\(\int \frac {(2+e x)^{9/2}}{(12-3 e^2 x^2)^{3/2}} \, dx\) [919]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F(-1)]
   Maxima [C] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 24, antiderivative size = 87 \[ \int \frac {(2+e x)^{9/2}}{\left (12-3 e^2 x^2\right )^{3/2}} \, dx=\frac {128}{3 \sqrt {3} e \sqrt {2-e x}}+\frac {32 \sqrt {2-e x}}{\sqrt {3} e}-\frac {8 (2-e x)^{3/2}}{3 \sqrt {3} e}+\frac {2 (2-e x)^{5/2}}{15 \sqrt {3} e} \]

[Out]

-8/9*(-e*x+2)^(3/2)/e*3^(1/2)+2/45*(-e*x+2)^(5/2)*3^(1/2)/e+128/9/e*3^(1/2)/(-e*x+2)^(1/2)+32/3*3^(1/2)*(-e*x+
2)^(1/2)/e

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {641, 45} \[ \int \frac {(2+e x)^{9/2}}{\left (12-3 e^2 x^2\right )^{3/2}} \, dx=\frac {2 (2-e x)^{5/2}}{15 \sqrt {3} e}-\frac {8 (2-e x)^{3/2}}{3 \sqrt {3} e}+\frac {32 \sqrt {2-e x}}{\sqrt {3} e}+\frac {128}{3 \sqrt {3} e \sqrt {2-e x}} \]

[In]

Int[(2 + e*x)^(9/2)/(12 - 3*e^2*x^2)^(3/2),x]

[Out]

128/(3*Sqrt[3]*e*Sqrt[2 - e*x]) + (32*Sqrt[2 - e*x])/(Sqrt[3]*e) - (8*(2 - e*x)^(3/2))/(3*Sqrt[3]*e) + (2*(2 -
 e*x)^(5/2))/(15*Sqrt[3]*e)

Rule 45

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 641

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c/e)*x)^
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*} \text {integral}& = \int \frac {(2+e x)^3}{(6-3 e x)^{3/2}} \, dx \\ & = \int \left (\frac {64}{(6-3 e x)^{3/2}}-\frac {16}{\sqrt {6-3 e x}}+\frac {4}{3} \sqrt {6-3 e x}-\frac {1}{27} (6-3 e x)^{3/2}\right ) \, dx \\ & = \frac {128}{3 \sqrt {3} e \sqrt {2-e x}}+\frac {32 \sqrt {2-e x}}{\sqrt {3} e}-\frac {8 (2-e x)^{3/2}}{3 \sqrt {3} e}+\frac {2 (2-e x)^{5/2}}{15 \sqrt {3} e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.68 \[ \int \frac {(2+e x)^{9/2}}{\left (12-3 e^2 x^2\right )^{3/2}} \, dx=\frac {2 \sqrt {4-e^2 x^2} \left (-728+172 e x+14 e^2 x^2+e^3 x^3\right )}{15 e (-2+e x) \sqrt {6+3 e x}} \]

[In]

Integrate[(2 + e*x)^(9/2)/(12 - 3*e^2*x^2)^(3/2),x]

[Out]

(2*Sqrt[4 - e^2*x^2]*(-728 + 172*e*x + 14*e^2*x^2 + e^3*x^3))/(15*e*(-2 + e*x)*Sqrt[6 + 3*e*x])

Maple [A] (verified)

Time = 2.22 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.59

method result size
gosper \(\frac {2 \left (e x -2\right ) \left (e^{3} x^{3}+14 x^{2} e^{2}+172 e x -728\right ) \left (e x +2\right )^{\frac {3}{2}}}{5 e \left (-3 x^{2} e^{2}+12\right )^{\frac {3}{2}}}\) \(51\)
default \(\frac {2 \sqrt {-3 x^{2} e^{2}+12}\, \left (e^{3} x^{3}+14 x^{2} e^{2}+172 e x -728\right )}{45 \sqrt {e x +2}\, \left (e x -2\right ) e}\) \(53\)
risch \(-\frac {2 \left (x^{2} e^{2}+16 e x +204\right ) \left (e x -2\right ) \sqrt {\frac {-3 x^{2} e^{2}+12}{e x +2}}\, \sqrt {e x +2}}{15 e \sqrt {-3 e x +6}\, \sqrt {-3 x^{2} e^{2}+12}}+\frac {128 \sqrt {\frac {-3 x^{2} e^{2}+12}{e x +2}}\, \sqrt {e x +2}}{3 e \sqrt {-3 e x +6}\, \sqrt {-3 x^{2} e^{2}+12}}\) \(124\)

[In]

int((e*x+2)^(9/2)/(-3*e^2*x^2+12)^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/5*(e*x-2)*(e^3*x^3+14*e^2*x^2+172*e*x-728)*(e*x+2)^(3/2)/e/(-3*e^2*x^2+12)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.63 \[ \int \frac {(2+e x)^{9/2}}{\left (12-3 e^2 x^2\right )^{3/2}} \, dx=\frac {2 \, {\left (e^{3} x^{3} + 14 \, e^{2} x^{2} + 172 \, e x - 728\right )} \sqrt {-3 \, e^{2} x^{2} + 12} \sqrt {e x + 2}}{45 \, {\left (e^{3} x^{2} - 4 \, e\right )}} \]

[In]

integrate((e*x+2)^(9/2)/(-3*e^2*x^2+12)^(3/2),x, algorithm="fricas")

[Out]

2/45*(e^3*x^3 + 14*e^2*x^2 + 172*e*x - 728)*sqrt(-3*e^2*x^2 + 12)*sqrt(e*x + 2)/(e^3*x^2 - 4*e)

Sympy [F(-1)]

Timed out. \[ \int \frac {(2+e x)^{9/2}}{\left (12-3 e^2 x^2\right )^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate((e*x+2)**(9/2)/(-3*e**2*x**2+12)**(3/2),x)

[Out]

Timed out

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.41 \[ \int \frac {(2+e x)^{9/2}}{\left (12-3 e^2 x^2\right )^{3/2}} \, dx=\frac {2 i \, \sqrt {3} {\left (e^{3} x^{3} + 14 \, e^{2} x^{2} + 172 \, e x - 728\right )}}{45 \, \sqrt {e x - 2} e} \]

[In]

integrate((e*x+2)^(9/2)/(-3*e^2*x^2+12)^(3/2),x, algorithm="maxima")

[Out]

2/45*I*sqrt(3)*(e^3*x^3 + 14*e^2*x^2 + 172*e*x - 728)/(sqrt(e*x - 2)*e)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.91 \[ \int \frac {(2+e x)^{9/2}}{\left (12-3 e^2 x^2\right )^{3/2}} \, dx=-\frac {1024 \, \sqrt {3}}{45 \, e} + \frac {128 \, \sqrt {3}}{9 \, \sqrt {-e x + 2} e} + \frac {2 \, \sqrt {3} {\left ({\left (e x - 2\right )}^{2} \sqrt {-e x + 2} e^{4} - 20 \, {\left (-e x + 2\right )}^{\frac {3}{2}} e^{4} + 240 \, \sqrt {-e x + 2} e^{4}\right )}}{45 \, e^{5}} \]

[In]

integrate((e*x+2)^(9/2)/(-3*e^2*x^2+12)^(3/2),x, algorithm="giac")

[Out]

-1024/45*sqrt(3)/e + 128/9*sqrt(3)/(sqrt(-e*x + 2)*e) + 2/45*sqrt(3)*((e*x - 2)^2*sqrt(-e*x + 2)*e^4 - 20*(-e*
x + 2)^(3/2)*e^4 + 240*sqrt(-e*x + 2)*e^4)/e^5

Mupad [B] (verification not implemented)

Time = 10.31 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.92 \[ \int \frac {(2+e x)^{9/2}}{\left (12-3 e^2 x^2\right )^{3/2}} \, dx=-\frac {\sqrt {12-3\,e^2\,x^2}\,\left (\frac {2\,x^3\,\sqrt {e\,x+2}}{45}-\frac {1456\,\sqrt {e\,x+2}}{45\,e^3}+\frac {344\,x\,\sqrt {e\,x+2}}{45\,e^2}+\frac {28\,x^2\,\sqrt {e\,x+2}}{45\,e}\right )}{\frac {4}{e^2}-x^2} \]

[In]

int((e*x + 2)^(9/2)/(12 - 3*e^2*x^2)^(3/2),x)

[Out]

-((12 - 3*e^2*x^2)^(1/2)*((2*x^3*(e*x + 2)^(1/2))/45 - (1456*(e*x + 2)^(1/2))/(45*e^3) + (344*x*(e*x + 2)^(1/2
))/(45*e^2) + (28*x^2*(e*x + 2)^(1/2))/(45*e)))/(4/e^2 - x^2)