\(\int \frac {1}{(3+2 x^2+2 x^4)^{3/2}} \, dx\) [252]

Optimal result

Integrand size = 16, antiderivative size = 254 \[ \int \frac {1}{\left (3+2 x^2+2 x^4\right )^{3/2}} \, dx=\frac {x \left (2-x^2\right )}{15 \sqrt {3+2 x^2+2 x^4}}+\frac {x \sqrt {3+2 x^2+2 x^4}}{15 \left (\sqrt {6}+2 x^2\right )}-\frac {\left (3+\sqrt {6} x^2\right ) \sqrt {\frac {3+2 x^2+2 x^4}{\left (3+\sqrt {6} x^2\right )^2}} E\left (2 \arctan \left (\sqrt [4]{\frac {2}{3}} x\right )|\frac {1}{12} \left (6-\sqrt {6}\right )\right )}{5\ 6^{3/4} \sqrt {3+2 x^2+2 x^4}}+\frac {\left (1+\sqrt {6}\right ) \left (3+\sqrt {6} x^2\right ) \sqrt {\frac {3+2 x^2+2 x^4}{\left (3+\sqrt {6} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{\frac {2}{3}} x\right ),\frac {1}{12} \left (6-\sqrt {6}\right )\right )}{10\ 6^{3/4} \sqrt {3+2 x^2+2 x^4}} \] Output:

1/15*x*(-x^2+2)/(2*x^4+2*x^2+3)^(1/2)+x*(2*x^4+2*x^2+3)^(1/2)/(15*6^(1/2)+ 
30*x^2)-1/30*(3+6^(1/2)*x^2)*((2*x^4+2*x^2+3)/(3+6^(1/2)*x^2)^2)^(1/2)*Ell 
ipticE(sin(2*arctan(1/3*2^(1/4)*3^(3/4)*x)),1/6*(18-3*6^(1/2))^(1/2))*6^(1 
/4)/(2*x^4+2*x^2+3)^(1/2)+1/60*(1+6^(1/2))*(3+6^(1/2)*x^2)*((2*x^4+2*x^2+3 
)/(3+6^(1/2)*x^2)^2)^(1/2)*InverseJacobiAM(2*arctan(1/3*2^(1/4)*3^(3/4)*x) 
,1/6*(18-3*6^(1/2))^(1/2))*6^(1/4)/(2*x^4+2*x^2+3)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 5.91 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.28 \[ \int \frac {1}{\left (3+2 x^2+2 x^4\right )^{3/2}} \, dx=\frac {-4 \sqrt {-\frac {i}{-i+\sqrt {5}}} x \left (-2+x^2\right )-\sqrt {2} \left (i+\sqrt {5}\right ) \sqrt {\frac {-i+\sqrt {5}-2 i x^2}{-i+\sqrt {5}}} \sqrt {\frac {i+\sqrt {5}+2 i x^2}{i+\sqrt {5}}} E\left (i \text {arcsinh}\left (\sqrt {-\frac {2 i}{-i+\sqrt {5}}} x\right )|\frac {i-\sqrt {5}}{i+\sqrt {5}}\right )+\sqrt {2} \left (-5 i+\sqrt {5}\right ) \sqrt {\frac {-i+\sqrt {5}-2 i x^2}{-i+\sqrt {5}}} \sqrt {\frac {i+\sqrt {5}+2 i x^2}{i+\sqrt {5}}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {2 i}{-i+\sqrt {5}}} x\right ),\frac {i-\sqrt {5}}{i+\sqrt {5}}\right )}{60 \sqrt {-\frac {i}{-i+\sqrt {5}}} \sqrt {3+2 x^2+2 x^4}} \] Input:

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

Output:

(-4*Sqrt[(-I)/(-I + Sqrt[5])]*x*(-2 + x^2) - Sqrt[2]*(I + Sqrt[5])*Sqrt[(- 
I + Sqrt[5] - (2*I)*x^2)/(-I + Sqrt[5])]*Sqrt[(I + Sqrt[5] + (2*I)*x^2)/(I 
 + Sqrt[5])]*EllipticE[I*ArcSinh[Sqrt[(-2*I)/(-I + Sqrt[5])]*x], (I - Sqrt 
[5])/(I + Sqrt[5])] + Sqrt[2]*(-5*I + Sqrt[5])*Sqrt[(-I + Sqrt[5] - (2*I)* 
x^2)/(-I + Sqrt[5])]*Sqrt[(I + Sqrt[5] + (2*I)*x^2)/(I + Sqrt[5])]*Ellipti 
cF[I*ArcSinh[Sqrt[(-2*I)/(-I + Sqrt[5])]*x], (I - Sqrt[5])/(I + Sqrt[5])]) 
/(60*Sqrt[(-I)/(-I + Sqrt[5])]*Sqrt[3 + 2*x^2 + 2*x^4])
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {1405, 27, 1511, 27, 1416, 1509}

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.

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

\(\Big \downarrow \) 1405

\(\displaystyle \frac {1}{60} \int \frac {4 \left (x^2+3\right )}{\sqrt {2 x^4+2 x^2+3}}dx+\frac {x \left (2-x^2\right )}{15 \sqrt {2 x^4+2 x^2+3}}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1511

\(\displaystyle \frac {1}{15} \left (\frac {1}{2} \left (6+\sqrt {6}\right ) \int \frac {1}{\sqrt {2 x^4+2 x^2+3}}dx-\sqrt {\frac {3}{2}} \int \frac {3-\sqrt {6} x^2}{3 \sqrt {2 x^4+2 x^2+3}}dx\right )+\frac {x \left (2-x^2\right )}{15 \sqrt {2 x^4+2 x^2+3}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{15} \left (\frac {1}{2} \left (6+\sqrt {6}\right ) \int \frac {1}{\sqrt {2 x^4+2 x^2+3}}dx-\frac {\int \frac {3-\sqrt {6} x^2}{\sqrt {2 x^4+2 x^2+3}}dx}{\sqrt {6}}\right )+\frac {x \left (2-x^2\right )}{15 \sqrt {2 x^4+2 x^2+3}}\)

\(\Big \downarrow \) 1416

\(\displaystyle \frac {1}{15} \left (\frac {\left (6+\sqrt {6}\right ) \left (\sqrt {6} x^2+3\right ) \sqrt {\frac {2 x^4+2 x^2+3}{\left (\sqrt {6} x^2+3\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{\frac {2}{3}} x\right ),\frac {1}{12} \left (6-\sqrt {6}\right )\right )}{4 \sqrt [4]{6} \sqrt {2 x^4+2 x^2+3}}-\frac {\int \frac {3-\sqrt {6} x^2}{\sqrt {2 x^4+2 x^2+3}}dx}{\sqrt {6}}\right )+\frac {x \left (2-x^2\right )}{15 \sqrt {2 x^4+2 x^2+3}}\)

\(\Big \downarrow \) 1509

\(\displaystyle \frac {1}{15} \left (\frac {\left (6+\sqrt {6}\right ) \left (\sqrt {6} x^2+3\right ) \sqrt {\frac {2 x^4+2 x^2+3}{\left (\sqrt {6} x^2+3\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{\frac {2}{3}} x\right ),\frac {1}{12} \left (6-\sqrt {6}\right )\right )}{4 \sqrt [4]{6} \sqrt {2 x^4+2 x^2+3}}-\frac {\frac {3^{3/4} \left (\sqrt {6} x^2+3\right ) \sqrt {\frac {2 x^4+2 x^2+3}{\left (\sqrt {6} x^2+3\right )^2}} E\left (2 \arctan \left (\sqrt [4]{\frac {2}{3}} x\right )|\frac {1}{12} \left (6-\sqrt {6}\right )\right )}{\sqrt [4]{2} \sqrt {2 x^4+2 x^2+3}}-\frac {3 x \sqrt {2 x^4+2 x^2+3}}{\sqrt {6} x^2+3}}{\sqrt {6}}\right )+\frac {x \left (2-x^2\right )}{15 \sqrt {2 x^4+2 x^2+3}}\)

Input:

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

Output:

(x*(2 - x^2))/(15*Sqrt[3 + 2*x^2 + 2*x^4]) + (-(((-3*x*Sqrt[3 + 2*x^2 + 2* 
x^4])/(3 + Sqrt[6]*x^2) + (3^(3/4)*(3 + Sqrt[6]*x^2)*Sqrt[(3 + 2*x^2 + 2*x 
^4)/(3 + Sqrt[6]*x^2)^2]*EllipticE[2*ArcTan[(2/3)^(1/4)*x], (6 - Sqrt[6])/ 
12])/(2^(1/4)*Sqrt[3 + 2*x^2 + 2*x^4]))/Sqrt[6]) + ((6 + Sqrt[6])*(3 + Sqr 
t[6]*x^2)*Sqrt[(3 + 2*x^2 + 2*x^4)/(3 + Sqrt[6]*x^2)^2]*EllipticF[2*ArcTan 
[(2/3)^(1/4)*x], (6 - Sqrt[6])/12])/(4*6^(1/4)*Sqrt[3 + 2*x^2 + 2*x^4]))/1 
5
 

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 1416

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

rule 1509

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo 
l] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + b*x^2 + c*x^4]/(a*(1 + q 
^2*x^2))), x] + Simp[d*(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q^2* 
x^2)^2)]/(q*Sqrt[a + b*x^2 + c*x^4]))*EllipticE[2*ArcTan[q*x], 1/2 - b*(q^2 
/(4*c))], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 
- 4*a*c, 0] && PosQ[c/a]
 

rule 1511

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo 
l] :> With[{q = Rt[c/a, 2]}, Simp[(e + d*q)/q   Int[1/Sqrt[a + b*x^2 + c*x^ 
4], x], x] - Simp[e/q   Int[(1 - q*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x] /; 
NeQ[e + d*q, 0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && Pos 
Q[c/a]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.80 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.93

method	result	size
risch	\(-\frac {x \left (x^{2}-2\right )}{15 \sqrt {2 x^{4}+2 x^{2}+3}}+\frac {3 \sqrt {1-\left (-\frac {1}{3}+\frac {i \sqrt {5}}{3}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{3}-\frac {i \sqrt {5}}{3}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-3+3 i \sqrt {5}}}{3}, \frac {\sqrt {-6+3 i \sqrt {5}}}{3}\right )}{5 \sqrt {-3+3 i \sqrt {5}}\, \sqrt {2 x^{4}+2 x^{2}+3}}-\frac {6 \sqrt {1-\left (-\frac {1}{3}+\frac {i \sqrt {5}}{3}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{3}-\frac {i \sqrt {5}}{3}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {-3+3 i \sqrt {5}}}{3}, \frac {\sqrt {-6+3 i \sqrt {5}}}{3}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {-3+3 i \sqrt {5}}}{3}, \frac {\sqrt {-6+3 i \sqrt {5}}}{3}\right )\right )}{5 \sqrt {-3+3 i \sqrt {5}}\, \sqrt {2 x^{4}+2 x^{2}+3}\, \left (2+2 i \sqrt {5}\right )}\)	\(235\)
default	\(-\frac {4 \left (-\frac {1}{30} x +\frac {1}{60} x^{3}\right )}{\sqrt {2 x^{4}+2 x^{2}+3}}+\frac {3 \sqrt {1-\left (-\frac {1}{3}+\frac {i \sqrt {5}}{3}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{3}-\frac {i \sqrt {5}}{3}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-3+3 i \sqrt {5}}}{3}, \frac {\sqrt {-6+3 i \sqrt {5}}}{3}\right )}{5 \sqrt {-3+3 i \sqrt {5}}\, \sqrt {2 x^{4}+2 x^{2}+3}}-\frac {6 \sqrt {1-\left (-\frac {1}{3}+\frac {i \sqrt {5}}{3}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{3}-\frac {i \sqrt {5}}{3}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {-3+3 i \sqrt {5}}}{3}, \frac {\sqrt {-6+3 i \sqrt {5}}}{3}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {-3+3 i \sqrt {5}}}{3}, \frac {\sqrt {-6+3 i \sqrt {5}}}{3}\right )\right )}{5 \sqrt {-3+3 i \sqrt {5}}\, \sqrt {2 x^{4}+2 x^{2}+3}\, \left (2+2 i \sqrt {5}\right )}\)	\(238\)
elliptic	\(-\frac {4 \left (-\frac {1}{30} x +\frac {1}{60} x^{3}\right )}{\sqrt {2 x^{4}+2 x^{2}+3}}+\frac {3 \sqrt {1-\left (-\frac {1}{3}+\frac {i \sqrt {5}}{3}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{3}-\frac {i \sqrt {5}}{3}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-3+3 i \sqrt {5}}}{3}, \frac {\sqrt {-6+3 i \sqrt {5}}}{3}\right )}{5 \sqrt {-3+3 i \sqrt {5}}\, \sqrt {2 x^{4}+2 x^{2}+3}}-\frac {6 \sqrt {1-\left (-\frac {1}{3}+\frac {i \sqrt {5}}{3}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{3}-\frac {i \sqrt {5}}{3}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {-3+3 i \sqrt {5}}}{3}, \frac {\sqrt {-6+3 i \sqrt {5}}}{3}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {-3+3 i \sqrt {5}}}{3}, \frac {\sqrt {-6+3 i \sqrt {5}}}{3}\right )\right )}{5 \sqrt {-3+3 i \sqrt {5}}\, \sqrt {2 x^{4}+2 x^{2}+3}\, \left (2+2 i \sqrt {5}\right )}\)	\(238\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.65 \[ \int \frac {1}{\left (3+2 x^2+2 x^4\right )^{3/2}} \, dx=\frac {\sqrt {3} {\left (2 \, x^{4} + 2 \, x^{2} - \sqrt {-5} {\left (2 \, x^{4} + 2 \, x^{2} + 3\right )} + 3\right )} \sqrt {\frac {1}{3} \, \sqrt {-5} - \frac {1}{3}} E(\arcsin \left (x \sqrt {\frac {1}{3} \, \sqrt {-5} - \frac {1}{3}}\right )\,|\,\frac {1}{3} \, \sqrt {-5} - \frac {2}{3}) - 2 \, \sqrt {3} {\left (4 \, x^{4} + 4 \, x^{2} + \sqrt {-5} {\left (2 \, x^{4} + 2 \, x^{2} + 3\right )} + 6\right )} \sqrt {\frac {1}{3} \, \sqrt {-5} - \frac {1}{3}} F(\arcsin \left (x \sqrt {\frac {1}{3} \, \sqrt {-5} - \frac {1}{3}}\right )\,|\,\frac {1}{3} \, \sqrt {-5} - \frac {2}{3}) - 6 \, \sqrt {2 \, x^{4} + 2 \, x^{2} + 3} {\left (x^{3} - 2 \, x\right )}}{90 \, {\left (2 \, x^{4} + 2 \, x^{2} + 3\right )}} \] Input:

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

Output:

1/90*(sqrt(3)*(2*x^4 + 2*x^2 - sqrt(-5)*(2*x^4 + 2*x^2 + 3) + 3)*sqrt(1/3* 
sqrt(-5) - 1/3)*elliptic_e(arcsin(x*sqrt(1/3*sqrt(-5) - 1/3)), 1/3*sqrt(-5 
) - 2/3) - 2*sqrt(3)*(4*x^4 + 4*x^2 + sqrt(-5)*(2*x^4 + 2*x^2 + 3) + 6)*sq 
rt(1/3*sqrt(-5) - 1/3)*elliptic_f(arcsin(x*sqrt(1/3*sqrt(-5) - 1/3)), 1/3* 
sqrt(-5) - 2/3) - 6*sqrt(2*x^4 + 2*x^2 + 3)*(x^3 - 2*x))/(2*x^4 + 2*x^2 + 
3)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\left (3+2 x^2+2 x^4\right )^{3/2}} \, dx=\int \frac {\sqrt {2 x^{4}+2 x^{2}+3}}{4 x^{8}+8 x^{6}+16 x^{4}+12 x^{2}+9}d x \] Input:

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

Output:

int(sqrt(2*x**4 + 2*x**2 + 3)/(4*x**8 + 8*x**6 + 16*x**4 + 12*x**2 + 9),x)