\(\int \frac {1}{(3+8 x^2+2 x^4)^{3/2}} \, dx\) [246]

Optimal result

Integrand size = 16, antiderivative size = 239 \[ \int \frac {1}{\left (3+8 x^2+2 x^4\right )^{3/2}} \, dx=\frac {x}{\sqrt {10} \left (4-\sqrt {10}\right ) \sqrt {3+8 x^2+2 x^4}}+\frac {2 \sqrt {3+8 x^2+2 x^4} E\left (\arctan \left (\sqrt {\frac {2}{4+\sqrt {10}}} x\right )|-\frac {2}{3} \left (5+2 \sqrt {10}\right )\right )}{5 \sqrt {3 \left (4+\sqrt {10}\right )} \sqrt {4-\sqrt {10}+2 x^2} \sqrt {4+\sqrt {10}+2 x^2}}-\frac {\sqrt {3+\left (4-\sqrt {10}\right ) x^2} \sqrt {3+\left (4+\sqrt {10}\right ) x^2} \operatorname {EllipticF}\left (\arctan \left (\sqrt {\frac {1}{3} \left (4-\sqrt {10}\right )} x\right ),-\frac {2}{3} \left (5+2 \sqrt {10}\right )\right )}{10 \sqrt {3 \left (4-\sqrt {10}\right )} \sqrt {3+8 x^2+2 x^4}} \] Output:

1/10*x*10^(1/2)/(4-10^(1/2))/(2*x^4+8*x^2+3)^(1/2)+2/5*(2*x^4+8*x^2+3)^(1/ 
2)*EllipticE(2^(1/2)/(4+10^(1/2))^(1/2)*x/(1+2/(4+10^(1/2))*x^2)^(1/2),1/3 
*(-30-12*10^(1/2))^(1/2))/(12+3*10^(1/2))^(1/2)/(4-10^(1/2)+2*x^2)^(1/2)/( 
4+10^(1/2)+2*x^2)^(1/2)-1/10*(3+(4-10^(1/2))*x^2)^(1/2)*(3+(4+10^(1/2))*x^ 
2)^(1/2)*InverseJacobiAM(arctan(1/3*(12-3*10^(1/2))^(1/2)*x),1/3*(-30-12*1 
0^(1/2))^(1/2))/(12-3*10^(1/2))^(1/2)/(2*x^4+8*x^2+3)^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 6.22 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.93 \[ \int \frac {1}{\left (3+8 x^2+2 x^4\right )^{3/2}} \, dx=\frac {26 x+8 x^3+4 i \left (2 \sqrt {2}-\sqrt {5}\right ) \sqrt {\frac {-4+\sqrt {10}-2 x^2}{-4+\sqrt {10}}} \sqrt {4+\sqrt {10}+2 x^2} E\left (i \text {arcsinh}\left (\sqrt {\frac {2}{4+\sqrt {10}}} x\right )|\frac {13}{3}+\frac {4 \sqrt {10}}{3}\right )-i \left (5 \sqrt {2}-4 \sqrt {5}\right ) \sqrt {\frac {-4+\sqrt {10}-2 x^2}{-4+\sqrt {10}}} \sqrt {4+\sqrt {10}+2 x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {2}{4+\sqrt {10}}} x\right ),\frac {13}{3}+\frac {4 \sqrt {10}}{3}\right )}{60 \sqrt {3+8 x^2+2 x^4}} \] Input:

Integrate[(3 + 8*x^2 + 2*x^4)^(-3/2),x]
 

Output:

(26*x + 8*x^3 + (4*I)*(2*Sqrt[2] - Sqrt[5])*Sqrt[(-4 + Sqrt[10] - 2*x^2)/( 
-4 + Sqrt[10])]*Sqrt[4 + Sqrt[10] + 2*x^2]*EllipticE[I*ArcSinh[Sqrt[2/(4 + 
 Sqrt[10])]*x], 13/3 + (4*Sqrt[10])/3] - I*(5*Sqrt[2] - 4*Sqrt[5])*Sqrt[(- 
4 + Sqrt[10] - 2*x^2)/(-4 + Sqrt[10])]*Sqrt[4 + Sqrt[10] + 2*x^2]*Elliptic 
F[I*ArcSinh[Sqrt[2/(4 + Sqrt[10])]*x], 13/3 + (4*Sqrt[10])/3])/(60*Sqrt[3 
+ 8*x^2 + 2*x^4])
 

Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.25, 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.

\(\displaystyle \int \frac {1}{\left (2 x^4+8 x^2+3\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1405

\(\displaystyle \frac {x \left (4 x^2+13\right )}{30 \sqrt {2 x^4+8 x^2+3}}-\frac {1}{120} \int \frac {4 \left (4 x^2+3\right )}{\sqrt {2 x^4+8 x^2+3}}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x \left (4 x^2+13\right )}{30 \sqrt {2 x^4+8 x^2+3}}-\frac {1}{30} \int \frac {4 x^2+3}{\sqrt {2 x^4+8 x^2+3}}dx\)

\(\Big \downarrow \) 1503

\(\displaystyle \frac {1}{30} \left (-3 \int \frac {1}{\sqrt {2 x^4+8 x^2+3}}dx-4 \int \frac {x^2}{\sqrt {2 x^4+8 x^2+3}}dx\right )+\frac {x \left (4 x^2+13\right )}{30 \sqrt {2 x^4+8 x^2+3}}\)

\(\Big \downarrow \) 1412

\(\displaystyle \frac {1}{30} \left (-4 \int \frac {x^2}{\sqrt {2 x^4+8 x^2+3}}dx-\frac {\sqrt {\frac {3}{4+\sqrt {10}}} \sqrt {\frac {\left (4-\sqrt {10}\right ) x^2+3}{\left (4+\sqrt {10}\right ) x^2+3}} \left (\left (4+\sqrt {10}\right ) x^2+3\right ) \operatorname {EllipticF}\left (\arctan \left (\sqrt {\frac {1}{3} \left (4+\sqrt {10}\right )} x\right ),-\frac {2}{3} \left (5-2 \sqrt {10}\right )\right )}{\sqrt {2 x^4+8 x^2+3}}\right )+\frac {x \left (4 x^2+13\right )}{30 \sqrt {2 x^4+8 x^2+3}}\)

\(\Big \downarrow \) 1455

\(\displaystyle \frac {1}{30} \left (-\frac {\sqrt {\frac {3}{4+\sqrt {10}}} \sqrt {\frac {\left (4-\sqrt {10}\right ) x^2+3}{\left (4+\sqrt {10}\right ) x^2+3}} \left (\left (4+\sqrt {10}\right ) x^2+3\right ) \operatorname {EllipticF}\left (\arctan \left (\sqrt {\frac {1}{3} \left (4+\sqrt {10}\right )} x\right ),-\frac {2}{3} \left (5-2 \sqrt {10}\right )\right )}{\sqrt {2 x^4+8 x^2+3}}-4 \left (\frac {x \left (2 x^2+\sqrt {10}+4\right )}{2 \sqrt {2 x^4+8 x^2+3}}-\frac {\sqrt {\frac {1}{3} \left (4+\sqrt {10}\right )} \sqrt {\frac {\left (4-\sqrt {10}\right ) x^2+3}{\left (4+\sqrt {10}\right ) x^2+3}} \left (\left (4+\sqrt {10}\right ) x^2+3\right ) E\left (\arctan \left (\sqrt {\frac {1}{3} \left (4+\sqrt {10}\right )} x\right )|-\frac {2}{3} \left (5-2 \sqrt {10}\right )\right )}{2 \sqrt {2 x^4+8 x^2+3}}\right )\right )+\frac {x \left (4 x^2+13\right )}{30 \sqrt {2 x^4+8 x^2+3}}\)

Input:

Int[(3 + 8*x^2 + 2*x^4)^(-3/2),x]
 

Output:

(x*(13 + 4*x^2))/(30*Sqrt[3 + 8*x^2 + 2*x^4]) + (-4*((x*(4 + Sqrt[10] + 2* 
x^2))/(2*Sqrt[3 + 8*x^2 + 2*x^4]) - (Sqrt[(4 + Sqrt[10])/3]*Sqrt[(3 + (4 - 
 Sqrt[10])*x^2)/(3 + (4 + Sqrt[10])*x^2)]*(3 + (4 + Sqrt[10])*x^2)*Ellipti 
cE[ArcTan[Sqrt[(4 + Sqrt[10])/3]*x], (-2*(5 - 2*Sqrt[10]))/3])/(2*Sqrt[3 + 
 8*x^2 + 2*x^4])) - (Sqrt[3/(4 + Sqrt[10])]*Sqrt[(3 + (4 - Sqrt[10])*x^2)/ 
(3 + (4 + Sqrt[10])*x^2)]*(3 + (4 + Sqrt[10])*x^2)*EllipticF[ArcTan[Sqrt[( 
4 + Sqrt[10])/3]*x], (-2*(5 - 2*Sqrt[10]))/3])/Sqrt[3 + 8*x^2 + 2*x^4])/30
 

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 = 2.60 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.94

method	result	size
risch	\(\frac {x \left (4 x^{2}+13\right )}{30 \sqrt {2 x^{4}+8 x^{2}+3}}-\frac {3 \sqrt {1-\left (-\frac {4}{3}+\frac {\sqrt {10}}{3}\right ) x^{2}}\, \sqrt {1-\left (-\frac {4}{3}-\frac {\sqrt {10}}{3}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-12+3 \sqrt {10}}}{3}, \frac {2 \sqrt {6}}{3}+\frac {\sqrt {15}}{3}\right )}{10 \sqrt {-12+3 \sqrt {10}}\, \sqrt {2 x^{4}+8 x^{2}+3}}+\frac {12 \sqrt {1-\left (-\frac {4}{3}+\frac {\sqrt {10}}{3}\right ) x^{2}}\, \sqrt {1-\left (-\frac {4}{3}-\frac {\sqrt {10}}{3}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {-12+3 \sqrt {10}}}{3}, \frac {2 \sqrt {6}}{3}+\frac {\sqrt {15}}{3}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {-12+3 \sqrt {10}}}{3}, \frac {2 \sqrt {6}}{3}+\frac {\sqrt {15}}{3}\right )\right )}{5 \sqrt {-12+3 \sqrt {10}}\, \sqrt {2 x^{4}+8 x^{2}+3}\, \left (8+2 \sqrt {10}\right )}\)	\(224\)
default	\(-\frac {4 \left (-\frac {13}{120} x -\frac {1}{30} x^{3}\right )}{\sqrt {2 x^{4}+8 x^{2}+3}}-\frac {3 \sqrt {1-\left (-\frac {4}{3}+\frac {\sqrt {10}}{3}\right ) x^{2}}\, \sqrt {1-\left (-\frac {4}{3}-\frac {\sqrt {10}}{3}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-12+3 \sqrt {10}}}{3}, \frac {2 \sqrt {6}}{3}+\frac {\sqrt {15}}{3}\right )}{10 \sqrt {-12+3 \sqrt {10}}\, \sqrt {2 x^{4}+8 x^{2}+3}}+\frac {12 \sqrt {1-\left (-\frac {4}{3}+\frac {\sqrt {10}}{3}\right ) x^{2}}\, \sqrt {1-\left (-\frac {4}{3}-\frac {\sqrt {10}}{3}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {-12+3 \sqrt {10}}}{3}, \frac {2 \sqrt {6}}{3}+\frac {\sqrt {15}}{3}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {-12+3 \sqrt {10}}}{3}, \frac {2 \sqrt {6}}{3}+\frac {\sqrt {15}}{3}\right )\right )}{5 \sqrt {-12+3 \sqrt {10}}\, \sqrt {2 x^{4}+8 x^{2}+3}\, \left (8+2 \sqrt {10}\right )}\)	\(225\)
elliptic	\(-\frac {4 \left (-\frac {13}{120} x -\frac {1}{30} x^{3}\right )}{\sqrt {2 x^{4}+8 x^{2}+3}}-\frac {3 \sqrt {1-\left (-\frac {4}{3}+\frac {\sqrt {10}}{3}\right ) x^{2}}\, \sqrt {1-\left (-\frac {4}{3}-\frac {\sqrt {10}}{3}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-12+3 \sqrt {10}}}{3}, \frac {2 \sqrt {6}}{3}+\frac {\sqrt {15}}{3}\right )}{10 \sqrt {-12+3 \sqrt {10}}\, \sqrt {2 x^{4}+8 x^{2}+3}}+\frac {12 \sqrt {1-\left (-\frac {4}{3}+\frac {\sqrt {10}}{3}\right ) x^{2}}\, \sqrt {1-\left (-\frac {4}{3}-\frac {\sqrt {10}}{3}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {-12+3 \sqrt {10}}}{3}, \frac {2 \sqrt {6}}{3}+\frac {\sqrt {15}}{3}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {-12+3 \sqrt {10}}}{3}, \frac {2 \sqrt {6}}{3}+\frac {\sqrt {15}}{3}\right )\right )}{5 \sqrt {-12+3 \sqrt {10}}\, \sqrt {2 x^{4}+8 x^{2}+3}\, \left (8+2 \sqrt {10}\right )}\)	\(225\)

Input:

int(1/(2*x^4+8*x^2+3)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/30*x*(4*x^2+13)/(2*x^4+8*x^2+3)^(1/2)-3/10/(-12+3*10^(1/2))^(1/2)*(1-(-4 
/3+1/3*10^(1/2))*x^2)^(1/2)*(1-(-4/3-1/3*10^(1/2))*x^2)^(1/2)/(2*x^4+8*x^2 
+3)^(1/2)*EllipticF(1/3*x*(-12+3*10^(1/2))^(1/2),2/3*6^(1/2)+1/3*15^(1/2)) 
+12/5/(-12+3*10^(1/2))^(1/2)*(1-(-4/3+1/3*10^(1/2))*x^2)^(1/2)*(1-(-4/3-1/ 
3*10^(1/2))*x^2)^(1/2)/(2*x^4+8*x^2+3)^(1/2)/(8+2*10^(1/2))*(EllipticF(1/3 
*x*(-12+3*10^(1/2))^(1/2),2/3*6^(1/2)+1/3*15^(1/2))-EllipticE(1/3*x*(-12+3 
*10^(1/2))^(1/2),2/3*6^(1/2)+1/3*15^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.74 \[ \int \frac {1}{\left (3+8 x^2+2 x^4\right )^{3/2}} \, dx=\frac {4 \, {\left (\sqrt {10} \sqrt {3} {\left (2 \, x^{4} + 8 \, x^{2} + 3\right )} - 4 \, \sqrt {3} {\left (2 \, x^{4} + 8 \, x^{2} + 3\right )}\right )} \sqrt {\frac {1}{3} \, \sqrt {10} - \frac {4}{3}} E(\arcsin \left (x \sqrt {\frac {1}{3} \, \sqrt {10} - \frac {4}{3}}\right )\,|\,\frac {4}{3} \, \sqrt {10} + \frac {13}{3}) - {\left (\sqrt {10} \sqrt {3} {\left (2 \, x^{4} + 8 \, x^{2} + 3\right )} - 28 \, \sqrt {3} {\left (2 \, x^{4} + 8 \, x^{2} + 3\right )}\right )} \sqrt {\frac {1}{3} \, \sqrt {10} - \frac {4}{3}} F(\arcsin \left (x \sqrt {\frac {1}{3} \, \sqrt {10} - \frac {4}{3}}\right )\,|\,\frac {4}{3} \, \sqrt {10} + \frac {13}{3}) + 6 \, \sqrt {2 \, x^{4} + 8 \, x^{2} + 3} {\left (4 \, x^{3} + 13 \, x\right )}}{180 \, {\left (2 \, x^{4} + 8 \, x^{2} + 3\right )}} \] Input:

integrate(1/(2*x^4+8*x^2+3)^(3/2),x, algorithm="fricas")
 

Output:

1/180*(4*(sqrt(10)*sqrt(3)*(2*x^4 + 8*x^2 + 3) - 4*sqrt(3)*(2*x^4 + 8*x^2 
+ 3))*sqrt(1/3*sqrt(10) - 4/3)*elliptic_e(arcsin(x*sqrt(1/3*sqrt(10) - 4/3 
)), 4/3*sqrt(10) + 13/3) - (sqrt(10)*sqrt(3)*(2*x^4 + 8*x^2 + 3) - 28*sqrt 
(3)*(2*x^4 + 8*x^2 + 3))*sqrt(1/3*sqrt(10) - 4/3)*elliptic_f(arcsin(x*sqrt 
(1/3*sqrt(10) - 4/3)), 4/3*sqrt(10) + 13/3) + 6*sqrt(2*x^4 + 8*x^2 + 3)*(4 
*x^3 + 13*x))/(2*x^4 + 8*x^2 + 3)
                                                                                    
                                                                                    
 

Sympy [F]

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

integrate(1/(2*x**4+8*x**2+3)**(3/2),x)
 

Output:

Integral((2*x**4 + 8*x**2 + 3)**(-3/2), x)
 

Maxima [F]

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

integrate(1/(2*x^4+8*x^2+3)^(3/2),x, algorithm="maxima")
 

Output:

integrate((2*x^4 + 8*x^2 + 3)^(-3/2), x)
 

Giac [F]

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

integrate(1/(2*x^4+8*x^2+3)^(3/2),x, algorithm="giac")
 

Output:

integrate((2*x^4 + 8*x^2 + 3)^(-3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (3+8 x^2+2 x^4\right )^{3/2}} \, dx=\int \frac {1}{{\left (2\,x^4+8\,x^2+3\right )}^{3/2}} \,d x \] Input:

int(1/(8*x^2 + 2*x^4 + 3)^(3/2),x)
 

Output:

int(1/(8*x^2 + 2*x^4 + 3)^(3/2), x)
 

Reduce [F]

\[ \int \frac {1}{\left (3+8 x^2+2 x^4\right )^{3/2}} \, dx=\int \frac {\sqrt {2 x^{4}+8 x^{2}+3}}{4 x^{8}+32 x^{6}+76 x^{4}+48 x^{2}+9}d x \] Input:

int(1/(2*x^4+8*x^2+3)^(3/2),x)
 

Output:

int(sqrt(2*x**4 + 8*x**2 + 3)/(4*x**8 + 32*x**6 + 76*x**4 + 48*x**2 + 9),x 
)