3.918 \(\int \frac {(2+e x)^{11/2}}{(12-3 e^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=111 \[ -\frac {2 (2-e x)^{7/2}}{21 \sqrt {3} e}+\frac {32 (2-e x)^{5/2}}{15 \sqrt {3} e}-\frac {64 (2-e x)^{3/2}}{3 \sqrt {3} e}+\frac {512 \sqrt {2-e x}}{3 \sqrt {3} e}+\frac {512}{3 \sqrt {3} e \sqrt {2-e x}} \]

[Out]

-64/9*(-e*x+2)^(3/2)/e*3^(1/2)+32/45*(-e*x+2)^(5/2)*3^(1/2)/e-2/63*(-e*x+2)^(7/2)*3^(1/2)/e+512/9/e*3^(1/2)/(-
e*x+2)^(1/2)+512/9*3^(1/2)*(-e*x+2)^(1/2)/e

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Rubi [A]  time = 0.03, antiderivative size = 111, 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 = {627, 43} \[ -\frac {2 (2-e x)^{7/2}}{21 \sqrt {3} e}+\frac {32 (2-e x)^{5/2}}{15 \sqrt {3} e}-\frac {64 (2-e x)^{3/2}}{3 \sqrt {3} e}+\frac {512 \sqrt {2-e x}}{3 \sqrt {3} e}+\frac {512}{3 \sqrt {3} e \sqrt {2-e x}} \]

Antiderivative was successfully verified.

[In]

Int[(2 + e*x)^(11/2)/(12 - 3*e^2*x^2)^(3/2),x]

[Out]

512/(3*Sqrt[3]*e*Sqrt[2 - e*x]) + (512*Sqrt[2 - e*x])/(3*Sqrt[3]*e) - (64*(2 - e*x)^(3/2))/(3*Sqrt[3]*e) + (32
*(2 - e*x)^(5/2))/(15*Sqrt[3]*e) - (2*(2 - e*x)^(7/2))/(21*Sqrt[3]*e)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rubi steps

\begin {align*} \int \frac {(2+e x)^{11/2}}{\left (12-3 e^2 x^2\right )^{3/2}} \, dx &=\int \frac {(2+e x)^4}{(6-3 e x)^{3/2}} \, dx\\ &=\int \left (\frac {256}{(6-3 e x)^{3/2}}-\frac {256}{3 \sqrt {6-3 e x}}+\frac {32}{3} \sqrt {6-3 e x}-\frac {16}{27} (6-3 e x)^{3/2}+\frac {1}{81} (6-3 e x)^{5/2}\right ) \, dx\\ &=\frac {512}{3 \sqrt {3} e \sqrt {2-e x}}+\frac {512 \sqrt {2-e x}}{3 \sqrt {3} e}-\frac {64 (2-e x)^{3/2}}{3 \sqrt {3} e}+\frac {32 (2-e x)^{5/2}}{15 \sqrt {3} e}-\frac {2 (2-e x)^{7/2}}{21 \sqrt {3} e}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 60, normalized size = 0.54 \[ -\frac {2 \sqrt {e x+2} \left (5 e^4 x^4+72 e^3 x^3+568 e^2 x^2+5664 e x-23216\right )}{105 e \sqrt {12-3 e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + e*x)^(11/2)/(12 - 3*e^2*x^2)^(3/2),x]

[Out]

(-2*Sqrt[2 + e*x]*(-23216 + 5664*e*x + 568*e^2*x^2 + 72*e^3*x^3 + 5*e^4*x^4))/(105*e*Sqrt[12 - 3*e^2*x^2])

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fricas [A]  time = 0.60, size = 64, normalized size = 0.58 \[ \frac {2 \, {\left (5 \, e^{4} x^{4} + 72 \, e^{3} x^{3} + 568 \, e^{2} x^{2} + 5664 \, e x - 23216\right )} \sqrt {-3 \, e^{2} x^{2} + 12} \sqrt {e x + 2}}{315 \, {\left (e^{3} x^{2} - 4 \, e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+2)^(11/2)/(-3*e^2*x^2+12)^(3/2),x, algorithm="fricas")

[Out]

2/315*(5*e^4*x^4 + 72*e^3*x^3 + 568*e^2*x^2 + 5664*e*x - 23216)*sqrt(-3*e^2*x^2 + 12)*sqrt(e*x + 2)/(e^3*x^2 -
4*e)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+2)^(11/2)/(-3*e^2*x^2+12)^(3/2),x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.05, size = 60, normalized size = 0.54 \[ \frac {2 \left (e x -2\right ) \left (5 e^{4} x^{4}+72 e^{3} x^{3}+568 e^{2} x^{2}+5664 e x -23216\right ) \left (e x +2\right )^{\frac {3}{2}}}{35 \left (-3 e^{2} x^{2}+12\right )^{\frac {3}{2}} e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+2)^(11/2)/(-3*e^2*x^2+12)^(3/2),x)

[Out]

2/35*(e*x-2)*(5*e^4*x^4+72*e^3*x^3+568*e^2*x^2+5664*e*x-23216)*(e*x+2)^(3/2)/e/(-3*e^2*x^2+12)^(3/2)

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maxima [C]  time = 3.00, size = 58, normalized size = 0.52 \[ \frac {10 i \, \sqrt {3} e^{4} x^{4} + 144 i \, \sqrt {3} e^{3} x^{3} + 1136 i \, \sqrt {3} e^{2} x^{2} + 11328 i \, \sqrt {3} e x - 46432 i \, \sqrt {3}}{315 \, \sqrt {e x - 2} e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+2)^(11/2)/(-3*e^2*x^2+12)^(3/2),x, algorithm="maxima")

[Out]

1/315*(10*I*sqrt(3)*e^4*x^4 + 144*I*sqrt(3)*e^3*x^3 + 1136*I*sqrt(3)*e^2*x^2 + 11328*I*sqrt(3)*e*x - 46432*I*s
qrt(3))/(sqrt(e*x - 2)*e)

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mupad [B]  time = 0.30, size = 93, normalized size = 0.84 \[ -\frac {\sqrt {12-3\,e^2\,x^2}\,\left (\frac {16\,x^3\,\sqrt {e\,x+2}}{35}-\frac {46432\,\sqrt {e\,x+2}}{315\,e^3}+\frac {3776\,x\,\sqrt {e\,x+2}}{105\,e^2}+\frac {2\,e\,x^4\,\sqrt {e\,x+2}}{63}+\frac {1136\,x^2\,\sqrt {e\,x+2}}{315\,e}\right )}{\frac {4}{e^2}-x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x + 2)^(11/2)/(12 - 3*e^2*x^2)^(3/2),x)

[Out]

-((12 - 3*e^2*x^2)^(1/2)*((16*x^3*(e*x + 2)^(1/2))/35 - (46432*(e*x + 2)^(1/2))/(315*e^3) + (3776*x*(e*x + 2)^
(1/2))/(105*e^2) + (2*e*x^4*(e*x + 2)^(1/2))/63 + (1136*x^2*(e*x + 2)^(1/2))/(315*e)))/(4/e^2 - x^2)

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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((e*x+2)**(11/2)/(-3*e**2*x**2+12)**(3/2),x)

[Out]

Timed out

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