\(\int \frac {1}{(2+3 x^2+3 x^4)^{3/2}} \, dx\) [235]

Optimal result

Integrand size = 16, antiderivative size = 261 \[ \int \frac {1}{\left (2+3 x^2+3 x^4\right )^{3/2}} \, dx=\frac {x \left (1-3 x^2\right )}{10 \sqrt {2+3 x^2+3 x^4}}+\frac {3 x \sqrt {2+3 x^2+3 x^4}}{10 \left (\sqrt {6}+3 x^2\right )}-\frac {\sqrt [4]{3} \left (2+\sqrt {6} x^2\right ) \sqrt {\frac {2+3 x^2+3 x^4}{\left (2+\sqrt {6} x^2\right )^2}} E\left (2 \arctan \left (\sqrt [4]{\frac {3}{2}} x\right )|\frac {1}{8} \left (4-\sqrt {6}\right )\right )}{5\ 2^{3/4} \sqrt {2+3 x^2+3 x^4}}+\frac {\left (3+2 \sqrt {6}\right ) \left (2+\sqrt {6} x^2\right ) \sqrt {\frac {2+3 x^2+3 x^4}{\left (2+\sqrt {6} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{\frac {3}{2}} x\right ),\frac {1}{8} \left (4-\sqrt {6}\right )\right )}{10\ 6^{3/4} \sqrt {2+3 x^2+3 x^4}} \] Output:

1/10*x*(-3*x^2+1)/(3*x^4+3*x^2+2)^(1/2)+3*x*(3*x^4+3*x^2+2)^(1/2)/(10*6^(1 
/2)+30*x^2)-1/10*3^(1/4)*(2+6^(1/2)*x^2)*((3*x^4+3*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)),1/4*(8-2*6^(1/2))^ 
(1/2))*2^(1/4)/(3*x^4+3*x^2+2)^(1/2)+1/60*(3+2*6^(1/2))*(2+6^(1/2)*x^2)*(( 
3*x^4+3*x^2+2)/(2+6^(1/2)*x^2)^2)^(1/2)*InverseJacobiAM(2*arctan(1/2*3^(1/ 
4)*2^(3/4)*x),1/4*(8-2*6^(1/2))^(1/2))*6^(1/4)/(3*x^4+3*x^2+2)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 6.24 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.32 \[ \int \frac {1}{\left (2+3 x^2+3 x^4\right )^{3/2}} \, dx=\frac {12 \sqrt {-\frac {i}{-3 i+\sqrt {15}}} x \left (1-3 x^2\right )-3 i \sqrt {2} \left (\sqrt {3}-i \sqrt {5}\right ) \sqrt {\frac {-3 i+\sqrt {15}-6 i x^2}{-3 i+\sqrt {15}}} \sqrt {\frac {3 i+\sqrt {15}+6 i x^2}{3 i+\sqrt {15}}} E\left (i \text {arcsinh}\left (\sqrt {-\frac {6 i}{-3 i+\sqrt {15}}} x\right )|\frac {3 i-\sqrt {15}}{3 i+\sqrt {15}}\right )+\sqrt {2} \left (-5 i \sqrt {3}+3 \sqrt {5}\right ) \sqrt {\frac {-3 i+\sqrt {15}-6 i x^2}{-3 i+\sqrt {15}}} \sqrt {\frac {3 i+\sqrt {15}+6 i x^2}{3 i+\sqrt {15}}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {6 i}{-3 i+\sqrt {15}}} x\right ),\frac {3 i-\sqrt {15}}{3 i+\sqrt {15}}\right )}{120 \sqrt {-\frac {i}{-3 i+\sqrt {15}}} \sqrt {2+3 x^2+3 x^4}} \] Input:

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

Output:

(12*Sqrt[(-I)/(-3*I + Sqrt[15])]*x*(1 - 3*x^2) - (3*I)*Sqrt[2]*(Sqrt[3] - 
I*Sqrt[5])*Sqrt[(-3*I + Sqrt[15] - (6*I)*x^2)/(-3*I + Sqrt[15])]*Sqrt[(3*I 
 + Sqrt[15] + (6*I)*x^2)/(3*I + Sqrt[15])]*EllipticE[I*ArcSinh[Sqrt[(-6*I) 
/(-3*I + Sqrt[15])]*x], (3*I - Sqrt[15])/(3*I + Sqrt[15])] + Sqrt[2]*((-5* 
I)*Sqrt[3] + 3*Sqrt[5])*Sqrt[(-3*I + Sqrt[15] - (6*I)*x^2)/(-3*I + Sqrt[15 
])]*Sqrt[(3*I + Sqrt[15] + (6*I)*x^2)/(3*I + Sqrt[15])]*EllipticF[I*ArcSin 
h[Sqrt[(-6*I)/(-3*I + Sqrt[15])]*x], (3*I - Sqrt[15])/(3*I + Sqrt[15])])/( 
120*Sqrt[(-I)/(-3*I + Sqrt[15])]*Sqrt[2 + 3*x^2 + 3*x^4])
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.03, 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.

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

\(\Big \downarrow \) 1405

\(\displaystyle \frac {1}{30} \int \frac {3 \left (3 x^2+4\right )}{\sqrt {3 x^4+3 x^2+2}}dx+\frac {x \left (1-3 x^2\right )}{10 \sqrt {3 x^4+3 x^2+2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{10} \int \frac {3 x^2+4}{\sqrt {3 x^4+3 x^2+2}}dx+\frac {x \left (1-3 x^2\right )}{10 \sqrt {3 x^4+3 x^2+2}}\)

\(\Big \downarrow \) 1511

\(\displaystyle \frac {1}{10} \left (\left (4+\sqrt {6}\right ) \int \frac {1}{\sqrt {3 x^4+3 x^2+2}}dx-\sqrt {6} \int \frac {2-\sqrt {6} x^2}{2 \sqrt {3 x^4+3 x^2+2}}dx\right )+\frac {x \left (1-3 x^2\right )}{10 \sqrt {3 x^4+3 x^2+2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{10} \left (\left (4+\sqrt {6}\right ) \int \frac {1}{\sqrt {3 x^4+3 x^2+2}}dx-\sqrt {\frac {3}{2}} \int \frac {2-\sqrt {6} x^2}{\sqrt {3 x^4+3 x^2+2}}dx\right )+\frac {x \left (1-3 x^2\right )}{10 \sqrt {3 x^4+3 x^2+2}}\)

\(\Big \downarrow \) 1416

\(\displaystyle \frac {1}{10} \left (\frac {\left (4+\sqrt {6}\right ) \left (\sqrt {6} x^2+2\right ) \sqrt {\frac {3 x^4+3 x^2+2}{\left (\sqrt {6} x^2+2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{\frac {3}{2}} x\right ),\frac {1}{8} \left (4-\sqrt {6}\right )\right )}{2 \sqrt [4]{6} \sqrt {3 x^4+3 x^2+2}}-\sqrt {\frac {3}{2}} \int \frac {2-\sqrt {6} x^2}{\sqrt {3 x^4+3 x^2+2}}dx\right )+\frac {x \left (1-3 x^2\right )}{10 \sqrt {3 x^4+3 x^2+2}}\)

\(\Big \downarrow \) 1509

\(\displaystyle \frac {1}{10} \left (\frac {\left (4+\sqrt {6}\right ) \left (\sqrt {6} x^2+2\right ) \sqrt {\frac {3 x^4+3 x^2+2}{\left (\sqrt {6} x^2+2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{\frac {3}{2}} x\right ),\frac {1}{8} \left (4-\sqrt {6}\right )\right )}{2 \sqrt [4]{6} \sqrt {3 x^4+3 x^2+2}}-\sqrt {\frac {3}{2}} \left (\frac {2^{3/4} \left (\sqrt {6} x^2+2\right ) \sqrt {\frac {3 x^4+3 x^2+2}{\left (\sqrt {6} x^2+2\right )^2}} E\left (2 \arctan \left (\sqrt [4]{\frac {3}{2}} x\right )|\frac {1}{8} \left (4-\sqrt {6}\right )\right )}{\sqrt [4]{3} \sqrt {3 x^4+3 x^2+2}}-\frac {2 x \sqrt {3 x^4+3 x^2+2}}{\sqrt {6} x^2+2}\right )\right )+\frac {x \left (1-3 x^2\right )}{10 \sqrt {3 x^4+3 x^2+2}}\)

Input:

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

Output:

(x*(1 - 3*x^2))/(10*Sqrt[2 + 3*x^2 + 3*x^4]) + (-(Sqrt[3/2]*((-2*x*Sqrt[2 
+ 3*x^2 + 3*x^4])/(2 + Sqrt[6]*x^2) + (2^(3/4)*(2 + Sqrt[6]*x^2)*Sqrt[(2 + 
 3*x^2 + 3*x^4)/(2 + Sqrt[6]*x^2)^2]*EllipticE[2*ArcTan[(3/2)^(1/4)*x], (4 
 - Sqrt[6])/8])/(3^(1/4)*Sqrt[2 + 3*x^2 + 3*x^4]))) + ((4 + Sqrt[6])*(2 + 
Sqrt[6]*x^2)*Sqrt[(2 + 3*x^2 + 3*x^4)/(2 + Sqrt[6]*x^2)^2]*EllipticF[2*Arc 
Tan[(3/2)^(1/4)*x], (4 - Sqrt[6])/8])/(2*6^(1/4)*Sqrt[2 + 3*x^2 + 3*x^4])) 
/10
 

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 = 2.26 (sec) , antiderivative size = 237, normalized size of antiderivative = 0.91

method	result	size
risch	\(-\frac {x \left (3 x^{2}-1\right )}{10 \sqrt {3 x^{4}+3 x^{2}+2}}+\frac {4 \sqrt {1-\left (-\frac {3}{4}+\frac {i \sqrt {15}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {3}{4}-\frac {i \sqrt {15}}{4}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-3+i \sqrt {15}}}{2}, \frac {\sqrt {-1+i \sqrt {15}}}{2}\right )}{5 \sqrt {-3+i \sqrt {15}}\, \sqrt {3 x^{4}+3 x^{2}+2}}-\frac {12 \sqrt {1-\left (-\frac {3}{4}+\frac {i \sqrt {15}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {3}{4}-\frac {i \sqrt {15}}{4}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {-3+i \sqrt {15}}}{2}, \frac {\sqrt {-1+i \sqrt {15}}}{2}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {-3+i \sqrt {15}}}{2}, \frac {\sqrt {-1+i \sqrt {15}}}{2}\right )\right )}{5 \sqrt {-3+i \sqrt {15}}\, \sqrt {3 x^{4}+3 x^{2}+2}\, \left (3+i \sqrt {15}\right )}\)	\(237\)
default	\(-\frac {6 \left (-\frac {1}{60} x +\frac {1}{20} x^{3}\right )}{\sqrt {3 x^{4}+3 x^{2}+2}}+\frac {4 \sqrt {1-\left (-\frac {3}{4}+\frac {i \sqrt {15}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {3}{4}-\frac {i \sqrt {15}}{4}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-3+i \sqrt {15}}}{2}, \frac {\sqrt {-1+i \sqrt {15}}}{2}\right )}{5 \sqrt {-3+i \sqrt {15}}\, \sqrt {3 x^{4}+3 x^{2}+2}}-\frac {12 \sqrt {1-\left (-\frac {3}{4}+\frac {i \sqrt {15}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {3}{4}-\frac {i \sqrt {15}}{4}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {-3+i \sqrt {15}}}{2}, \frac {\sqrt {-1+i \sqrt {15}}}{2}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {-3+i \sqrt {15}}}{2}, \frac {\sqrt {-1+i \sqrt {15}}}{2}\right )\right )}{5 \sqrt {-3+i \sqrt {15}}\, \sqrt {3 x^{4}+3 x^{2}+2}\, \left (3+i \sqrt {15}\right )}\)	\(238\)
elliptic	\(-\frac {6 \left (-\frac {1}{60} x +\frac {1}{20} x^{3}\right )}{\sqrt {3 x^{4}+3 x^{2}+2}}+\frac {4 \sqrt {1-\left (-\frac {3}{4}+\frac {i \sqrt {15}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {3}{4}-\frac {i \sqrt {15}}{4}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-3+i \sqrt {15}}}{2}, \frac {\sqrt {-1+i \sqrt {15}}}{2}\right )}{5 \sqrt {-3+i \sqrt {15}}\, \sqrt {3 x^{4}+3 x^{2}+2}}-\frac {12 \sqrt {1-\left (-\frac {3}{4}+\frac {i \sqrt {15}}{4}\right ) x^{2}}\, \sqrt {1-\left (-\frac {3}{4}-\frac {i \sqrt {15}}{4}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {-3+i \sqrt {15}}}{2}, \frac {\sqrt {-1+i \sqrt {15}}}{2}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {-3+i \sqrt {15}}}{2}, \frac {\sqrt {-1+i \sqrt {15}}}{2}\right )\right )}{5 \sqrt {-3+i \sqrt {15}}\, \sqrt {3 x^{4}+3 x^{2}+2}\, \left (3+i \sqrt {15}\right )}\)	\(238\)

Input:

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

Output:

-1/10*x*(3*x^2-1)/(3*x^4+3*x^2+2)^(1/2)+4/5/(-3+I*15^(1/2))^(1/2)*(1-(-3/4 
+1/4*I*15^(1/2))*x^2)^(1/2)*(1-(-3/4-1/4*I*15^(1/2))*x^2)^(1/2)/(3*x^4+3*x 
^2+2)^(1/2)*EllipticF(1/2*x*(-3+I*15^(1/2))^(1/2),1/2*(-1+I*15^(1/2))^(1/2 
))-12/5/(-3+I*15^(1/2))^(1/2)*(1-(-3/4+1/4*I*15^(1/2))*x^2)^(1/2)*(1-(-3/4 
-1/4*I*15^(1/2))*x^2)^(1/2)/(3*x^4+3*x^2+2)^(1/2)/(3+I*15^(1/2))*(Elliptic 
F(1/2*x*(-3+I*15^(1/2))^(1/2),1/2*(-1+I*15^(1/2))^(1/2))-EllipticE(1/2*x*( 
-3+I*15^(1/2))^(1/2),1/2*(-1+I*15^(1/2))^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.62 \[ \int \frac {1}{\left (2+3 x^2+3 x^4\right )^{3/2}} \, dx=\frac {3 \, \sqrt {2} {\left (9 \, x^{4} + 9 \, x^{2} - \sqrt {-15} {\left (3 \, x^{4} + 3 \, x^{2} + 2\right )} + 6\right )} \sqrt {\sqrt {-15} - 3} E(\arcsin \left (\frac {1}{2} \, x \sqrt {\sqrt {-15} - 3}\right )\,|\,\frac {1}{4} \, \sqrt {-15} - \frac {1}{4}) - \sqrt {2} {\left (63 \, x^{4} + 63 \, x^{2} + \sqrt {-15} {\left (3 \, x^{4} + 3 \, x^{2} + 2\right )} + 42\right )} \sqrt {\sqrt {-15} - 3} F(\arcsin \left (\frac {1}{2} \, x \sqrt {\sqrt {-15} - 3}\right )\,|\,\frac {1}{4} \, \sqrt {-15} - \frac {1}{4}) - 24 \, \sqrt {3 \, x^{4} + 3 \, x^{2} + 2} {\left (3 \, x^{3} - x\right )}}{240 \, {\left (3 \, x^{4} + 3 \, x^{2} + 2\right )}} \] Input:

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

Output:

1/240*(3*sqrt(2)*(9*x^4 + 9*x^2 - sqrt(-15)*(3*x^4 + 3*x^2 + 2) + 6)*sqrt( 
sqrt(-15) - 3)*elliptic_e(arcsin(1/2*x*sqrt(sqrt(-15) - 3)), 1/4*sqrt(-15) 
 - 1/4) - sqrt(2)*(63*x^4 + 63*x^2 + sqrt(-15)*(3*x^4 + 3*x^2 + 2) + 42)*s 
qrt(sqrt(-15) - 3)*elliptic_f(arcsin(1/2*x*sqrt(sqrt(-15) - 3)), 1/4*sqrt( 
-15) - 1/4) - 24*sqrt(3*x^4 + 3*x^2 + 2)*(3*x^3 - x))/(3*x^4 + 3*x^2 + 2)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\left (2+3 x^2+3 x^4\right )^{3/2}} \, dx=\int \frac {\sqrt {3 x^{4}+3 x^{2}+2}}{9 x^{8}+18 x^{6}+21 x^{4}+12 x^{2}+4}d x \] Input:

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

Output:

int(sqrt(3*x**4 + 3*x**2 + 2)/(9*x**8 + 18*x**6 + 21*x**4 + 12*x**2 + 4),x 
)