\(\int \frac {1}{(-3+5 x^2-2 x^4)^{3/2}} \, dx\) [265]

Optimal result
Mathematica [A] (verified)
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 = 61 \[ \int \frac {1}{\left (-3+5 x^2-2 x^4\right )^{3/2}} \, dx=-\frac {x \left (13-10 x^2\right )}{3 \sqrt {-3+5 x^2-2 x^4}}+\frac {5}{3} E\left (\left .\arccos \left (\sqrt {\frac {2}{3}} x\right )\right |3\right )-\frac {2}{3} \operatorname {EllipticF}\left (\arccos \left (\sqrt {\frac {2}{3}} x\right ),3\right ) \] Output:

-1/3*x*(-10*x^2+13)/(-2*x^4+5*x^2-3)^(1/2)+5/3*EllipticE(1/3*(-6*x^2+9)^(1 
/2),3^(1/2))-2/3*InverseJacobiAM(arccos(1/3*x*6^(1/2)),3^(1/2))
 

Mathematica [A] (verified)

Time = 6.36 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.70 \[ \int \frac {1}{\left (-3+5 x^2-2 x^4\right )^{3/2}} \, dx=\frac {-13 x+10 x^3+5 \sqrt {6-4 x^2} \sqrt {1-x^2} E\left (\arcsin \left (\sqrt {\frac {2}{3}} x\right )|\frac {3}{2}\right )+\sqrt {6-4 x^2} \sqrt {1-x^2} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {2}{3}} x\right ),\frac {3}{2}\right )}{3 \sqrt {-3+5 x^2-2 x^4}} \] Input:

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

Output:

(-13*x + 10*x^3 + 5*Sqrt[6 - 4*x^2]*Sqrt[1 - x^2]*EllipticE[ArcSin[Sqrt[2/ 
3]*x], 3/2] + Sqrt[6 - 4*x^2]*Sqrt[1 - x^2]*EllipticF[ArcSin[Sqrt[2/3]*x], 
 3/2])/(3*Sqrt[-3 + 5*x^2 - 2*x^4])
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.05, 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, 322, 328}

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.

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

\(\Big \downarrow \) 1405

\(\displaystyle \frac {1}{3} \int \frac {2 \left (6-5 x^2\right )}{\sqrt {-2 x^4+5 x^2-3}}dx-\frac {x \left (13-10 x^2\right )}{3 \sqrt {-2 x^4+5 x^2-3}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2}{3} \int \frac {6-5 x^2}{\sqrt {-2 x^4+5 x^2-3}}dx-\frac {x \left (13-10 x^2\right )}{3 \sqrt {-2 x^4+5 x^2-3}}\)

\(\Big \downarrow \) 1494

\(\displaystyle \frac {4}{3} \sqrt {2} \int \frac {6-5 x^2}{2 \sqrt {2} \sqrt {3-2 x^2} \sqrt {x^2-1}}dx-\frac {x \left (13-10 x^2\right )}{3 \sqrt {-2 x^4+5 x^2-3}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2}{3} \int \frac {6-5 x^2}{\sqrt {3-2 x^2} \sqrt {x^2-1}}dx-\frac {x \left (13-10 x^2\right )}{3 \sqrt {-2 x^4+5 x^2-3}}\)

\(\Big \downarrow \) 399

\(\displaystyle \frac {2}{3} \left (\int \frac {1}{\sqrt {3-2 x^2} \sqrt {x^2-1}}dx-5 \int \frac {\sqrt {x^2-1}}{\sqrt {3-2 x^2}}dx\right )-\frac {x \left (13-10 x^2\right )}{3 \sqrt {-2 x^4+5 x^2-3}}\)

\(\Big \downarrow \) 322

\(\displaystyle \frac {2}{3} \left (-5 \int \frac {\sqrt {x^2-1}}{\sqrt {3-2 x^2}}dx-\operatorname {EllipticF}\left (\arccos \left (\sqrt {\frac {2}{3}} x\right ),3\right )\right )-\frac {x \left (13-10 x^2\right )}{3 \sqrt {-2 x^4+5 x^2-3}}\)

\(\Big \downarrow \) 328

\(\displaystyle \frac {2}{3} \left (\frac {5}{2} E\left (\left .\arccos \left (\sqrt {\frac {2}{3}} x\right )\right |3\right )-\operatorname {EllipticF}\left (\arccos \left (\sqrt {\frac {2}{3}} x\right ),3\right )\right )-\frac {x \left (13-10 x^2\right )}{3 \sqrt {-2 x^4+5 x^2-3}}\)

Input:

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

Output:

-1/3*(x*(13 - 10*x^2))/Sqrt[-3 + 5*x^2 - 2*x^4] + (2*((5*EllipticE[ArcCos[ 
Sqrt[2/3]*x], 3])/2 - EllipticF[ArcCos[Sqrt[2/3]*x], 3]))/3
 

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 322
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S 
imp[(-(Sqrt[c]*Rt[-d/c, 2]*Sqrt[a - b*(c/d)])^(-1))*EllipticF[ArcCos[Rt[-d/ 
c, 2]*x], b*(c/(b*c - a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && 
GtQ[c, 0] && GtQ[a - b*(c/d), 0]
 

rule 328
Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ 
(-Sqrt[a - b*(c/d)]/(Sqrt[c]*Rt[-d/c, 2]))*EllipticE[ArcCos[Rt[-d/c, 2]*x], 
 b*(c/(b*c - a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] 
 && GtQ[a - b*(c/d), 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)

Leaf count of result is larger than twice the leaf count of optimal. \(138\) vs. \(2(55)=110\).

Time = 2.67 (sec) , antiderivative size = 139, normalized size of antiderivative = 2.28

method result size
risch \(\frac {x \left (10 x^{2}-13\right )}{3 \sqrt {-2 x^{4}+5 x^{2}-3}}+\frac {2 \sqrt {6}\, \sqrt {-6 x^{2}+9}\, \sqrt {-x^{2}+1}\, \operatorname {EllipticF}\left (\frac {x \sqrt {6}}{3}, \frac {\sqrt {6}}{2}\right )}{3 \sqrt {-2 x^{4}+5 x^{2}-3}}-\frac {5 \sqrt {6}\, \sqrt {-6 x^{2}+9}\, \sqrt {-x^{2}+1}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {6}}{3}, \frac {\sqrt {6}}{2}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {6}}{3}, \frac {\sqrt {6}}{2}\right )\right )}{9 \sqrt {-2 x^{4}+5 x^{2}-3}}\) \(139\)
default \(\frac {\frac {10}{3} x^{3}-\frac {13}{3} x}{\sqrt {-2 x^{4}+5 x^{2}-3}}+\frac {2 \sqrt {6}\, \sqrt {-6 x^{2}+9}\, \sqrt {-x^{2}+1}\, \operatorname {EllipticF}\left (\frac {x \sqrt {6}}{3}, \frac {\sqrt {6}}{2}\right )}{3 \sqrt {-2 x^{4}+5 x^{2}-3}}-\frac {5 \sqrt {6}\, \sqrt {-6 x^{2}+9}\, \sqrt {-x^{2}+1}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {6}}{3}, \frac {\sqrt {6}}{2}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {6}}{3}, \frac {\sqrt {6}}{2}\right )\right )}{9 \sqrt {-2 x^{4}+5 x^{2}-3}}\) \(140\)
elliptic \(\frac {\frac {10}{3} x^{3}-\frac {13}{3} x}{\sqrt {-2 x^{4}+5 x^{2}-3}}+\frac {2 \sqrt {6}\, \sqrt {-6 x^{2}+9}\, \sqrt {-x^{2}+1}\, \operatorname {EllipticF}\left (\frac {x \sqrt {6}}{3}, \frac {\sqrt {6}}{2}\right )}{3 \sqrt {-2 x^{4}+5 x^{2}-3}}-\frac {5 \sqrt {6}\, \sqrt {-6 x^{2}+9}\, \sqrt {-x^{2}+1}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {6}}{3}, \frac {\sqrt {6}}{2}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {6}}{3}, \frac {\sqrt {6}}{2}\right )\right )}{9 \sqrt {-2 x^{4}+5 x^{2}-3}}\) \(140\)

Input:

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

Output:

1/3*x*(10*x^2-13)/(-2*x^4+5*x^2-3)^(1/2)+2/3*6^(1/2)*(-6*x^2+9)^(1/2)*(-x^ 
2+1)^(1/2)/(-2*x^4+5*x^2-3)^(1/2)*EllipticF(1/3*x*6^(1/2),1/2*6^(1/2))-5/9 
*6^(1/2)*(-6*x^2+9)^(1/2)*(-x^2+1)^(1/2)/(-2*x^4+5*x^2-3)^(1/2)*(EllipticF 
(1/3*x*6^(1/2),1/2*6^(1/2))-EllipticE(1/3*x*6^(1/2),1/2*6^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.36 \[ \int \frac {1}{\left (-3+5 x^2-2 x^4\right )^{3/2}} \, dx=-\frac {5 \, \sqrt {-3} {\left (2 \, x^{4} - 5 \, x^{2} + 3\right )} E(\arcsin \left (x\right )\,|\,\frac {2}{3}) - \sqrt {-3} {\left (2 \, x^{4} - 5 \, x^{2} + 3\right )} F(\arcsin \left (x\right )\,|\,\frac {2}{3}) + \sqrt {-2 \, x^{4} + 5 \, x^{2} - 3} {\left (10 \, x^{3} - 13 \, x\right )}}{3 \, {\left (2 \, x^{4} - 5 \, x^{2} + 3\right )}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

Integral((-2*x**4 + 5*x**2 - 3)**(-3/2), x)
 

Maxima [F]

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

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

Output:

integrate((-2*x^4 + 5*x^2 - 3)^(-3/2), x)
 

Giac [F]

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

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

Output:

integrate((-2*x^4 + 5*x^2 - 3)^(-3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (-3+5 x^2-2 x^4\right )^{3/2}} \, dx=\int \frac {1}{{\left (-2\,x^4+5\,x^2-3\right )}^{3/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\left (-3+5 x^2-2 x^4\right )^{3/2}} \, dx=\int \frac {\sqrt {-2 x^{4}+5 x^{2}-3}}{4 x^{8}-20 x^{6}+37 x^{4}-30 x^{2}+9}d x \] Input:

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

Output:

int(sqrt( - 2*x**4 + 5*x**2 - 3)/(4*x**8 - 20*x**6 + 37*x**4 - 30*x**2 + 9 
),x)