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

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

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Rubi [A]  time = 0.02, antiderivative size = 55, 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 = {627, 47, 63, 206} \begin {gather*} \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)} \end {gather*}

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 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])

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 {\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*}

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Mathematica [A]  time = 0.09, size = 74, normalized size = 1.35 \begin {gather*} -\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [B]  time = 0.33, size = 161, normalized size = 2.93 \begin {gather*} \frac {\frac {\sqrt {3} \sqrt {4 (e x+2)-(e x+2)^2}}{e \sqrt {e x+2}}-\frac {\sqrt {3} (e x+2) \tanh ^{-1}\left (\frac {2 \sqrt {e x+2}}{\sqrt {4 (e x+2)-(e x+2)^2}}\right )}{2 e}}{\left (\frac {\sqrt {4 (e x+2)-(e x+2)^2}}{\sqrt {e x+2}}-2\right ) \left (\frac {\sqrt {4 (e x+2)-(e x+2)^2}}{\sqrt {e x+2}}+2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.39, size = 116, normalized size = 2.11 \begin {gather*} \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 {gather*}

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {-3 \, e^{2} x^{2} + 12}}{{\left (e x + 2\right )}^{\frac {5}{2}}}\,{d x} \end {gather*}

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)

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maple [B]  time = 0.10, size = 88, normalized size = 1.60 \begin {gather*} \frac {\sqrt {-e^{2} x^{2}+4}\, \left (\sqrt {3}\, e x \arctanh \left (\frac {\sqrt {3}\, \sqrt {-3 e x +6}}{6}\right )+2 \sqrt {3}\, \arctanh \left (\frac {\sqrt {3}\, \sqrt {-3 e x +6}}{6}\right )-2 \sqrt {-3 e x +6}\right ) \sqrt {3}}{2 \sqrt {\left (e x +2\right )^{3}}\, \sqrt {-3 e x +6}\, e} \end {gather*}

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)*(arctanh(1/6*3^(1/2)*(-3*e*x+6)^(1/2))*3^(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {-3 \, e^{2} x^{2} + 12}}{{\left (e x + 2\right )}^{\frac {5}{2}}}\,{d x} \end {gather*}

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\sqrt {12-3\,e^2\,x^2}}{{\left (e\,x+2\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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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