\(\int \frac {1}{(2-4 x^2-3 x^4)^{3/2}} \, dx\) [182]

Optimal result

Integrand size = 16, antiderivative size = 127 \[ \int \frac {1}{\left (2-4 x^2-3 x^4\right )^{3/2}} \, dx=\frac {x \left (7+3 x^2\right )}{20 \sqrt {2-4 x^2-3 x^4}}-\frac {1}{20} \sqrt {2+\sqrt {10}} E\left (\arcsin \left (\sqrt {\frac {1}{2} \left (2+\sqrt {10}\right )} x\right )|\frac {1}{3} \left (-7+2 \sqrt {10}\right )\right )+\frac {1}{4} \sqrt {\frac {1}{10} \left (2+\sqrt {10}\right )} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {1}{2} \left (2+\sqrt {10}\right )} x\right ),\frac {1}{3} \left (-7+2 \sqrt {10}\right )\right ) \] Output:

1/20*x*(3*x^2+7)/(-3*x^4-4*x^2+2)^(1/2)-1/20*(2+10^(1/2))^(1/2)*EllipticE( 
1/2*(4+2*10^(1/2))^(1/2)*x,1/3*I*15^(1/2)-1/3*I*6^(1/2))+1/40*(20+10*10^(1 
/2))^(1/2)*EllipticF(1/2*(4+2*10^(1/2))^(1/2)*x,1/3*I*15^(1/2)-1/3*I*6^(1/ 
2))
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 9.32 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.06 \[ \int \frac {1}{\left (2-4 x^2-3 x^4\right )^{3/2}} \, dx=\frac {1}{20} \left (\frac {x \left (7+3 x^2\right )}{\sqrt {2-4 x^2-3 x^4}}-i \sqrt {-2+\sqrt {10}} E\left (i \text {arcsinh}\left (\sqrt {-1+\sqrt {\frac {5}{2}}} x\right )|\frac {1}{3} \left (-7-2 \sqrt {10}\right )\right )+\frac {i \left (-5+\sqrt {10}\right ) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-1+\sqrt {\frac {5}{2}}} x\right ),\frac {1}{3} \left (-7-2 \sqrt {10}\right )\right )}{\sqrt {-2+\sqrt {10}}}\right ) \] Input:

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

Output:

((x*(7 + 3*x^2))/Sqrt[2 - 4*x^2 - 3*x^4] - I*Sqrt[-2 + Sqrt[10]]*EllipticE 
[I*ArcSinh[Sqrt[-1 + Sqrt[5/2]]*x], (-7 - 2*Sqrt[10])/3] + (I*(-5 + Sqrt[1 
0])*EllipticF[I*ArcSinh[Sqrt[-1 + Sqrt[5/2]]*x], (-7 - 2*Sqrt[10])/3])/Sqr 
t[-2 + Sqrt[10]])/20
 

Rubi [A] (warning: unable to verify)

Time = 0.60 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.17, 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.

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

\(\Big \downarrow \) 1405

\(\displaystyle \frac {x \left (3 x^2+7\right )}{20 \sqrt {-3 x^4-4 x^2+2}}-\frac {1}{80} \int -\frac {12 \left (1-x^2\right )}{\sqrt {-3 x^4-4 x^2+2}}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3}{20} \int \frac {1-x^2}{\sqrt {-3 x^4-4 x^2+2}}dx+\frac {x \left (3 x^2+7\right )}{20 \sqrt {-3 x^4-4 x^2+2}}\)

\(\Big \downarrow \) 1494

\(\displaystyle \frac {3}{10} \sqrt {3} \int \frac {1-x^2}{2 \sqrt {-3 x^2+\sqrt {10}-2} \sqrt {3 x^2+\sqrt {10}+2}}dx+\frac {x \left (3 x^2+7\right )}{20 \sqrt {-3 x^4-4 x^2+2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3}{20} \sqrt {3} \int \frac {1-x^2}{\sqrt {-3 x^2+\sqrt {10}-2} \sqrt {3 x^2+\sqrt {10}+2}}dx+\frac {x \left (3 x^2+7\right )}{20 \sqrt {-3 x^4-4 x^2+2}}\)

\(\Big \downarrow \) 399

\(\displaystyle \frac {3}{20} \sqrt {3} \left (\frac {1}{3} \left (5+\sqrt {10}\right ) \int \frac {1}{\sqrt {-3 x^2+\sqrt {10}-2} \sqrt {3 x^2+\sqrt {10}+2}}dx-\frac {1}{3} \int \frac {\sqrt {3 x^2+\sqrt {10}+2}}{\sqrt {-3 x^2+\sqrt {10}-2}}dx\right )+\frac {x \left (3 x^2+7\right )}{20 \sqrt {-3 x^4-4 x^2+2}}\)

\(\Big \downarrow \) 321

\(\displaystyle \frac {3}{20} \sqrt {3} \left (\frac {1}{9} \sqrt {\frac {1}{2} \left (\sqrt {10}-2\right )} \left (5+\sqrt {10}\right ) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {1}{2} \left (2+\sqrt {10}\right )} x\right ),\frac {1}{3} \left (-7+2 \sqrt {10}\right )\right )-\frac {1}{3} \int \frac {\sqrt {3 x^2+\sqrt {10}+2}}{\sqrt {-3 x^2+\sqrt {10}-2}}dx\right )+\frac {x \left (3 x^2+7\right )}{20 \sqrt {-3 x^4-4 x^2+2}}\)

\(\Big \downarrow \) 327

\(\displaystyle \frac {3}{20} \sqrt {3} \left (\frac {1}{9} \sqrt {\frac {1}{2} \left (\sqrt {10}-2\right )} \left (5+\sqrt {10}\right ) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {1}{2} \left (2+\sqrt {10}\right )} x\right ),\frac {1}{3} \left (-7+2 \sqrt {10}\right )\right )-\frac {1}{3} \sqrt {\frac {2}{\sqrt {10}-2}} E\left (\arcsin \left (\sqrt {\frac {1}{2} \left (2+\sqrt {10}\right )} x\right )|\frac {1}{3} \left (-7+2 \sqrt {10}\right )\right )\right )+\frac {x \left (3 x^2+7\right )}{20 \sqrt {-3 x^4-4 x^2+2}}\)

Input:

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

Output:

(x*(7 + 3*x^2))/(20*Sqrt[2 - 4*x^2 - 3*x^4]) + (3*Sqrt[3]*(-1/3*(Sqrt[2/(- 
2 + Sqrt[10])]*EllipticE[ArcSin[Sqrt[(2 + Sqrt[10])/2]*x], (-7 + 2*Sqrt[10 
])/3]) + (Sqrt[(-2 + Sqrt[10])/2]*(5 + Sqrt[10])*EllipticF[ArcSin[Sqrt[(2 
+ Sqrt[10])/2]*x], (-7 + 2*Sqrt[10])/3])/9))/20
 

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. 229 vs. \(2 (97 ) = 194\).

Time = 2.24 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.81

method	result	size
risch	\(\frac {x \left (3 x^{2}+7\right )}{20 \sqrt {-3 x^{4}-4 x^{2}+2}}+\frac {3 \sqrt {1-\left (1+\frac {\sqrt {10}}{2}\right ) x^{2}}\, \sqrt {1-\left (1-\frac {\sqrt {10}}{2}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {\sqrt {4+2 \sqrt {10}}\, x}{2}, \frac {i \sqrt {15}}{3}-\frac {i \sqrt {6}}{3}\right )}{10 \sqrt {4+2 \sqrt {10}}\, \sqrt {-3 x^{4}-4 x^{2}+2}}+\frac {6 \sqrt {1-\left (1+\frac {\sqrt {10}}{2}\right ) x^{2}}\, \sqrt {1-\left (1-\frac {\sqrt {10}}{2}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {\sqrt {4+2 \sqrt {10}}\, x}{2}, \frac {i \sqrt {15}}{3}-\frac {i \sqrt {6}}{3}\right )-\operatorname {EllipticE}\left (\frac {\sqrt {4+2 \sqrt {10}}\, x}{2}, \frac {i \sqrt {15}}{3}-\frac {i \sqrt {6}}{3}\right )\right )}{5 \sqrt {4+2 \sqrt {10}}\, \sqrt {-3 x^{4}-4 x^{2}+2}\, \left (-4+2 \sqrt {10}\right )}\)	\(230\)
default	\(\frac {\frac {7}{20} x +\frac {3}{20} x^{3}}{\sqrt {-3 x^{4}-4 x^{2}+2}}+\frac {3 \sqrt {1-\left (1+\frac {\sqrt {10}}{2}\right ) x^{2}}\, \sqrt {1-\left (1-\frac {\sqrt {10}}{2}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {\sqrt {4+2 \sqrt {10}}\, x}{2}, \frac {i \sqrt {15}}{3}-\frac {i \sqrt {6}}{3}\right )}{10 \sqrt {4+2 \sqrt {10}}\, \sqrt {-3 x^{4}-4 x^{2}+2}}+\frac {6 \sqrt {1-\left (1+\frac {\sqrt {10}}{2}\right ) x^{2}}\, \sqrt {1-\left (1-\frac {\sqrt {10}}{2}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {\sqrt {4+2 \sqrt {10}}\, x}{2}, \frac {i \sqrt {15}}{3}-\frac {i \sqrt {6}}{3}\right )-\operatorname {EllipticE}\left (\frac {\sqrt {4+2 \sqrt {10}}\, x}{2}, \frac {i \sqrt {15}}{3}-\frac {i \sqrt {6}}{3}\right )\right )}{5 \sqrt {4+2 \sqrt {10}}\, \sqrt {-3 x^{4}-4 x^{2}+2}\, \left (-4+2 \sqrt {10}\right )}\)	\(231\)
elliptic	\(\frac {\frac {7}{20} x +\frac {3}{20} x^{3}}{\sqrt {-3 x^{4}-4 x^{2}+2}}+\frac {3 \sqrt {1-\left (1+\frac {\sqrt {10}}{2}\right ) x^{2}}\, \sqrt {1-\left (1-\frac {\sqrt {10}}{2}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {\sqrt {4+2 \sqrt {10}}\, x}{2}, \frac {i \sqrt {15}}{3}-\frac {i \sqrt {6}}{3}\right )}{10 \sqrt {4+2 \sqrt {10}}\, \sqrt {-3 x^{4}-4 x^{2}+2}}+\frac {6 \sqrt {1-\left (1+\frac {\sqrt {10}}{2}\right ) x^{2}}\, \sqrt {1-\left (1-\frac {\sqrt {10}}{2}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {\sqrt {4+2 \sqrt {10}}\, x}{2}, \frac {i \sqrt {15}}{3}-\frac {i \sqrt {6}}{3}\right )-\operatorname {EllipticE}\left (\frac {\sqrt {4+2 \sqrt {10}}\, x}{2}, \frac {i \sqrt {15}}{3}-\frac {i \sqrt {6}}{3}\right )\right )}{5 \sqrt {4+2 \sqrt {10}}\, \sqrt {-3 x^{4}-4 x^{2}+2}\, \left (-4+2 \sqrt {10}\right )}\)	\(231\)

Input:

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

Output:

1/20*x*(3*x^2+7)/(-3*x^4-4*x^2+2)^(1/2)+3/10/(4+2*10^(1/2))^(1/2)*(1-(1+1/ 
2*10^(1/2))*x^2)^(1/2)*(1-(1-1/2*10^(1/2))*x^2)^(1/2)/(-3*x^4-4*x^2+2)^(1/ 
2)*EllipticF(1/2*(4+2*10^(1/2))^(1/2)*x,1/3*I*15^(1/2)-1/3*I*6^(1/2))+6/5/ 
(4+2*10^(1/2))^(1/2)*(1-(1+1/2*10^(1/2))*x^2)^(1/2)*(1-(1-1/2*10^(1/2))*x^ 
2)^(1/2)/(-3*x^4-4*x^2+2)^(1/2)/(-4+2*10^(1/2))*(EllipticF(1/2*(4+2*10^(1/ 
2))^(1/2)*x,1/3*I*15^(1/2)-1/3*I*6^(1/2))-EllipticE(1/2*(4+2*10^(1/2))^(1/ 
2)*x,1/3*I*15^(1/2)-1/3*I*6^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.25 \[ \int \frac {1}{\left (2-4 x^2-3 x^4\right )^{3/2}} \, dx=\frac {2 \, \sqrt {10} \sqrt {2} {\left (3 \, x^{4} + 4 \, x^{2} - 2\right )} \sqrt {\frac {1}{2} \, \sqrt {10} + 1} F(\arcsin \left (x \sqrt {\frac {1}{2} \, \sqrt {10} + 1}\right )\,|\,\frac {2}{3} \, \sqrt {10} - \frac {7}{3}) - {\left (\sqrt {10} \sqrt {2} {\left (3 \, x^{4} + 4 \, x^{2} - 2\right )} + 2 \, \sqrt {2} {\left (3 \, x^{4} + 4 \, x^{2} - 2\right )}\right )} \sqrt {\frac {1}{2} \, \sqrt {10} + 1} E(\arcsin \left (x \sqrt {\frac {1}{2} \, \sqrt {10} + 1}\right )\,|\,\frac {2}{3} \, \sqrt {10} - \frac {7}{3}) - 2 \, \sqrt {-3 \, x^{4} - 4 \, x^{2} + 2} {\left (3 \, x^{3} + 7 \, x\right )}}{40 \, {\left (3 \, x^{4} + 4 \, x^{2} - 2\right )}} \] Input:

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

Output:

1/40*(2*sqrt(10)*sqrt(2)*(3*x^4 + 4*x^2 - 2)*sqrt(1/2*sqrt(10) + 1)*ellipt 
ic_f(arcsin(x*sqrt(1/2*sqrt(10) + 1)), 2/3*sqrt(10) - 7/3) - (sqrt(10)*sqr 
t(2)*(3*x^4 + 4*x^2 - 2) + 2*sqrt(2)*(3*x^4 + 4*x^2 - 2))*sqrt(1/2*sqrt(10 
) + 1)*elliptic_e(arcsin(x*sqrt(1/2*sqrt(10) + 1)), 2/3*sqrt(10) - 7/3) - 
2*sqrt(-3*x^4 - 4*x^2 + 2)*(3*x^3 + 7*x))/(3*x^4 + 4*x^2 - 2)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\left (2-4 x^2-3 x^4\right )^{3/2}} \, dx=\int \frac {\sqrt {-3 x^{4}-4 x^{2}+2}}{9 x^{8}+24 x^{6}+4 x^{4}-16 x^{2}+4}d x \] Input:

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

Output:

int(sqrt( - 3*x**4 - 4*x**2 + 2)/(9*x**8 + 24*x**6 + 4*x**4 - 16*x**2 + 4) 
,x)