\(\int \frac {1}{(2+5 x^2+x^4)^{3/2}} \, dx\) [301]

Optimal result

Integrand size = 14, antiderivative size = 234 \[ \int \frac {1}{\left (2+5 x^2+x^4\right )^{3/2}} \, dx=\frac {2 x}{\sqrt {17} \left (5-\sqrt {17}\right ) \sqrt {2+5 x^2+x^4}}+\frac {5 \sqrt {\frac {2}{5+\sqrt {17}}} \sqrt {2+5 x^2+x^4} E\left (\arctan \left (\sqrt {\frac {2}{5+\sqrt {17}}} x\right )|\frac {1}{4} \left (-17-5 \sqrt {17}\right )\right )}{17 \sqrt {\frac {4}{5+\sqrt {17}}+x^2} \sqrt {5+\sqrt {17}+2 x^2}}-\frac {\sqrt {4+\left (5-\sqrt {17}\right ) x^2} \sqrt {4+\left (5+\sqrt {17}\right ) x^2} \operatorname {EllipticF}\left (\arctan \left (\frac {1}{2} \sqrt {5-\sqrt {17}} x\right ),\frac {1}{4} \left (-17-5 \sqrt {17}\right )\right )}{17 \sqrt {5-\sqrt {17}} \sqrt {2+5 x^2+x^4}} \] Output:

2/17*x*17^(1/2)/(5-17^(1/2))/(x^4+5*x^2+2)^(1/2)+5/17*2^(1/2)/(5+17^(1/2)) 
^(1/2)*(x^4+5*x^2+2)^(1/2)*EllipticE(2^(1/2)/(5+17^(1/2))^(1/2)*x/(1+2/(5+ 
17^(1/2))*x^2)^(1/2),1/2*(-17-5*17^(1/2))^(1/2))/(4/(5+17^(1/2))+x^2)^(1/2 
)/(5+17^(1/2)+2*x^2)^(1/2)-1/17*(4+(5-17^(1/2))*x^2)^(1/2)*(4+(5+17^(1/2)) 
*x^2)^(1/2)*InverseJacobiAM(arctan(1/2*(5-17^(1/2))^(1/2)*x),1/2*(-17-5*17 
^(1/2))^(1/2))/(5-17^(1/2))^(1/2)/(x^4+5*x^2+2)^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 5.53 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.94 \[ \int \frac {1}{\left (2+5 x^2+x^4\right )^{3/2}} \, dx=\frac {4 x \left (21+5 x^2\right )-5 i \sqrt {2} \left (-5+\sqrt {17}\right ) \sqrt {\frac {-5+\sqrt {17}-2 x^2}{-5+\sqrt {17}}} \sqrt {5+\sqrt {17}+2 x^2} E\left (i \text {arcsinh}\left (\sqrt {\frac {2}{5+\sqrt {17}}} x\right )|\frac {21}{4}+\frac {5 \sqrt {17}}{4}\right )+i \sqrt {2} \left (-17+5 \sqrt {17}\right ) \sqrt {\frac {-5+\sqrt {17}-2 x^2}{-5+\sqrt {17}}} \sqrt {5+\sqrt {17}+2 x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {2}{5+\sqrt {17}}} x\right ),\frac {21}{4}+\frac {5 \sqrt {17}}{4}\right )}{136 \sqrt {2+5 x^2+x^4}} \] Input:

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

Output:

(4*x*(21 + 5*x^2) - (5*I)*Sqrt[2]*(-5 + Sqrt[17])*Sqrt[(-5 + Sqrt[17] - 2* 
x^2)/(-5 + Sqrt[17])]*Sqrt[5 + Sqrt[17] + 2*x^2]*EllipticE[I*ArcSinh[Sqrt[ 
2/(5 + Sqrt[17])]*x], 21/4 + (5*Sqrt[17])/4] + I*Sqrt[2]*(-17 + 5*Sqrt[17] 
)*Sqrt[(-5 + Sqrt[17] - 2*x^2)/(-5 + Sqrt[17])]*Sqrt[5 + Sqrt[17] + 2*x^2] 
*EllipticF[I*ArcSinh[Sqrt[2/(5 + Sqrt[17])]*x], 21/4 + (5*Sqrt[17])/4])/(1 
36*Sqrt[2 + 5*x^2 + x^4])
 

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.20, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1405, 1503, 1412, 1455}

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

\(\displaystyle \frac {x \left (5 x^2+21\right )}{34 \sqrt {x^4+5 x^2+2}}-\frac {1}{34} \int \frac {5 x^2+4}{\sqrt {x^4+5 x^2+2}}dx\)

\(\Big \downarrow \) 1503

\(\displaystyle \frac {1}{34} \left (-4 \int \frac {1}{\sqrt {x^4+5 x^2+2}}dx-5 \int \frac {x^2}{\sqrt {x^4+5 x^2+2}}dx\right )+\frac {x \left (5 x^2+21\right )}{34 \sqrt {x^4+5 x^2+2}}\)

\(\Big \downarrow \) 1412

\(\displaystyle \frac {1}{34} \left (-5 \int \frac {x^2}{\sqrt {x^4+5 x^2+2}}dx-\frac {2 \sqrt {\frac {\left (5-\sqrt {17}\right ) x^2+4}{\left (5+\sqrt {17}\right ) x^2+4}} \left (\left (5+\sqrt {17}\right ) x^2+4\right ) \operatorname {EllipticF}\left (\arctan \left (\frac {1}{2} \sqrt {5+\sqrt {17}} x\right ),\frac {1}{4} \left (-17+5 \sqrt {17}\right )\right )}{\sqrt {5+\sqrt {17}} \sqrt {x^4+5 x^2+2}}\right )+\frac {x \left (5 x^2+21\right )}{34 \sqrt {x^4+5 x^2+2}}\)

\(\Big \downarrow \) 1455

\(\displaystyle \frac {1}{34} \left (-\frac {2 \sqrt {\frac {\left (5-\sqrt {17}\right ) x^2+4}{\left (5+\sqrt {17}\right ) x^2+4}} \left (\left (5+\sqrt {17}\right ) x^2+4\right ) \operatorname {EllipticF}\left (\arctan \left (\frac {1}{2} \sqrt {5+\sqrt {17}} x\right ),\frac {1}{4} \left (-17+5 \sqrt {17}\right )\right )}{\sqrt {5+\sqrt {17}} \sqrt {x^4+5 x^2+2}}-5 \left (\frac {x \left (2 x^2+\sqrt {17}+5\right )}{2 \sqrt {x^4+5 x^2+2}}-\frac {\sqrt {5+\sqrt {17}} \sqrt {\frac {\left (5-\sqrt {17}\right ) x^2+4}{\left (5+\sqrt {17}\right ) x^2+4}} \left (\left (5+\sqrt {17}\right ) x^2+4\right ) E\left (\arctan \left (\frac {1}{2} \sqrt {5+\sqrt {17}} x\right )|\frac {1}{4} \left (-17+5 \sqrt {17}\right )\right )}{4 \sqrt {x^4+5 x^2+2}}\right )\right )+\frac {x \left (5 x^2+21\right )}{34 \sqrt {x^4+5 x^2+2}}\)

Input:

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

Output:

(x*(21 + 5*x^2))/(34*Sqrt[2 + 5*x^2 + x^4]) + (-5*((x*(5 + Sqrt[17] + 2*x^ 
2))/(2*Sqrt[2 + 5*x^2 + x^4]) - (Sqrt[5 + Sqrt[17]]*Sqrt[(4 + (5 - Sqrt[17 
])*x^2)/(4 + (5 + Sqrt[17])*x^2)]*(4 + (5 + Sqrt[17])*x^2)*EllipticE[ArcTa 
n[(Sqrt[5 + Sqrt[17]]*x)/2], (-17 + 5*Sqrt[17])/4])/(4*Sqrt[2 + 5*x^2 + x^ 
4])) - (2*Sqrt[(4 + (5 - Sqrt[17])*x^2)/(4 + (5 + Sqrt[17])*x^2)]*(4 + (5 
+ Sqrt[17])*x^2)*EllipticF[ArcTan[(Sqrt[5 + Sqrt[17]]*x)/2], (-17 + 5*Sqrt 
[17])/4])/(Sqrt[5 + Sqrt[17]]*Sqrt[2 + 5*x^2 + x^4]))/34
 

Defintions of rubi rules used

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 1412

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b 
^2 - 4*a*c, 2]}, Simp[(2*a + (b + q)*x^2)*(Sqrt[(2*a + (b - q)*x^2)/(2*a + 
(b + q)*x^2)]/(2*a*Rt[(b + q)/(2*a), 2]*Sqrt[a + b*x^2 + c*x^4]))*EllipticF 
[ArcTan[Rt[(b + q)/(2*a), 2]*x], 2*(q/(b + q))], x] /; PosQ[(b + q)/a] && 
!(PosQ[(b - q)/a] && SimplerSqrtQ[(b - q)/(2*a), (b + q)/(2*a)])] /; FreeQ[ 
{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]
 

rule 1455

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[b^2 - 4*a*c, 2]}, Simp[x*((b + q + 2*c*x^2)/(2*c*Sqrt[a + b*x^2 + c*x^4 
])), x] - Simp[Rt[(b + q)/(2*a), 2]*(2*a + (b + q)*x^2)*(Sqrt[(2*a + (b - q 
)*x^2)/(2*a + (b + q)*x^2)]/(2*c*Sqrt[a + b*x^2 + c*x^4]))*EllipticE[ArcTan 
[Rt[(b + q)/(2*a), 2]*x], 2*(q/(b + q))], x] /; PosQ[(b + q)/a] &&  !(PosQ[ 
(b - q)/a] && SimplerSqrtQ[(b - q)/(2*a), (b + q)/(2*a)])] /; FreeQ[{a, b, 
c}, x] && GtQ[b^2 - 4*a*c, 0]
 

rule 1503

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[d   Int[1/Sqrt[a + b*x^2 + c*x^4] 
, x], x] + Simp[e   Int[x^2/Sqrt[a + b*x^2 + c*x^4], x], x] /; PosQ[(b + q) 
/a] || PosQ[(b - q)/a]] /; FreeQ[{a, b, c, d, e}, x] && GtQ[b^2 - 4*a*c, 0]
 

Maple [A] (verified)

Time = 2.18 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.88

method	result	size
risch	\(\frac {x \left (5 x^{2}+21\right )}{34 \sqrt {x^{4}+5 x^{2}+2}}-\frac {4 \sqrt {1-\left (-\frac {5}{4}+\frac {\sqrt {17}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{4}-\frac {\sqrt {17}}{4}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-5+\sqrt {17}}}{2}, \frac {5 \sqrt {2}}{4}+\frac {\sqrt {34}}{4}\right )}{17 \sqrt {-5+\sqrt {17}}\, \sqrt {x^{4}+5 x^{2}+2}}+\frac {20 \sqrt {1-\left (-\frac {5}{4}+\frac {\sqrt {17}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{4}-\frac {\sqrt {17}}{4}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {-5+\sqrt {17}}}{2}, \frac {5 \sqrt {2}}{4}+\frac {\sqrt {34}}{4}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {-5+\sqrt {17}}}{2}, \frac {5 \sqrt {2}}{4}+\frac {\sqrt {34}}{4}\right )\right )}{17 \sqrt {-5+\sqrt {17}}\, \sqrt {x^{4}+5 x^{2}+2}\, \left (5+\sqrt {17}\right )}\)	\(206\)
default	\(-\frac {2 \left (-\frac {21}{68} x -\frac {5}{68} x^{3}\right )}{\sqrt {x^{4}+5 x^{2}+2}}-\frac {4 \sqrt {1-\left (-\frac {5}{4}+\frac {\sqrt {17}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{4}-\frac {\sqrt {17}}{4}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-5+\sqrt {17}}}{2}, \frac {5 \sqrt {2}}{4}+\frac {\sqrt {34}}{4}\right )}{17 \sqrt {-5+\sqrt {17}}\, \sqrt {x^{4}+5 x^{2}+2}}+\frac {20 \sqrt {1-\left (-\frac {5}{4}+\frac {\sqrt {17}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{4}-\frac {\sqrt {17}}{4}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {-5+\sqrt {17}}}{2}, \frac {5 \sqrt {2}}{4}+\frac {\sqrt {34}}{4}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {-5+\sqrt {17}}}{2}, \frac {5 \sqrt {2}}{4}+\frac {\sqrt {34}}{4}\right )\right )}{17 \sqrt {-5+\sqrt {17}}\, \sqrt {x^{4}+5 x^{2}+2}\, \left (5+\sqrt {17}\right )}\)	\(207\)
elliptic	\(-\frac {2 \left (-\frac {21}{68} x -\frac {5}{68} x^{3}\right )}{\sqrt {x^{4}+5 x^{2}+2}}-\frac {4 \sqrt {1-\left (-\frac {5}{4}+\frac {\sqrt {17}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{4}-\frac {\sqrt {17}}{4}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-5+\sqrt {17}}}{2}, \frac {5 \sqrt {2}}{4}+\frac {\sqrt {34}}{4}\right )}{17 \sqrt {-5+\sqrt {17}}\, \sqrt {x^{4}+5 x^{2}+2}}+\frac {20 \sqrt {1-\left (-\frac {5}{4}+\frac {\sqrt {17}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {5}{4}-\frac {\sqrt {17}}{4}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {-5+\sqrt {17}}}{2}, \frac {5 \sqrt {2}}{4}+\frac {\sqrt {34}}{4}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {-5+\sqrt {17}}}{2}, \frac {5 \sqrt {2}}{4}+\frac {\sqrt {34}}{4}\right )\right )}{17 \sqrt {-5+\sqrt {17}}\, \sqrt {x^{4}+5 x^{2}+2}\, \left (5+\sqrt {17}\right )}\)	\(207\)

Input:

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

Output:

1/34*x*(5*x^2+21)/(x^4+5*x^2+2)^(1/2)-4/17/(-5+17^(1/2))^(1/2)*(1-(-5/4+1/ 
4*17^(1/2))*x^2)^(1/2)*(1-(-5/4-1/4*17^(1/2))*x^2)^(1/2)/(x^4+5*x^2+2)^(1/ 
2)*EllipticF(1/2*x*(-5+17^(1/2))^(1/2),5/4*2^(1/2)+1/4*34^(1/2))+20/17/(-5 
+17^(1/2))^(1/2)*(1-(-5/4+1/4*17^(1/2))*x^2)^(1/2)*(1-(-5/4-1/4*17^(1/2))* 
x^2)^(1/2)/(x^4+5*x^2+2)^(1/2)/(5+17^(1/2))*(EllipticF(1/2*x*(-5+17^(1/2)) 
^(1/2),5/4*2^(1/2)+1/4*34^(1/2))-EllipticE(1/2*x*(-5+17^(1/2))^(1/2),5/4*2 
^(1/2)+1/4*34^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.68 \[ \int \frac {1}{\left (2+5 x^2+x^4\right )^{3/2}} \, dx=\frac {5 \, {\left (\sqrt {17} \sqrt {2} {\left (x^{4} + 5 \, x^{2} + 2\right )} - 5 \, \sqrt {2} {\left (x^{4} + 5 \, x^{2} + 2\right )}\right )} \sqrt {\sqrt {17} - 5} E(\arcsin \left (\frac {1}{2} \, x \sqrt {\sqrt {17} - 5}\right )\,|\,\frac {5}{4} \, \sqrt {17} + \frac {21}{4}) - {\left (\sqrt {17} \sqrt {2} {\left (x^{4} + 5 \, x^{2} + 2\right )} - 45 \, \sqrt {2} {\left (x^{4} + 5 \, x^{2} + 2\right )}\right )} \sqrt {\sqrt {17} - 5} F(\arcsin \left (\frac {1}{2} \, x \sqrt {\sqrt {17} - 5}\right )\,|\,\frac {5}{4} \, \sqrt {17} + \frac {21}{4}) + 8 \, \sqrt {x^{4} + 5 \, x^{2} + 2} {\left (5 \, x^{3} + 21 \, x\right )}}{272 \, {\left (x^{4} + 5 \, x^{2} + 2\right )}} \] Input:

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

Output:

1/272*(5*(sqrt(17)*sqrt(2)*(x^4 + 5*x^2 + 2) - 5*sqrt(2)*(x^4 + 5*x^2 + 2) 
)*sqrt(sqrt(17) - 5)*elliptic_e(arcsin(1/2*x*sqrt(sqrt(17) - 5)), 5/4*sqrt 
(17) + 21/4) - (sqrt(17)*sqrt(2)*(x^4 + 5*x^2 + 2) - 45*sqrt(2)*(x^4 + 5*x 
^2 + 2))*sqrt(sqrt(17) - 5)*elliptic_f(arcsin(1/2*x*sqrt(sqrt(17) - 5)), 5 
/4*sqrt(17) + 21/4) + 8*sqrt(x^4 + 5*x^2 + 2)*(5*x^3 + 21*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]

\[ \int \frac {1}{\left (2+5 x^2+x^4\right )^{3/2}} \, dx=\int \frac {\sqrt {x^{4}+5 x^{2}+2}}{x^{8}+10 x^{6}+29 x^{4}+20 x^{2}+4}d x \] 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 + 29*x**4 + 20*x**2 + 4),x)