\(\int \frac {1}{(3+4 x^2-2 x^4)^{3/2}} \, dx\) [187]

Optimal result
Mathematica [C] (warning: unable to verify)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 16, antiderivative size = 127 \[ \int \frac {1}{\left (3+4 x^2-2 x^4\right )^{3/2}} \, dx=\frac {x \left (7-2 x^2\right )}{30 \sqrt {3+4 x^2-2 x^4}}+\frac {1}{30} \sqrt {-2+\sqrt {10}} E\left (\arcsin \left (\sqrt {\frac {2}{2+\sqrt {10}}} x\right )|\frac {1}{3} \left (-7-2 \sqrt {10}\right )\right )+\frac {1}{6} \sqrt {\frac {1}{10} \left (-2+\sqrt {10}\right )} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {2}{2+\sqrt {10}}} x\right ),\frac {1}{3} \left (-7-2 \sqrt {10}\right )\right ) \] Output:

1/30*x*(-2*x^2+7)/(-2*x^4+4*x^2+3)^(1/2)+1/30*(-2+10^(1/2))^(1/2)*Elliptic 
E(2^(1/2)/(2+10^(1/2))^(1/2)*x,1/3*I*6^(1/2)+1/3*I*15^(1/2))+1/60*(-20+10* 
10^(1/2))^(1/2)*EllipticF(2^(1/2)/(2+10^(1/2))^(1/2)*x,1/3*I*6^(1/2)+1/3*I 
*15^(1/2))
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 9.51 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.09 \[ \int \frac {1}{\left (3+4 x^2-2 x^4\right )^{3/2}} \, dx=\frac {1}{30} \left (\frac {x \left (7-2 x^2\right )}{\sqrt {3+4 x^2-2 x^4}}+i \sqrt {2+\sqrt {10}} E\left (i \text {arcsinh}\left (\sqrt {\frac {2}{-2+\sqrt {10}}} x\right )|-\frac {7}{3}+\frac {2 \sqrt {10}}{3}\right )-\frac {i \left (5+\sqrt {10}\right ) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {2}{-2+\sqrt {10}}} x\right ),-\frac {7}{3}+\frac {2 \sqrt {10}}{3}\right )}{\sqrt {2+\sqrt {10}}}\right ) \] Input:

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

Output:

((x*(7 - 2*x^2))/Sqrt[3 + 4*x^2 - 2*x^4] + I*Sqrt[2 + Sqrt[10]]*EllipticE[ 
I*ArcSinh[Sqrt[2/(-2 + Sqrt[10])]*x], -7/3 + (2*Sqrt[10])/3] - (I*(5 + Sqr 
t[10])*EllipticF[I*ArcSinh[Sqrt[2/(-2 + Sqrt[10])]*x], -7/3 + (2*Sqrt[10]) 
/3])/Sqrt[2 + Sqrt[10]])/30
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.12, 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.

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

\(\Big \downarrow \) 1405

\(\displaystyle \frac {x \left (7-2 x^2\right )}{30 \sqrt {-2 x^4+4 x^2+3}}-\frac {1}{120} \int -\frac {4 \left (2 x^2+3\right )}{\sqrt {-2 x^4+4 x^2+3}}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{30} \int \frac {2 x^2+3}{\sqrt {-2 x^4+4 x^2+3}}dx+\frac {x \left (7-2 x^2\right )}{30 \sqrt {-2 x^4+4 x^2+3}}\)

\(\Big \downarrow \) 1494

\(\displaystyle \frac {1}{15} \sqrt {2} \int \frac {2 x^2+3}{2 \sqrt {-2 x^2+\sqrt {10}+2} \sqrt {2 x^2+\sqrt {10}-2}}dx+\frac {x \left (7-2 x^2\right )}{30 \sqrt {-2 x^4+4 x^2+3}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {2 x^2+3}{\sqrt {-2 x^2+\sqrt {10}+2} \sqrt {2 x^2+\sqrt {10}-2}}dx}{15 \sqrt {2}}+\frac {x \left (7-2 x^2\right )}{30 \sqrt {-2 x^4+4 x^2+3}}\)

\(\Big \downarrow \) 399

\(\displaystyle \frac {\left (5-\sqrt {10}\right ) \int \frac {1}{\sqrt {-2 x^2+\sqrt {10}+2} \sqrt {2 x^2+\sqrt {10}-2}}dx+\int \frac {\sqrt {2 x^2+\sqrt {10}-2}}{\sqrt {-2 x^2+\sqrt {10}+2}}dx}{15 \sqrt {2}}+\frac {x \left (7-2 x^2\right )}{30 \sqrt {-2 x^4+4 x^2+3}}\)

\(\Big \downarrow \) 321

\(\displaystyle \frac {\int \frac {\sqrt {2 x^2+\sqrt {10}-2}}{\sqrt {-2 x^2+\sqrt {10}+2}}dx+\frac {\left (5-\sqrt {10}\right ) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {2}{2+\sqrt {10}}} x\right ),\frac {1}{3} \left (-7-2 \sqrt {10}\right )\right )}{\sqrt {2 \left (\sqrt {10}-2\right )}}}{15 \sqrt {2}}+\frac {x \left (7-2 x^2\right )}{30 \sqrt {-2 x^4+4 x^2+3}}\)

\(\Big \downarrow \) 327

\(\displaystyle \frac {\frac {\left (5-\sqrt {10}\right ) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {2}{2+\sqrt {10}}} x\right ),\frac {1}{3} \left (-7-2 \sqrt {10}\right )\right )}{\sqrt {2 \left (\sqrt {10}-2\right )}}+\sqrt {\frac {1}{2} \left (\sqrt {10}-2\right )} E\left (\arcsin \left (\sqrt {\frac {2}{2+\sqrt {10}}} x\right )|\frac {1}{3} \left (-7-2 \sqrt {10}\right )\right )}{15 \sqrt {2}}+\frac {x \left (7-2 x^2\right )}{30 \sqrt {-2 x^4+4 x^2+3}}\)

Input:

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

Output:

(x*(7 - 2*x^2))/(30*Sqrt[3 + 4*x^2 - 2*x^4]) + (Sqrt[(-2 + Sqrt[10])/2]*El 
lipticE[ArcSin[Sqrt[2/(2 + Sqrt[10])]*x], (-7 - 2*Sqrt[10])/3] + ((5 - Sqr 
t[10])*EllipticF[ArcSin[Sqrt[2/(2 + Sqrt[10])]*x], (-7 - 2*Sqrt[10])/3])/S 
qrt[2*(-2 + Sqrt[10])])/(15*Sqrt[2])
 

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 = 2.30 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.81

method result size
risch \(-\frac {x \left (2 x^{2}-7\right )}{30 \sqrt {-2 x^{4}+4 x^{2}+3}}+\frac {3 \sqrt {1-\left (-\frac {2}{3}+\frac {\sqrt {10}}{3}\right ) x^{2}}\, \sqrt {1-\left (-\frac {2}{3}-\frac {\sqrt {10}}{3}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {\sqrt {-6+3 \sqrt {10}}\, x}{3}, \frac {i \sqrt {6}}{3}+\frac {i \sqrt {15}}{3}\right )}{10 \sqrt {-6+3 \sqrt {10}}\, \sqrt {-2 x^{4}+4 x^{2}+3}}-\frac {6 \sqrt {1-\left (-\frac {2}{3}+\frac {\sqrt {10}}{3}\right ) x^{2}}\, \sqrt {1-\left (-\frac {2}{3}-\frac {\sqrt {10}}{3}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {\sqrt {-6+3 \sqrt {10}}\, x}{3}, \frac {i \sqrt {6}}{3}+\frac {i \sqrt {15}}{3}\right )-\operatorname {EllipticE}\left (\frac {\sqrt {-6+3 \sqrt {10}}\, x}{3}, \frac {i \sqrt {6}}{3}+\frac {i \sqrt {15}}{3}\right )\right )}{5 \sqrt {-6+3 \sqrt {10}}\, \sqrt {-2 x^{4}+4 x^{2}+3}\, \left (4+2 \sqrt {10}\right )}\) \(230\)
default \(\frac {\frac {7}{30} x -\frac {1}{15} x^{3}}{\sqrt {-2 x^{4}+4 x^{2}+3}}+\frac {3 \sqrt {1-\left (-\frac {2}{3}+\frac {\sqrt {10}}{3}\right ) x^{2}}\, \sqrt {1-\left (-\frac {2}{3}-\frac {\sqrt {10}}{3}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {\sqrt {-6+3 \sqrt {10}}\, x}{3}, \frac {i \sqrt {6}}{3}+\frac {i \sqrt {15}}{3}\right )}{10 \sqrt {-6+3 \sqrt {10}}\, \sqrt {-2 x^{4}+4 x^{2}+3}}-\frac {6 \sqrt {1-\left (-\frac {2}{3}+\frac {\sqrt {10}}{3}\right ) x^{2}}\, \sqrt {1-\left (-\frac {2}{3}-\frac {\sqrt {10}}{3}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {\sqrt {-6+3 \sqrt {10}}\, x}{3}, \frac {i \sqrt {6}}{3}+\frac {i \sqrt {15}}{3}\right )-\operatorname {EllipticE}\left (\frac {\sqrt {-6+3 \sqrt {10}}\, x}{3}, \frac {i \sqrt {6}}{3}+\frac {i \sqrt {15}}{3}\right )\right )}{5 \sqrt {-6+3 \sqrt {10}}\, \sqrt {-2 x^{4}+4 x^{2}+3}\, \left (4+2 \sqrt {10}\right )}\) \(231\)
elliptic \(\frac {\frac {7}{30} x -\frac {1}{15} x^{3}}{\sqrt {-2 x^{4}+4 x^{2}+3}}+\frac {3 \sqrt {1-\left (-\frac {2}{3}+\frac {\sqrt {10}}{3}\right ) x^{2}}\, \sqrt {1-\left (-\frac {2}{3}-\frac {\sqrt {10}}{3}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {\sqrt {-6+3 \sqrt {10}}\, x}{3}, \frac {i \sqrt {6}}{3}+\frac {i \sqrt {15}}{3}\right )}{10 \sqrt {-6+3 \sqrt {10}}\, \sqrt {-2 x^{4}+4 x^{2}+3}}-\frac {6 \sqrt {1-\left (-\frac {2}{3}+\frac {\sqrt {10}}{3}\right ) x^{2}}\, \sqrt {1-\left (-\frac {2}{3}-\frac {\sqrt {10}}{3}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {\sqrt {-6+3 \sqrt {10}}\, x}{3}, \frac {i \sqrt {6}}{3}+\frac {i \sqrt {15}}{3}\right )-\operatorname {EllipticE}\left (\frac {\sqrt {-6+3 \sqrt {10}}\, x}{3}, \frac {i \sqrt {6}}{3}+\frac {i \sqrt {15}}{3}\right )\right )}{5 \sqrt {-6+3 \sqrt {10}}\, \sqrt {-2 x^{4}+4 x^{2}+3}\, \left (4+2 \sqrt {10}\right )}\) \(231\)

Input:

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

Output:

-1/30*x*(2*x^2-7)/(-2*x^4+4*x^2+3)^(1/2)+3/10/(-6+3*10^(1/2))^(1/2)*(1-(-2 
/3+1/3*10^(1/2))*x^2)^(1/2)*(1-(-2/3-1/3*10^(1/2))*x^2)^(1/2)/(-2*x^4+4*x^ 
2+3)^(1/2)*EllipticF(1/3*(-6+3*10^(1/2))^(1/2)*x,1/3*I*6^(1/2)+1/3*I*15^(1 
/2))-6/5/(-6+3*10^(1/2))^(1/2)*(1-(-2/3+1/3*10^(1/2))*x^2)^(1/2)*(1-(-2/3- 
1/3*10^(1/2))*x^2)^(1/2)/(-2*x^4+4*x^2+3)^(1/2)/(4+2*10^(1/2))*(EllipticF( 
1/3*(-6+3*10^(1/2))^(1/2)*x,1/3*I*6^(1/2)+1/3*I*15^(1/2))-EllipticE(1/3*(- 
6+3*10^(1/2))^(1/2)*x,1/3*I*6^(1/2)+1/3*I*15^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.39 \[ \int \frac {1}{\left (3+4 x^2-2 x^4\right )^{3/2}} \, dx=\frac {2 \, {\left (\sqrt {10} \sqrt {3} {\left (2 \, x^{4} - 4 \, x^{2} - 3\right )} - 2 \, \sqrt {3} {\left (2 \, x^{4} - 4 \, x^{2} - 3\right )}\right )} \sqrt {\frac {1}{3} \, \sqrt {10} - \frac {2}{3}} E(\arcsin \left (x \sqrt {\frac {1}{3} \, \sqrt {10} - \frac {2}{3}}\right )\,|\,-\frac {2}{3} \, \sqrt {10} - \frac {7}{3}) + {\left (\sqrt {10} \sqrt {3} {\left (2 \, x^{4} - 4 \, x^{2} - 3\right )} + 10 \, \sqrt {3} {\left (2 \, x^{4} - 4 \, x^{2} - 3\right )}\right )} \sqrt {\frac {1}{3} \, \sqrt {10} - \frac {2}{3}} F(\arcsin \left (x \sqrt {\frac {1}{3} \, \sqrt {10} - \frac {2}{3}}\right )\,|\,-\frac {2}{3} \, \sqrt {10} - \frac {7}{3}) + 6 \, \sqrt {-2 \, x^{4} + 4 \, x^{2} + 3} {\left (2 \, x^{3} - 7 \, x\right )}}{180 \, {\left (2 \, x^{4} - 4 \, x^{2} - 3\right )}} \] Input:

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

Output:

1/180*(2*(sqrt(10)*sqrt(3)*(2*x^4 - 4*x^2 - 3) - 2*sqrt(3)*(2*x^4 - 4*x^2 
- 3))*sqrt(1/3*sqrt(10) - 2/3)*elliptic_e(arcsin(x*sqrt(1/3*sqrt(10) - 2/3 
)), -2/3*sqrt(10) - 7/3) + (sqrt(10)*sqrt(3)*(2*x^4 - 4*x^2 - 3) + 10*sqrt 
(3)*(2*x^4 - 4*x^2 - 3))*sqrt(1/3*sqrt(10) - 2/3)*elliptic_f(arcsin(x*sqrt 
(1/3*sqrt(10) - 2/3)), -2/3*sqrt(10) - 7/3) + 6*sqrt(-2*x^4 + 4*x^2 + 3)*( 
2*x^3 - 7*x))/(2*x^4 - 4*x^2 - 3)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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