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

Optimal. Leaf size=45 \[ \frac{6 \sqrt{3} (2-e x)^{7/2}}{7 e}-\frac{24 \sqrt{3} (2-e x)^{5/2}}{5 e} \]

[Out]

(-24*Sqrt[3]*(2 - e*x)^(5/2))/(5*e) + (6*Sqrt[3]*(2 - e*x)^(7/2))/(7*e)

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Rubi [A]  time = 0.0156231, antiderivative size = 45, 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} \[ \frac{6 \sqrt{3} (2-e x)^{7/2}}{7 e}-\frac{24 \sqrt{3} (2-e x)^{5/2}}{5 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-24*Sqrt[3]*(2 - e*x)^(5/2))/(5*e) + (6*Sqrt[3]*(2 - e*x)^(7/2))/(7*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{\left (12-3 e^2 x^2\right )^{3/2}}{\sqrt{2+e x}} \, dx &=\int (6-3 e x)^{3/2} (2+e x) \, dx\\ &=\int \left (4 (6-3 e x)^{3/2}-\frac{1}{3} (6-3 e x)^{5/2}\right ) \, dx\\ &=-\frac{24 \sqrt{3} (2-e x)^{5/2}}{5 e}+\frac{6 \sqrt{3} (2-e x)^{7/2}}{7 e}\\ \end{align*}

Mathematica [A]  time = 0.0545767, size = 43, normalized size = 0.96 \[ -\frac{6 (e x-2)^2 (5 e x+18) \sqrt{12-3 e^2 x^2}}{35 e \sqrt{e x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*(-2 + e*x)^2*(18 + 5*e*x)*Sqrt[12 - 3*e^2*x^2])/(35*e*Sqrt[2 + e*x])

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Maple [A]  time = 0.04, size = 36, normalized size = 0.8 \begin{align*}{\frac{ \left ( 2\,ex-4 \right ) \left ( 5\,ex+18 \right ) }{35\,e} \left ( -3\,{e}^{2}{x}^{2}+12 \right ) ^{{\frac{3}{2}}} \left ( ex+2 \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(e*x-2)*(5*e*x+18)*(-3*e^2*x^2+12)^(3/2)/e/(e*x+2)^(3/2)

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Maxima [C]  time = 1.78863, size = 63, normalized size = 1.4 \begin{align*} -\frac{{\left (30 i \, \sqrt{3} e^{3} x^{3} - 12 i \, \sqrt{3} e^{2} x^{2} - 312 i \, \sqrt{3} e x + 432 i \, \sqrt{3}\right )} \sqrt{e x - 2}}{35 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/35*(30*I*sqrt(3)*e^3*x^3 - 12*I*sqrt(3)*e^2*x^2 - 312*I*sqrt(3)*e*x + 432*I*sqrt(3))*sqrt(e*x - 2)/e

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Fricas [A]  time = 1.77618, size = 128, normalized size = 2.84 \begin{align*} -\frac{6 \,{\left (5 \, e^{3} x^{3} - 2 \, e^{2} x^{2} - 52 \, e x + 72\right )} \sqrt{-3 \, e^{2} x^{2} + 12} \sqrt{e x + 2}}{35 \,{\left (e^{2} x + 2 \, e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6/35*(5*e^3*x^3 - 2*e^2*x^2 - 52*e*x + 72)*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*} 3 \sqrt{3} \left (\int \frac{4 \sqrt{- e^{2} x^{2} + 4}}{\sqrt{e x + 2}}\, dx + \int - \frac{e^{2} x^{2} \sqrt{- e^{2} x^{2} + 4}}{\sqrt{e x + 2}}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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