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

Optimal. Leaf size=25 \[ -\frac {\tanh ^{-1}\left (\frac {1}{2} \sqrt {2-e x}\right )}{\sqrt {3} e} \]

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {627, 63, 206} \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {1}{2} \sqrt {2-e x}\right )}{\sqrt {3} e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 50, normalized size = 2.00 \begin {gather*} \frac {\sqrt {e x-2} \sqrt {e x+2} \tan ^{-1}\left (\frac {1}{2} \sqrt {e x-2}\right )}{e \sqrt {12-3 e^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [B]  time = 0.30, size = 96, normalized size = 3.84 \begin {gather*} \frac {\log \left (\sqrt {4 (e x+2)-(e x+2)^2}-2 \sqrt {e x+2}\right )}{2 \sqrt {3} e}-\frac {\log \left (2 e \sqrt {e x+2}+e \sqrt {4 (e x+2)-(e x+2)^2}\right )}{2 \sqrt {3} e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.41, size = 64, normalized size = 2.56 \begin {gather*} \frac {\sqrt {3} \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 )}{6 \, e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*sqrt(3)*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))/e

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {-3 \, e^{2} x^{2} + 12} \sqrt {e x + 2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.07, size = 50, normalized size = 2.00 \begin {gather*} -\frac {\sqrt {-e^{2} x^{2}+4}\, \sqrt {3}\, \arctanh \left (\frac {\sqrt {3}\, \sqrt {-3 e x +6}}{6}\right )}{3 \sqrt {e x +2}\, \sqrt {-e x +2}\, e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {-3 \, e^{2} x^{2} + 12} \sqrt {e x + 2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {1}{\sqrt {12-3\,e^2\,x^2}\,\sqrt {e\,x+2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________