\(\int \frac {1}{(2+4 x^2+3 x^4)^{3/2}} \, dx\) [234]

Optimal result

Integrand size = 16, antiderivative size = 260 \[ \int \frac {1}{\left (2+4 x^2+3 x^4\right )^{3/2}} \, dx=-\frac {x \left (1+3 x^2\right )}{4 \sqrt {2+4 x^2+3 x^4}}+\frac {3 x \sqrt {2+4 x^2+3 x^4}}{4 \left (\sqrt {6}+3 x^2\right )}-\frac {\sqrt [4]{3} \left (2+\sqrt {6} x^2\right ) \sqrt {\frac {2+4 x^2+3 x^4}{\left (2+\sqrt {6} x^2\right )^2}} E\left (2 \arctan \left (\sqrt [4]{\frac {3}{2}} x\right )|\frac {1}{2}-\frac {1}{\sqrt {6}}\right )}{2\ 2^{3/4} \sqrt {2+4 x^2+3 x^4}}+\frac {\sqrt [4]{3} \left (2+\sqrt {6}\right ) \left (2+\sqrt {6} x^2\right ) \sqrt {\frac {2+4 x^2+3 x^4}{\left (2+\sqrt {6} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{\frac {3}{2}} x\right ),\frac {1}{2}-\frac {1}{\sqrt {6}}\right )}{8\ 2^{3/4} \sqrt {2+4 x^2+3 x^4}} \] Output:

-1/4*x*(3*x^2+1)/(3*x^4+4*x^2+2)^(1/2)+3*x*(3*x^4+4*x^2+2)^(1/2)/(4*6^(1/2 
)+12*x^2)-1/4*3^(1/4)*(2+6^(1/2)*x^2)*((3*x^4+4*x^2+2)/(2+6^(1/2)*x^2)^2)^ 
(1/2)*EllipticE(sin(2*arctan(1/2*3^(1/4)*2^(3/4)*x)),1/6*(18-6*6^(1/2))^(1 
/2))*2^(1/4)/(3*x^4+4*x^2+2)^(1/2)+1/16*3^(1/4)*(2+6^(1/2))*(2+6^(1/2)*x^2 
)*((3*x^4+4*x^2+2)/(2+6^(1/2)*x^2)^2)^(1/2)*InverseJacobiAM(2*arctan(1/2*3 
^(1/4)*2^(3/4)*x),1/6*(18-6*6^(1/2))^(1/2))*2^(1/4)/(3*x^4+4*x^2+2)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 5.99 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.26 \[ \int \frac {1}{\left (2+4 x^2+3 x^4\right )^{3/2}} \, dx=\frac {-3 \sqrt {-\frac {i}{-2 i+\sqrt {2}}} x \left (1+3 x^2\right )-\sqrt {3} \left (2 i+\sqrt {2}\right ) \sqrt {\frac {-2 i+\sqrt {2}-3 i x^2}{-2 i+\sqrt {2}}} \sqrt {\frac {2 i+\sqrt {2}+3 i x^2}{2 i+\sqrt {2}}} E\left (i \text {arcsinh}\left (\sqrt {-\frac {3 i}{-2 i+\sqrt {2}}} x\right )|\frac {2 i-\sqrt {2}}{2 i+\sqrt {2}}\right )+\sqrt {3} \left (-i+\sqrt {2}\right ) \sqrt {\frac {-2 i+\sqrt {2}-3 i x^2}{-2 i+\sqrt {2}}} \sqrt {\frac {2 i+\sqrt {2}+3 i x^2}{2 i+\sqrt {2}}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {3 i}{-2 i+\sqrt {2}}} x\right ),\frac {2 i-\sqrt {2}}{2 i+\sqrt {2}}\right )}{12 \sqrt {-\frac {i}{-2 i+\sqrt {2}}} \sqrt {2+4 x^2+3 x^4}} \] Input:

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

Output:

(-3*Sqrt[(-I)/(-2*I + Sqrt[2])]*x*(1 + 3*x^2) - Sqrt[3]*(2*I + Sqrt[2])*Sq 
rt[(-2*I + Sqrt[2] - (3*I)*x^2)/(-2*I + Sqrt[2])]*Sqrt[(2*I + Sqrt[2] + (3 
*I)*x^2)/(2*I + Sqrt[2])]*EllipticE[I*ArcSinh[Sqrt[(-3*I)/(-2*I + Sqrt[2]) 
]*x], (2*I - Sqrt[2])/(2*I + Sqrt[2])] + Sqrt[3]*(-I + Sqrt[2])*Sqrt[(-2*I 
 + Sqrt[2] - (3*I)*x^2)/(-2*I + Sqrt[2])]*Sqrt[(2*I + Sqrt[2] + (3*I)*x^2) 
/(2*I + Sqrt[2])]*EllipticF[I*ArcSinh[Sqrt[(-3*I)/(-2*I + Sqrt[2])]*x], (2 
*I - Sqrt[2])/(2*I + Sqrt[2])])/(12*Sqrt[(-I)/(-2*I + Sqrt[2])]*Sqrt[2 + 4 
*x^2 + 3*x^4])
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.01, 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.

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

\(\Big \downarrow \) 1405

\(\displaystyle \frac {1}{16} \int \frac {12 \left (x^2+1\right )}{\sqrt {3 x^4+4 x^2+2}}dx-\frac {x \left (3 x^2+1\right )}{4 \sqrt {3 x^4+4 x^2+2}}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1511

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1416

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

\(\Big \downarrow \) 1509

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

Input:

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

Output:

-1/4*(x*(1 + 3*x^2))/Sqrt[2 + 4*x^2 + 3*x^4] + (3*(-(((-2*x*Sqrt[2 + 4*x^2 
 + 3*x^4])/(2 + Sqrt[6]*x^2) + (2^(3/4)*(2 + Sqrt[6]*x^2)*Sqrt[(2 + 4*x^2 
+ 3*x^4)/(2 + Sqrt[6]*x^2)^2]*EllipticE[2*ArcTan[(3/2)^(1/4)*x], 1/2 - 1/S 
qrt[6]])/(3^(1/4)*Sqrt[2 + 4*x^2 + 3*x^4]))/Sqrt[6]) + ((3 + Sqrt[6])*(2 + 
 Sqrt[6]*x^2)*Sqrt[(2 + 4*x^2 + 3*x^4)/(2 + Sqrt[6]*x^2)^2]*EllipticF[2*Ar 
cTan[(3/2)^(1/4)*x], 1/2 - 1/Sqrt[6]])/(6*6^(1/4)*Sqrt[2 + 4*x^2 + 3*x^4]) 
))/4
 

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 = 1.61 (sec) , antiderivative size = 237, normalized size of antiderivative = 0.91

method	result	size
risch	\(-\frac {x \left (3 x^{2}+1\right )}{4 \sqrt {3 x^{4}+4 x^{2}+2}}+\frac {3 \sqrt {1-\left (-1+\frac {i \sqrt {2}}{2}\right ) x^{2}}\, \sqrt {1-\left (-1-\frac {i \sqrt {2}}{2}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-4+2 i \sqrt {2}}}{2}, \frac {\sqrt {3+6 i \sqrt {2}}}{3}\right )}{2 \sqrt {-4+2 i \sqrt {2}}\, \sqrt {3 x^{4}+4 x^{2}+2}}-\frac {6 \sqrt {1-\left (-1+\frac {i \sqrt {2}}{2}\right ) x^{2}}\, \sqrt {1-\left (-1-\frac {i \sqrt {2}}{2}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {-4+2 i \sqrt {2}}}{2}, \frac {\sqrt {3+6 i \sqrt {2}}}{3}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {-4+2 i \sqrt {2}}}{2}, \frac {\sqrt {3+6 i \sqrt {2}}}{3}\right )\right )}{\sqrt {-4+2 i \sqrt {2}}\, \sqrt {3 x^{4}+4 x^{2}+2}\, \left (4+2 i \sqrt {2}\right )}\)	\(237\)
default	\(-\frac {6 \left (\frac {1}{24} x +\frac {1}{8} x^{3}\right )}{\sqrt {3 x^{4}+4 x^{2}+2}}+\frac {3 \sqrt {1-\left (-1+\frac {i \sqrt {2}}{2}\right ) x^{2}}\, \sqrt {1-\left (-1-\frac {i \sqrt {2}}{2}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-4+2 i \sqrt {2}}}{2}, \frac {\sqrt {3+6 i \sqrt {2}}}{3}\right )}{2 \sqrt {-4+2 i \sqrt {2}}\, \sqrt {3 x^{4}+4 x^{2}+2}}-\frac {6 \sqrt {1-\left (-1+\frac {i \sqrt {2}}{2}\right ) x^{2}}\, \sqrt {1-\left (-1-\frac {i \sqrt {2}}{2}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {-4+2 i \sqrt {2}}}{2}, \frac {\sqrt {3+6 i \sqrt {2}}}{3}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {-4+2 i \sqrt {2}}}{2}, \frac {\sqrt {3+6 i \sqrt {2}}}{3}\right )\right )}{\sqrt {-4+2 i \sqrt {2}}\, \sqrt {3 x^{4}+4 x^{2}+2}\, \left (4+2 i \sqrt {2}\right )}\)	\(238\)
elliptic	\(-\frac {6 \left (\frac {1}{24} x +\frac {1}{8} x^{3}\right )}{\sqrt {3 x^{4}+4 x^{2}+2}}+\frac {3 \sqrt {1-\left (-1+\frac {i \sqrt {2}}{2}\right ) x^{2}}\, \sqrt {1-\left (-1-\frac {i \sqrt {2}}{2}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-4+2 i \sqrt {2}}}{2}, \frac {\sqrt {3+6 i \sqrt {2}}}{3}\right )}{2 \sqrt {-4+2 i \sqrt {2}}\, \sqrt {3 x^{4}+4 x^{2}+2}}-\frac {6 \sqrt {1-\left (-1+\frac {i \sqrt {2}}{2}\right ) x^{2}}\, \sqrt {1-\left (-1-\frac {i \sqrt {2}}{2}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {-4+2 i \sqrt {2}}}{2}, \frac {\sqrt {3+6 i \sqrt {2}}}{3}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {-4+2 i \sqrt {2}}}{2}, \frac {\sqrt {3+6 i \sqrt {2}}}{3}\right )\right )}{\sqrt {-4+2 i \sqrt {2}}\, \sqrt {3 x^{4}+4 x^{2}+2}\, \left (4+2 i \sqrt {2}\right )}\)	\(238\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.57 \[ \int \frac {1}{\left (2+4 x^2+3 x^4\right )^{3/2}} \, dx=\frac {\sqrt {2} {\left (6 \, x^{4} + 8 \, x^{2} - \sqrt {-2} {\left (3 \, x^{4} + 4 \, x^{2} + 2\right )} + 4\right )} \sqrt {\frac {1}{2} \, \sqrt {-2} - 1} E(\arcsin \left (x \sqrt {\frac {1}{2} \, \sqrt {-2} - 1}\right )\,|\,\frac {2}{3} \, \sqrt {-2} + \frac {1}{3}) - 4 \, \sqrt {2} {\left (3 \, x^{4} + 4 \, x^{2} + 2\right )} \sqrt {\frac {1}{2} \, \sqrt {-2} - 1} F(\arcsin \left (x \sqrt {\frac {1}{2} \, \sqrt {-2} - 1}\right )\,|\,\frac {2}{3} \, \sqrt {-2} + \frac {1}{3}) - 2 \, \sqrt {3 \, x^{4} + 4 \, x^{2} + 2} {\left (3 \, x^{3} + x\right )}}{8 \, {\left (3 \, x^{4} + 4 \, x^{2} + 2\right )}} \] Input:

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

Output:

1/8*(sqrt(2)*(6*x^4 + 8*x^2 - sqrt(-2)*(3*x^4 + 4*x^2 + 2) + 4)*sqrt(1/2*s 
qrt(-2) - 1)*elliptic_e(arcsin(x*sqrt(1/2*sqrt(-2) - 1)), 2/3*sqrt(-2) + 1 
/3) - 4*sqrt(2)*(3*x^4 + 4*x^2 + 2)*sqrt(1/2*sqrt(-2) - 1)*elliptic_f(arcs 
in(x*sqrt(1/2*sqrt(-2) - 1)), 2/3*sqrt(-2) + 1/3) - 2*sqrt(3*x^4 + 4*x^2 + 
 2)*(3*x^3 + x))/(3*x^4 + 4*x^2 + 2)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\left (2+4 x^2+3 x^4\right )^{3/2}} \, dx=\int \frac {\sqrt {3 x^{4}+4 x^{2}+2}}{9 x^{8}+24 x^{6}+28 x^{4}+16 x^{2}+4}d x \] Input:

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

Output:

int(sqrt(3*x**4 + 4*x**2 + 2)/(9*x**8 + 24*x**6 + 28*x**4 + 16*x**2 + 4),x 
)