\(\int \frac {1}{(2+5 x^2-x^4)^{3/2}} \, dx\) [302]

Optimal result

Integrand size = 16, antiderivative size = 131 \[ \int \frac {1}{\left (2+5 x^2-x^4\right )^{3/2}} \, dx=\frac {x \left (29-5 x^2\right )}{66 \sqrt {2+5 x^2-x^4}}+\frac {5}{66} \sqrt {\frac {1}{2} \left (-5+\sqrt {33}\right )} E\left (\arcsin \left (\sqrt {\frac {2}{5+\sqrt {33}}} x\right )|\frac {1}{4} \left (-29-5 \sqrt {33}\right )\right )+\frac {1}{2} \sqrt {\frac {1}{66} \left (-5+\sqrt {33}\right )} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {2}{5+\sqrt {33}}} x\right ),\frac {1}{4} \left (-29-5 \sqrt {33}\right )\right ) \] Output:

1/66*x*(-5*x^2+29)/(-x^4+5*x^2+2)^(1/2)+5/132*(-10+2*33^(1/2))^(1/2)*Ellip 
ticE(2^(1/2)/(5+33^(1/2))^(1/2)*x,5/4*I*2^(1/2)+1/4*I*66^(1/2))+1/132*(-33 
0+66*33^(1/2))^(1/2)*EllipticF(2^(1/2)/(5+33^(1/2))^(1/2)*x,5/4*I*2^(1/2)+ 
1/4*I*66^(1/2))
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 6.46 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.34 \[ \int \frac {1}{\left (2+5 x^2-x^4\right )^{3/2}} \, dx=\frac {4 x \left (29-5 x^2\right )+10 i \sqrt {2 \left (5+\sqrt {33}\right )} \sqrt {2+5 x^2-x^4} E\left (i \text {arcsinh}\left (\sqrt {\frac {2}{-5+\sqrt {33}}} x\right )|-\frac {29}{4}+\frac {5 \sqrt {33}}{4}\right )-\frac {2 i \left (33+5 \sqrt {33}\right ) \sqrt {4+10 x^2-2 x^4} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {2}{-5+\sqrt {33}}} x\right ),-\frac {29}{4}+\frac {5 \sqrt {33}}{4}\right )}{\sqrt {5+\sqrt {33}}}}{264 \sqrt {2+5 x^2-x^4}} \] Input:

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

Output:

(4*x*(29 - 5*x^2) + (10*I)*Sqrt[2*(5 + Sqrt[33])]*Sqrt[2 + 5*x^2 - x^4]*El 
lipticE[I*ArcSinh[Sqrt[2/(-5 + Sqrt[33])]*x], -29/4 + (5*Sqrt[33])/4] - (( 
2*I)*(33 + 5*Sqrt[33])*Sqrt[4 + 10*x^2 - 2*x^4]*EllipticF[I*ArcSinh[Sqrt[2 
/(-5 + Sqrt[33])]*x], -29/4 + (5*Sqrt[33])/4])/Sqrt[5 + Sqrt[33]])/(264*Sq 
rt[2 + 5*x^2 - x^4])
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {1405, 25, 1494, 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.

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

\(\Big \downarrow \) 1405

\(\displaystyle \frac {x \left (29-5 x^2\right )}{66 \sqrt {-x^4+5 x^2+2}}-\frac {1}{66} \int -\frac {5 x^2+4}{\sqrt {-x^4+5 x^2+2}}dx\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{66} \int \frac {5 x^2+4}{\sqrt {-x^4+5 x^2+2}}dx+\frac {x \left (29-5 x^2\right )}{66 \sqrt {-x^4+5 x^2+2}}\)

\(\Big \downarrow \) 1494

\(\displaystyle \frac {1}{33} \int \frac {5 x^2+4}{\sqrt {-2 x^2+\sqrt {33}+5} \sqrt {2 x^2+\sqrt {33}-5}}dx+\frac {x \left (29-5 x^2\right )}{66 \sqrt {-x^4+5 x^2+2}}\)

\(\Big \downarrow \) 399

\(\displaystyle \frac {1}{33} \left (\frac {1}{2} \left (33-5 \sqrt {33}\right ) \int \frac {1}{\sqrt {-2 x^2+\sqrt {33}+5} \sqrt {2 x^2+\sqrt {33}-5}}dx+\frac {5}{2} \int \frac {\sqrt {2 x^2+\sqrt {33}-5}}{\sqrt {-2 x^2+\sqrt {33}+5}}dx\right )+\frac {x \left (29-5 x^2\right )}{66 \sqrt {-x^4+5 x^2+2}}\)

\(\Big \downarrow \) 321

\(\displaystyle \frac {1}{33} \left (\frac {5}{2} \int \frac {\sqrt {2 x^2+\sqrt {33}-5}}{\sqrt {-2 x^2+\sqrt {33}+5}}dx+\frac {\left (33-5 \sqrt {33}\right ) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {2}{5+\sqrt {33}}} x\right ),\frac {1}{4} \left (-29-5 \sqrt {33}\right )\right )}{2 \sqrt {2 \left (\sqrt {33}-5\right )}}\right )+\frac {x \left (29-5 x^2\right )}{66 \sqrt {-x^4+5 x^2+2}}\)

\(\Big \downarrow \) 327

\(\displaystyle \frac {1}{33} \left (\frac {\left (33-5 \sqrt {33}\right ) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {2}{5+\sqrt {33}}} x\right ),\frac {1}{4} \left (-29-5 \sqrt {33}\right )\right )}{2 \sqrt {2 \left (\sqrt {33}-5\right )}}+\frac {5}{2} \sqrt {\frac {1}{2} \left (\sqrt {33}-5\right )} E\left (\arcsin \left (\sqrt {\frac {2}{5+\sqrt {33}}} x\right )|\frac {1}{4} \left (-29-5 \sqrt {33}\right )\right )\right )+\frac {x \left (29-5 x^2\right )}{66 \sqrt {-x^4+5 x^2+2}}\)

Input:

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

Output:

(x*(29 - 5*x^2))/(66*Sqrt[2 + 5*x^2 - x^4]) + ((5*Sqrt[(-5 + Sqrt[33])/2]* 
EllipticE[ArcSin[Sqrt[2/(5 + Sqrt[33])]*x], (-29 - 5*Sqrt[33])/4])/2 + ((3 
3 - 5*Sqrt[33])*EllipticF[ArcSin[Sqrt[2/(5 + Sqrt[33])]*x], (-29 - 5*Sqrt[ 
33])/4])/(2*Sqrt[2*(-5 + Sqrt[33])]))/33
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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. 217 vs. \(2 (99 ) = 198\).

Time = 1.98 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.66

method	result	size
risch	\(-\frac {x \left (5 x^{2}-29\right )}{66 \sqrt {-x^{4}+5 x^{2}+2}}+\frac {4 \sqrt {1-\left (-\frac {5}{4}+\frac {\sqrt {33}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{4}-\frac {\sqrt {33}}{4}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-5+\sqrt {33}}}{2}, \frac {5 i \sqrt {2}}{4}+\frac {i \sqrt {66}}{4}\right )}{33 \sqrt {-5+\sqrt {33}}\, \sqrt {-x^{4}+5 x^{2}+2}}-\frac {20 \sqrt {1-\left (-\frac {5}{4}+\frac {\sqrt {33}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{4}-\frac {\sqrt {33}}{4}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {-5+\sqrt {33}}}{2}, \frac {5 i \sqrt {2}}{4}+\frac {i \sqrt {66}}{4}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {-5+\sqrt {33}}}{2}, \frac {5 i \sqrt {2}}{4}+\frac {i \sqrt {66}}{4}\right )\right )}{33 \sqrt {-5+\sqrt {33}}\, \sqrt {-x^{4}+5 x^{2}+2}\, \left (\sqrt {33}+5\right )}\)	\(218\)
default	\(\frac {\frac {29}{66} x -\frac {5}{66} x^{3}}{\sqrt {-x^{4}+5 x^{2}+2}}+\frac {4 \sqrt {1-\left (-\frac {5}{4}+\frac {\sqrt {33}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{4}-\frac {\sqrt {33}}{4}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-5+\sqrt {33}}}{2}, \frac {5 i \sqrt {2}}{4}+\frac {i \sqrt {66}}{4}\right )}{33 \sqrt {-5+\sqrt {33}}\, \sqrt {-x^{4}+5 x^{2}+2}}-\frac {20 \sqrt {1-\left (-\frac {5}{4}+\frac {\sqrt {33}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{4}-\frac {\sqrt {33}}{4}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {-5+\sqrt {33}}}{2}, \frac {5 i \sqrt {2}}{4}+\frac {i \sqrt {66}}{4}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {-5+\sqrt {33}}}{2}, \frac {5 i \sqrt {2}}{4}+\frac {i \sqrt {66}}{4}\right )\right )}{33 \sqrt {-5+\sqrt {33}}\, \sqrt {-x^{4}+5 x^{2}+2}\, \left (\sqrt {33}+5\right )}\)	\(219\)
elliptic	\(\frac {\frac {29}{66} x -\frac {5}{66} x^{3}}{\sqrt {-x^{4}+5 x^{2}+2}}+\frac {4 \sqrt {1-\left (-\frac {5}{4}+\frac {\sqrt {33}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{4}-\frac {\sqrt {33}}{4}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-5+\sqrt {33}}}{2}, \frac {5 i \sqrt {2}}{4}+\frac {i \sqrt {66}}{4}\right )}{33 \sqrt {-5+\sqrt {33}}\, \sqrt {-x^{4}+5 x^{2}+2}}-\frac {20 \sqrt {1-\left (-\frac {5}{4}+\frac {\sqrt {33}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{4}-\frac {\sqrt {33}}{4}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {-5+\sqrt {33}}}{2}, \frac {5 i \sqrt {2}}{4}+\frac {i \sqrt {66}}{4}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {-5+\sqrt {33}}}{2}, \frac {5 i \sqrt {2}}{4}+\frac {i \sqrt {66}}{4}\right )\right )}{33 \sqrt {-5+\sqrt {33}}\, \sqrt {-x^{4}+5 x^{2}+2}\, \left (\sqrt {33}+5\right )}\)	\(219\)

Input:

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

Output:

-1/66*x*(5*x^2-29)/(-x^4+5*x^2+2)^(1/2)+4/33/(-5+33^(1/2))^(1/2)*(1-(-5/4+ 
1/4*33^(1/2))*x^2)^(1/2)*(1-(-5/4-1/4*33^(1/2))*x^2)^(1/2)/(-x^4+5*x^2+2)^ 
(1/2)*EllipticF(1/2*x*(-5+33^(1/2))^(1/2),5/4*I*2^(1/2)+1/4*I*66^(1/2))-20 
/33/(-5+33^(1/2))^(1/2)*(1-(-5/4+1/4*33^(1/2))*x^2)^(1/2)*(1-(-5/4-1/4*33^ 
(1/2))*x^2)^(1/2)/(-x^4+5*x^2+2)^(1/2)/(33^(1/2)+5)*(EllipticF(1/2*x*(-5+3 
3^(1/2))^(1/2),5/4*I*2^(1/2)+1/4*I*66^(1/2))-EllipticE(1/2*x*(-5+33^(1/2)) 
^(1/2),5/4*I*2^(1/2)+1/4*I*66^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.24 \[ \int \frac {1}{\left (2+5 x^2-x^4\right )^{3/2}} \, dx=\frac {5 \, {\left (\sqrt {33} \sqrt {2} {\left (x^{4} - 5 \, x^{2} - 2\right )} - 5 \, \sqrt {2} {\left (x^{4} - 5 \, x^{2} - 2\right )}\right )} \sqrt {\sqrt {33} - 5} E(\arcsin \left (\frac {1}{2} \, x \sqrt {\sqrt {33} - 5}\right )\,|\,-\frac {5}{4} \, \sqrt {33} - \frac {29}{4}) - {\left (\sqrt {33} \sqrt {2} {\left (x^{4} - 5 \, x^{2} - 2\right )} - 45 \, \sqrt {2} {\left (x^{4} - 5 \, x^{2} - 2\right )}\right )} \sqrt {\sqrt {33} - 5} F(\arcsin \left (\frac {1}{2} \, x \sqrt {\sqrt {33} - 5}\right )\,|\,-\frac {5}{4} \, \sqrt {33} - \frac {29}{4}) + 8 \, \sqrt {-x^{4} + 5 \, x^{2} + 2} {\left (5 \, x^{3} - 29 \, x\right )}}{528 \, {\left (x^{4} - 5 \, x^{2} - 2\right )}} \] Input:

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

Output:

1/528*(5*(sqrt(33)*sqrt(2)*(x^4 - 5*x^2 - 2) - 5*sqrt(2)*(x^4 - 5*x^2 - 2) 
)*sqrt(sqrt(33) - 5)*elliptic_e(arcsin(1/2*x*sqrt(sqrt(33) - 5)), -5/4*sqr 
t(33) - 29/4) - (sqrt(33)*sqrt(2)*(x^4 - 5*x^2 - 2) - 45*sqrt(2)*(x^4 - 5* 
x^2 - 2))*sqrt(sqrt(33) - 5)*elliptic_f(arcsin(1/2*x*sqrt(sqrt(33) - 5)), 
-5/4*sqrt(33) - 29/4) + 8*sqrt(-x^4 + 5*x^2 + 2)*(5*x^3 - 29*x))/(x^4 - 5* 
x^2 - 2)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\left (2+5 x^2-x^4\right )^{3/2}} \, dx=\int \frac {\sqrt {-x^{4}+5 x^{2}+2}}{x^{8}-10 x^{6}+21 x^{4}+20 x^{2}+4}d x \] Input:

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

Output:

int(sqrt( - x**4 + 5*x**2 + 2)/(x**8 - 10*x**6 + 21*x**4 + 20*x**2 + 4),x)