^{2} + 4} + 2 \sqrt{e x + 2} \sqrt{- e^{2} x^{2} + 4}}\, dx}{3} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+2)**(3/2)/(-3*e**2*x**2+12)**(1/2),x)

[Out]

sqrt(3)*Integral(1/(e*x*sqrt(e*x + 2)*sqrt(-e**2*x**2 + 4) + 2*sqrt(e*x + 2)*sqrt(-e**2*x**2 + 4)), x)/3

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-3 \, e^{2} x^{2} + 12}{\left (e x + 2\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+2)^(3/2)/(-3*e^2*x^2+12)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(-3*e^2*x^2 + 12)*(e*x + 2)^(3/2)), x)