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

Optimal. Leaf size=79 \[ \frac {1}{16 \sqrt {3} e \sqrt {2-e x}}-\frac {1}{12 \sqrt {3} e \sqrt {2-e x} (e x+2)}-\frac {\tanh ^{-1}\left (\frac {1}{2} \sqrt {2-e x}\right )}{32 \sqrt {3} e} \]

[Out]

-1/96*arctanh(1/2*(-e*x+2)^(1/2))*3^(1/2)/e+1/48/e*3^(1/2)/(-e*x+2)^(1/2)-1/36/e/(e*x+2)*3^(1/2)/(-e*x+2)^(1/2
)

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Rubi [A]  time = 0.03, antiderivative size = 86, normalized size of antiderivative = 1.09, 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} \[ -\frac {\sqrt {2-e x}}{16 \sqrt {3} e (e x+2)}+\frac {1}{6 \sqrt {3} e \sqrt {2-e x} (e x+2)}-\frac {\tanh ^{-1}\left (\frac {1}{2} \sqrt {2-e x}\right )}{32 \sqrt {3} e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

1/(6*Sqrt[3]*e*Sqrt[2 - e*x]*(2 + e*x)) - Sqrt[2 - e*x]/(16*Sqrt[3]*e*(2 + e*x)) - ArcTanh[Sqrt[2 - e*x]/2]/(3
2*Sqrt[3]*e)

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])

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]))

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {2+e x} \left (12-3 e^2 x^2\right )^{3/2}} \, dx &=\int \frac {1}{(6-3 e x)^{3/2} (2+e x)^2} \, dx\\ &=\frac {1}{6 \sqrt {3} e \sqrt {2-e x} (2+e x)}+\frac {1}{4} \int \frac {1}{\sqrt {6-3 e x} (2+e x)^2} \, dx\\ &=\frac {1}{6 \sqrt {3} e \sqrt {2-e x} (2+e x)}-\frac {\sqrt {2-e x}}{16 \sqrt {3} e (2+e x)}+\frac {1}{32} \int \frac {1}{\sqrt {6-3 e x} (2+e x)} \, dx\\ &=\frac {1}{6 \sqrt {3} e \sqrt {2-e x} (2+e x)}-\frac {\sqrt {2-e x}}{16 \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 )}{48 e}\\ &=\frac {1}{6 \sqrt {3} e \sqrt {2-e x} (2+e x)}-\frac {\sqrt {2-e x}}{16 \sqrt {3} e (2+e x)}-\frac {\tanh ^{-1}\left (\frac {1}{2} \sqrt {2-e x}\right )}{32 \sqrt {3} e}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 48, normalized size = 0.61 \[ \frac {\sqrt {e x+2} \, _2F_1\left (-\frac {1}{2},2;\frac {1}{2};\frac {1}{2}-\frac {e x}{4}\right )}{24 e \sqrt {12-3 e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2 + e*x]*Hypergeometric2F1[-1/2, 2, 1/2, 1/2 - (e*x)/4])/(24*e*Sqrt[12 - 3*e^2*x^2])

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fricas [B]  time = 0.99, size = 139, normalized size = 1.76 \[ \frac {3 \, \sqrt {3} {\left (e^{3} x^{3} + 2 \, e^{2} x^{2} - 4 \, e x - 8\right )} \log \left (-\frac {3 \, e^{2} x^{2} - 12 \, e x + 4 \, \sqrt {3} \sqrt {-3 \, e^{2} x^{2} + 12} \sqrt {e x + 2} - 36}{e^{2} x^{2} + 4 \, e x + 4}\right ) - 4 \, \sqrt {-3 \, e^{2} x^{2} + 12} {\left (3 \, e x + 2\right )} \sqrt {e x + 2}}{576 \, {\left (e^{4} x^{3} + 2 \, e^{3} x^{2} - 4 \, e^{2} x - 8 \, e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/576*(3*sqrt(3)*(e^3*x^3 + 2*e^2*x^2 - 4*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 + 2)*sqrt(e*x + 2))/(e^4*x^3 + 2*e
^3*x^2 - 4*e^2*x - 8*e)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.07, size = 93, normalized size = 1.18 \[ \frac {\sqrt {-3 e^{2} x^{2}+12}\, \left (\sqrt {3}\, \sqrt {-3 e x +6}\, e x \arctanh \left (\frac {\sqrt {3}\, \sqrt {-3 e x +6}}{6}\right )-6 e x +2 \sqrt {3}\, \sqrt {-3 e x +6}\, \arctanh \left (\frac {\sqrt {3}\, \sqrt {-3 e x +6}}{6}\right )-4\right )}{288 \left (e x +2\right )^{\frac {3}{2}} \left (e x -2\right ) e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/288/(e*x+2)^(3/2)*(-3*e^2*x^2+12)^(1/2)*(3^(1/2)*(-3*e*x+6)^(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))*(-3*e*x+6)^(1/2)-6*e*x-4)/(e*x-2)/e

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {3}{2}} \sqrt {e x + 2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (12-3\,e^2\,x^2\right )}^{3/2}\,\sqrt {e\,x+2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\sqrt {3} \int \frac {1}{- e^{2} x^{2} \sqrt {e x + 2} \sqrt {- e^{2} x^{2} + 4} + 4 \sqrt {e x + 2} \sqrt {- e^{2} x^{2} + 4}}\, dx}{9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(3)*Integral(1/(-e**2*x**2*sqrt(e*x + 2)*sqrt(-e**2*x**2 + 4) + 4*sqrt(e*x + 2)*sqrt(-e**2*x**2 + 4)), x)/
9

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