### 3.899 $$\int \frac{\sqrt{12-3 e^2 x^2}}{(2+e x)^{5/2}} \, dx$$

Optimal. Leaf size=55 $\frac{\sqrt{3} \tanh ^{-1}\left (\frac{1}{2} \sqrt{2-e x}\right )}{2 e}-\frac{\sqrt{3} \sqrt{2-e x}}{e (e x+2)}$

[Out]

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

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Rubi [A]  time = 0.020318, antiderivative size = 55, 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, 47, 63, 206} $\frac{\sqrt{3} \tanh ^{-1}\left (\frac{1}{2} \sqrt{2-e x}\right )}{2 e}-\frac{\sqrt{3} \sqrt{2-e x}}{e (e x+2)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-((Sqrt[3]*Sqrt[2 - e*x])/(e*(2 + e*x))) + (Sqrt[3]*ArcTanh[Sqrt[2 - e*x]/2])/(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 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 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)^{5/2}} \, dx &=\int \frac{\sqrt{6-3 e x}}{(2+e x)^2} \, dx\\ &=-\frac{\sqrt{3} \sqrt{2-e x}}{e (2+e x)}-\frac{3}{2} \int \frac{1}{\sqrt{6-3 e x} (2+e x)} \, dx\\ &=-\frac{\sqrt{3} \sqrt{2-e x}}{e (2+e x)}+\frac{\operatorname{Subst}\left (\int \frac{1}{4-\frac{x^2}{3}} \, dx,x,\sqrt{6-3 e x}\right )}{e}\\ &=-\frac{\sqrt{3} \sqrt{2-e x}}{e (2+e x)}+\frac{\sqrt{3} \tanh ^{-1}\left (\frac{1}{2} \sqrt{2-e x}\right )}{2 e}\\ \end{align*}

Mathematica [A]  time = 0.0887889, size = 74, normalized size = 1.35 $-\frac{\sqrt{12-3 e^2 x^2} \left (2 e x+\sqrt{2-e x} (e x+2) \tanh ^{-1}\left (\frac{1}{2} \sqrt{2-e x}\right )-4\right )}{2 e (e x-2) (e x+2)^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(Sqrt[12 - 3*e^2*x^2]*(-4 + 2*e*x + Sqrt[2 - e*x]*(2 + e*x)*ArcTanh[Sqrt[2 - e*x]/2]))/(2*e*(-2 + e*x)*(2 + e
*x)^(3/2))

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Maple [B]  time = 0.149, size = 88, normalized size = 1.6 \begin{align*}{\frac{\sqrt{3}}{2\,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((-3*e^2*x^2+12)^(1/2)/(e*x+2)^(5/2),x)

[Out]

1/2*(-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{\sqrt{-3 \, e^{2} x^{2} + 12}}{{\left (e x + 2\right )}^{\frac{5}{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)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.88265, size = 277, normalized size = 5.04 \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}}{4 \,{\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((-3*e^2*x^2+12)^(1/2)/(e*x+2)^(5/2),x, algorithm="fricas")

[Out]

1/4*(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*} \sqrt{3} \int \frac{\sqrt{- e^{2} x^{2} + 4}}{e^{2} x^{2} \sqrt{e x + 2} + 4 e x \sqrt{e x + 2} + 4 \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)**(5/2),x)

[Out]

sqrt(3)*Integral(sqrt(-e**2*x**2 + 4)/(e**2*x**2*sqrt(e*x + 2) + 4*e*x*sqrt(e*x + 2) + 4*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{5}{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)^(5/2),x, algorithm="giac")

[Out]

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