\(\int \frac {1}{(-2+5 x^2-3 x^4)^{3/2}} \, dx\) [280]

Optimal result

Integrand size = 16, antiderivative size = 43 \[ \int \frac {1}{\left (-2+5 x^2-3 x^4\right )^{3/2}} \, dx=-\frac {x \left (13-15 x^2\right )}{2 \sqrt {-2+5 x^2-3 x^4}}+\frac {5 E(\arccos (x)|3)}{2}-\operatorname {EllipticF}(\arccos (x),3) \] Output:

-1/2*x*(-15*x^2+13)/(-3*x^4+5*x^2-2)^(1/2)+5/2*EllipticE((-x^2+1)^(1/2),3^ 
(1/2))-InverseJacobiAM(arccos(x),3^(1/2))
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(105\) vs. \(2(43)=86\).

Time = 6.45 (sec) , antiderivative size = 105, normalized size of antiderivative = 2.44 \[ \int \frac {1}{\left (-2+5 x^2-3 x^4\right )^{3/2}} \, dx=\frac {-13 x+15 x^3+5 \sqrt {6-9 x^2} \sqrt {1-x^2} E\left (\arcsin \left (\sqrt {\frac {3}{2}} x\right )|\frac {2}{3}\right )-\sqrt {6-9 x^2} \sqrt {1-x^2} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {3}{2}} x\right ),\frac {2}{3}\right )}{2 \sqrt {-2+5 x^2-3 x^4}} \] Input:

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

Output:

(-13*x + 15*x^3 + 5*Sqrt[6 - 9*x^2]*Sqrt[1 - x^2]*EllipticE[ArcSin[Sqrt[3/ 
2]*x], 2/3] - Sqrt[6 - 9*x^2]*Sqrt[1 - x^2]*EllipticF[ArcSin[Sqrt[3/2]*x], 
 2/3])/(2*Sqrt[-2 + 5*x^2 - 3*x^4])
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.16, 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.

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

\(\Big \downarrow \) 1405

\(\displaystyle \frac {1}{2} \int \frac {3 \left (4-5 x^2\right )}{\sqrt {-3 x^4+5 x^2-2}}dx-\frac {x \left (13-15 x^2\right )}{2 \sqrt {-3 x^4+5 x^2-2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3}{2} \int \frac {4-5 x^2}{\sqrt {-3 x^4+5 x^2-2}}dx-\frac {x \left (13-15 x^2\right )}{2 \sqrt {-3 x^4+5 x^2-2}}\)

\(\Big \downarrow \) 1494

\(\displaystyle 3 \sqrt {3} \int \frac {4-5 x^2}{2 \sqrt {3} \sqrt {1-x^2} \sqrt {3 x^2-2}}dx-\frac {x \left (13-15 x^2\right )}{2 \sqrt {-3 x^4+5 x^2-2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3}{2} \int \frac {4-5 x^2}{\sqrt {1-x^2} \sqrt {3 x^2-2}}dx-\frac {x \left (13-15 x^2\right )}{2 \sqrt {-3 x^4+5 x^2-2}}\)

\(\Big \downarrow \) 399

\(\displaystyle \frac {3}{2} \left (\frac {2}{3} \int \frac {1}{\sqrt {1-x^2} \sqrt {3 x^2-2}}dx-\frac {5}{3} \int \frac {\sqrt {3 x^2-2}}{\sqrt {1-x^2}}dx\right )-\frac {x \left (13-15 x^2\right )}{2 \sqrt {-3 x^4+5 x^2-2}}\)

\(\Big \downarrow \) 322

\(\displaystyle \frac {3}{2} \left (-\frac {5}{3} \int \frac {\sqrt {3 x^2-2}}{\sqrt {1-x^2}}dx-\frac {2}{3} \operatorname {EllipticF}(\arccos (x),3)\right )-\frac {x \left (13-15 x^2\right )}{2 \sqrt {-3 x^4+5 x^2-2}}\)

\(\Big \downarrow \) 328

\(\displaystyle \frac {3}{2} \left (\frac {5 E(\arccos (x)|3)}{3}-\frac {2 \operatorname {EllipticF}(\arccos (x),3)}{3}\right )-\frac {x \left (13-15 x^2\right )}{2 \sqrt {-3 x^4+5 x^2-2}}\)

Input:

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

Output:

-1/2*(x*(13 - 15*x^2))/Sqrt[-2 + 5*x^2 - 3*x^4] + (3*((5*EllipticE[ArcCos[ 
x], 3])/3 - (2*EllipticF[ArcCos[x], 3])/3))/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 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. \(117\) vs. \(2(48)=96\).

Time = 2.63 (sec) , antiderivative size = 118, normalized size of antiderivative = 2.74

method	result	size
risch	\(\frac {x \left (15 x^{2}-13\right )}{2 \sqrt {-3 x^{4}+5 x^{2}-2}}+\frac {3 \sqrt {-x^{2}+1}\, \sqrt {-6 x^{2}+4}\, \operatorname {EllipticF}\left (x , \frac {\sqrt {6}}{2}\right )}{\sqrt {-3 x^{4}+5 x^{2}-2}}-\frac {5 \sqrt {-x^{2}+1}\, \sqrt {-6 x^{2}+4}\, \left (\operatorname {EllipticF}\left (x , \frac {\sqrt {6}}{2}\right )-\operatorname {EllipticE}\left (x , \frac {\sqrt {6}}{2}\right )\right )}{2 \sqrt {-3 x^{4}+5 x^{2}-2}}\)	\(118\)
default	\(\frac {\frac {15}{2} x^{3}-\frac {13}{2} x}{\sqrt {-3 x^{4}+5 x^{2}-2}}+\frac {3 \sqrt {-x^{2}+1}\, \sqrt {-6 x^{2}+4}\, \operatorname {EllipticF}\left (x , \frac {\sqrt {6}}{2}\right )}{\sqrt {-3 x^{4}+5 x^{2}-2}}-\frac {5 \sqrt {-x^{2}+1}\, \sqrt {-6 x^{2}+4}\, \left (\operatorname {EllipticF}\left (x , \frac {\sqrt {6}}{2}\right )-\operatorname {EllipticE}\left (x , \frac {\sqrt {6}}{2}\right )\right )}{2 \sqrt {-3 x^{4}+5 x^{2}-2}}\)	\(119\)
elliptic	\(\frac {\frac {15}{2} x^{3}-\frac {13}{2} x}{\sqrt {-3 x^{4}+5 x^{2}-2}}+\frac {3 \sqrt {-x^{2}+1}\, \sqrt {-6 x^{2}+4}\, \operatorname {EllipticF}\left (x , \frac {\sqrt {6}}{2}\right )}{\sqrt {-3 x^{4}+5 x^{2}-2}}-\frac {5 \sqrt {-x^{2}+1}\, \sqrt {-6 x^{2}+4}\, \left (\operatorname {EllipticF}\left (x , \frac {\sqrt {6}}{2}\right )-\operatorname {EllipticE}\left (x , \frac {\sqrt {6}}{2}\right )\right )}{2 \sqrt {-3 x^{4}+5 x^{2}-2}}\)	\(119\)

Input:

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

Output:

1/2*x*(15*x^2-13)/(-3*x^4+5*x^2-2)^(1/2)+3*(-x^2+1)^(1/2)*(-6*x^2+4)^(1/2) 
/(-3*x^4+5*x^2-2)^(1/2)*EllipticF(x,1/2*6^(1/2))-5/2*(-x^2+1)^(1/2)*(-6*x^ 
2+4)^(1/2)/(-3*x^4+5*x^2-2)^(1/2)*(EllipticF(x,1/2*6^(1/2))-EllipticE(x,1/ 
2*6^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.91 \[ \int \frac {1}{\left (-2+5 x^2-3 x^4\right )^{3/2}} \, dx=-\frac {5 \, \sqrt {-2} {\left (3 \, x^{4} - 5 \, x^{2} + 2\right )} E(\arcsin \left (x\right )\,|\,\frac {3}{2}) + \sqrt {-2} {\left (3 \, x^{4} - 5 \, x^{2} + 2\right )} F(\arcsin \left (x\right )\,|\,\frac {3}{2}) + \sqrt {-3 \, x^{4} + 5 \, x^{2} - 2} {\left (15 \, x^{3} - 13 \, x\right )}}{2 \, {\left (3 \, x^{4} - 5 \, x^{2} + 2\right )}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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