3.10.33 \(\int \frac {\sqrt [4]{12-3 e^2 x^2}}{(2+e x)^{5/2}} \, dx\) [933]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 35, normalized size of antiderivative = 1.00, 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 = {665} \begin {gather*} -\frac {\sqrt [4]{3} \left (4-e^2 x^2\right )^{5/4}}{5 e (e x+2)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 665

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*}

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Mathematica [A]
time = 0.16, size = 35, normalized size = 1.00 \begin {gather*} -\frac {\sqrt [4]{3} \left (4-e^2 x^2\right )^{5/4}}{5 e (2+e x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.55, size = 30, normalized size = 0.86

method	result	size
gosper	\(\frac {\left (e x -2\right ) \left (-3 e^{2} x^{2}+12\right )^{\frac {1}{4}}}{5 \left (e x +2\right )^{\frac {3}{2}} e}\)	\(30\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

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Fricas [A]
time = 2.80, size = 45, normalized size = 1.29 \begin {gather*} \frac {{\left (-3 \, x^{2} e^{2} + 12\right )}^{\frac {1}{4}} \sqrt {x e + 2} {\left (x e - 2\right )}}{5 \, {\left (x^{2} e^{3} + 4 \, x e^{2} + 4 \, e\right )}} \end {gather*}

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

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

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]

3**(1/4)*Integral((-e**2*x**2 + 4)**(1/4)/(e**2*x**2*sqrt(e*x + 2) + 4*e*x*sqrt(e*x + 2) + 4*sqrt(e*x + 2)), x
)

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Giac [A]
time = 1.22, size = 21, normalized size = 0.60 \begin {gather*} -\frac {1}{5} \cdot 3^{\frac {1}{4}} {\left (\frac {4}{x e + 2} - 1\right )}^{\frac {5}{4}} e^{\left (-1\right )} \end {gather*}

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)*(4/(x*e + 2) - 1)^(5/4)*e^(-1)

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Mupad [B]
time = 0.64, size = 49, normalized size = 1.40 \begin {gather*} \frac {\left (\frac {x}{5\,e}-\frac {2}{5\,e^2}\right )\,{\left (12-3\,e^2\,x^2\right )}^{1/4}}{\frac {2\,\sqrt {e\,x+2}}{e}+x\,\sqrt {e\,x+2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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