3.917 \(\int \frac{1}{(2+e x)^{5/2} \sqrt{12-3 e^2 x^2}} \, dx\)

Optimal. Leaf size=86 \[ -\frac{\sqrt{3} \sqrt{2-e x}}{64 e (e x+2)}-\frac{\sqrt{2-e x}}{8 \sqrt{3} e (e x+2)^2}-\frac{\sqrt{3} \tanh ^{-1}\left (\frac{1}{2} \sqrt{2-e x}\right )}{128 e} \]

[Out]

-Sqrt[2 - e*x]/(8*Sqrt[3]*e*(2 + e*x)^2) - (Sqrt[3]*Sqrt[2 - e*x])/(64*e*(2 + e*x)) - (Sqrt[3]*ArcTanh[Sqrt[2
- e*x]/2])/(128*e)

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Rubi [A]  time = 0.0278343, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {627, 51, 63, 206} \[ -\frac{\sqrt{3} \sqrt{2-e x}}{64 e (e x+2)}-\frac{\sqrt{2-e x}}{8 \sqrt{3} e (e x+2)^2}-\frac{\sqrt{3} \tanh ^{-1}\left (\frac{1}{2} \sqrt{2-e x}\right )}{128 e} \]

Antiderivative was successfully verified.

[In]

Int[1/((2 + e*x)^(5/2)*Sqrt[12 - 3*e^2*x^2]),x]

[Out]

-Sqrt[2 - e*x]/(8*Sqrt[3]*e*(2 + e*x)^2) - (Sqrt[3]*Sqrt[2 - e*x])/(64*e*(2 + e*x)) - (Sqrt[3]*ArcTanh[Sqrt[2
- e*x]/2])/(128*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)^{5/2} \sqrt{12-3 e^2 x^2}} \, dx &=\int \frac{1}{\sqrt{6-3 e x} (2+e x)^3} \, dx\\ &=-\frac{\sqrt{2-e x}}{8 \sqrt{3} e (2+e x)^2}+\frac{3}{16} \int \frac{1}{\sqrt{6-3 e x} (2+e x)^2} \, dx\\ &=-\frac{\sqrt{2-e x}}{8 \sqrt{3} e (2+e x)^2}-\frac{\sqrt{3} \sqrt{2-e x}}{64 e (2+e x)}+\frac{3}{128} \int \frac{1}{\sqrt{6-3 e x} (2+e x)} \, dx\\ &=-\frac{\sqrt{2-e x}}{8 \sqrt{3} e (2+e x)^2}-\frac{\sqrt{3} \sqrt{2-e x}}{64 e (2+e x)}-\frac{\operatorname{Subst}\left (\int \frac{1}{4-\frac{x^2}{3}} \, dx,x,\sqrt{6-3 e x}\right )}{64 e}\\ &=-\frac{\sqrt{2-e x}}{8 \sqrt{3} e (2+e x)^2}-\frac{\sqrt{3} \sqrt{2-e x}}{64 e (2+e x)}-\frac{\sqrt{3} \tanh ^{-1}\left (\frac{1}{2} \sqrt{2-e x}\right )}{128 e}\\ \end{align*}

Mathematica [C]  time = 0.0608295, size = 53, normalized size = 0.62 \[ \frac{(e x-2) \sqrt{e x+2} \, _2F_1\left (\frac{1}{2},3;\frac{3}{2};\frac{1}{2}-\frac{e x}{4}\right )}{32 e \sqrt{12-3 e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((2 + e*x)^(5/2)*Sqrt[12 - 3*e^2*x^2]),x]

[Out]

((-2 + e*x)*Sqrt[2 + e*x]*Hypergeometric2F1[1/2, 3, 3/2, 1/2 - (e*x)/4])/(32*e*Sqrt[12 - 3*e^2*x^2])

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Maple [A]  time = 0.128, size = 126, normalized size = 1.5 \begin{align*} -{\frac{\sqrt{3}}{384\,e}\sqrt{-{e}^{2}{x}^{2}+4} \left ( 3\,\sqrt{3}{\it Artanh} \left ( 1/6\,\sqrt{3}\sqrt{-3\,ex+6} \right ){x}^{2}{e}^{2}+12\,\sqrt{3}{\it Artanh} \left ( 1/6\,\sqrt{3}\sqrt{-3\,ex+6} \right ) xe+6\,xe\sqrt{-3\,ex+6}+12\,\sqrt{3}{\it Artanh} \left ( 1/6\,\sqrt{3}\sqrt{-3\,ex+6} \right ) +28\,\sqrt{-3\,ex+6} \right ){\frac{1}{\sqrt{ \left ( ex+2 \right ) ^{5}}}}{\frac{1}{\sqrt{-3\,ex+6}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+2)^(5/2)/(-3*e^2*x^2+12)^(1/2),x)

[Out]

-1/384*(-e^2*x^2+4)^(1/2)*(3*3^(1/2)*arctanh(1/6*3^(1/2)*(-3*e*x+6)^(1/2))*x^2*e^2+12*3^(1/2)*arctanh(1/6*3^(1
/2)*(-3*e*x+6)^(1/2))*x*e+6*x*e*(-3*e*x+6)^(1/2)+12*3^(1/2)*arctanh(1/6*3^(1/2)*(-3*e*x+6)^(1/2))+28*(-3*e*x+6
)^(1/2))/((e*x+2)^5)^(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{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+2)^(5/2)/(-3*e^2*x^2+12)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(-3*e^2*x^2 + 12)*(e*x + 2)^(5/2)), x)

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Fricas [B]  time = 1.90118, size = 335, normalized size = 3.9 \begin{align*} \frac{3 \, \sqrt{3}{\left (e^{3} x^{3} + 6 \, e^{2} x^{2} + 12 \, e x + 8\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}{\left (3 \, e x + 14\right )} \sqrt{e x + 2}}{768 \,{\left (e^{4} x^{3} + 6 \, e^{3} x^{2} + 12 \, e^{2} x + 8 \, e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+2)^(5/2)/(-3*e^2*x^2+12)^(1/2),x, algorithm="fricas")

[Out]

1/768*(3*sqrt(3)*(e^3*x^3 + 6*e^2*x^2 + 12*e*x + 8)*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)*(3*e*x + 14)*sqrt(e*x + 2))/(e^4*x^3 + 6
*e^3*x^2 + 12*e^2*x + 8*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(1/(e*x+2)**(5/2)/(-3*e**2*x**2+12)**(1/2),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(1/(e*x+2)^(5/2)/(-3*e^2*x^2+12)^(1/2),x, algorithm="giac")

[Out]

sage0*x