3.944 \(\int \frac{1}{(2+e x)^{9/2} \sqrt [4]{12-3 e^2 x^2}} \, dx\)

Optimal. Leaf size=141 \[ -\frac{2 \left (4-e^2 x^2\right )^{3/4}}{1155 \sqrt [4]{3} e (e x+2)^{3/2}}-\frac{2 \left (4-e^2 x^2\right )^{3/4}}{385 \sqrt [4]{3} e (e x+2)^{5/2}}-\frac{\left (4-e^2 x^2\right )^{3/4}}{55 \sqrt [4]{3} e (e x+2)^{7/2}}-\frac{\left (4-e^2 x^2\right )^{3/4}}{15 \sqrt [4]{3} e (e x+2)^{9/2}} \]

[Out]

-(4 - e^2*x^2)^(3/4)/(15*3^(1/4)*e*(2 + e*x)^(9/2)) - (4 - e^2*x^2)^(3/4)/(55*3^(1/4)*e*(2 + e*x)^(7/2)) - (2*
(4 - e^2*x^2)^(3/4))/(385*3^(1/4)*e*(2 + e*x)^(5/2)) - (2*(4 - e^2*x^2)^(3/4))/(1155*3^(1/4)*e*(2 + e*x)^(3/2)
)

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Rubi [A]  time = 0.0729962, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {659, 651} \[ -\frac{2 \left (4-e^2 x^2\right )^{3/4}}{1155 \sqrt [4]{3} e (e x+2)^{3/2}}-\frac{2 \left (4-e^2 x^2\right )^{3/4}}{385 \sqrt [4]{3} e (e x+2)^{5/2}}-\frac{\left (4-e^2 x^2\right )^{3/4}}{55 \sqrt [4]{3} e (e x+2)^{7/2}}-\frac{\left (4-e^2 x^2\right )^{3/4}}{15 \sqrt [4]{3} e (e x+2)^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((2 + e*x)^(9/2)*(12 - 3*e^2*x^2)^(1/4)),x]

[Out]

-(4 - e^2*x^2)^(3/4)/(15*3^(1/4)*e*(2 + e*x)^(9/2)) - (4 - e^2*x^2)^(3/4)/(55*3^(1/4)*e*(2 + e*x)^(7/2)) - (2*
(4 - e^2*x^2)^(3/4))/(385*3^(1/4)*e*(2 + e*x)^(5/2)) - (2*(4 - e^2*x^2)^(3/4))/(1155*3^(1/4)*e*(2 + e*x)^(3/2)
)

Rule 659

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a + c*x^2)^(p + 1)
)/(2*c*d*(m + p + 1)), x] + Dist[Simplify[m + 2*p + 2]/(2*d*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p,
x], x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p + 2
], 0]

Rule 651

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^m*(a + c*x^2)^(p + 1))
/(2*c*d*(p + 1)), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p
+ 2, 0]

Rubi steps

\begin{align*} \int \frac{1}{(2+e x)^{9/2} \sqrt [4]{12-3 e^2 x^2}} \, dx &=-\frac{\left (4-e^2 x^2\right )^{3/4}}{15 \sqrt [4]{3} e (2+e x)^{9/2}}+\frac{1}{5} \int \frac{1}{(2+e x)^{7/2} \sqrt [4]{12-3 e^2 x^2}} \, dx\\ &=-\frac{\left (4-e^2 x^2\right )^{3/4}}{15 \sqrt [4]{3} e (2+e x)^{9/2}}-\frac{\left (4-e^2 x^2\right )^{3/4}}{55 \sqrt [4]{3} e (2+e x)^{7/2}}+\frac{2}{55} \int \frac{1}{(2+e x)^{5/2} \sqrt [4]{12-3 e^2 x^2}} \, dx\\ &=-\frac{\left (4-e^2 x^2\right )^{3/4}}{15 \sqrt [4]{3} e (2+e x)^{9/2}}-\frac{\left (4-e^2 x^2\right )^{3/4}}{55 \sqrt [4]{3} e (2+e x)^{7/2}}-\frac{2 \left (4-e^2 x^2\right )^{3/4}}{385 \sqrt [4]{3} e (2+e x)^{5/2}}+\frac{2}{385} \int \frac{1}{(2+e x)^{3/2} \sqrt [4]{12-3 e^2 x^2}} \, dx\\ &=-\frac{\left (4-e^2 x^2\right )^{3/4}}{15 \sqrt [4]{3} e (2+e x)^{9/2}}-\frac{\left (4-e^2 x^2\right )^{3/4}}{55 \sqrt [4]{3} e (2+e x)^{7/2}}-\frac{2 \left (4-e^2 x^2\right )^{3/4}}{385 \sqrt [4]{3} e (2+e x)^{5/2}}-\frac{2 \left (4-e^2 x^2\right )^{3/4}}{1155 \sqrt [4]{3} e (2+e x)^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0704015, size = 57, normalized size = 0.4 \[ \frac{(e x-2) \left (2 e^3 x^3+18 e^2 x^2+69 e x+159\right )}{1155 e (e x+2)^{7/2} \sqrt [4]{12-3 e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((2 + e*x)^(9/2)*(12 - 3*e^2*x^2)^(1/4)),x]

[Out]

((-2 + e*x)*(159 + 69*e*x + 18*e^2*x^2 + 2*e^3*x^3))/(1155*e*(2 + e*x)^(7/2)*(12 - 3*e^2*x^2)^(1/4))

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Maple [A]  time = 0.04, size = 52, normalized size = 0.4 \begin{align*}{\frac{ \left ( ex-2 \right ) \left ( 2\,{e}^{3}{x}^{3}+18\,{e}^{2}{x}^{2}+69\,ex+159 \right ) }{1155\,e} \left ( ex+2 \right ) ^{-{\frac{7}{2}}}{\frac{1}{\sqrt [4]{-3\,{e}^{2}{x}^{2}+12}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+2)^(9/2)/(-3*e^2*x^2+12)^(1/4),x)

[Out]

1/1155*(e*x-2)*(2*e^3*x^3+18*e^2*x^2+69*e*x+159)/(e*x+2)^(7/2)/e/(-3*e^2*x^2+12)^(1/4)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac{1}{4}}{\left (e x + 2\right )}^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+2)^(9/2)/(-3*e^2*x^2+12)^(1/4),x, algorithm="maxima")

[Out]

integrate(1/((-3*e^2*x^2 + 12)^(1/4)*(e*x + 2)^(9/2)), x)

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Fricas [A]  time = 1.93047, size = 208, normalized size = 1.48 \begin{align*} -\frac{{\left (2 \, e^{3} x^{3} + 18 \, e^{2} x^{2} + 69 \, e x + 159\right )}{\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac{3}{4}} \sqrt{e x + 2}}{3465 \,{\left (e^{6} x^{5} + 10 \, e^{5} x^{4} + 40 \, e^{4} x^{3} + 80 \, e^{3} x^{2} + 80 \, e^{2} x + 32 \, e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+2)^(9/2)/(-3*e^2*x^2+12)^(1/4),x, algorithm="fricas")

[Out]

-1/3465*(2*e^3*x^3 + 18*e^2*x^2 + 69*e*x + 159)*(-3*e^2*x^2 + 12)^(3/4)*sqrt(e*x + 2)/(e^6*x^5 + 10*e^5*x^4 +
40*e^4*x^3 + 80*e^3*x^2 + 80*e^2*x + 32*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(1/(e*x+2)**(9/2)/(-3*e**2*x**2+12)**(1/4),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac{1}{4}}{\left (e x + 2\right )}^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+2)^(9/2)/(-3*e^2*x^2+12)^(1/4),x, algorithm="giac")

[Out]

integrate(1/((-3*e^2*x^2 + 12)^(1/4)*(e*x + 2)^(9/2)), x)