\(\int \frac {1}{(2+e x)^{9/2} \sqrt [4]{12-3 e^2 x^2}} \, dx\) [944]

Optimal result

Integrand size = 24, antiderivative size = 141 \[ \int \frac {1}{(2+e x)^{9/2} \sqrt [4]{12-3 e^2 x^2}} \, dx=-\frac {\left (4-e^2 x^2\right )^{3/4}}{15 \sqrt [4]{3} e (2+e x)^{9/2}}-\frac {\left (4-e^2 x^2\right )^{3/4}}{55 \sqrt [4]{3} e (2+e x)^{7/2}}-\frac {2 \left (4-e^2 x^2\right )^{3/4}}{385 \sqrt [4]{3} e (2+e x)^{5/2}}-\frac {2 \left (4-e^2 x^2\right )^{3/4}}{1155 \sqrt [4]{3} e (2+e x)^{3/2}} \]

[Out]

-1/45*(-e^2*x^2+4)^(3/4)*3^(3/4)/e/(e*x+2)^(9/2)-1/165*(-e^2*x^2+4)^(3/4)*3^(3/4)/e/(e*x+2)^(7/2)-2/1155*(-e^2
*x^2+4)^(3/4)*3^(3/4)/e/(e*x+2)^(5/2)-2/3465*(-e^2*x^2+4)^(3/4)*3^(3/4)/e/(e*x+2)^(3/2)

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} \[ \int \frac {1}{(2+e x)^{9/2} \sqrt [4]{12-3 e^2 x^2}} \, dx=-\frac {2 \left (4-e^2 x^2\right )^{3/4}}{1155 \sqrt [4]{3} e (e x+2)^{3/2}}-\frac {2 \left (4-e^2 x^2\right )^{3/4}}{385 \sqrt [4]{3} e (e x+2)^{5/2}}-\frac {\left (4-e^2 x^2\right )^{3/4}}{55 \sqrt [4]{3} e (e x+2)^{7/2}}-\frac {\left (4-e^2 x^2\right )^{3/4}}{15 \sqrt [4]{3} e (e x+2)^{9/2}} \]

[In]

Int[1/((2 + e*x)^(9/2)*(12 - 3*e^2*x^2)^(1/4)),x]

[Out]

-1/15*(4 - e^2*x^2)^(3/4)/(3^(1/4)*e*(2 + e*x)^(9/2)) - (4 - e^2*x^2)^(3/4)/(55*3^(1/4)*e*(2 + e*x)^(7/2)) - (
2*(4 - e^2*x^2)^(3/4))/(385*3^(1/4)*e*(2 + e*x)^(5/2)) - (2*(4 - e^2*x^2)^(3/4))/(1155*3^(1/4)*e*(2 + e*x)^(3/
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 {\left (4-e^2 x^2\right )^{3/4}}{15 \sqrt [4]{3} e (2+e x)^{9/2}}+\frac {1}{5} \int \frac {1}{(2+e x)^{7/2} \sqrt [4]{12-3 e^2 x^2}} \, dx \\ & = -\frac {\left (4-e^2 x^2\right )^{3/4}}{15 \sqrt [4]{3} e (2+e x)^{9/2}}-\frac {\left (4-e^2 x^2\right )^{3/4}}{55 \sqrt [4]{3} e (2+e x)^{7/2}}+\frac {2}{55} \int \frac {1}{(2+e x)^{5/2} \sqrt [4]{12-3 e^2 x^2}} \, dx \\ & = -\frac {\left (4-e^2 x^2\right )^{3/4}}{15 \sqrt [4]{3} e (2+e x)^{9/2}}-\frac {\left (4-e^2 x^2\right )^{3/4}}{55 \sqrt [4]{3} e (2+e x)^{7/2}}-\frac {2 \left (4-e^2 x^2\right )^{3/4}}{385 \sqrt [4]{3} e (2+e x)^{5/2}}+\frac {2}{385} \int \frac {1}{(2+e x)^{3/2} \sqrt [4]{12-3 e^2 x^2}} \, dx \\ & = -\frac {\left (4-e^2 x^2\right )^{3/4}}{15 \sqrt [4]{3} e (2+e x)^{9/2}}-\frac {\left (4-e^2 x^2\right )^{3/4}}{55 \sqrt [4]{3} e (2+e x)^{7/2}}-\frac {2 \left (4-e^2 x^2\right )^{3/4}}{385 \sqrt [4]{3} e (2+e x)^{5/2}}-\frac {2 \left (4-e^2 x^2\right )^{3/4}}{1155 \sqrt [4]{3} e (2+e x)^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.79 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.40 \[ \int \frac {1}{(2+e x)^{9/2} \sqrt [4]{12-3 e^2 x^2}} \, dx=-\frac {\left (4-e^2 x^2\right )^{3/4} \left (159+69 e x+18 e^2 x^2+2 e^3 x^3\right )}{1155 \sqrt [4]{3} e (2+e x)^{9/2}} \]

[In]

Integrate[1/((2 + e*x)^(9/2)*(12 - 3*e^2*x^2)^(1/4)),x]

[Out]

-1/1155*((4 - e^2*x^2)^(3/4)*(159 + 69*e*x + 18*e^2*x^2 + 2*e^3*x^3))/(3^(1/4)*e*(2 + e*x)^(9/2))

Maple [A] (verified)

Time = 2.25 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.37

method	result	size
gosper	\(\frac {\left (e x -2\right ) \left (2 e^{3} x^{3}+18 x^{2} e^{2}+69 e x +159\right )}{1155 \left (e x +2\right )^{\frac {7}{2}} e \left (-3 x^{2} e^{2}+12\right )^{\frac {1}{4}}}\)	\(52\)

[In]

int(1/(e*x+2)^(9/2)/(-3*e^2*x^2+12)^(1/4),x,method=_RETURNVERBOSE)

[Out]

1/1155*(e*x-2)*(2*e^3*x^3+18*e^2*x^2+69*e*x+159)/(e*x+2)^(7/2)/e/(-3*e^2*x^2+12)^(1/4)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.61 \[ \int \frac {1}{(2+e x)^{9/2} \sqrt [4]{12-3 e^2 x^2}} \, dx=-\frac {{\left (2 \, e^{3} x^{3} + 18 \, e^{2} x^{2} + 69 \, e x + 159\right )} {\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {3}{4}} \sqrt {e x + 2}}{3465 \, {\left (e^{6} x^{5} + 10 \, e^{5} x^{4} + 40 \, e^{4} x^{3} + 80 \, e^{3} x^{2} + 80 \, e^{2} x + 32 \, e\right )}} \]

[In]

integrate(1/(e*x+2)^(9/2)/(-3*e^2*x^2+12)^(1/4),x, algorithm="fricas")

[Out]

-1/3465*(2*e^3*x^3 + 18*e^2*x^2 + 69*e*x + 159)*(-3*e^2*x^2 + 12)^(3/4)*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. \[ \int \frac {1}{(2+e x)^{9/2} \sqrt [4]{12-3 e^2 x^2}} \, dx=\text {Timed out} \]

[In]

integrate(1/(e*x+2)**(9/2)/(-3*e**2*x**2+12)**(1/4),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {1}{(2+e x)^{9/2} \sqrt [4]{12-3 e^2 x^2}} \, dx=\int { \frac {1}{{\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {1}{4}} {\left (e x + 2\right )}^{\frac {9}{2}}} \,d x } \]

[In]

integrate(1/(e*x+2)^(9/2)/(-3*e^2*x^2+12)^(1/4),x, algorithm="maxima")

[Out]

integrate(1/((-3*e^2*x^2 + 12)^(1/4)*(e*x + 2)^(9/2)), x)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.49 \[ \int \frac {1}{(2+e x)^{9/2} \sqrt [4]{12-3 e^2 x^2}} \, dx=-\frac {3^{\frac {3}{4}} {\left (77 \, {\left (\frac {4}{e x + 2} - 1\right )}^{\frac {15}{4}} + 315 \, {\left (\frac {4}{e x + 2} - 1\right )}^{\frac {11}{4}} + 495 \, {\left (\frac {4}{e x + 2} - 1\right )}^{\frac {7}{4}} + 385 \, {\left (\frac {4}{e x + 2} - 1\right )}^{\frac {3}{4}}\right )}}{221760 \, e} \]

[In]

integrate(1/(e*x+2)^(9/2)/(-3*e^2*x^2+12)^(1/4),x, algorithm="giac")

[Out]

-1/221760*3^(3/4)*(77*(4/(e*x + 2) - 1)^(15/4) + 315*(4/(e*x + 2) - 1)^(11/4) + 495*(4/(e*x + 2) - 1)^(7/4) +
385*(4/(e*x + 2) - 1)^(3/4))/e

Mupad [B] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.79 \[ \int \frac {1}{(2+e x)^{9/2} \sqrt [4]{12-3 e^2 x^2}} \, dx=-\frac {{\left (12-3\,e^2\,x^2\right )}^{3/4}\,\left (\frac {23\,x}{1155\,e^4}+\frac {53}{1155\,e^5}+\frac {2\,x^3}{3465\,e^2}+\frac {2\,x^2}{385\,e^3}\right )}{\frac {16\,\sqrt {e\,x+2}}{e^4}+x^4\,\sqrt {e\,x+2}+\frac {32\,x\,\sqrt {e\,x+2}}{e^3}+\frac {8\,x^3\,\sqrt {e\,x+2}}{e}+\frac {24\,x^2\,\sqrt {e\,x+2}}{e^2}} \]

[In]

int(1/((12 - 3*e^2*x^2)^(1/4)*(e*x + 2)^(9/2)),x)

[Out]

-((12 - 3*e^2*x^2)^(3/4)*((23*x)/(1155*e^4) + 53/(1155*e^5) + (2*x^3)/(3465*e^2) + (2*x^2)/(385*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)