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

Optimal. Leaf size=43 \[ \frac{2 (2-e x)^{3/2}}{3 \sqrt{3} e}-\frac{8 \sqrt{2-e x}}{\sqrt{3} e} \]

[Out]

(-8*Sqrt[2 - e*x])/(Sqrt[3]*e) + (2*(2 - e*x)^(3/2))/(3*Sqrt[3]*e)

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Rubi [A]  time = 0.0167946, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {627, 43} \[ \frac{2 (2-e x)^{3/2}}{3 \sqrt{3} e}-\frac{8 \sqrt{2-e x}}{\sqrt{3} e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*Sqrt[2 - e*x])/(Sqrt[3]*e) + (2*(2 - e*x)^(3/2))/(3*Sqrt[3]*e)

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]))

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{(2+e x)^{3/2}}{\sqrt{12-3 e^2 x^2}} \, dx &=\int \frac{2+e x}{\sqrt{6-3 e x}} \, dx\\ &=\int \left (\frac{4}{\sqrt{6-3 e x}}-\frac{1}{3} \sqrt{6-3 e x}\right ) \, dx\\ &=-\frac{8 \sqrt{2-e x}}{\sqrt{3} e}+\frac{2 (2-e x)^{3/2}}{3 \sqrt{3} e}\\ \end{align*}

Mathematica [A]  time = 0.0506008, size = 40, normalized size = 0.93 \[ \frac{2 (e x-2) \sqrt{e x+2} (e x+10)}{3 e \sqrt{12-3 e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-2 + e*x)*Sqrt[2 + e*x]*(10 + e*x))/(3*e*Sqrt[12 - 3*e^2*x^2])

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Maple [A]  time = 0.04, size = 35, normalized size = 0.8 \begin{align*}{\frac{ \left ( 2\,ex-4 \right ) \left ( ex+10 \right ) }{3\,e}\sqrt{ex+2}{\frac{1}{\sqrt{-3\,{e}^{2}{x}^{2}+12}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(e*x-2)*(e*x+10)*(e*x+2)^(1/2)/e/(-3*e^2*x^2+12)^(1/2)

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Maxima [C]  time = 1.96245, size = 38, normalized size = 0.88 \begin{align*} -\frac{2 i \, \sqrt{3}{\left (e^{2} x^{2} + 8 \, e x - 20\right )}}{9 \, \sqrt{e x - 2} e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/9*I*sqrt(3)*(e^2*x^2 + 8*e*x - 20)/(sqrt(e*x - 2)*e)

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Fricas [A]  time = 1.72781, size = 90, normalized size = 2.09 \begin{align*} -\frac{2 \, \sqrt{-3 \, e^{2} x^{2} + 12}{\left (e x + 10\right )} \sqrt{e x + 2}}{9 \,{\left (e^{2} x + 2 \, e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/9*sqrt(-3*e^2*x^2 + 12)*(e*x + 10)*sqrt(e*x + 2)/(e^2*x + 2*e)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\sqrt{3} \left (\int \frac{2 \sqrt{e x + 2}}{\sqrt{- e^{2} x^{2} + 4}}\, dx + \int \frac{e x \sqrt{e x + 2}}{\sqrt{- e^{2} x^{2} + 4}}\, dx\right )}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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