3.943 \(\int \frac {1}{(2+e x)^{7/2} \sqrt [4]{12-3 e^2 x^2}} \, dx\)

Optimal. Leaf size=106 \[ -\frac {2 \left (4-e^2 x^2\right )^{3/4}}{231 \sqrt [4]{3} e (e x+2)^{3/2}}-\frac {2 \left (4-e^2 x^2\right )^{3/4}}{77 \sqrt [4]{3} e (e x+2)^{5/2}}-\frac {\left (4-e^2 x^2\right )^{3/4}}{11 \sqrt [4]{3} e (e x+2)^{7/2}} \]

[Out]

-1/33*(-e^2*x^2+4)^(3/4)*3^(3/4)/e/(e*x+2)^(7/2)-2/231*(-e^2*x^2+4)^(3/4)*3^(3/4)/e/(e*x+2)^(5/2)-2/693*(-e^2*
x^2+4)^(3/4)*3^(3/4)/e/(e*x+2)^(3/2)

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Rubi [A]  time = 0.05, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {659, 651} \[ -\frac {2 \left (4-e^2 x^2\right )^{3/4}}{231 \sqrt [4]{3} e (e x+2)^{3/2}}-\frac {2 \left (4-e^2 x^2\right )^{3/4}}{77 \sqrt [4]{3} e (e x+2)^{5/2}}-\frac {\left (4-e^2 x^2\right )^{3/4}}{11 \sqrt [4]{3} e (e x+2)^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(4 - e^2*x^2)^(3/4)/(11*3^(1/4)*e*(2 + e*x)^(7/2)) - (2*(4 - e^2*x^2)^(3/4))/(77*3^(1/4)*e*(2 + e*x)^(5/2)) -
(2*(4 - e^2*x^2)^(3/4))/(231*3^(1/4)*e*(2 + e*x)^(3/2))

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 {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}}{11 \sqrt [4]{3} e (2+e x)^{7/2}}+\frac {2}{11} \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}}{11 \sqrt [4]{3} e (2+e x)^{7/2}}-\frac {2 \left (4-e^2 x^2\right )^{3/4}}{77 \sqrt [4]{3} e (2+e x)^{5/2}}+\frac {2}{77} \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}}{11 \sqrt [4]{3} e (2+e x)^{7/2}}-\frac {2 \left (4-e^2 x^2\right )^{3/4}}{77 \sqrt [4]{3} e (2+e x)^{5/2}}-\frac {2 \left (4-e^2 x^2\right )^{3/4}}{231 \sqrt [4]{3} e (2+e x)^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 49, normalized size = 0.46 \[ \frac {(e x-2) \left (2 e^2 x^2+14 e x+41\right )}{231 e (e x+2)^{5/2} \sqrt [4]{12-3 e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2 + e*x)*(41 + 14*e*x + 2*e^2*x^2))/(231*e*(2 + e*x)^(5/2)*(12 - 3*e^2*x^2)^(1/4))

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fricas [A]  time = 0.83, size = 70, normalized size = 0.66 \[ -\frac {{\left (2 \, e^{2} x^{2} + 14 \, e x + 41\right )} {\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {3}{4}} \sqrt {e x + 2}}{693 \, {\left (e^{5} x^{4} + 8 \, e^{4} x^{3} + 24 \, e^{3} x^{2} + 32 \, e^{2} x + 16 \, e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/693*(2*e^2*x^2 + 14*e*x + 41)*(-3*e^2*x^2 + 12)^(3/4)*sqrt(e*x + 2)/(e^5*x^4 + 8*e^4*x^3 + 24*e^3*x^2 + 32*
e^2*x + 16*e)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+2)^(7/2)/(-3*e^2*x^2+12)^(1/4),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 = 0.05, size = 44, normalized size = 0.42 \[ \frac {\left (e x -2\right ) \left (2 e^{2} x^{2}+14 e x +41\right )}{231 \left (e x +2\right )^{\frac {5}{2}} \left (-3 e^{2} x^{2}+12\right )^{\frac {1}{4}} e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/231*(e*x-2)*(2*e^2*x^2+14*e*x+41)/(e*x+2)^(5/2)/e/(-3*e^2*x^2+12)^(1/4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 0.34, size = 88, normalized size = 0.83 \[ -\frac {{\left (12-3\,e^2\,x^2\right )}^{3/4}\,\left (\frac {2\,x}{99\,e^3}+\frac {41}{693\,e^4}+\frac {2\,x^2}{693\,e^2}\right )}{\frac {8\,\sqrt {e\,x+2}}{e^3}+x^3\,\sqrt {e\,x+2}+\frac {12\,x\,\sqrt {e\,x+2}}{e^2}+\frac {6\,x^2\,\sqrt {e\,x+2}}{e}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((12 - 3*e^2*x^2)^(3/4)*((2*x)/(99*e^3) + 41/(693*e^4) + (2*x^2)/(693*e^2)))/((8*(e*x + 2)^(1/2))/e^3 + x^3*(
e*x + 2)^(1/2) + (12*x*(e*x + 2)^(1/2))/e^2 + (6*x^2*(e*x + 2)^(1/2))/e)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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