\(\int \frac {1}{(2+5 x^2+2 x^4)^{3/2}} \, dx\) [300]

Optimal result

Integrand size = 16, antiderivative size = 122 \[ \int \frac {1}{\left (2+5 x^2+2 x^4\right )^{3/2}} \, dx=\frac {2 x}{3 \sqrt {2+5 x^2+2 x^4}}+\frac {5 \sqrt {2+5 x^2+2 x^4} E\left (\left .\arctan \left (\frac {x}{\sqrt {2}}\right )\right |-3\right )}{18 \sqrt {2+x^2} \sqrt {1+2 x^2}}-\frac {4 \sqrt {2+x^2} \sqrt {1+2 x^2} \operatorname {EllipticF}\left (\arctan \left (\frac {x}{\sqrt {2}}\right ),-3\right )}{9 \sqrt {2+5 x^2+2 x^4}} \] Output:

2/3*x/(2*x^4+5*x^2+2)^(1/2)+5/18*(2*x^4+5*x^2+2)^(1/2)*EllipticE(x*2^(1/2) 
/(2*x^2+4)^(1/2),I*3^(1/2))/(x^2+2)^(1/2)/(2*x^2+1)^(1/2)-4/9*(x^2+2)^(1/2 
)*(2*x^2+1)^(1/2)*InverseJacobiAM(arctan(1/2*x*2^(1/2)),I*3^(1/2))/(2*x^4+ 
5*x^2+2)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 4.67 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.89 \[ \int \frac {1}{\left (2+5 x^2+2 x^4\right )^{3/2}} \, dx=\frac {17 x+10 x^3+10 i \sqrt {2+x^2} \sqrt {1+2 x^2} E\left (i \text {arcsinh}\left (\sqrt {2} x\right )|\frac {1}{4}\right )-6 i \sqrt {2+x^2} \sqrt {1+2 x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {2} x\right ),\frac {1}{4}\right )}{18 \sqrt {2+5 x^2+2 x^4}} \] Input:

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

Output:

(17*x + 10*x^3 + (10*I)*Sqrt[2 + x^2]*Sqrt[1 + 2*x^2]*EllipticE[I*ArcSinh[ 
Sqrt[2]*x], 1/4] - (6*I)*Sqrt[2 + x^2]*Sqrt[1 + 2*x^2]*EllipticF[I*ArcSinh 
[Sqrt[2]*x], 1/4])/(18*Sqrt[2 + 5*x^2 + 2*x^4])
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.41, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {1405, 27, 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.

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

\(\Big \downarrow \) 1405

\(\displaystyle \frac {x \left (10 x^2+17\right )}{18 \sqrt {2 x^4+5 x^2+2}}-\frac {1}{18} \int \frac {2 \left (5 x^2+4\right )}{\sqrt {2 x^4+5 x^2+2}}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x \left (10 x^2+17\right )}{18 \sqrt {2 x^4+5 x^2+2}}-\frac {1}{9} \int \frac {5 x^2+4}{\sqrt {2 x^4+5 x^2+2}}dx\)

\(\Big \downarrow \) 1503

\(\displaystyle \frac {1}{9} \left (-4 \int \frac {1}{\sqrt {2 x^4+5 x^2+2}}dx-5 \int \frac {x^2}{\sqrt {2 x^4+5 x^2+2}}dx\right )+\frac {x \left (10 x^2+17\right )}{18 \sqrt {2 x^4+5 x^2+2}}\)

\(\Big \downarrow \) 1412

\(\displaystyle \frac {1}{9} \left (-5 \int \frac {x^2}{\sqrt {2 x^4+5 x^2+2}}dx-\frac {2 \sqrt {\frac {x^2+2}{2 x^2+1}} \left (2 x^2+1\right ) \operatorname {EllipticF}\left (\arctan \left (\sqrt {2} x\right ),\frac {3}{4}\right )}{\sqrt {2 x^4+5 x^2+2}}\right )+\frac {x \left (10 x^2+17\right )}{18 \sqrt {2 x^4+5 x^2+2}}\)

\(\Big \downarrow \) 1455

\(\displaystyle \frac {1}{9} \left (-\frac {2 \sqrt {\frac {x^2+2}{2 x^2+1}} \left (2 x^2+1\right ) \operatorname {EllipticF}\left (\arctan \left (\sqrt {2} x\right ),\frac {3}{4}\right )}{\sqrt {2 x^4+5 x^2+2}}-5 \left (\frac {x \left (x^2+2\right )}{\sqrt {2 x^4+5 x^2+2}}-\frac {\sqrt {\frac {x^2+2}{2 x^2+1}} \left (2 x^2+1\right ) E\left (\arctan \left (\sqrt {2} x\right )|\frac {3}{4}\right )}{\sqrt {2 x^4+5 x^2+2}}\right )\right )+\frac {x \left (10 x^2+17\right )}{18 \sqrt {2 x^4+5 x^2+2}}\)

Input:

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

Output:

(x*(17 + 10*x^2))/(18*Sqrt[2 + 5*x^2 + 2*x^4]) + (-5*((x*(2 + x^2))/Sqrt[2 
 + 5*x^2 + 2*x^4] - (Sqrt[(2 + x^2)/(1 + 2*x^2)]*(1 + 2*x^2)*EllipticE[Arc 
Tan[Sqrt[2]*x], 3/4])/Sqrt[2 + 5*x^2 + 2*x^4]) - (2*Sqrt[(2 + x^2)/(1 + 2* 
x^2)]*(1 + 2*x^2)*EllipticF[ArcTan[Sqrt[2]*x], 3/4])/Sqrt[2 + 5*x^2 + 2*x^ 
4])/9
 

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 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 = 1.78 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.08

method	result	size
risch	\(\frac {x \left (10 x^{2}+17\right )}{18 \sqrt {2 x^{4}+5 x^{2}+2}}+\frac {2 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {2 x^{2}+1}\, \operatorname {EllipticF}\left (\frac {i x \sqrt {2}}{2}, 2\right )}{9 \sqrt {2 x^{4}+5 x^{2}+2}}-\frac {5 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {2 x^{2}+1}\, \left (\operatorname {EllipticF}\left (\frac {i x \sqrt {2}}{2}, 2\right )-\operatorname {EllipticE}\left (\frac {i x \sqrt {2}}{2}, 2\right )\right )}{36 \sqrt {2 x^{4}+5 x^{2}+2}}\)	\(132\)
default	\(-\frac {4 \left (-\frac {17}{72} x -\frac {5}{36} x^{3}\right )}{\sqrt {2 x^{4}+5 x^{2}+2}}+\frac {2 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {2 x^{2}+1}\, \operatorname {EllipticF}\left (\frac {i x \sqrt {2}}{2}, 2\right )}{9 \sqrt {2 x^{4}+5 x^{2}+2}}-\frac {5 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {2 x^{2}+1}\, \left (\operatorname {EllipticF}\left (\frac {i x \sqrt {2}}{2}, 2\right )-\operatorname {EllipticE}\left (\frac {i x \sqrt {2}}{2}, 2\right )\right )}{36 \sqrt {2 x^{4}+5 x^{2}+2}}\)	\(133\)
elliptic	\(-\frac {4 \left (-\frac {17}{72} x -\frac {5}{36} x^{3}\right )}{\sqrt {2 x^{4}+5 x^{2}+2}}+\frac {2 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {2 x^{2}+1}\, \operatorname {EllipticF}\left (\frac {i x \sqrt {2}}{2}, 2\right )}{9 \sqrt {2 x^{4}+5 x^{2}+2}}-\frac {5 i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {2 x^{2}+1}\, \left (\operatorname {EllipticF}\left (\frac {i x \sqrt {2}}{2}, 2\right )-\operatorname {EllipticE}\left (\frac {i x \sqrt {2}}{2}, 2\right )\right )}{36 \sqrt {2 x^{4}+5 x^{2}+2}}\)	\(133\)

Input:

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

Output:

1/18*x*(10*x^2+17)/(2*x^4+5*x^2+2)^(1/2)+2/9*I*2^(1/2)*(2*x^2+4)^(1/2)*(2* 
x^2+1)^(1/2)/(2*x^4+5*x^2+2)^(1/2)*EllipticF(1/2*I*x*2^(1/2),2)-5/36*I*2^( 
1/2)*(2*x^2+4)^(1/2)*(2*x^2+1)^(1/2)/(2*x^4+5*x^2+2)^(1/2)*(EllipticF(1/2* 
I*x*2^(1/2),2)-EllipticE(1/2*I*x*2^(1/2),2))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.80 \[ \int \frac {1}{\left (2+5 x^2+2 x^4\right )^{3/2}} \, dx=-\frac {5 \, \sqrt {2} \sqrt {-\frac {1}{2}} {\left (2 \, x^{4} + 5 \, x^{2} + 2\right )} E(\arcsin \left (\sqrt {-\frac {1}{2}} x\right )\,|\,4) - 21 \, \sqrt {2} \sqrt {-\frac {1}{2}} {\left (2 \, x^{4} + 5 \, x^{2} + 2\right )} F(\arcsin \left (\sqrt {-\frac {1}{2}} x\right )\,|\,4) - 2 \, \sqrt {2 \, x^{4} + 5 \, x^{2} + 2} {\left (10 \, x^{3} + 17 \, x\right )}}{36 \, {\left (2 \, x^{4} + 5 \, x^{2} + 2\right )}} \] Input:

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

Output:

-1/36*(5*sqrt(2)*sqrt(-1/2)*(2*x^4 + 5*x^2 + 2)*elliptic_e(arcsin(sqrt(-1/ 
2)*x), 4) - 21*sqrt(2)*sqrt(-1/2)*(2*x^4 + 5*x^2 + 2)*elliptic_f(arcsin(sq 
rt(-1/2)*x), 4) - 2*sqrt(2*x^4 + 5*x^2 + 2)*(10*x^3 + 17*x))/(2*x^4 + 5*x^ 
2 + 2)
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\left (2+5 x^2+2 x^4\right )^{3/2}} \, dx=\int \frac {\sqrt {2 x^{4}+5 x^{2}+2}}{4 x^{8}+20 x^{6}+33 x^{4}+20 x^{2}+4}d x \] Input:

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

Output:

int(sqrt(2*x**4 + 5*x**2 + 2)/(4*x**8 + 20*x**6 + 33*x**4 + 20*x**2 + 4),x 
)