\(\int (2+e x)^{5/2} (12-3 e^2 x^2)^{3/2} \, dx\) [901]

Optimal result

Integrand size = 24, antiderivative size = 109 \[ \int (2+e x)^{5/2} \left (12-3 e^2 x^2\right )^{3/2} \, dx=-\frac {1536 \sqrt {3} (2-e x)^{5/2}}{5 e}+\frac {1536 \sqrt {3} (2-e x)^{7/2}}{7 e}-\frac {64 \sqrt {3} (2-e x)^{9/2}}{e}+\frac {96 \sqrt {3} (2-e x)^{11/2}}{11 e}-\frac {6 \sqrt {3} (2-e x)^{13/2}}{13 e} \]

[Out]

-1536/5*(-e*x+2)^(5/2)*3^(1/2)/e+1536/7*(-e*x+2)^(7/2)*3^(1/2)/e-64*(-e*x+2)^(9/2)/e*3^(1/2)+96/11*(-e*x+2)^(1
1/2)*3^(1/2)/e-6/13*(-e*x+2)^(13/2)*3^(1/2)/e

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 109, 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} \[ \int (2+e x)^{5/2} \left (12-3 e^2 x^2\right )^{3/2} \, dx=-\frac {6 \sqrt {3} (2-e x)^{13/2}}{13 e}+\frac {96 \sqrt {3} (2-e x)^{11/2}}{11 e}-\frac {64 \sqrt {3} (2-e x)^{9/2}}{e}+\frac {1536 \sqrt {3} (2-e x)^{7/2}}{7 e}-\frac {1536 \sqrt {3} (2-e x)^{5/2}}{5 e} \]

[In]

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

[Out]

(-1536*Sqrt[3]*(2 - e*x)^(5/2))/(5*e) + (1536*Sqrt[3]*(2 - e*x)^(7/2))/(7*e) - (64*Sqrt[3]*(2 - e*x)^(9/2))/e
+ (96*Sqrt[3]*(2 - e*x)^(11/2))/(11*e) - (6*Sqrt[3]*(2 - e*x)^(13/2))/(13*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 (6-3 e x)^{3/2} (2+e x)^4 \, dx \\ & = \int \left (256 (6-3 e x)^{3/2}-\frac {256}{3} (6-3 e x)^{5/2}+\frac {32}{3} (6-3 e x)^{7/2}-\frac {16}{27} (6-3 e x)^{9/2}+\frac {1}{81} (6-3 e x)^{11/2}\right ) \, dx \\ & = -\frac {1536 \sqrt {3} (2-e x)^{5/2}}{5 e}+\frac {1536 \sqrt {3} (2-e x)^{7/2}}{7 e}-\frac {64 \sqrt {3} (2-e x)^{9/2}}{e}+\frac {96 \sqrt {3} (2-e x)^{11/2}}{11 e}-\frac {6 \sqrt {3} (2-e x)^{13/2}}{13 e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.61 \[ \int (2+e x)^{5/2} \left (12-3 e^2 x^2\right )^{3/2} \, dx=-\frac {2 (-2+e x)^2 \sqrt {12-3 e^2 x^2} \left (154928+133600 e x+56840 e^2 x^2+12600 e^3 x^3+1155 e^4 x^4\right )}{5005 e \sqrt {2+e x}} \]

[In]

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

[Out]

(-2*(-2 + e*x)^2*Sqrt[12 - 3*e^2*x^2]*(154928 + 133600*e*x + 56840*e^2*x^2 + 12600*e^3*x^3 + 1155*e^4*x^4))/(5
005*e*Sqrt[2 + e*x])

Maple [A] (verified)

Time = 2.26 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.55

method	result	size
gosper	\(\frac {2 \left (e x -2\right ) \left (1155 e^{4} x^{4}+12600 e^{3} x^{3}+56840 x^{2} e^{2}+133600 e x +154928\right ) \left (-3 x^{2} e^{2}+12\right )^{\frac {3}{2}}}{15015 e \left (e x +2\right )^{\frac {3}{2}}}\)	\(60\)
default	\(-\frac {2 \sqrt {-3 x^{2} e^{2}+12}\, \left (e x -2\right )^{2} \left (1155 e^{4} x^{4}+12600 e^{3} x^{3}+56840 x^{2} e^{2}+133600 e x +154928\right )}{5005 \sqrt {e x +2}\, e}\)	\(62\)
risch	\(\frac {6 \sqrt {\frac {-3 x^{2} e^{2}+12}{e x +2}}\, \sqrt {e x +2}\, \left (1155 e^{6} x^{6}+7980 e^{5} x^{5}+11060 e^{4} x^{4}-43360 e^{3} x^{3}-152112 x^{2} e^{2}-85312 e x +619712\right ) \left (e x -2\right )}{5005 \sqrt {-3 x^{2} e^{2}+12}\, e \sqrt {-3 e x +6}}\)	\(104\)

[In]

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

[Out]

2/15015*(e*x-2)*(1155*e^4*x^4+12600*e^3*x^3+56840*e^2*x^2+133600*e*x+154928)*(-3*e^2*x^2+12)^(3/2)/e/(e*x+2)^(
3/2)

Fricas [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.72 \[ \int (2+e x)^{5/2} \left (12-3 e^2 x^2\right )^{3/2} \, dx=-\frac {2 \, {\left (1155 \, e^{6} x^{6} + 7980 \, e^{5} x^{5} + 11060 \, e^{4} x^{4} - 43360 \, e^{3} x^{3} - 152112 \, e^{2} x^{2} - 85312 \, e x + 619712\right )} \sqrt {-3 \, e^{2} x^{2} + 12} \sqrt {e x + 2}}{5005 \, {\left (e^{2} x + 2 \, e\right )}} \]

[In]

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

[Out]

-2/5005*(1155*e^6*x^6 + 7980*e^5*x^5 + 11060*e^4*x^4 - 43360*e^3*x^3 - 152112*e^2*x^2 - 85312*e*x + 619712)*sq
rt(-3*e^2*x^2 + 12)*sqrt(e*x + 2)/(e^2*x + 2*e)

Sympy [F(-1)]

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

[In]

integrate((e*x+2)**(5/2)*(-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 = 0.28 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.85 \[ \int (2+e x)^{5/2} \left (12-3 e^2 x^2\right )^{3/2} \, dx=\frac {2 \, {\left (-1155 i \, \sqrt {3} e^{6} x^{6} - 7980 i \, \sqrt {3} e^{5} x^{5} - 11060 i \, \sqrt {3} e^{4} x^{4} + 43360 i \, \sqrt {3} e^{3} x^{3} + 152112 i \, \sqrt {3} e^{2} x^{2} + 85312 i \, \sqrt {3} e x - 619712 i \, \sqrt {3}\right )} {\left (e x + 2\right )} \sqrt {e x - 2}}{5005 \, {\left (e^{2} x + 2 \, e\right )}} \]

[In]

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

[Out]

2/5005*(-1155*I*sqrt(3)*e^6*x^6 - 7980*I*sqrt(3)*e^5*x^5 - 11060*I*sqrt(3)*e^4*x^4 + 43360*I*sqrt(3)*e^3*x^3 +
152112*I*sqrt(3)*e^2*x^2 + 85312*I*sqrt(3)*e*x - 619712*I*sqrt(3))*(e*x + 2)*sqrt(e*x - 2)/(e^2*x + 2*e)

Giac [F(-2)]

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

[In]

integrate((e*x+2)^(5/2)*(-3*e^2*x^2+12)^(3/2),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 = 0.26 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.80 \[ \int (2+e x)^{5/2} \left (12-3 e^2 x^2\right )^{3/2} \, dx=\frac {2\,\sqrt {12-3\,e^2\,x^2}\,\sqrt {e\,x+2}\,\left (-1155\,e^5\,x^5-5670\,e^4\,x^4+280\,e^3\,x^3+42800\,e^2\,x^2+66512\,e\,x-47712\right )}{5005\,e}-\frac {1048576\,\sqrt {12-3\,e^2\,x^2}}{5005\,e\,\sqrt {e\,x+2}} \]

[In]

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

[Out]

(2*(12 - 3*e^2*x^2)^(1/2)*(e*x + 2)^(1/2)*(66512*e*x + 42800*e^2*x^2 + 280*e^3*x^3 - 5670*e^4*x^4 - 1155*e^5*x
^5 - 47712))/(5005*e) - (1048576*(12 - 3*e^2*x^2)^(1/2))/(5005*e*(e*x + 2)^(1/2))