3.9.98 \(\int \frac {\sqrt {12-3 e^2 x^2}}{(2+e x)^{3/2}} \, dx\) [898]

Optimal. Leaf size=46 \[ \frac {2 \sqrt {3} \sqrt {2-e x}}{e}-\frac {4 \sqrt {3} \tanh ^{-1}\left (\frac {1}{2} \sqrt {2-e x}\right )}{e} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 46, 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, 52, 65, 212} \begin {gather*} \frac {2 \sqrt {3} \sqrt {2-e x}}{e}-\frac {4 \sqrt {3} \tanh ^{-1}\left (\frac {1}{2} \sqrt {2-e x}\right )}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[3]*Sqrt[2 - e*x])/e - (4*Sqrt[3]*ArcTanh[Sqrt[2 - e*x]/2])/e

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 {12-3 e^2 x^2}}{(2+e x)^{3/2}} \, dx &=\int \frac {\sqrt {6-3 e x}}{2+e x} \, dx\\ &=\frac {2 \sqrt {3} \sqrt {2-e x}}{e}+12 \int \frac {1}{\sqrt {6-3 e x} (2+e x)} \, dx\\ &=\frac {2 \sqrt {3} \sqrt {2-e x}}{e}-\frac {8 \text {Subst}\left (\int \frac {1}{4-\frac {x^2}{3}} \, dx,x,\sqrt {6-3 e x}\right )}{e}\\ &=\frac {2 \sqrt {3} \sqrt {2-e x}}{e}-\frac {4 \sqrt {3} \tanh ^{-1}\left (\frac {1}{2} \sqrt {2-e x}\right )}{e}\\ \end {align*}

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Mathematica [A]
time = 0.21, size = 63, normalized size = 1.37 \begin {gather*} \frac {2 \sqrt {3} \left (\frac {\sqrt {4-e^2 x^2}}{\sqrt {2+e x}}-2 \tanh ^{-1}\left (\frac {2 \sqrt {2+e x}}{\sqrt {4-e^2 x^2}}\right )\right )}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.58, size = 66, normalized size = 1.43

method	result	size
default	\(-\frac {2 \sqrt {-e^{2} x^{2}+4}\, \left (2 \sqrt {3}\, \arctanh \left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right )-\sqrt {-3 e x +6}\right ) \sqrt {3}}{\sqrt {e x +2}\, \sqrt {-3 e x +6}\, e}\)	\(66\)
risch	\(-\frac {6 \left (e x -2\right ) \sqrt {\frac {-3 e^{2} x^{2}+12}{e x +2}}\, \sqrt {e x +2}}{e \sqrt {-3 e x +6}\, \sqrt {-3 e^{2} x^{2}+12}}-\frac {4 \sqrt {3}\, \arctanh \left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right ) \sqrt {\frac {-3 e^{2} x^{2}+12}{e x +2}}\, \sqrt {e x +2}}{e \sqrt {-3 e^{2} x^{2}+12}}\)	\(120\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (36) = 72\).
time = 2.20, size = 100, normalized size = 2.17 \begin {gather*} \frac {2 \, {\left (\sqrt {3} {\left (x e + 2\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 ) + \sqrt {-3 \, x^{2} e^{2} + 12} \sqrt {x e + 2}\right )}}{x e^{2} + 2 \, e} \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)^(3/2),x, algorithm="fricas")

[Out]

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

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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 x \sqrt {e x + 2} + 2 \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)**(3/2),x)

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \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)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Error index.cc index_gcd Error: Bad Argument ValueError index.cc index_gcd Error: Bad Argument Value

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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 )}^{3/2}} \,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)^(3/2),x)

[Out]

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

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