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

Optimal. Leaf size=43 \[ \frac{2 \sqrt{3} (2-e x)^{5/2}}{5 e}-\frac{8 (2-e x)^{3/2}}{\sqrt{3} e} \]

[Out]

(-8*(2 - e*x)^(3/2))/(Sqrt[3]*e) + (2*Sqrt[3]*(2 - e*x)^(5/2))/(5*e)

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Rubi [A]  time = 0.0153062, 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} \[ \frac{2 \sqrt{3} (2-e x)^{5/2}}{5 e}-\frac{8 (2-e x)^{3/2}}{\sqrt{3} e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*(2 - e*x)^(3/2))/(Sqrt[3]*e) + (2*Sqrt[3]*(2 - e*x)^(5/2))/(5*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 \sqrt{2+e x} \sqrt{12-3 e^2 x^2} \, dx &=\int \sqrt{6-3 e x} (2+e x) \, dx\\ &=\int \left (4 \sqrt{6-3 e x}-\frac{1}{3} (6-3 e x)^{3/2}\right ) \, dx\\ &=-\frac{8 (2-e x)^{3/2}}{\sqrt{3} e}+\frac{2 \sqrt{3} (2-e x)^{5/2}}{5 e}\\ \end{align*}

Mathematica [A]  time = 0.0439205, size = 42, normalized size = 0.98 \[ \frac{2 (e x-2) (3 e x+14) \sqrt{4-e^2 x^2}}{5 e \sqrt{3 e x+6}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-2 + e*x)*(14 + 3*e*x)*Sqrt[4 - e^2*x^2])/(5*e*Sqrt[6 + 3*e*x])

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Maple [A]  time = 0.039, size = 36, normalized size = 0.8 \begin{align*}{\frac{ \left ( 2\,ex-4 \right ) \left ( 3\,ex+14 \right ) }{15\,e}\sqrt{-3\,{e}^{2}{x}^{2}+12}{\frac{1}{\sqrt{ex+2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(e*x-2)*(3*e*x+14)*(-3*e^2*x^2+12)^(1/2)/e/(e*x+2)^(1/2)

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Maxima [C]  time = 2.1625, size = 66, normalized size = 1.53 \begin{align*} \frac{{\left (6 i \, \sqrt{3} e^{2} x^{2} + 16 i \, \sqrt{3} e x - 56 i \, \sqrt{3}\right )}{\left (e x + 2\right )} \sqrt{e x - 2}}{15 \,{\left (e^{2} x + 2 \, e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(6*I*sqrt(3)*e^2*x^2 + 16*I*sqrt(3)*e*x - 56*I*sqrt(3))*(e*x + 2)*sqrt(e*x - 2)/(e^2*x + 2*e)

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Fricas [A]  time = 2.11975, size = 109, normalized size = 2.53 \begin{align*} \frac{2 \,{\left (3 \, e^{2} x^{2} + 8 \, e x - 28\right )} \sqrt{-3 \, e^{2} x^{2} + 12} \sqrt{e x + 2}}{15 \,{\left (e^{2} x + 2 \, e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*e^2*x^2 + 8*e*x - 28)*sqrt(-3*e^2*x^2 + 12)*sqrt(e*x + 2)/(e^2*x + 2*e)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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