3.10.17 \(\int \frac {1}{(2+e x)^{5/2} \sqrt {12-3 e^2 x^2}} \, dx\) [917]

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

[Out]

-1/128*arctanh(1/2*(-e*x+2)^(1/2))*3^(1/2)/e-1/24*3^(1/2)*(-e*x+2)^(1/2)/e/(e*x+2)^2-1/64*3^(1/2)*(-e*x+2)^(1/
2)/e/(e*x+2)

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Rubi [A]
time = 0.02, antiderivative size = 86, normalized size of antiderivative = 1.00, 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 = {641, 44, 65, 212} \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*Sqrt[2 - e*x]/(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 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

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 {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 {\text {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*}

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Mathematica [A]
time = 0.28, size = 81, normalized size = 0.94 \begin {gather*} \frac {-2 (14+3 e x) \sqrt {4-e^2 x^2}-3 (2+e x)^{5/2} \tanh ^{-1}\left (\frac {2 \sqrt {2+e x}}{\sqrt {4-e^2 x^2}}\right )}{128 \sqrt {3} e (2+e x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(14 + 3*e*x)*Sqrt[4 - e^2*x^2] - 3*(2 + e*x)^(5/2)*ArcTanh[(2*Sqrt[2 + e*x])/Sqrt[4 - e^2*x^2]])/(128*Sqrt
[3]*e*(2 + e*x)^(5/2))

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Maple [A]
time = 0.48, size = 126, normalized size = 1.47

method	result	size
default	\(-\frac {\sqrt {-e^{2} x^{2}+4}\, \left (3 \sqrt {3}\, \arctanh \left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right ) e^{2} x^{2}+12 \sqrt {3}\, \arctanh \left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right ) e x +6 e x \sqrt {-3 e x +6}+12 \sqrt {3}\, \arctanh \left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right )+28 \sqrt {-3 e x +6}\right ) \sqrt {3}}{384 \sqrt {\left (e x +2\right )^{5}}\, \sqrt {-3 e x +6}\, e}\)	\(126\)

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,method=_RETURNVERBOSE)

[Out]

-1/384*(-e^2*x^2+4)^(1/2)*(3*3^(1/2)*arctanh(1/6*(-3*e*x+6)^(1/2)*3^(1/2))*e^2*x^2+12*3^(1/2)*arctanh(1/6*(-3*
e*x+6)^(1/2)*3^(1/2))*e*x+6*e*x*(-3*e*x+6)^(1/2)+12*3^(1/2)*arctanh(1/6*(-3*e*x+6)^(1/2)*3^(1/2))+28*(-3*e*x+6
)^(1/2))/((e*x+2)^5)^(1/2)/(-3*e*x+6)^(1/2)*3^(1/2)/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(1/(e*x+2)^(5/2)/(-3*e^2*x^2+12)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 137 vs. \(2 (68) = 136\).
time = 2.96, size = 137, normalized size = 1.59 \begin {gather*} \frac {3 \, \sqrt {3} {\left (x^{3} e^{3} + 6 \, x^{2} e^{2} + 12 \, x e + 8\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} {\left (3 \, x e + 14\right )} \sqrt {x e + 2}}{768 \, {\left (x^{3} e^{4} + 6 \, x^{2} e^{3} + 12 \, x e^{2} + 8 \, e\right )}} \end {gather*}

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)*(x^3*e^3 + 6*x^2*e^2 + 12*x*e + 8)*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)*(3*x*e + 14)*sqrt(x*e + 2))/(x^3*e^4 + 6
*x^2*e^3 + 12*x*e^2 + 8*e)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\sqrt {3} \int \frac {1}{e^{2} x^{2} \sqrt {e x + 2} \sqrt {- e^{2} x^{2} + 4} + 4 e x \sqrt {e x + 2} \sqrt {- e^{2} x^{2} + 4} + 4 \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)**(5/2)/(-3*e**2*x**2+12)**(1/2),x)

[Out]

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

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Giac [A]
time = 4.28, size = 71, normalized size = 0.83 \begin {gather*} \frac {1}{768} \, \sqrt {3} {\left (\frac {4 \, {\left (3 \, {\left (-x e + 2\right )}^{\frac {3}{2}} - 20 \, \sqrt {-x e + 2}\right )}}{{\left (x e + 2\right )}^{2}} - 3 \, \log \left (\sqrt {-x e + 2} + 2\right ) + 3 \, \log \left (-\sqrt {-x e + 2} + 2\right )\right )} e^{\left (-1\right )} \end {gather*}

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]

1/768*sqrt(3)*(4*(3*(-x*e + 2)^(3/2) - 20*sqrt(-x*e + 2))/(x*e + 2)^2 - 3*log(sqrt(-x*e + 2) + 2) + 3*log(-sqr
t(-x*e + 2) + 2))*e^(-1)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\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(1/((12 - 3*e^2*x^2)^(1/2)*(e*x + 2)^(5/2)),x)

[Out]

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

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