3.918 \(\int \frac{(2+e x)^{11/2}}{(12-3 e^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=111 \[ -\frac{2 (2-e x)^{7/2}}{21 \sqrt{3} e}+\frac{32 (2-e x)^{5/2}}{15 \sqrt{3} e}-\frac{64 (2-e x)^{3/2}}{3 \sqrt{3} e}+\frac{512 \sqrt{2-e x}}{3 \sqrt{3} e}+\frac{512}{3 \sqrt{3} e \sqrt{2-e x}} \]

[Out]

512/(3*Sqrt[3]*e*Sqrt[2 - e*x]) + (512*Sqrt[2 - e*x])/(3*Sqrt[3]*e) - (64*(2 - e*x)^(3/2))/(3*Sqrt[3]*e) + (32
*(2 - e*x)^(5/2))/(15*Sqrt[3]*e) - (2*(2 - e*x)^(7/2))/(21*Sqrt[3]*e)

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Rubi [A]  time = 0.0285124, antiderivative size = 111, 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} \[ -\frac{2 (2-e x)^{7/2}}{21 \sqrt{3} e}+\frac{32 (2-e x)^{5/2}}{15 \sqrt{3} e}-\frac{64 (2-e x)^{3/2}}{3 \sqrt{3} e}+\frac{512 \sqrt{2-e x}}{3 \sqrt{3} e}+\frac{512}{3 \sqrt{3} e \sqrt{2-e x}} \]

Antiderivative was successfully verified.

[In]

Int[(2 + e*x)^(11/2)/(12 - 3*e^2*x^2)^(3/2),x]

[Out]

512/(3*Sqrt[3]*e*Sqrt[2 - e*x]) + (512*Sqrt[2 - e*x])/(3*Sqrt[3]*e) - (64*(2 - e*x)^(3/2))/(3*Sqrt[3]*e) + (32
*(2 - e*x)^(5/2))/(15*Sqrt[3]*e) - (2*(2 - e*x)^(7/2))/(21*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)^{11/2}}{\left (12-3 e^2 x^2\right )^{3/2}} \, dx &=\int \frac{(2+e x)^4}{(6-3 e x)^{3/2}} \, dx\\ &=\int \left (\frac{256}{(6-3 e x)^{3/2}}-\frac{256}{3 \sqrt{6-3 e x}}+\frac{32}{3} \sqrt{6-3 e x}-\frac{16}{27} (6-3 e x)^{3/2}+\frac{1}{81} (6-3 e x)^{5/2}\right ) \, dx\\ &=\frac{512}{3 \sqrt{3} e \sqrt{2-e x}}+\frac{512 \sqrt{2-e x}}{3 \sqrt{3} e}-\frac{64 (2-e x)^{3/2}}{3 \sqrt{3} e}+\frac{32 (2-e x)^{5/2}}{15 \sqrt{3} e}-\frac{2 (2-e x)^{7/2}}{21 \sqrt{3} e}\\ \end{align*}

Mathematica [A]  time = 0.101524, size = 60, normalized size = 0.54 \[ -\frac{2 \sqrt{e x+2} \left (5 e^4 x^4+72 e^3 x^3+568 e^2 x^2+5664 e x-23216\right )}{105 e \sqrt{12-3 e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + e*x)^(11/2)/(12 - 3*e^2*x^2)^(3/2),x]

[Out]

(-2*Sqrt[2 + e*x]*(-23216 + 5664*e*x + 568*e^2*x^2 + 72*e^3*x^3 + 5*e^4*x^4))/(105*e*Sqrt[12 - 3*e^2*x^2])

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Maple [A]  time = 0.043, size = 60, normalized size = 0.5 \begin{align*}{\frac{ \left ( 2\,ex-4 \right ) \left ( 5\,{e}^{4}{x}^{4}+72\,{e}^{3}{x}^{3}+568\,{e}^{2}{x}^{2}+5664\,ex-23216 \right ) }{35\,e} \left ( ex+2 \right ) ^{{\frac{3}{2}}} \left ( -3\,{e}^{2}{x}^{2}+12 \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+2)^(11/2)/(-3*e^2*x^2+12)^(3/2),x)

[Out]

2/35*(e*x-2)*(5*e^4*x^4+72*e^3*x^3+568*e^2*x^2+5664*e*x-23216)*(e*x+2)^(3/2)/e/(-3*e^2*x^2+12)^(3/2)

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Maxima [C]  time = 1.67088, size = 78, normalized size = 0.7 \begin{align*} \frac{10 i \, \sqrt{3} e^{4} x^{4} + 144 i \, \sqrt{3} e^{3} x^{3} + 1136 i \, \sqrt{3} e^{2} x^{2} + 11328 i \, \sqrt{3} e x - 46432 i \, \sqrt{3}}{315 \, \sqrt{e x - 2} e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+2)^(11/2)/(-3*e^2*x^2+12)^(3/2),x, algorithm="maxima")

[Out]

1/315*(10*I*sqrt(3)*e^4*x^4 + 144*I*sqrt(3)*e^3*x^3 + 1136*I*sqrt(3)*e^2*x^2 + 11328*I*sqrt(3)*e*x - 46432*I*s
qrt(3))/(sqrt(e*x - 2)*e)

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Fricas [A]  time = 1.84925, size = 158, normalized size = 1.42 \begin{align*} \frac{2 \,{\left (5 \, e^{4} x^{4} + 72 \, e^{3} x^{3} + 568 \, e^{2} x^{2} + 5664 \, e x - 23216\right )} \sqrt{-3 \, e^{2} x^{2} + 12} \sqrt{e x + 2}}{315 \,{\left (e^{3} x^{2} - 4 \, e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+2)^(11/2)/(-3*e^2*x^2+12)^(3/2),x, algorithm="fricas")

[Out]

2/315*(5*e^4*x^4 + 72*e^3*x^3 + 568*e^2*x^2 + 5664*e*x - 23216)*sqrt(-3*e^2*x^2 + 12)*sqrt(e*x + 2)/(e^3*x^2 -
4*e)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+2)**(11/2)/(-3*e**2*x**2+12)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+2)^(11/2)/(-3*e^2*x^2+12)^(3/2),x, algorithm="giac")

[Out]

sage0*x