\(\int \frac {1}{(-2-x^2+3 x^4)^{3/2}} \, dx\) [210]

Optimal result

Integrand size = 16, antiderivative size = 135 \[ \int \frac {1}{\left (-2-x^2+3 x^4\right )^{3/2}} \, dx=-\frac {x \left (13-3 x^2\right )}{50 \sqrt {-2-x^2+3 x^4}}-\frac {\sqrt {1-x^2} \sqrt {2+3 x^2} E\left (\arcsin (x)\left |-\frac {3}{2}\right .\right )}{25 \sqrt {2} \sqrt {-2-x^2+3 x^4}}-\frac {\sqrt {1-x^2} \sqrt {2+3 x^2} \operatorname {EllipticF}\left (\arcsin (x),-\frac {3}{2}\right )}{5 \sqrt {2} \sqrt {-2-x^2+3 x^4}} \] Output:

-1/50*x*(-3*x^2+13)/(3*x^4-x^2-2)^(1/2)-1/50*(-x^2+1)^(1/2)*(3*x^2+2)^(1/2 
)*EllipticE(x,1/2*I*6^(1/2))*2^(1/2)/(3*x^4-x^2-2)^(1/2)-1/10*(-x^2+1)^(1/ 
2)*(3*x^2+2)^(1/2)*EllipticF(x,1/2*I*6^(1/2))*2^(1/2)/(3*x^4-x^2-2)^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 6.81 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.78 \[ \int \frac {1}{\left (-2-x^2+3 x^4\right )^{3/2}} \, dx=\frac {-13 x+3 x^3-i \sqrt {6+3 x^2-9 x^4} E\left (i \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right )|-\frac {2}{3}\right )+5 i \sqrt {6+3 x^2-9 x^4} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right ),-\frac {2}{3}\right )}{50 \sqrt {-2-x^2+3 x^4}} \] Input:

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

Output:

(-13*x + 3*x^3 - I*Sqrt[6 + 3*x^2 - 9*x^4]*EllipticE[I*ArcSinh[Sqrt[3/2]*x 
], -2/3] + (5*I)*Sqrt[6 + 3*x^2 - 9*x^4]*EllipticF[I*ArcSinh[Sqrt[3/2]*x], 
 -2/3])/(50*Sqrt[-2 - x^2 + 3*x^4])
 

Rubi [A] (warning: unable to verify)

Time = 0.60 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.61, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {1405, 27, 1501, 27, 1410, 1498}

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.

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

\(\Big \downarrow \) 1405

\(\displaystyle \frac {1}{50} \int -\frac {3 \left (x^2+4\right )}{\sqrt {3 x^4-x^2-2}}dx-\frac {x \left (13-3 x^2\right )}{50 \sqrt {3 x^4-x^2-2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {3}{50} \int \frac {x^2+4}{\sqrt {3 x^4-x^2-2}}dx-\frac {x \left (13-3 x^2\right )}{50 \sqrt {3 x^4-x^2-2}}\)

\(\Big \downarrow \) 1501

\(\displaystyle -\frac {3}{50} \left (5 \int \frac {1}{\sqrt {3 x^4-x^2-2}}dx+\frac {1}{6} \int -\frac {6 \left (1-x^2\right )}{\sqrt {3 x^4-x^2-2}}dx\right )-\frac {x \left (13-3 x^2\right )}{50 \sqrt {3 x^4-x^2-2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {3}{50} \left (5 \int \frac {1}{\sqrt {3 x^4-x^2-2}}dx-\int \frac {1-x^2}{\sqrt {3 x^4-x^2-2}}dx\right )-\frac {x \left (13-3 x^2\right )}{50 \sqrt {3 x^4-x^2-2}}\)

\(\Big \downarrow \) 1410

\(\displaystyle -\frac {3}{50} \left (\frac {\sqrt {5} \sqrt {x^2-1} \sqrt {3 x^2+2} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {5}{2}} x}{\sqrt {x^2-1}}\right ),\frac {2}{5}\right )}{\sqrt {3 x^4-x^2-2}}-\int \frac {1-x^2}{\sqrt {3 x^4-x^2-2}}dx\right )-\frac {x \left (13-3 x^2\right )}{50 \sqrt {3 x^4-x^2-2}}\)

\(\Big \downarrow \) 1498

\(\displaystyle -\frac {3}{50} \left (\frac {\sqrt {5} \sqrt {x^2-1} \sqrt {3 x^2+2} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {5}{2}} x}{\sqrt {x^2-1}}\right ),\frac {2}{5}\right )}{\sqrt {3 x^4-x^2-2}}-\frac {\sqrt {5} \sqrt {x^2-1} \sqrt {\frac {3 x^2+2}{1-x^2}} E\left (\arcsin \left (\frac {\sqrt {\frac {5}{2}} x}{\sqrt {x^2-1}}\right )|\frac {2}{5}\right )}{3 \sqrt {\frac {1}{1-x^2}} \sqrt {3 x^4-x^2-2}}+\frac {x \left (3 x^2+2\right )}{3 \sqrt {3 x^4-x^2-2}}\right )-\frac {x \left (13-3 x^2\right )}{50 \sqrt {3 x^4-x^2-2}}\)

Input:

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

Output:

-1/50*(x*(13 - 3*x^2))/Sqrt[-2 - x^2 + 3*x^4] - (3*((x*(2 + 3*x^2))/(3*Sqr 
t[-2 - x^2 + 3*x^4]) - (Sqrt[5]*Sqrt[-1 + x^2]*Sqrt[(2 + 3*x^2)/(1 - x^2)] 
*EllipticE[ArcSin[(Sqrt[5/2]*x)/Sqrt[-1 + x^2]], 2/5])/(3*Sqrt[(1 - x^2)^( 
-1)]*Sqrt[-2 - x^2 + 3*x^4]) + (Sqrt[5]*Sqrt[-1 + x^2]*Sqrt[2 + 3*x^2]*Ell 
ipticF[ArcSin[(Sqrt[5/2]*x)/Sqrt[-1 + x^2]], 2/5])/Sqrt[-2 - x^2 + 3*x^4]) 
)/50
 

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 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 1410

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b 
^2 - 4*a*c, 2]}, Simp[Sqrt[-2*a - (b - q)*x^2]*(Sqrt[(2*a + (b + q)*x^2)/q] 
/(2*Sqrt[-a]*Sqrt[a + b*x^2 + c*x^4]))*EllipticF[ArcSin[x/Sqrt[(2*a + (b + 
q)*x^2)/(2*q)]], (b + q)/(2*q)], x] /; IntegerQ[q]] /; FreeQ[{a, b, c}, x] 
&& GtQ[b^2 - 4*a*c, 0] && LtQ[a, 0] && GtQ[c, 0]
 

rule 1498

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[e*x*((b + q + 2*c*x^2)/(2*c*Sqrt[ 
a + b*x^2 + c*x^4])), x] - Simp[e*q*Sqrt[(2*a + (b - q)*x^2)/(2*a + (b + q) 
*x^2)]*(Sqrt[(2*a + (b + q)*x^2)/q]/(2*c*Sqrt[a + b*x^2 + c*x^4]*Sqrt[a/(2* 
a + (b + q)*x^2)]))*EllipticE[ArcSin[x/Sqrt[(2*a + (b + q)*x^2)/(2*q)]], (b 
 + q)/(2*q)], x] /; EqQ[2*c*d - e*(b - q), 0]] /; FreeQ[{a, b, c, d, e}, x] 
 && GtQ[b^2 - 4*a*c, 0] && LtQ[a, 0] && GtQ[c, 0]
 

rule 1501

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*c*d - e*(b - q))/(2*c)   Int[1 
/Sqrt[a + b*x^2 + c*x^4], x], x] + Simp[e/(2*c)   Int[(b - q + 2*c*x^2)/Sqr 
t[a + b*x^2 + c*x^4], x], x] /; NeQ[2*c*d - e*(b - q), 0]] /; FreeQ[{a, b, 
c, d, e}, x] && GtQ[b^2 - 4*a*c, 0] && LtQ[a, 0] && GtQ[c, 0]
 

Maple [A] (verified)

Time = 2.22 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.09

method	result	size
risch	\(\frac {x \left (3 x^{2}-13\right )}{50 \sqrt {3 x^{4}-x^{2}-2}}+\frac {i \sqrt {6}\, \sqrt {6 x^{2}+4}\, \sqrt {-x^{2}+1}\, \operatorname {EllipticF}\left (\frac {i x \sqrt {6}}{2}, \frac {i \sqrt {6}}{3}\right )}{25 \sqrt {3 x^{4}-x^{2}-2}}+\frac {i \sqrt {6}\, \sqrt {6 x^{2}+4}\, \sqrt {-x^{2}+1}\, \left (\operatorname {EllipticF}\left (\frac {i x \sqrt {6}}{2}, \frac {i \sqrt {6}}{3}\right )-\operatorname {EllipticE}\left (\frac {i x \sqrt {6}}{2}, \frac {i \sqrt {6}}{3}\right )\right )}{100 \sqrt {3 x^{4}-x^{2}-2}}\)	\(147\)
default	\(-\frac {6 \left (\frac {13}{300} x -\frac {1}{100} x^{3}\right )}{\sqrt {3 x^{4}-x^{2}-2}}+\frac {i \sqrt {6}\, \sqrt {6 x^{2}+4}\, \sqrt {-x^{2}+1}\, \operatorname {EllipticF}\left (\frac {i x \sqrt {6}}{2}, \frac {i \sqrt {6}}{3}\right )}{25 \sqrt {3 x^{4}-x^{2}-2}}+\frac {i \sqrt {6}\, \sqrt {6 x^{2}+4}\, \sqrt {-x^{2}+1}\, \left (\operatorname {EllipticF}\left (\frac {i x \sqrt {6}}{2}, \frac {i \sqrt {6}}{3}\right )-\operatorname {EllipticE}\left (\frac {i x \sqrt {6}}{2}, \frac {i \sqrt {6}}{3}\right )\right )}{100 \sqrt {3 x^{4}-x^{2}-2}}\)	\(148\)
elliptic	\(-\frac {6 \left (\frac {13}{300} x -\frac {1}{100} x^{3}\right )}{\sqrt {3 x^{4}-x^{2}-2}}+\frac {i \sqrt {6}\, \sqrt {6 x^{2}+4}\, \sqrt {-x^{2}+1}\, \operatorname {EllipticF}\left (\frac {i x \sqrt {6}}{2}, \frac {i \sqrt {6}}{3}\right )}{25 \sqrt {3 x^{4}-x^{2}-2}}+\frac {i \sqrt {6}\, \sqrt {6 x^{2}+4}\, \sqrt {-x^{2}+1}\, \left (\operatorname {EllipticF}\left (\frac {i x \sqrt {6}}{2}, \frac {i \sqrt {6}}{3}\right )-\operatorname {EllipticE}\left (\frac {i x \sqrt {6}}{2}, \frac {i \sqrt {6}}{3}\right )\right )}{100 \sqrt {3 x^{4}-x^{2}-2}}\)	\(148\)

Input:

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

Output:

1/50*x*(3*x^2-13)/(3*x^4-x^2-2)^(1/2)+1/25*I*6^(1/2)*(6*x^2+4)^(1/2)*(-x^2 
+1)^(1/2)/(3*x^4-x^2-2)^(1/2)*EllipticF(1/2*I*x*6^(1/2),1/3*I*6^(1/2))+1/1 
00*I*6^(1/2)*(6*x^2+4)^(1/2)*(-x^2+1)^(1/2)/(3*x^4-x^2-2)^(1/2)*(EllipticF 
(1/2*I*x*6^(1/2),1/3*I*6^(1/2))-EllipticE(1/2*I*x*6^(1/2),1/3*I*6^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.61 \[ \int \frac {1}{\left (-2-x^2+3 x^4\right )^{3/2}} \, dx=\frac {\sqrt {-2} {\left (3 \, x^{4} - x^{2} - 2\right )} E(\arcsin \left (x\right )\,|\,-\frac {3}{2}) + 5 \, \sqrt {-2} {\left (3 \, x^{4} - x^{2} - 2\right )} F(\arcsin \left (x\right )\,|\,-\frac {3}{2}) + \sqrt {3 \, x^{4} - x^{2} - 2} {\left (3 \, x^{3} - 13 \, x\right )}}{50 \, {\left (3 \, x^{4} - x^{2} - 2\right )}} \] Input:

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

Output:

1/50*(sqrt(-2)*(3*x^4 - x^2 - 2)*elliptic_e(arcsin(x), -3/2) + 5*sqrt(-2)* 
(3*x^4 - x^2 - 2)*elliptic_f(arcsin(x), -3/2) + sqrt(3*x^4 - x^2 - 2)*(3*x 
^3 - 13*x))/(3*x^4 - x^2 - 2)
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\left (-2-x^2+3 x^4\right )^{3/2}} \, dx=\int \frac {\sqrt {3 x^{4}-x^{2}-2}}{9 x^{8}-6 x^{6}-11 x^{4}+4 x^{2}+4}d x \] Input:

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

Output:

int(sqrt(3*x**4 - x**2 - 2)/(9*x**8 - 6*x**6 - 11*x**4 + 4*x**2 + 4),x)