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

Optimal. Leaf size=35 \[ -\frac{\sqrt [4]{3} \left (4-e^2 x^2\right )^{5/4}}{5 e (e x+2)^{5/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0100162, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.042, Rules used = {651} \[ -\frac{\sqrt [4]{3} \left (4-e^2 x^2\right )^{5/4}}{5 e (e x+2)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0533383, size = 35, normalized size = 1. \[ \frac{(e x-2) \sqrt [4]{12-3 e^2 x^2}}{5 e (e x+2)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.049, size = 30, normalized size = 0.9 \begin{align*}{\frac{ex-2}{5\,e}\sqrt [4]{-3\,{e}^{2}{x}^{2}+12} \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)^(1/4)/(e*x+2)^(5/2),x)

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\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((-3*e^2*x^2+12)^(1/4)/(e*x+2)^(5/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.7725, size = 107, normalized size = 3.06 \begin{align*} \frac{{\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac{1}{4}} \sqrt{e x + 2}{\left (e x - 2\right )}}{5 \,{\left (e^{3} x^{2} + 4 \, e^{2} x + 4 \, e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.25996, size = 62, normalized size = 1.77 \begin{align*} -\frac{3^{\frac{1}{4}}{\left (-{\left (x e + 2\right )}^{2} + 4 \, x e + 8\right )}^{\frac{1}{4}}{\left (\frac{4}{x e + 2} - 1\right )} e^{\left (-1\right )}}{5 \, \sqrt{x e + 2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*3^(1/4)*(-(x*e + 2)^2 + 4*x*e + 8)^(1/4)*(4/(x*e + 2) - 1)*e^(-1)/sqrt(x*e + 2)