\(\int \frac {1}{(3+6 x^2-2 x^4)^{3/2}} \, dx\) [185]

Optimal result

Integrand size = 16, antiderivative size = 119 \[ \int \frac {1}{\left (3+6 x^2-2 x^4\right )^{3/2}} \, dx=\frac {x \left (4-x^2\right )}{15 \sqrt {3+6 x^2-2 x^4}}+\frac {1}{30} \sqrt {-3+\sqrt {15}} E\left (\arcsin \left (\sqrt {\frac {1}{3} \left (-3+\sqrt {15}\right )} x\right )|-4-\sqrt {15}\right )+\frac {1}{6} \sqrt {\frac {1}{15} \left (-3+\sqrt {15}\right )} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {1}{3} \left (-3+\sqrt {15}\right )} x\right ),-4-\sqrt {15}\right ) \] Output:

1/15*x*(-x^2+4)/(-2*x^4+6*x^2+3)^(1/2)+1/30*(-3+15^(1/2))^(1/2)*EllipticE( 
1/3*(-9+3*15^(1/2))^(1/2)*x,1/2*I*6^(1/2)+1/2*I*10^(1/2))+1/90*(-45+15*15^ 
(1/2))^(1/2)*EllipticF(1/3*(-9+3*15^(1/2))^(1/2)*x,1/2*I*6^(1/2)+1/2*I*10^ 
(1/2))
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 9.17 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.03 \[ \int \frac {1}{\left (3+6 x^2-2 x^4\right )^{3/2}} \, dx=\frac {1}{30} \left (-\frac {2 x \left (-4+x^2\right )}{\sqrt {3+6 x^2-2 x^4}}+i \sqrt {3+\sqrt {15}} E\left (i \text {arcsinh}\left (\sqrt {1+\sqrt {\frac {5}{3}}} x\right )|-4+\sqrt {15}\right )-\frac {i \left (5+\sqrt {15}\right ) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {1+\sqrt {\frac {5}{3}}} x\right ),-4+\sqrt {15}\right )}{\sqrt {3+\sqrt {15}}}\right ) \] Input:

Integrate[(3 + 6*x^2 - 2*x^4)^(-3/2),x]
 

Output:

((-2*x*(-4 + x^2))/Sqrt[3 + 6*x^2 - 2*x^4] + I*Sqrt[3 + Sqrt[15]]*Elliptic 
E[I*ArcSinh[Sqrt[1 + Sqrt[5/3]]*x], -4 + Sqrt[15]] - (I*(5 + Sqrt[15])*Ell 
ipticF[I*ArcSinh[Sqrt[1 + Sqrt[5/3]]*x], -4 + Sqrt[15]])/Sqrt[3 + Sqrt[15] 
])/30
 

Rubi [A] (warning: unable to verify)

Time = 0.60 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.19, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {1405, 27, 1494, 27, 399, 321, 327}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{\left (-2 x^4+6 x^2+3\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1405

\(\displaystyle \frac {x \left (4-x^2\right )}{15 \sqrt {-2 x^4+6 x^2+3}}-\frac {1}{180} \int -\frac {12 \left (x^2+1\right )}{\sqrt {-2 x^4+6 x^2+3}}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{15} \int \frac {x^2+1}{\sqrt {-2 x^4+6 x^2+3}}dx+\frac {x \left (4-x^2\right )}{15 \sqrt {-2 x^4+6 x^2+3}}\)

\(\Big \downarrow \) 1494

\(\displaystyle \frac {2}{15} \sqrt {2} \int \frac {x^2+1}{2 \sqrt {-2 x^2+\sqrt {15}+3} \sqrt {2 x^2+\sqrt {15}-3}}dx+\frac {x \left (4-x^2\right )}{15 \sqrt {-2 x^4+6 x^2+3}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{15} \sqrt {2} \int \frac {x^2+1}{\sqrt {-2 x^2+\sqrt {15}+3} \sqrt {2 x^2+\sqrt {15}-3}}dx+\frac {x \left (4-x^2\right )}{15 \sqrt {-2 x^4+6 x^2+3}}\)

\(\Big \downarrow \) 399

\(\displaystyle \frac {1}{15} \sqrt {2} \left (\frac {1}{2} \left (5-\sqrt {15}\right ) \int \frac {1}{\sqrt {-2 x^2+\sqrt {15}+3} \sqrt {2 x^2+\sqrt {15}-3}}dx+\frac {1}{2} \int \frac {\sqrt {2 x^2+\sqrt {15}-3}}{\sqrt {-2 x^2+\sqrt {15}+3}}dx\right )+\frac {x \left (4-x^2\right )}{15 \sqrt {-2 x^4+6 x^2+3}}\)

\(\Big \downarrow \) 321

\(\displaystyle \frac {1}{15} \sqrt {2} \left (\frac {1}{2} \int \frac {\sqrt {2 x^2+\sqrt {15}-3}}{\sqrt {-2 x^2+\sqrt {15}+3}}dx+\frac {1}{4} \left (5-\sqrt {15}\right ) \sqrt {\frac {1}{3} \left (3+\sqrt {15}\right )} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {1}{3} \left (-3+\sqrt {15}\right )} x\right ),-4-\sqrt {15}\right )\right )+\frac {x \left (4-x^2\right )}{15 \sqrt {-2 x^4+6 x^2+3}}\)

\(\Big \downarrow \) 327

\(\displaystyle \frac {1}{15} \sqrt {2} \left (\frac {1}{4} \left (5-\sqrt {15}\right ) \sqrt {\frac {1}{3} \left (3+\sqrt {15}\right )} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {1}{3} \left (-3+\sqrt {15}\right )} x\right ),-4-\sqrt {15}\right )+\frac {1}{2} \sqrt {\frac {3}{3+\sqrt {15}}} E\left (\arcsin \left (\sqrt {\frac {1}{3} \left (-3+\sqrt {15}\right )} x\right )|-4-\sqrt {15}\right )\right )+\frac {x \left (4-x^2\right )}{15 \sqrt {-2 x^4+6 x^2+3}}\)

Input:

Int[(3 + 6*x^2 - 2*x^4)^(-3/2),x]
 

Output:

(x*(4 - x^2))/(15*Sqrt[3 + 6*x^2 - 2*x^4]) + (Sqrt[2]*((Sqrt[3/(3 + Sqrt[1 
5])]*EllipticE[ArcSin[Sqrt[(-3 + Sqrt[15])/3]*x], -4 - Sqrt[15]])/2 + ((5 
- Sqrt[15])*Sqrt[(3 + Sqrt[15])/3]*EllipticF[ArcSin[Sqrt[(-3 + Sqrt[15])/3 
]*x], -4 - Sqrt[15]])/4))/15
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 321

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S 
imp[(1/(Sqrt[a]*Sqrt[c]*Rt[-d/c, 2]))*EllipticF[ArcSin[Rt[-d/c, 2]*x], b*(c 
/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 
0] &&  !(NegQ[b/a] && SimplerSqrtQ[-b/a, -d/c])
 

rule 327

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ 
(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*EllipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d) 
)], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0]
 

rule 399

Int[((e_) + (f_.)*(x_)^2)/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_) 
^2]), x_Symbol] :> Simp[f/b   Int[Sqrt[a + b*x^2]/Sqrt[c + d*x^2], x], x] + 
 Simp[(b*e - a*f)/b   Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] /; Fr 
eeQ[{a, b, c, d, e, f}, x] &&  !((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && 
(PosQ[d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-b/a, -d/c])))))
 

rule 1405

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(-x)*(b^2 
- 2*a*c + b*c*x^2)*((a + b*x^2 + c*x^4)^(p + 1)/(2*a*(p + 1)*(b^2 - 4*a*c)) 
), x] + Simp[1/(2*a*(p + 1)*(b^2 - 4*a*c))   Int[(b^2 - 2*a*c + 2*(p + 1)*( 
b^2 - 4*a*c) + b*c*(4*p + 7)*x^2)*(a + b*x^2 + c*x^4)^(p + 1), x], x] /; Fr 
eeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && IntegerQ[2*p]
 

rule 1494

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo 
l] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[2*Sqrt[-c]   Int[(d + e*x^2)/(Sqr 
t[b + q + 2*c*x^2]*Sqrt[-b + q - 2*c*x^2]), x], x]] /; FreeQ[{a, b, c, d, e 
}, x] && GtQ[b^2 - 4*a*c, 0] && LtQ[c, 0]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 227 vs. \(2 (97 ) = 194\).

Time = 1.63 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.92

method	result	size
risch	\(-\frac {x \left (x^{2}-4\right )}{15 \sqrt {-2 x^{4}+6 x^{2}+3}}+\frac {\sqrt {1-\left (-1+\frac {\sqrt {15}}{3}\right ) x^{2}}\, \sqrt {1-\left (-1-\frac {\sqrt {15}}{3}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {\sqrt {-9+3 \sqrt {15}}\, x}{3}, \frac {i \sqrt {6}}{2}+\frac {i \sqrt {10}}{2}\right )}{5 \sqrt {-9+3 \sqrt {15}}\, \sqrt {-2 x^{4}+6 x^{2}+3}}-\frac {6 \sqrt {1-\left (-1+\frac {\sqrt {15}}{3}\right ) x^{2}}\, \sqrt {1-\left (-1-\frac {\sqrt {15}}{3}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {\sqrt {-9+3 \sqrt {15}}\, x}{3}, \frac {i \sqrt {6}}{2}+\frac {i \sqrt {10}}{2}\right )-\operatorname {EllipticE}\left (\frac {\sqrt {-9+3 \sqrt {15}}\, x}{3}, \frac {i \sqrt {6}}{2}+\frac {i \sqrt {10}}{2}\right )\right )}{5 \sqrt {-9+3 \sqrt {15}}\, \sqrt {-2 x^{4}+6 x^{2}+3}\, \left (6+2 \sqrt {15}\right )}\)	\(228\)
default	\(\frac {\frac {4}{15} x -\frac {1}{15} x^{3}}{\sqrt {-2 x^{4}+6 x^{2}+3}}+\frac {\sqrt {1-\left (-1+\frac {\sqrt {15}}{3}\right ) x^{2}}\, \sqrt {1-\left (-1-\frac {\sqrt {15}}{3}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {\sqrt {-9+3 \sqrt {15}}\, x}{3}, \frac {i \sqrt {6}}{2}+\frac {i \sqrt {10}}{2}\right )}{5 \sqrt {-9+3 \sqrt {15}}\, \sqrt {-2 x^{4}+6 x^{2}+3}}-\frac {6 \sqrt {1-\left (-1+\frac {\sqrt {15}}{3}\right ) x^{2}}\, \sqrt {1-\left (-1-\frac {\sqrt {15}}{3}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {\sqrt {-9+3 \sqrt {15}}\, x}{3}, \frac {i \sqrt {6}}{2}+\frac {i \sqrt {10}}{2}\right )-\operatorname {EllipticE}\left (\frac {\sqrt {-9+3 \sqrt {15}}\, x}{3}, \frac {i \sqrt {6}}{2}+\frac {i \sqrt {10}}{2}\right )\right )}{5 \sqrt {-9+3 \sqrt {15}}\, \sqrt {-2 x^{4}+6 x^{2}+3}\, \left (6+2 \sqrt {15}\right )}\)	\(231\)
elliptic	\(\frac {\frac {4}{15} x -\frac {1}{15} x^{3}}{\sqrt {-2 x^{4}+6 x^{2}+3}}+\frac {\sqrt {1-\left (-1+\frac {\sqrt {15}}{3}\right ) x^{2}}\, \sqrt {1-\left (-1-\frac {\sqrt {15}}{3}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {\sqrt {-9+3 \sqrt {15}}\, x}{3}, \frac {i \sqrt {6}}{2}+\frac {i \sqrt {10}}{2}\right )}{5 \sqrt {-9+3 \sqrt {15}}\, \sqrt {-2 x^{4}+6 x^{2}+3}}-\frac {6 \sqrt {1-\left (-1+\frac {\sqrt {15}}{3}\right ) x^{2}}\, \sqrt {1-\left (-1-\frac {\sqrt {15}}{3}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {\sqrt {-9+3 \sqrt {15}}\, x}{3}, \frac {i \sqrt {6}}{2}+\frac {i \sqrt {10}}{2}\right )-\operatorname {EllipticE}\left (\frac {\sqrt {-9+3 \sqrt {15}}\, x}{3}, \frac {i \sqrt {6}}{2}+\frac {i \sqrt {10}}{2}\right )\right )}{5 \sqrt {-9+3 \sqrt {15}}\, \sqrt {-2 x^{4}+6 x^{2}+3}\, \left (6+2 \sqrt {15}\right )}\)	\(231\)

Input:

int(1/(-2*x^4+6*x^2+3)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/15*x*(x^2-4)/(-2*x^4+6*x^2+3)^(1/2)+1/5/(-9+3*15^(1/2))^(1/2)*(1-(-1+1/ 
3*15^(1/2))*x^2)^(1/2)*(1-(-1-1/3*15^(1/2))*x^2)^(1/2)/(-2*x^4+6*x^2+3)^(1 
/2)*EllipticF(1/3*(-9+3*15^(1/2))^(1/2)*x,1/2*I*6^(1/2)+1/2*I*10^(1/2))-6/ 
5/(-9+3*15^(1/2))^(1/2)*(1-(-1+1/3*15^(1/2))*x^2)^(1/2)*(1-(-1-1/3*15^(1/2 
))*x^2)^(1/2)/(-2*x^4+6*x^2+3)^(1/2)/(6+2*15^(1/2))*(EllipticF(1/3*(-9+3*1 
5^(1/2))^(1/2)*x,1/2*I*6^(1/2)+1/2*I*10^(1/2))-EllipticE(1/3*(-9+3*15^(1/2 
))^(1/2)*x,1/2*I*6^(1/2)+1/2*I*10^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.18 \[ \int \frac {1}{\left (3+6 x^2-2 x^4\right )^{3/2}} \, dx=-\frac {\sqrt {3} {\left (2 \, x^{4} - 6 \, x^{2} - \sqrt {\frac {5}{3}} {\left (2 \, x^{4} - 6 \, x^{2} - 3\right )} - 3\right )} \sqrt {\sqrt {\frac {5}{3}} - 1} E(\arcsin \left (x \sqrt {\sqrt {\frac {5}{3}} - 1}\right )\,|\,-3 \, \sqrt {\frac {5}{3}} - 4) - 2 \, \sqrt {3} {\left (2 \, x^{4} - 6 \, x^{2} - 3\right )} \sqrt {\sqrt {\frac {5}{3}} - 1} F(\arcsin \left (x \sqrt {\sqrt {\frac {5}{3}} - 1}\right )\,|\,-3 \, \sqrt {\frac {5}{3}} - 4) - 2 \, \sqrt {-2 \, x^{4} + 6 \, x^{2} + 3} {\left (x^{3} - 4 \, x\right )}}{30 \, {\left (2 \, x^{4} - 6 \, x^{2} - 3\right )}} \] Input:

integrate(1/(-2*x^4+6*x^2+3)^(3/2),x, algorithm="fricas")
 

Output:

-1/30*(sqrt(3)*(2*x^4 - 6*x^2 - sqrt(5/3)*(2*x^4 - 6*x^2 - 3) - 3)*sqrt(sq 
rt(5/3) - 1)*elliptic_e(arcsin(x*sqrt(sqrt(5/3) - 1)), -3*sqrt(5/3) - 4) - 
 2*sqrt(3)*(2*x^4 - 6*x^2 - 3)*sqrt(sqrt(5/3) - 1)*elliptic_f(arcsin(x*sqr 
t(sqrt(5/3) - 1)), -3*sqrt(5/3) - 4) - 2*sqrt(-2*x^4 + 6*x^2 + 3)*(x^3 - 4 
*x))/(2*x^4 - 6*x^2 - 3)
 

Sympy [F]

\[ \int \frac {1}{\left (3+6 x^2-2 x^4\right )^{3/2}} \, dx=\int \frac {1}{\left (- 2 x^{4} + 6 x^{2} + 3\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(1/(-2*x**4+6*x**2+3)**(3/2),x)
 

Output:

Integral((-2*x**4 + 6*x**2 + 3)**(-3/2), x)
 

Maxima [F]

\[ \int \frac {1}{\left (3+6 x^2-2 x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (-2 \, x^{4} + 6 \, x^{2} + 3\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/(-2*x^4+6*x^2+3)^(3/2),x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

integrate((-2*x^4 + 6*x^2 + 3)^(-3/2), x)
 

Giac [F]

\[ \int \frac {1}{\left (3+6 x^2-2 x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (-2 \, x^{4} + 6 \, x^{2} + 3\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/(-2*x^4+6*x^2+3)^(3/2),x, algorithm="giac")
 

Output:

integrate((-2*x^4 + 6*x^2 + 3)^(-3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (3+6 x^2-2 x^4\right )^{3/2}} \, dx=\int \frac {1}{{\left (-2\,x^4+6\,x^2+3\right )}^{3/2}} \,d x \] Input:

int(1/(6*x^2 - 2*x^4 + 3)^(3/2),x)
 

Output:

int(1/(6*x^2 - 2*x^4 + 3)^(3/2), x)
 

Reduce [F]

\[ \int \frac {1}{\left (3+6 x^2-2 x^4\right )^{3/2}} \, dx=\int \frac {\sqrt {-2 x^{4}+6 x^{2}+3}}{4 x^{8}-24 x^{6}+24 x^{4}+36 x^{2}+9}d x \] Input:

int(1/(-2*x^4+6*x^2+3)^(3/2),x)
 

Output:

int(sqrt( - 2*x**4 + 6*x**2 + 3)/(4*x**8 - 24*x**6 + 24*x**4 + 36*x**2 + 9 
),x)