\(\int \frac {1}{(2+5 x^2-3 x^4)^{3/2}} \, dx\) [173]

Optimal result

Integrand size = 16, antiderivative size = 57 \[ \int \frac {1}{\left (2+5 x^2-3 x^4\right )^{3/2}} \, dx=\frac {x \left (37-15 x^2\right )}{98 \sqrt {2+5 x^2-3 x^4}}+\frac {5}{98} E\left (\left .\arcsin \left (\frac {x}{\sqrt {2}}\right )\right |-6\right )+\frac {1}{14} \operatorname {EllipticF}\left (\arcsin \left (\frac {x}{\sqrt {2}}\right ),-6\right ) \] Output:

1/98*x*(-15*x^2+37)/(-3*x^4+5*x^2+2)^(1/2)+5/98*EllipticE(1/2*x*2^(1/2),I* 
6^(1/2))+1/14*EllipticF(1/2*x*2^(1/2),I*6^(1/2))
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 9.29 (sec) , antiderivative size = 123, normalized size of antiderivative = 2.16 \[ \int \frac {1}{\left (2+5 x^2-3 x^4\right )^{3/2}} \, dx=\frac {37 x-15 x^3+5 i \sqrt {6} \sqrt {2-x^2} \sqrt {1+3 x^2} E\left (i \text {arcsinh}\left (\sqrt {3} x\right )|-\frac {1}{6}\right )-7 i \sqrt {6} \sqrt {2-x^2} \sqrt {1+3 x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {3} x\right ),-\frac {1}{6}\right )}{98 \sqrt {2+5 x^2-3 x^4}} \] Input:

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

Output:

(37*x - 15*x^3 + (5*I)*Sqrt[6]*Sqrt[2 - x^2]*Sqrt[1 + 3*x^2]*EllipticE[I*A 
rcSinh[Sqrt[3]*x], -1/6] - (7*I)*Sqrt[6]*Sqrt[2 - x^2]*Sqrt[1 + 3*x^2]*Ell 
ipticF[I*ArcSinh[Sqrt[3]*x], -1/6])/(98*Sqrt[2 + 5*x^2 - 3*x^4])
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.09, 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.

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

\(\Big \downarrow \) 1405

\(\displaystyle \frac {x \left (37-15 x^2\right )}{98 \sqrt {-3 x^4+5 x^2+2}}-\frac {1}{98} \int -\frac {3 \left (5 x^2+4\right )}{\sqrt {-3 x^4+5 x^2+2}}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3}{98} \int \frac {5 x^2+4}{\sqrt {-3 x^4+5 x^2+2}}dx+\frac {x \left (37-15 x^2\right )}{98 \sqrt {-3 x^4+5 x^2+2}}\)

\(\Big \downarrow \) 1494

\(\displaystyle \frac {3}{49} \sqrt {3} \int \frac {5 x^2+4}{2 \sqrt {3} \sqrt {2-x^2} \sqrt {3 x^2+1}}dx+\frac {x \left (37-15 x^2\right )}{98 \sqrt {-3 x^4+5 x^2+2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3}{98} \int \frac {5 x^2+4}{\sqrt {2-x^2} \sqrt {3 x^2+1}}dx+\frac {x \left (37-15 x^2\right )}{98 \sqrt {-3 x^4+5 x^2+2}}\)

\(\Big \downarrow \) 399

\(\displaystyle \frac {3}{98} \left (\frac {7}{3} \int \frac {1}{\sqrt {2-x^2} \sqrt {3 x^2+1}}dx+\frac {5}{3} \int \frac {\sqrt {3 x^2+1}}{\sqrt {2-x^2}}dx\right )+\frac {x \left (37-15 x^2\right )}{98 \sqrt {-3 x^4+5 x^2+2}}\)

\(\Big \downarrow \) 321

\(\displaystyle \frac {3}{98} \left (\frac {5}{3} \int \frac {\sqrt {3 x^2+1}}{\sqrt {2-x^2}}dx+\frac {7}{3} \operatorname {EllipticF}\left (\arcsin \left (\frac {x}{\sqrt {2}}\right ),-6\right )\right )+\frac {x \left (37-15 x^2\right )}{98 \sqrt {-3 x^4+5 x^2+2}}\)

\(\Big \downarrow \) 327

\(\displaystyle \frac {3}{98} \left (\frac {7}{3} \operatorname {EllipticF}\left (\arcsin \left (\frac {x}{\sqrt {2}}\right ),-6\right )+\frac {5}{3} E\left (\left .\arcsin \left (\frac {x}{\sqrt {2}}\right )\right |-6\right )\right )+\frac {x \left (37-15 x^2\right )}{98 \sqrt {-3 x^4+5 x^2+2}}\)

Input:

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

Output:

(x*(37 - 15*x^2))/(98*Sqrt[2 + 5*x^2 - 3*x^4]) + (3*((5*EllipticE[ArcSin[x 
/Sqrt[2]], -6])/3 + (7*EllipticF[ArcSin[x/Sqrt[2]], -6])/3))/98
 

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. 141 vs. \(2 (55 ) = 110\).

Time = 2.38 (sec) , antiderivative size = 142, normalized size of antiderivative = 2.49

method	result	size
risch	\(-\frac {x \left (15 x^{2}-37\right )}{98 \sqrt {-3 x^{4}+5 x^{2}+2}}+\frac {3 \sqrt {2}\, \sqrt {-2 x^{2}+4}\, \sqrt {3 x^{2}+1}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}}{2}, i \sqrt {6}\right )}{49 \sqrt {-3 x^{4}+5 x^{2}+2}}-\frac {5 \sqrt {2}\, \sqrt {-2 x^{2}+4}\, \sqrt {3 x^{2}+1}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {2}}{2}, i \sqrt {6}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {2}}{2}, i \sqrt {6}\right )\right )}{196 \sqrt {-3 x^{4}+5 x^{2}+2}}\)	\(142\)
default	\(\frac {\frac {37}{98} x -\frac {15}{98} x^{3}}{\sqrt {-3 x^{4}+5 x^{2}+2}}+\frac {3 \sqrt {2}\, \sqrt {-2 x^{2}+4}\, \sqrt {3 x^{2}+1}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}}{2}, i \sqrt {6}\right )}{49 \sqrt {-3 x^{4}+5 x^{2}+2}}-\frac {5 \sqrt {2}\, \sqrt {-2 x^{2}+4}\, \sqrt {3 x^{2}+1}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {2}}{2}, i \sqrt {6}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {2}}{2}, i \sqrt {6}\right )\right )}{196 \sqrt {-3 x^{4}+5 x^{2}+2}}\)	\(143\)
elliptic	\(\frac {\frac {37}{98} x -\frac {15}{98} x^{3}}{\sqrt {-3 x^{4}+5 x^{2}+2}}+\frac {3 \sqrt {2}\, \sqrt {-2 x^{2}+4}\, \sqrt {3 x^{2}+1}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}}{2}, i \sqrt {6}\right )}{49 \sqrt {-3 x^{4}+5 x^{2}+2}}-\frac {5 \sqrt {2}\, \sqrt {-2 x^{2}+4}\, \sqrt {3 x^{2}+1}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {2}}{2}, i \sqrt {6}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {2}}{2}, i \sqrt {6}\right )\right )}{196 \sqrt {-3 x^{4}+5 x^{2}+2}}\)	\(143\)

Input:

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

Output:

-1/98*x*(15*x^2-37)/(-3*x^4+5*x^2+2)^(1/2)+3/49*2^(1/2)*(-2*x^2+4)^(1/2)*( 
3*x^2+1)^(1/2)/(-3*x^4+5*x^2+2)^(1/2)*EllipticF(1/2*x*2^(1/2),I*6^(1/2))-5 
/196*2^(1/2)*(-2*x^2+4)^(1/2)*(3*x^2+1)^(1/2)/(-3*x^4+5*x^2+2)^(1/2)*(Elli 
pticF(1/2*x*2^(1/2),I*6^(1/2))-EllipticE(1/2*x*2^(1/2),I*6^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.54 \[ \int \frac {1}{\left (2+5 x^2-3 x^4\right )^{3/2}} \, dx=\frac {5 \, {\left (3 \, x^{4} - 5 \, x^{2} - 2\right )} E(\arcsin \left (\frac {1}{2} \, \sqrt {2} x\right )\,|\,-6) + 19 \, {\left (3 \, x^{4} - 5 \, x^{2} - 2\right )} F(\arcsin \left (\frac {1}{2} \, \sqrt {2} x\right )\,|\,-6) + 2 \, \sqrt {-3 \, x^{4} + 5 \, x^{2} + 2} {\left (15 \, x^{3} - 37 \, x\right )}}{196 \, {\left (3 \, x^{4} - 5 \, x^{2} - 2\right )}} \] Input:

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

Output:

1/196*(5*(3*x^4 - 5*x^2 - 2)*elliptic_e(arcsin(1/2*sqrt(2)*x), -6) + 19*(3 
*x^4 - 5*x^2 - 2)*elliptic_f(arcsin(1/2*sqrt(2)*x), -6) + 2*sqrt(-3*x^4 + 
5*x^2 + 2)*(15*x^3 - 37*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]

\[ \int \frac {1}{\left (2+5 x^2-3 x^4\right )^{3/2}} \, dx=\int \frac {\sqrt {-3 x^{4}+5 x^{2}+2}}{9 x^{8}-30 x^{6}+13 x^{4}+20 x^{2}+4}d x \] 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 + 13*x**4 + 20*x**2 + 4 
),x)