\(\int \frac {\sqrt [4]{12-3 e^2 x^2}}{(2+e x)^{11/2}} \, dx\) [936]

Optimal result

Integrand size = 24, antiderivative size = 141 \[ \int \frac {\sqrt [4]{12-3 e^2 x^2}}{(2+e x)^{11/2}} \, dx=-\frac {\sqrt [4]{3} \left (4-e^2 x^2\right )^{5/4}}{17 e (2+e x)^{11/2}}-\frac {3 \sqrt [4]{3} \left (4-e^2 x^2\right )^{5/4}}{221 e (2+e x)^{9/2}}-\frac {2 \left (4-e^2 x^2\right )^{5/4}}{221\ 3^{3/4} e (2+e x)^{7/2}}-\frac {2 \left (4-e^2 x^2\right )^{5/4}}{1105\ 3^{3/4} e (2+e x)^{5/2}} \]

[Out]

-1/17*3^(1/4)*(-e^2*x^2+4)^(5/4)/e/(e*x+2)^(11/2)-3/221*3^(1/4)*(-e^2*x^2+4)^(5/4)/e/(e*x+2)^(9/2)-2/663*(-e^2
*x^2+4)^(5/4)*3^(1/4)/e/(e*x+2)^(7/2)-2/3315*3^(1/4)*(-e^2*x^2+4)^(5/4)/e/(e*x+2)^(5/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {673, 665} \[ \int \frac {\sqrt [4]{12-3 e^2 x^2}}{(2+e x)^{11/2}} \, dx=-\frac {2 \left (4-e^2 x^2\right )^{5/4}}{1105\ 3^{3/4} e (e x+2)^{5/2}}-\frac {2 \left (4-e^2 x^2\right )^{5/4}}{221\ 3^{3/4} e (e x+2)^{7/2}}-\frac {3 \sqrt [4]{3} \left (4-e^2 x^2\right )^{5/4}}{221 e (e x+2)^{9/2}}-\frac {\sqrt [4]{3} \left (4-e^2 x^2\right )^{5/4}}{17 e (e x+2)^{11/2}} \]

[In]

Int[(12 - 3*e^2*x^2)^(1/4)/(2 + e*x)^(11/2),x]

[Out]

-1/17*(3^(1/4)*(4 - e^2*x^2)^(5/4))/(e*(2 + e*x)^(11/2)) - (3*3^(1/4)*(4 - e^2*x^2)^(5/4))/(221*e*(2 + e*x)^(9
/2)) - (2*(4 - e^2*x^2)^(5/4))/(221*3^(3/4)*e*(2 + e*x)^(7/2)) - (2*(4 - e^2*x^2)^(5/4))/(1105*3^(3/4)*e*(2 +
e*x)^(5/2))

Rule 665

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^m*((a + c*x^2)^(p + 1)/
(2*c*d*(p + 1))), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p
+ 2, 0]

Rule 673

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-e)*(d + e*x)^m*((a + c*x^2)^(p +
1)/(2*c*d*(m + p + 1))), x] + Dist[Simplify[m + 2*p + 2]/(2*d*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^
p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p +
2], 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {\sqrt [4]{3} \left (4-e^2 x^2\right )^{5/4}}{17 e (2+e x)^{11/2}}+\frac {3}{17} \int \frac {\sqrt [4]{12-3 e^2 x^2}}{(2+e x)^{9/2}} \, dx \\ & = -\frac {\sqrt [4]{3} \left (4-e^2 x^2\right )^{5/4}}{17 e (2+e x)^{11/2}}-\frac {3 \sqrt [4]{3} \left (4-e^2 x^2\right )^{5/4}}{221 e (2+e x)^{9/2}}+\frac {6}{221} \int \frac {\sqrt [4]{12-3 e^2 x^2}}{(2+e x)^{7/2}} \, dx \\ & = -\frac {\sqrt [4]{3} \left (4-e^2 x^2\right )^{5/4}}{17 e (2+e x)^{11/2}}-\frac {3 \sqrt [4]{3} \left (4-e^2 x^2\right )^{5/4}}{221 e (2+e x)^{9/2}}-\frac {2 \left (4-e^2 x^2\right )^{5/4}}{221\ 3^{3/4} e (2+e x)^{7/2}}+\frac {2}{663} \int \frac {\sqrt [4]{12-3 e^2 x^2}}{(2+e x)^{5/2}} \, dx \\ & = -\frac {\sqrt [4]{3} \left (4-e^2 x^2\right )^{5/4}}{17 e (2+e x)^{11/2}}-\frac {3 \sqrt [4]{3} \left (4-e^2 x^2\right )^{5/4}}{221 e (2+e x)^{9/2}}-\frac {2 \left (4-e^2 x^2\right )^{5/4}}{221\ 3^{3/4} e (2+e x)^{7/2}}-\frac {2 \left (4-e^2 x^2\right )^{5/4}}{1105\ 3^{3/4} e (2+e x)^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.44 \[ \int \frac {\sqrt [4]{12-3 e^2 x^2}}{(2+e x)^{11/2}} \, dx=\frac {(-2+e x) \sqrt [4]{4-e^2 x^2} \left (341+109 e x+22 e^2 x^2+2 e^3 x^3\right )}{1105\ 3^{3/4} e (2+e x)^{9/2}} \]

[In]

Integrate[(12 - 3*e^2*x^2)^(1/4)/(2 + e*x)^(11/2),x]

[Out]

((-2 + e*x)*(4 - e^2*x^2)^(1/4)*(341 + 109*e*x + 22*e^2*x^2 + 2*e^3*x^3))/(1105*3^(3/4)*e*(2 + e*x)^(9/2))

Maple [A] (verified)

Time = 2.40 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.37

method	result	size
gosper	\(\frac {\left (e x -2\right ) \left (2 e^{3} x^{3}+22 x^{2} e^{2}+109 e x +341\right ) \left (-3 x^{2} e^{2}+12\right )^{\frac {1}{4}}}{3315 \left (e x +2\right )^{\frac {9}{2}} e}\)	\(52\)

[In]

int((-3*e^2*x^2+12)^(1/4)/(e*x+2)^(11/2),x,method=_RETURNVERBOSE)

[Out]

1/3315*(e*x-2)*(2*e^3*x^3+22*e^2*x^2+109*e*x+341)*(-3*e^2*x^2+12)^(1/4)/(e*x+2)^(9/2)/e

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.67 \[ \int \frac {\sqrt [4]{12-3 e^2 x^2}}{(2+e x)^{11/2}} \, dx=\frac {{\left (2 \, e^{4} x^{4} + 18 \, e^{3} x^{3} + 65 \, e^{2} x^{2} + 123 \, e x - 682\right )} {\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {1}{4}} \sqrt {e x + 2}}{3315 \, {\left (e^{6} x^{5} + 10 \, e^{5} x^{4} + 40 \, e^{4} x^{3} + 80 \, e^{3} x^{2} + 80 \, e^{2} x + 32 \, e\right )}} \]

[In]

integrate((-3*e^2*x^2+12)^(1/4)/(e*x+2)^(11/2),x, algorithm="fricas")

[Out]

1/3315*(2*e^4*x^4 + 18*e^3*x^3 + 65*e^2*x^2 + 123*e*x - 682)*(-3*e^2*x^2 + 12)^(1/4)*sqrt(e*x + 2)/(e^6*x^5 +
10*e^5*x^4 + 40*e^4*x^3 + 80*e^3*x^2 + 80*e^2*x + 32*e)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt [4]{12-3 e^2 x^2}}{(2+e x)^{11/2}} \, dx=\text {Timed out} \]

[In]

integrate((-3*e**2*x**2+12)**(1/4)/(e*x+2)**(11/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\sqrt [4]{12-3 e^2 x^2}}{(2+e x)^{11/2}} \, dx=\int { \frac {{\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {1}{4}}}{{\left (e x + 2\right )}^{\frac {11}{2}}} \,d x } \]

[In]

integrate((-3*e^2*x^2+12)^(1/4)/(e*x+2)^(11/2),x, algorithm="maxima")

[Out]

integrate((-3*e^2*x^2 + 12)^(1/4)/(e*x + 2)^(11/2), x)

Giac [F(-2)]

Exception generated. \[ \int \frac {\sqrt [4]{12-3 e^2 x^2}}{(2+e x)^{11/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((-3*e^2*x^2+12)^(1/4)/(e*x+2)^(11/2),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 = 10.51 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.84 \[ \int \frac {\sqrt [4]{12-3 e^2 x^2}}{(2+e x)^{11/2}} \, dx=\frac {{\left (12-3\,e^2\,x^2\right )}^{1/4}\,\left (\frac {41\,x}{1105\,e^4}-\frac {682}{3315\,e^5}+\frac {2\,x^4}{3315\,e}+\frac {6\,x^3}{1105\,e^2}+\frac {x^2}{51\,e^3}\right )}{\frac {16\,\sqrt {e\,x+2}}{e^4}+x^4\,\sqrt {e\,x+2}+\frac {32\,x\,\sqrt {e\,x+2}}{e^3}+\frac {8\,x^3\,\sqrt {e\,x+2}}{e}+\frac {24\,x^2\,\sqrt {e\,x+2}}{e^2}} \]

[In]

int((12 - 3*e^2*x^2)^(1/4)/(e*x + 2)^(11/2),x)

[Out]

((12 - 3*e^2*x^2)^(1/4)*((41*x)/(1105*e^4) - 682/(3315*e^5) + (2*x^4)/(3315*e) + (6*x^3)/(1105*e^2) + x^2/(51*
e^3)))/((16*(e*x + 2)^(1/2))/e^4 + x^4*(e*x + 2)^(1/2) + (32*x*(e*x + 2)^(1/2))/e^3 + (8*x^3*(e*x + 2)^(1/2))/
e + (24*x^2*(e*x + 2)^(1/2))/e^2)