3.8.75 \(\int \frac {\sqrt [4]{12-3 e^2 x^2}}{(2+e x)^{7/2}} \, dx\)

Optimal. Leaf size=71 \[ -\frac {\left (4-e^2 x^2\right )^{5/4}}{15\ 3^{3/4} e (e x+2)^{5/2}}-\frac {\left (4-e^2 x^2\right )^{5/4}}{3\ 3^{3/4} e (e x+2)^{7/2}} \]

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Rubi [A]  time = 0.03, antiderivative size = 71, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\frac {\left (4-e^2 x^2\right )^{5/4}}{15\ 3^{3/4} e (e x+2)^{5/2}}-\frac {\left (4-e^2 x^2\right )^{5/4}}{3\ 3^{3/4} e (e x+2)^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(12 - 3*e^2*x^2)^(1/4)/(2 + e*x)^(7/2),x]

[Out]

-(4 - e^2*x^2)^(5/4)/(3*3^(3/4)*e*(2 + e*x)^(7/2)) - (4 - e^2*x^2)^(5/4)/(15*3^(3/4)*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]

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]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 45, normalized size = 0.63 \begin {gather*} \frac {(e x-2) (e x+7) \sqrt [4]{4-e^2 x^2}}{15\ 3^{3/4} e (e x+2)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(12 - 3*e^2*x^2)^(1/4)/(2 + e*x)^(7/2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.30, size = 47, normalized size = 0.66 \begin {gather*} -\frac {(e x+7) \left (4 (e x+2)-(e x+2)^2\right )^{5/4}}{15\ 3^{3/4} e (e x+2)^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(12 - 3*e^2*x^2)^(1/4)/(2 + e*x)^(7/2),x]

[Out]

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

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fricas [A]  time = 0.40, size = 61, normalized size = 0.86 \begin {gather*} \frac {{\left (e^{2} x^{2} + 5 \, e x - 14\right )} {\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {1}{4}} \sqrt {e x + 2}}{45 \, {\left (e^{4} x^{3} + 6 \, e^{3} x^{2} + 12 \, e^{2} x + 8 \, 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)^(7/2),x, algorithm="fricas")

[Out]

1/45*(e^2*x^2 + 5*e*x - 14)*(-3*e^2*x^2 + 12)^(1/4)*sqrt(e*x + 2)/(e^4*x^3 + 6*e^3*x^2 + 12*e^2*x + 8*e)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \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)^(7/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Erro
r index.cc index_gcd Error: Bad Argument ValueError index.cc index_gcd Error: Bad Argument Value

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maple [A]  time = 0.06, size = 35, normalized size = 0.49 \begin {gather*} \frac {\left (e x -2\right ) \left (e x +7\right ) \left (-3 e^{2} x^{2}+12\right )^{\frac {1}{4}}}{45 \left (e x +2\right )^{\frac {5}{2}} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3*e^2*x^2+12)^(1/4)/(e*x+2)^(7/2),x)

[Out]

1/45*(e*x-2)*(e*x+7)*(-3*e^2*x^2+12)^(1/4)/(e*x+2)^(5/2)/e

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {1}{4}}}{{\left (e x + 2\right )}^{\frac {7}{2}}}\,{d x} \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)^(7/2),x, algorithm="maxima")

[Out]

integrate((-3*e^2*x^2 + 12)^(1/4)/(e*x + 2)^(7/2), x)

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mupad [B]  time = 0.29, size = 37, normalized size = 0.52 \begin {gather*} \frac {{\left (12-3\,e^2\,x^2\right )}^{1/4}\,\left (e^2\,x^2+5\,e\,x-14\right )}{45\,e\,{\left (e\,x+2\right )}^{5/2}} \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)^(7/2),x)

[Out]

((12 - 3*e^2*x^2)^(1/4)*(5*e*x + e^2*x^2 - 14))/(45*e*(e*x + 2)^(5/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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)**(7/2),x)

[Out]

Timed out

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