\(\int (2+e x)^{5/2} \sqrt {12-3 e^2 x^2} \, dx\) [894]

Optimal result

Integrand size = 24, antiderivative size = 87 \[ \int (2+e x)^{5/2} \sqrt {12-3 e^2 x^2} \, dx=-\frac {128 (2-e x)^{3/2}}{\sqrt {3} e}+\frac {96 \sqrt {3} (2-e x)^{5/2}}{5 e}-\frac {24 \sqrt {3} (2-e x)^{7/2}}{7 e}+\frac {2 (2-e x)^{9/2}}{3 \sqrt {3} e} \]

[Out]

-128/3*(-e*x+2)^(3/2)/e*3^(1/2)+2/9*(-e*x+2)^(9/2)/e*3^(1/2)+96/5*(-e*x+2)^(5/2)*3^(1/2)/e-24/7*(-e*x+2)^(7/2)
*3^(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} \[ \int (2+e x)^{5/2} \sqrt {12-3 e^2 x^2} \, dx=\frac {2 (2-e x)^{9/2}}{3 \sqrt {3} e}-\frac {24 \sqrt {3} (2-e x)^{7/2}}{7 e}+\frac {96 \sqrt {3} (2-e x)^{5/2}}{5 e}-\frac {128 (2-e x)^{3/2}}{\sqrt {3} e} \]

[In]

Int[(2 + e*x)^(5/2)*Sqrt[12 - 3*e^2*x^2],x]

[Out]

(-128*(2 - e*x)^(3/2))/(Sqrt[3]*e) + (96*Sqrt[3]*(2 - e*x)^(5/2))/(5*e) - (24*Sqrt[3]*(2 - e*x)^(7/2))/(7*e) +
(2*(2 - e*x)^(9/2))/(3*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 \sqrt {6-3 e x} (2+e x)^3 \, dx \\ & = \int \left (64 \sqrt {6-3 e x}-16 (6-3 e x)^{3/2}+\frac {4}{3} (6-3 e x)^{5/2}-\frac {1}{27} (6-3 e x)^{7/2}\right ) \, dx \\ & = -\frac {128 (2-e x)^{3/2}}{\sqrt {3} e}+\frac {96 \sqrt {3} (2-e x)^{5/2}}{5 e}-\frac {24 \sqrt {3} (2-e x)^{7/2}}{7 e}+\frac {2 (2-e x)^{9/2}}{3 \sqrt {3} e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.67 \[ \int (2+e x)^{5/2} \sqrt {12-3 e^2 x^2} \, dx=\frac {2 (-2+e x) \sqrt {4-e^2 x^2} \left (2552+1284 e x+330 e^2 x^2+35 e^3 x^3\right )}{105 e \sqrt {6+3 e x}} \]

[In]

Integrate[(2 + e*x)^(5/2)*Sqrt[12 - 3*e^2*x^2],x]

[Out]

(2*(-2 + e*x)*Sqrt[4 - e^2*x^2]*(2552 + 1284*e*x + 330*e^2*x^2 + 35*e^3*x^3))/(105*e*Sqrt[6 + 3*e*x])

Maple [A] (verified)

Time = 2.20 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.60

method	result	size
gosper	\(\frac {2 \left (e x -2\right ) \left (35 e^{3} x^{3}+330 x^{2} e^{2}+1284 e x +2552\right ) \sqrt {-3 x^{2} e^{2}+12}}{315 e \sqrt {e x +2}}\)	\(52\)
default	\(\frac {2 \left (e x -2\right ) \left (35 e^{3} x^{3}+330 x^{2} e^{2}+1284 e x +2552\right ) \sqrt {-3 x^{2} e^{2}+12}}{315 e \sqrt {e x +2}}\)	\(52\)
risch	\(-\frac {2 \sqrt {\frac {-3 x^{2} e^{2}+12}{e x +2}}\, \sqrt {e x +2}\, \left (35 e^{4} x^{4}+260 e^{3} x^{3}+624 x^{2} e^{2}-16 e x -5104\right ) \left (e x -2\right )}{105 \sqrt {-3 x^{2} e^{2}+12}\, e \sqrt {-3 e x +6}}\)	\(88\)

[In]

int((e*x+2)^(5/2)*(-3*e^2*x^2+12)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/315*(e*x-2)*(35*e^3*x^3+330*e^2*x^2+1284*e*x+2552)*(-3*e^2*x^2+12)^(1/2)/e/(e*x+2)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.71 \[ \int (2+e x)^{5/2} \sqrt {12-3 e^2 x^2} \, dx=\frac {2 \, {\left (35 \, e^{4} x^{4} + 260 \, e^{3} x^{3} + 624 \, e^{2} x^{2} - 16 \, e x - 5104\right )} \sqrt {-3 \, e^{2} x^{2} + 12} \sqrt {e x + 2}}{315 \, {\left (e^{2} x + 2 \, e\right )}} \]

[In]

integrate((e*x+2)^(5/2)*(-3*e^2*x^2+12)^(1/2),x, algorithm="fricas")

[Out]

2/315*(35*e^4*x^4 + 260*e^3*x^3 + 624*e^2*x^2 - 16*e*x - 5104)*sqrt(-3*e^2*x^2 + 12)*sqrt(e*x + 2)/(e^2*x + 2*
e)

Sympy [F(-1)]

Timed out. \[ \int (2+e x)^{5/2} \sqrt {12-3 e^2 x^2} \, dx=\text {Timed out} \]

[In]

integrate((e*x+2)**(5/2)*(-3*e**2*x**2+12)**(1/2),x)

[Out]

Timed out

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.82 \[ \int (2+e x)^{5/2} \sqrt {12-3 e^2 x^2} \, dx=-\frac {2 \, {\left (-35 i \, \sqrt {3} e^{4} x^{4} - 260 i \, \sqrt {3} e^{3} x^{3} - 624 i \, \sqrt {3} e^{2} x^{2} + 16 i \, \sqrt {3} e x + 5104 i \, \sqrt {3}\right )} {\left (e x + 2\right )} \sqrt {e x - 2}}{315 \, {\left (e^{2} x + 2 \, e\right )}} \]

[In]

integrate((e*x+2)^(5/2)*(-3*e^2*x^2+12)^(1/2),x, algorithm="maxima")

[Out]

-2/315*(-35*I*sqrt(3)*e^4*x^4 - 260*I*sqrt(3)*e^3*x^3 - 624*I*sqrt(3)*e^2*x^2 + 16*I*sqrt(3)*e*x + 5104*I*sqrt
(3))*(e*x + 2)*sqrt(e*x - 2)/(e^2*x + 2*e)

Giac [F(-2)]

Exception generated. \[ \int (2+e x)^{5/2} \sqrt {12-3 e^2 x^2} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((e*x+2)^(5/2)*(-3*e^2*x^2+12)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [B] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.62 \[ \int (2+e x)^{5/2} \sqrt {12-3 e^2 x^2} \, dx=\frac {2\,\sqrt {12-3\,e^2\,x^2}\,\left (35\,e^4\,x^4+260\,e^3\,x^3+624\,e^2\,x^2-16\,e\,x-5104\right )}{315\,e\,\sqrt {e\,x+2}} \]

[In]

int((12 - 3*e^2*x^2)^(1/2)*(e*x + 2)^(5/2),x)

[Out]

(2*(12 - 3*e^2*x^2)^(1/2)*(624*e^2*x^2 - 16*e*x + 260*e^3*x^3 + 35*e^4*x^4 - 5104))/(315*e*(e*x + 2)^(1/2))