\(\int (2+e x)^{3/2} \sqrt {12-3 e^2 x^2} \, dx\) [895]

Optimal result

Integrand size = 24, antiderivative size = 65 \[ \int (2+e x)^{3/2} \sqrt {12-3 e^2 x^2} \, dx=-\frac {32 (2-e x)^{3/2}}{\sqrt {3} e}+\frac {16 \sqrt {3} (2-e x)^{5/2}}{5 e}-\frac {2 \sqrt {3} (2-e x)^{7/2}}{7 e} \]

[Out]

-32/3*(-e*x+2)^(3/2)/e*3^(1/2)+16/5*(-e*x+2)^(5/2)*3^(1/2)/e-2/7*(-e*x+2)^(7/2)*3^(1/2)/e

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 65, 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} \[ \int (2+e x)^{3/2} \sqrt {12-3 e^2 x^2} \, dx=-\frac {2 \sqrt {3} (2-e x)^{7/2}}{7 e}+\frac {16 \sqrt {3} (2-e x)^{5/2}}{5 e}-\frac {32 (2-e x)^{3/2}}{\sqrt {3} e} \]

[In]

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

[Out]

(-32*(2 - e*x)^(3/2))/(Sqrt[3]*e) + (16*Sqrt[3]*(2 - e*x)^(5/2))/(5*e) - (2*Sqrt[3]*(2 - e*x)^(7/2))/(7*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 \sqrt {6-3 e x} (2+e x)^2 \, dx \\ & = \int \left (16 \sqrt {6-3 e x}-\frac {8}{3} (6-3 e x)^{3/2}+\frac {1}{9} (6-3 e x)^{5/2}\right ) \, dx \\ & = -\frac {32 (2-e x)^{3/2}}{\sqrt {3} e}+\frac {16 \sqrt {3} (2-e x)^{5/2}}{5 e}-\frac {2 \sqrt {3} (2-e x)^{7/2}}{7 e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.77 \[ \int (2+e x)^{3/2} \sqrt {12-3 e^2 x^2} \, dx=\frac {2 (-2+e x) \sqrt {4-e^2 x^2} \left (284+108 e x+15 e^2 x^2\right )}{35 e \sqrt {6+3 e x}} \]

[In]

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

[Out]

(2*(-2 + e*x)*Sqrt[4 - e^2*x^2]*(284 + 108*e*x + 15*e^2*x^2))/(35*e*Sqrt[6 + 3*e*x])

Maple [A] (verified)

Time = 2.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.68

method	result	size
gosper	\(\frac {2 \left (e x -2\right ) \left (15 x^{2} e^{2}+108 e x +284\right ) \sqrt {-3 x^{2} e^{2}+12}}{105 \sqrt {e x +2}\, e}\)	\(44\)
default	\(\frac {2 \left (e x -2\right ) \left (15 x^{2} e^{2}+108 e x +284\right ) \sqrt {-3 x^{2} e^{2}+12}}{105 \sqrt {e x +2}\, e}\)	\(44\)
risch	\(-\frac {2 \sqrt {\frac {-3 x^{2} e^{2}+12}{e x +2}}\, \sqrt {e x +2}\, \left (15 e^{3} x^{3}+78 x^{2} e^{2}+68 e x -568\right ) \left (e x -2\right )}{35 \sqrt {-3 x^{2} e^{2}+12}\, e \sqrt {-3 e x +6}}\)	\(80\)

[In]

int((e*x+2)^(3/2)*(-3*e^2*x^2+12)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/105*(e*x-2)*(15*e^2*x^2+108*e*x+284)*(-3*e^2*x^2+12)^(1/2)/(e*x+2)^(1/2)/e

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.83 \[ \int (2+e x)^{3/2} \sqrt {12-3 e^2 x^2} \, dx=\frac {2 \, {\left (15 \, e^{3} x^{3} + 78 \, e^{2} x^{2} + 68 \, e x - 568\right )} \sqrt {-3 \, e^{2} x^{2} + 12} \sqrt {e x + 2}}{105 \, {\left (e^{2} x + 2 \, e\right )}} \]

[In]

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

[Out]

2/105*(15*e^3*x^3 + 78*e^2*x^2 + 68*e*x - 568)*sqrt(-3*e^2*x^2 + 12)*sqrt(e*x + 2)/(e^2*x + 2*e)

Sympy [F]

\[ \int (2+e x)^{3/2} \sqrt {12-3 e^2 x^2} \, dx=\sqrt {3} \left (\int 2 \sqrt {e x + 2} \sqrt {- e^{2} x^{2} + 4}\, dx + \int e x \sqrt {e x + 2} \sqrt {- e^{2} x^{2} + 4}\, dx\right ) \]

[In]

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

[Out]

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.92 \[ \int (2+e x)^{3/2} \sqrt {12-3 e^2 x^2} \, dx=-\frac {2 \, {\left (-15 i \, \sqrt {3} e^{3} x^{3} - 78 i \, \sqrt {3} e^{2} x^{2} - 68 i \, \sqrt {3} e x + 568 i \, \sqrt {3}\right )} {\left (e x + 2\right )} \sqrt {e x - 2}}{105 \, {\left (e^{2} x + 2 \, e\right )}} \]

[In]

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

[Out]

-2/105*(-15*I*sqrt(3)*e^3*x^3 - 78*I*sqrt(3)*e^2*x^2 - 68*I*sqrt(3)*e*x + 568*I*sqrt(3))*(e*x + 2)*sqrt(e*x -
2)/(e^2*x + 2*e)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.48 \[ \int (2+e x)^{3/2} \sqrt {12-3 e^2 x^2} \, dx=\frac {2 \, \sqrt {3} {\left (e^{2} {\left (\frac {15 \, {\left (e x - 2\right )}^{3} \sqrt {-e x + 2} + 84 \, {\left (e x - 2\right )}^{2} \sqrt {-e x + 2} - 140 \, {\left (-e x + 2\right )}^{\frac {3}{2}}}{e^{2}} + \frac {352}{e^{2}}\right )} + 84 \, {\left (e x - 2\right )}^{2} \sqrt {-e x + 2} - 420 \, {\left (-e x + 2\right )}^{\frac {3}{2}} + 672\right )}}{105 \, e} \]

[In]

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

[Out]

2/105*sqrt(3)*(e^2*((15*(e*x - 2)^3*sqrt(-e*x + 2) + 84*(e*x - 2)^2*sqrt(-e*x + 2) - 140*(-e*x + 2)^(3/2))/e^2
+ 352/e^2) + 84*(e*x - 2)^2*sqrt(-e*x + 2) - 420*(-e*x + 2)^(3/2) + 672)/e

Mupad [B] (verification not implemented)

Time = 9.89 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.12 \[ \int (2+e x)^{3/2} \sqrt {12-3 e^2 x^2} \, dx=\frac {\sqrt {12-3\,e^2\,x^2}\,\left (\frac {52\,x^2\,\sqrt {e\,x+2}}{35}-\frac {1136\,\sqrt {e\,x+2}}{105\,e^2}+\frac {136\,x\,\sqrt {e\,x+2}}{105\,e}+\frac {2\,e\,x^3\,\sqrt {e\,x+2}}{7}\right )}{x+\frac {2}{e}} \]

[In]

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

[Out]

((12 - 3*e^2*x^2)^(1/2)*((52*x^2*(e*x + 2)^(1/2))/35 - (1136*(e*x + 2)^(1/2))/(105*e^2) + (136*x*(e*x + 2)^(1/
2))/(105*e) + (2*e*x^3*(e*x + 2)^(1/2))/7))/(x + 2/e)