3.898 \(\int \frac{\sqrt{12-3 e^2 x^2}}{(2+e x)^{3/2}} \, dx\)

Optimal. Leaf size=46 \[ \frac{2 \sqrt{3} \sqrt{2-e x}}{e}-\frac{4 \sqrt{3} \tanh ^{-1}\left (\frac{1}{2} \sqrt{2-e x}\right )}{e} \]

[Out]

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

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Rubi [A]  time = 0.0202946, antiderivative size = 46, 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, 50, 63, 206} \[ \frac{2 \sqrt{3} \sqrt{2-e x}}{e}-\frac{4 \sqrt{3} \tanh ^{-1}\left (\frac{1}{2} \sqrt{2-e x}\right )}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[3]*Sqrt[2 - e*x])/e - (4*Sqrt[3]*ArcTanh[Sqrt[2 - e*x]/2])/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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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{\sqrt{12-3 e^2 x^2}}{(2+e x)^{3/2}} \, dx &=\int \frac{\sqrt{6-3 e x}}{2+e x} \, dx\\ &=\frac{2 \sqrt{3} \sqrt{2-e x}}{e}+12 \int \frac{1}{\sqrt{6-3 e x} (2+e x)} \, dx\\ &=\frac{2 \sqrt{3} \sqrt{2-e x}}{e}-\frac{8 \operatorname{Subst}\left (\int \frac{1}{4-\frac{x^2}{3}} \, dx,x,\sqrt{6-3 e x}\right )}{e}\\ &=\frac{2 \sqrt{3} \sqrt{2-e x}}{e}-\frac{4 \sqrt{3} \tanh ^{-1}\left (\frac{1}{2} \sqrt{2-e x}\right )}{e}\\ \end{align*}

Mathematica [A]  time = 0.0602468, size = 63, normalized size = 1.37 \[ \frac{2 \sqrt{12-3 e^2 x^2} \left (\sqrt{e x-2}-2 \tan ^{-1}\left (\frac{1}{2} \sqrt{e x-2}\right )\right )}{e \sqrt{e x-2} \sqrt{e x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.143, size = 66, normalized size = 1.4 \begin{align*} -2\,{\frac{\sqrt{-{e}^{2}{x}^{2}+4} \left ( 2\,\sqrt{3}{\it Artanh} \left ( 1/6\,\sqrt{3}\sqrt{-3\,ex+6} \right ) -\sqrt{-3\,ex+6} \right ) \sqrt{3}}{\sqrt{ex+2}\sqrt{-3\,ex+6}e}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(-e^2*x^2+4)^(1/2)*(2*3^(1/2)*arctanh(1/6*3^(1/2)*(-3*e*x+6)^(1/2))-(-3*e*x+6)^(1/2))/(e*x+2)^(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{\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((-3*e^2*x^2+12)^(1/2)/(e*x+2)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.13564, size = 239, normalized size = 5.2 \begin{align*} \frac{2 \,{\left (\sqrt{3}{\left (e x + 2\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 ) + \sqrt{-3 \, e^{2} x^{2} + 12} \sqrt{e x + 2}\right )}}{e^{2} x + 2 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\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((-3*e^2*x^2+12)^(1/2)/(e*x+2)^(3/2),x, algorithm="giac")

[Out]

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