\(\int \frac {(2+e x)^{7/2}}{\sqrt {12-3 e^2 x^2}} \, dx\) [911]

Optimal result

Integrand size = 24, antiderivative size = 85 \[ \int \frac {(2+e x)^{7/2}}{\sqrt {12-3 e^2 x^2}} \, dx=-\frac {128 \sqrt {2-e x}}{\sqrt {3} e}+\frac {32 (2-e x)^{3/2}}{\sqrt {3} e}-\frac {8 \sqrt {3} (2-e x)^{5/2}}{5 e}+\frac {2 (2-e x)^{7/2}}{7 \sqrt {3} e} \]

[Out]

32/3*(-e*x+2)^(3/2)/e*3^(1/2)+2/21*(-e*x+2)^(7/2)*3^(1/2)/e-8/5*(-e*x+2)^(5/2)*3^(1/2)/e-128/3*3^(1/2)*(-e*x+2
)^(1/2)/e

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 85, 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} \[ \int \frac {(2+e x)^{7/2}}{\sqrt {12-3 e^2 x^2}} \, dx=\frac {2 (2-e x)^{7/2}}{7 \sqrt {3} e}-\frac {8 \sqrt {3} (2-e x)^{5/2}}{5 e}+\frac {32 (2-e x)^{3/2}}{\sqrt {3} e}-\frac {128 \sqrt {2-e x}}{\sqrt {3} e} \]

[In]

Int[(2 + e*x)^(7/2)/Sqrt[12 - 3*e^2*x^2],x]

[Out]

(-128*Sqrt[2 - e*x])/(Sqrt[3]*e) + (32*(2 - e*x)^(3/2))/(Sqrt[3]*e) - (8*Sqrt[3]*(2 - e*x)^(5/2))/(5*e) + (2*(
2 - e*x)^(7/2))/(7*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}{\sqrt {6-3 e x}} \, dx \\ & = \int \left (\frac {64}{\sqrt {6-3 e x}}-16 \sqrt {6-3 e x}+\frac {4}{3} (6-3 e x)^{3/2}-\frac {1}{27} (6-3 e x)^{5/2}\right ) \, dx \\ & = -\frac {128 \sqrt {2-e x}}{\sqrt {3} e}+\frac {32 (2-e x)^{3/2}}{\sqrt {3} e}-\frac {8 \sqrt {3} (2-e x)^{5/2}}{5 e}+\frac {2 (2-e x)^{7/2}}{7 \sqrt {3} e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.62 \[ \int \frac {(2+e x)^{7/2}}{\sqrt {12-3 e^2 x^2}} \, dx=-\frac {2 \sqrt {4-e^2 x^2} \left (1416+284 e x+54 e^2 x^2+5 e^3 x^3\right )}{35 e \sqrt {6+3 e x}} \]

[In]

Integrate[(2 + e*x)^(7/2)/Sqrt[12 - 3*e^2*x^2],x]

[Out]

(-2*Sqrt[4 - e^2*x^2]*(1416 + 284*e*x + 54*e^2*x^2 + 5*e^3*x^3))/(35*e*Sqrt[6 + 3*e*x])

Maple [A] (verified)

Time = 2.26 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.55

method	result	size
default	\(-\frac {2 \sqrt {-3 x^{2} e^{2}+12}\, \left (5 e^{3} x^{3}+54 x^{2} e^{2}+284 e x +1416\right )}{105 \sqrt {e x +2}\, e}\)	\(47\)
gosper	\(\frac {2 \left (e x -2\right ) \left (5 e^{3} x^{3}+54 x^{2} e^{2}+284 e x +1416\right ) \sqrt {e x +2}}{35 e \sqrt {-3 x^{2} e^{2}+12}}\)	\(52\)
risch	\(\frac {2 \sqrt {\frac {-3 x^{2} e^{2}+12}{e x +2}}\, \sqrt {e x +2}\, \left (5 e^{3} x^{3}+54 x^{2} e^{2}+284 e x +1416\right ) \left (e x -2\right )}{35 \sqrt {-3 x^{2} e^{2}+12}\, e \sqrt {-3 e x +6}}\)	\(80\)

[In]

int((e*x+2)^(7/2)/(-3*e^2*x^2+12)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/105/(e*x+2)^(1/2)*(-3*e^2*x^2+12)^(1/2)*(5*e^3*x^3+54*e^2*x^2+284*e*x+1416)/e

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.64 \[ \int \frac {(2+e x)^{7/2}}{\sqrt {12-3 e^2 x^2}} \, dx=-\frac {2 \, {\left (5 \, e^{3} x^{3} + 54 \, e^{2} x^{2} + 284 \, e x + 1416\right )} \sqrt {-3 \, e^{2} x^{2} + 12} \sqrt {e x + 2}}{105 \, {\left (e^{2} x + 2 \, e\right )}} \]

[In]

integrate((e*x+2)^(7/2)/(-3*e^2*x^2+12)^(1/2),x, algorithm="fricas")

[Out]

-2/105*(5*e^3*x^3 + 54*e^2*x^2 + 284*e*x + 1416)*sqrt(-3*e^2*x^2 + 12)*sqrt(e*x + 2)/(e^2*x + 2*e)

Sympy [F(-1)]

Timed out. \[ \int \frac {(2+e x)^{7/2}}{\sqrt {12-3 e^2 x^2}} \, dx=\text {Timed out} \]

[In]

integrate((e*x+2)**(7/2)/(-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 = 0.28 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.53 \[ \int \frac {(2+e x)^{7/2}}{\sqrt {12-3 e^2 x^2}} \, dx=-\frac {2 i \, \sqrt {3} {\left (5 \, e^{4} x^{4} + 44 \, e^{3} x^{3} + 176 \, e^{2} x^{2} + 848 \, e x - 2832\right )}}{105 \, \sqrt {e x - 2} e} \]

[In]

integrate((e*x+2)^(7/2)/(-3*e^2*x^2+12)^(1/2),x, algorithm="maxima")

[Out]

-2/105*I*sqrt(3)*(5*e^4*x^4 + 44*e^3*x^3 + 176*e^2*x^2 + 848*e*x - 2832)/(sqrt(e*x - 2)*e)

Giac [F(-2)]

Exception generated. \[ \int \frac {(2+e x)^{7/2}}{\sqrt {12-3 e^2 x^2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((e*x+2)^(7/2)/(-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 = 0.21 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.87 \[ \int \frac {(2+e x)^{7/2}}{\sqrt {12-3 e^2 x^2}} \, dx=-\frac {\sqrt {12-3\,e^2\,x^2}\,\left (\frac {944\,\sqrt {e\,x+2}}{35\,e^2}+\frac {36\,x^2\,\sqrt {e\,x+2}}{35}+\frac {568\,x\,\sqrt {e\,x+2}}{105\,e}+\frac {2\,e\,x^3\,\sqrt {e\,x+2}}{21}\right )}{x+\frac {2}{e}} \]

[In]

int((e*x + 2)^(7/2)/(12 - 3*e^2*x^2)^(1/2),x)

[Out]

-((12 - 3*e^2*x^2)^(1/2)*((944*(e*x + 2)^(1/2))/(35*e^2) + (36*x^2*(e*x + 2)^(1/2))/35 + (568*x*(e*x + 2)^(1/2
))/(105*e) + (2*e*x^3*(e*x + 2)^(1/2))/21))/(x + 2/e)