\(\int \frac {1}{(2+e x)^{5/2} \sqrt [4]{12-3 e^2 x^2}} \, dx\) [942]

Optimal result

Integrand size = 24, antiderivative size = 71 \[ \int \frac {1}{(2+e x)^{5/2} \sqrt [4]{12-3 e^2 x^2}} \, dx=-\frac {\left (4-e^2 x^2\right )^{3/4}}{7 \sqrt [4]{3} e (2+e x)^{5/2}}-\frac {\left (4-e^2 x^2\right )^{3/4}}{21 \sqrt [4]{3} e (2+e x)^{3/2}} \]

[Out]

-1/21*(-e^2*x^2+4)^(3/4)*3^(3/4)/e/(e*x+2)^(5/2)-1/63*(-e^2*x^2+4)^(3/4)*3^(3/4)/e/(e*x+2)^(3/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {673, 665} \[ \int \frac {1}{(2+e x)^{5/2} \sqrt [4]{12-3 e^2 x^2}} \, dx=-\frac {\left (4-e^2 x^2\right )^{3/4}}{21 \sqrt [4]{3} e (e x+2)^{3/2}}-\frac {\left (4-e^2 x^2\right )^{3/4}}{7 \sqrt [4]{3} e (e x+2)^{5/2}} \]

[In]

Int[1/((2 + e*x)^(5/2)*(12 - 3*e^2*x^2)^(1/4)),x]

[Out]

-1/7*(4 - e^2*x^2)^(3/4)/(3^(1/4)*e*(2 + e*x)^(5/2)) - (4 - e^2*x^2)^(3/4)/(21*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}}{7 \sqrt [4]{3} e (2+e x)^{5/2}}+\frac {1}{7} \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}}{7 \sqrt [4]{3} e (2+e x)^{5/2}}-\frac {\left (4-e^2 x^2\right )^{3/4}}{21 \sqrt [4]{3} e (2+e x)^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.56 \[ \int \frac {1}{(2+e x)^{5/2} \sqrt [4]{12-3 e^2 x^2}} \, dx=-\frac {(5+e x) \left (4-e^2 x^2\right )^{3/4}}{21 \sqrt [4]{3} e (2+e x)^{5/2}} \]

[In]

Integrate[1/((2 + e*x)^(5/2)*(12 - 3*e^2*x^2)^(1/4)),x]

[Out]

-1/21*((5 + e*x)*(4 - e^2*x^2)^(3/4))/(3^(1/4)*e*(2 + e*x)^(5/2))

Maple [A] (verified)

Time = 2.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.49

method	result	size
gosper	\(\frac {\left (e x -2\right ) \left (e x +5\right )}{21 \left (e x +2\right )^{\frac {3}{2}} e \left (-3 x^{2} e^{2}+12\right )^{\frac {1}{4}}}\)	\(35\)

[In]

int(1/(e*x+2)^(5/2)/(-3*e^2*x^2+12)^(1/4),x,method=_RETURNVERBOSE)

[Out]

1/21*(e*x-2)*(e*x+5)/(e*x+2)^(3/2)/e/(-3*e^2*x^2+12)^(1/4)

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.75 \[ \int \frac {1}{(2+e x)^{5/2} \sqrt [4]{12-3 e^2 x^2}} \, dx=-\frac {{\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {3}{4}} {\left (e x + 5\right )} \sqrt {e x + 2}}{63 \, {\left (e^{4} x^{3} + 6 \, e^{3} x^{2} + 12 \, e^{2} x + 8 \, e\right )}} \]

[In]

integrate(1/(e*x+2)^(5/2)/(-3*e^2*x^2+12)^(1/4),x, algorithm="fricas")

[Out]

-1/63*(-3*e^2*x^2 + 12)^(3/4)*(e*x + 5)*sqrt(e*x + 2)/(e^4*x^3 + 6*e^3*x^2 + 12*e^2*x + 8*e)

Sympy [F]

\[ \int \frac {1}{(2+e x)^{5/2} \sqrt [4]{12-3 e^2 x^2}} \, dx=\frac {3^{\frac {3}{4}} \int \frac {1}{e^{2} x^{2} \sqrt {e x + 2} \sqrt [4]{- e^{2} x^{2} + 4} + 4 e x \sqrt {e x + 2} \sqrt [4]{- e^{2} x^{2} + 4} + 4 \sqrt {e x + 2} \sqrt [4]{- e^{2} x^{2} + 4}}\, dx}{3} \]

[In]

integrate(1/(e*x+2)**(5/2)/(-3*e**2*x**2+12)**(1/4),x)

[Out]

3**(3/4)*Integral(1/(e**2*x**2*sqrt(e*x + 2)*(-e**2*x**2 + 4)**(1/4) + 4*e*x*sqrt(e*x + 2)*(-e**2*x**2 + 4)**(
1/4) + 4*sqrt(e*x + 2)*(-e**2*x**2 + 4)**(1/4)), x)/3

Maxima [F]

\[ \int \frac {1}{(2+e x)^{5/2} \sqrt [4]{12-3 e^2 x^2}} \, dx=\int { \frac {1}{{\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {1}{4}} {\left (e x + 2\right )}^{\frac {5}{2}}} \,d x } \]

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.55 \[ \int \frac {1}{(2+e x)^{5/2} \sqrt [4]{12-3 e^2 x^2}} \, dx=-\frac {3^{\frac {3}{4}} {\left (3 \, {\left (\frac {4}{e x + 2} - 1\right )}^{\frac {7}{4}} + 7 \, {\left (\frac {4}{e x + 2} - 1\right )}^{\frac {3}{4}}\right )}}{252 \, e} \]

[In]

integrate(1/(e*x+2)^(5/2)/(-3*e^2*x^2+12)^(1/4),x, algorithm="giac")

[Out]

-1/252*3^(3/4)*(3*(4/(e*x + 2) - 1)^(7/4) + 7*(4/(e*x + 2) - 1)^(3/4))/e

Mupad [B] (verification not implemented)

Time = 10.65 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.92 \[ \int \frac {1}{(2+e x)^{5/2} \sqrt [4]{12-3 e^2 x^2}} \, dx=-\frac {\left (\frac {x}{63\,e^2}+\frac {5}{63\,e^3}\right )\,{\left (12-3\,e^2\,x^2\right )}^{3/4}}{\frac {4\,\sqrt {e\,x+2}}{e^2}+x^2\,\sqrt {e\,x+2}+\frac {4\,x\,\sqrt {e\,x+2}}{e}} \]

[In]

int(1/((12 - 3*e^2*x^2)^(1/4)*(e*x + 2)^(5/2)),x)

[Out]

-((x/(63*e^2) + 5/(63*e^3))*(12 - 3*e^2*x^2)^(3/4))/((4*(e*x + 2)^(1/2))/e^2 + x^2*(e*x + 2)^(1/2) + (4*x*(e*x
+ 2)^(1/2))/e)