3.10.23 \(\int \frac {\sqrt {2+e x}}{(12-3 e^2 x^2)^{3/2}} \, dx\) [923]

Optimal. Leaf size=50 \[ \frac {1}{6 \sqrt {3} e \sqrt {2-e x}}-\frac {\tanh ^{-1}\left (\frac {1}{2} \sqrt {2-e x}\right )}{12 \sqrt {3} e} \]

[Out]

-1/36*arctanh(1/2*(-e*x+2)^(1/2))*3^(1/2)/e+1/18/e*3^(1/2)/(-e*x+2)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 50, normalized size of antiderivative = 1.00, 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 = {641, 53, 65, 212} \begin {gather*} \frac {1}{6 \sqrt {3} e \sqrt {2-e x}}-\frac {\tanh ^{-1}\left (\frac {1}{2} \sqrt {2-e x}\right )}{12 \sqrt {3} e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

1/(6*Sqrt[3]*e*Sqrt[2 - e*x]) - ArcTanh[Sqrt[2 - e*x]/2]/(12*Sqrt[3]*e)

Rule 53

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 65

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 212

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

Rule 641

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c/e)*x)^
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 {\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)} \, dx\\ &=\frac {1}{6 \sqrt {3} e \sqrt {2-e x}}+\frac {1}{12} \int \frac {1}{\sqrt {6-3 e x} (2+e x)} \, dx\\ &=\frac {1}{6 \sqrt {3} e \sqrt {2-e x}}-\frac {\text {Subst}\left (\int \frac {1}{4-\frac {x^2}{3}} \, dx,x,\sqrt {6-3 e x}\right )}{18 e}\\ &=\frac {1}{6 \sqrt {3} e \sqrt {2-e x}}-\frac {\tanh ^{-1}\left (\frac {1}{2} \sqrt {2-e x}\right )}{12 \sqrt {3} e}\\ \end {align*}

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Mathematica [A]
time = 0.24, size = 83, normalized size = 1.66 \begin {gather*} \frac {-2 \sqrt {4-e^2 x^2}-(-2+e x) \sqrt {2+e x} \tanh ^{-1}\left (\frac {2 \sqrt {2+e x}}{\sqrt {4-e^2 x^2}}\right )}{12 e (-2+e x) \sqrt {6+3 e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[4 - e^2*x^2] - (-2 + e*x)*Sqrt[2 + e*x]*ArcTanh[(2*Sqrt[2 + e*x])/Sqrt[4 - e^2*x^2]])/(12*e*(-2 + e*x
)*Sqrt[6 + 3*e*x])

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Maple [A]
time = 0.49, size = 60, normalized size = 1.20

method	result	size
default	\(\frac {\sqrt {-3 e^{2} x^{2}+12}\, \left (\sqrt {3}\, \arctanh \left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right ) \sqrt {-3 e x +6}-6\right )}{108 \sqrt {e x +2}\, \left (e x -2\right ) e}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/108/(e*x+2)^(1/2)*(-3*e^2*x^2+12)^(1/2)*(3^(1/2)*arctanh(1/6*(-3*e*x+6)^(1/2)*3^(1/2))*(-3*e*x+6)^(1/2)-6)/(
e*x-2)/e

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 105 vs. \(2 (36) = 72\).
time = 3.78, size = 105, normalized size = 2.10 \begin {gather*} \frac {\sqrt {3} {\left (x^{2} e^{2} - 4\right )} \log \left (-\frac {3 \, x^{2} e^{2} - 12 \, x e + 4 \, \sqrt {3} \sqrt {-3 \, x^{2} e^{2} + 12} \sqrt {x e + 2} - 36}{x^{2} e^{2} + 4 \, x e + 4}\right ) - 4 \, \sqrt {-3 \, x^{2} e^{2} + 12} \sqrt {x e + 2}}{72 \, {\left (x^{2} e^{3} - 4 \, e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/72*(sqrt(3)*(x^2*e^2 - 4)*log(-(3*x^2*e^2 - 12*x*e + 4*sqrt(3)*sqrt(-3*x^2*e^2 + 12)*sqrt(x*e + 2) - 36)/(x^
2*e^2 + 4*x*e + 4)) - 4*sqrt(-3*x^2*e^2 + 12)*sqrt(x*e + 2))/(x^2*e^3 - 4*e)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\sqrt {3} \int \frac {\sqrt {e x + 2}}{- e^{2} x^{2} \sqrt {- e^{2} x^{2} + 4} + 4 \sqrt {- e^{2} x^{2} + 4}}\, dx}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.73, size = 57, normalized size = 1.14 \begin {gather*} -\frac {1}{72} \, \sqrt {3} e^{\left (-1\right )} \log \left (\sqrt {-x e + 2} + 2\right ) + \frac {1}{72} \, \sqrt {3} e^{\left (-1\right )} \log \left (-\sqrt {-x e + 2} + 2\right ) + \frac {\sqrt {3} e^{\left (-1\right )}}{18 \, \sqrt {-x e + 2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/72*sqrt(3)*e^(-1)*log(sqrt(-x*e + 2) + 2) + 1/72*sqrt(3)*e^(-1)*log(-sqrt(-x*e + 2) + 2) + 1/18*sqrt(3)*e^(
-1)/sqrt(-x*e + 2)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\sqrt {e\,x+2}}{{\left (12-3\,e^2\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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