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

Optimal. Leaf size=65 \[ -\frac {2 \sqrt {3} (2-e x)^{7/2}}{7 e}+\frac {16 \sqrt {3} (2-e x)^{5/2}}{5 e}-\frac {32 (2-e x)^{3/2}}{\sqrt {3} e} \]

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Rubi [A]  time = 0.02, antiderivative size = 65, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\frac {2 \sqrt {3} (2-e x)^{7/2}}{7 e}+\frac {16 \sqrt {3} (2-e x)^{5/2}}{5 e}-\frac {32 (2-e x)^{3/2}}{\sqrt {3} e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-32*(2 - e*x)^(3/2))/(Sqrt[3]*e) + (16*Sqrt[3]*(2 - e*x)^(5/2))/(5*e) - (2*Sqrt[3]*(2 - e*x)^(7/2))/(7*e)

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])

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]))

Rubi steps

\begin {align*} \int (2+e x)^{3/2} \sqrt {12-3 e^2 x^2} \, dx &=\int \sqrt {6-3 e x} (2+e x)^2 \, dx\\ &=\int \left (16 \sqrt {6-3 e x}-\frac {8}{3} (6-3 e x)^{3/2}+\frac {1}{9} (6-3 e x)^{5/2}\right ) \, dx\\ &=-\frac {32 (2-e x)^{3/2}}{\sqrt {3} e}+\frac {16 \sqrt {3} (2-e x)^{5/2}}{5 e}-\frac {2 \sqrt {3} (2-e x)^{7/2}}{7 e}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 50, normalized size = 0.77 \begin {gather*} \frac {2 (e x-2) \sqrt {4-e^2 x^2} \left (15 e^2 x^2+108 e x+284\right )}{35 e \sqrt {3 e x+6}} \end {gather*}

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[4 - e^2*x^2]*(284 + 108*e*x + 15*e^2*x^2))/(35*e*Sqrt[6 + 3*e*x])

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IntegrateAlgebraic [A]  time = 0.23, size = 60, normalized size = 0.92 \begin {gather*} -\frac {2 \left (4 (e x+2)-(e x+2)^2\right )^{3/2} \left (15 (e x+2)^2+48 (e x+2)+128\right )}{35 \sqrt {3} e (e x+2)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(4*(2 + e*x) - (2 + e*x)^2)^(3/2)*(128 + 48*(2 + e*x) + 15*(2 + e*x)^2))/(35*Sqrt[3]*e*(2 + e*x)^(3/2))

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fricas [A]  time = 0.40, size = 54, normalized size = 0.83 \begin {gather*} \frac {2 \, {\left (15 \, e^{3} x^{3} + 78 \, e^{2} x^{2} + 68 \, e x - 568\right )} \sqrt {-3 \, e^{2} x^{2} + 12} \sqrt {e x + 2}}{105 \, {\left (e^{2} x + 2 \, e\right )}} \end {gather*}

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]

2/105*(15*e^3*x^3 + 78*e^2*x^2 + 68*e*x - 568)*sqrt(-3*e^2*x^2 + 12)*sqrt(e*x + 2)/(e^2*x + 2*e)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {-3 \, e^{2} x^{2} + 12} {\left (e x + 2\right )}^{\frac {3}{2}}\,{d x} \end {gather*}

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

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maple [A]  time = 0.05, size = 44, normalized size = 0.68 \begin {gather*} \frac {2 \left (e x -2\right ) \left (15 e^{2} x^{2}+108 e x +284\right ) \sqrt {-3 e^{2} x^{2}+12}}{105 \sqrt {e x +2}\, e} \end {gather*}

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]

2/105*(e*x-2)*(15*e^2*x^2+108*e*x+284)*(-3*e^2*x^2+12)^(1/2)/(e*x+2)^(1/2)/e

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maxima [C]  time = 3.14, size = 60, normalized size = 0.92 \begin {gather*} \frac {{\left (30 i \, \sqrt {3} e^{3} x^{3} + 156 i \, \sqrt {3} e^{2} x^{2} + 136 i \, \sqrt {3} e x - 1136 i \, \sqrt {3}\right )} {\left (e x + 2\right )} \sqrt {e x - 2}}{105 \, {\left (e^{2} x + 2 \, e\right )}} \end {gather*}

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]

1/105*(30*I*sqrt(3)*e^3*x^3 + 156*I*sqrt(3)*e^2*x^2 + 136*I*sqrt(3)*e*x - 1136*I*sqrt(3))*(e*x + 2)*sqrt(e*x -
2)/(e^2*x + 2*e)

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mupad [B]  time = 0.52, size = 73, normalized size = 1.12 \begin {gather*} \frac {\sqrt {12-3\,e^2\,x^2}\,\left (\frac {52\,x^2\,\sqrt {e\,x+2}}{35}-\frac {1136\,\sqrt {e\,x+2}}{105\,e^2}+\frac {136\,x\,\sqrt {e\,x+2}}{105\,e}+\frac {2\,e\,x^3\,\sqrt {e\,x+2}}{7}\right )}{x+\frac {2}{e}} \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]

((12 - 3*e^2*x^2)^(1/2)*((52*x^2*(e*x + 2)^(1/2))/35 - (1136*(e*x + 2)^(1/2))/(105*e^2) + (136*x*(e*x + 2)^(1/
2))/(105*e) + (2*e*x^3*(e*x + 2)^(1/2))/7))/(x + 2/e)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \sqrt {3} \left (\int 2 \sqrt {e x + 2} \sqrt {- e^{2} x^{2} + 4}\, dx + \int e x \sqrt {e x + 2} \sqrt {- e^{2} x^{2} + 4}\, dx\right ) \end {gather*}

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))

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