3.942 \(\int \frac{1}{(2+e x)^{5/2} \sqrt [4]{12-3 e^2 x^2}} \, dx\)

Optimal. Leaf size=71 \[ -\frac{\left (4-e^2 x^2\right )^{3/4}}{21 \sqrt [4]{3} e (e x+2)^{3/2}}-\frac{\left (4-e^2 x^2\right )^{3/4}}{7 \sqrt [4]{3} e (e x+2)^{5/2}} \]

[Out]

-(4 - e^2*x^2)^(3/4)/(7*3^(1/4)*e*(2 + e*x)^(5/2)) - (4 - e^2*x^2)^(3/4)/(21*3^(1/4)*e*(2 + e*x)^(3/2))

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Rubi [A]  time = 0.0264299, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {659, 651} \[ -\frac{\left (4-e^2 x^2\right )^{3/4}}{21 \sqrt [4]{3} e (e x+2)^{3/2}}-\frac{\left (4-e^2 x^2\right )^{3/4}}{7 \sqrt [4]{3} e (e x+2)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((2 + e*x)^(5/2)*(12 - 3*e^2*x^2)^(1/4)),x]

[Out]

-(4 - e^2*x^2)^(3/4)/(7*3^(1/4)*e*(2 + e*x)^(5/2)) - (4 - e^2*x^2)^(3/4)/(21*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)^{5/2} \sqrt [4]{12-3 e^2 x^2}} \, dx &=-\frac{\left (4-e^2 x^2\right )^{3/4}}{7 \sqrt [4]{3} e (2+e x)^{5/2}}+\frac{1}{7} \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}}{7 \sqrt [4]{3} e (2+e x)^{5/2}}-\frac{\left (4-e^2 x^2\right )^{3/4}}{21 \sqrt [4]{3} e (2+e x)^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0572088, size = 40, normalized size = 0.56 \[ \frac{(e x-2) (e x+5)}{21 e (e x+2)^{3/2} \sqrt [4]{12-3 e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((2 + e*x)^(5/2)*(12 - 3*e^2*x^2)^(1/4)),x]

[Out]

((-2 + e*x)*(5 + e*x))/(21*e*(2 + e*x)^(3/2)*(12 - 3*e^2*x^2)^(1/4))

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Maple [A]  time = 0.043, size = 35, normalized size = 0.5 \begin{align*}{\frac{ \left ( ex-2 \right ) \left ( ex+5 \right ) }{21\,e} \left ( ex+2 \right ) ^{-{\frac{3}{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)^(5/2)/(-3*e^2*x^2+12)^(1/4),x)

[Out]

1/21*(e*x-2)*(e*x+5)/(e*x+2)^(3/2)/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{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+2)^(5/2)/(-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)^(5/2)), x)

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Fricas [A]  time = 1.81542, size = 127, normalized size = 1.79 \begin{align*} -\frac{{\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac{3}{4}}{\left (e x + 5\right )} \sqrt{e x + 2}}{63 \,{\left (e^{4} x^{3} + 6 \, e^{3} x^{2} + 12 \, e^{2} x + 8 \, e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+2)^(5/2)/(-3*e^2*x^2+12)^(1/4),x, algorithm="fricas")

[Out]

-1/63*(-3*e^2*x^2 + 12)^(3/4)*(e*x + 5)*sqrt(e*x + 2)/(e^4*x^3 + 6*e^3*x^2 + 12*e^2*x + 8*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)**(5/2)/(-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{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+2)^(5/2)/(-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)^(5/2)), x)