\(\int \frac {1}{(3+3 x^2+2 x^4)^{3/2}} \, dx\) [251]

Optimal result

Integrand size = 16, antiderivative size = 261 \[ \int \frac {1}{\left (3+3 x^2+2 x^4\right )^{3/2}} \, dx=\frac {x \left (1-2 x^2\right )}{15 \sqrt {3+3 x^2+2 x^4}}+\frac {2 x \sqrt {3+3 x^2+2 x^4}}{15 \left (\sqrt {6}+2 x^2\right )}-\frac {\sqrt [4]{2} \left (3+\sqrt {6} x^2\right ) \sqrt {\frac {3+3 x^2+2 x^4}{\left (3+\sqrt {6} x^2\right )^2}} E\left (2 \arctan \left (\sqrt [4]{\frac {2}{3}} x\right )|\frac {1}{8} \left (4-\sqrt {6}\right )\right )}{5\ 3^{3/4} \sqrt {3+3 x^2+2 x^4}}+\frac {\left (3+2 \sqrt {6}\right ) \left (3+\sqrt {6} x^2\right ) \sqrt {\frac {3+3 x^2+2 x^4}{\left (3+\sqrt {6} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{\frac {2}{3}} x\right ),\frac {1}{8} \left (4-\sqrt {6}\right )\right )}{15\ 6^{3/4} \sqrt {3+3 x^2+2 x^4}} \] Output:

1/15*x*(-2*x^2+1)/(2*x^4+3*x^2+3)^(1/2)+2*x*(2*x^4+3*x^2+3)^(1/2)/(15*6^(1 
/2)+30*x^2)-1/15*2^(1/4)*(3+6^(1/2)*x^2)*((2*x^4+3*x^2+3)/(3+6^(1/2)*x^2)^ 
2)^(1/2)*EllipticE(sin(2*arctan(1/3*2^(1/4)*3^(3/4)*x)),1/4*(8-2*6^(1/2))^ 
(1/2))*3^(1/4)/(2*x^4+3*x^2+3)^(1/2)+1/90*(3+2*6^(1/2))*(3+6^(1/2)*x^2)*(( 
2*x^4+3*x^2+3)/(3+6^(1/2)*x^2)^2)^(1/2)*InverseJacobiAM(2*arctan(1/3*2^(1/ 
4)*3^(3/4)*x),1/4*(8-2*6^(1/2))^(1/2))*6^(1/4)/(2*x^4+3*x^2+3)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 6.04 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.23 \[ \int \frac {1}{\left (3+3 x^2+2 x^4\right )^{3/2}} \, dx=\frac {4 \sqrt {-\frac {i}{-3 i+\sqrt {15}}} x \left (1-2 x^2\right )-\left (3 i+\sqrt {15}\right ) \sqrt {\frac {-3 i+\sqrt {15}-4 i x^2}{-3 i+\sqrt {15}}} \sqrt {\frac {3 i+\sqrt {15}+4 i x^2}{3 i+\sqrt {15}}} E\left (i \text {arcsinh}\left (2 \sqrt {-\frac {i}{-3 i+\sqrt {15}}} x\right )|\frac {3 i-\sqrt {15}}{3 i+\sqrt {15}}\right )+\left (-5 i+\sqrt {15}\right ) \sqrt {\frac {-3 i+\sqrt {15}-4 i x^2}{-3 i+\sqrt {15}}} \sqrt {\frac {3 i+\sqrt {15}+4 i x^2}{3 i+\sqrt {15}}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (2 \sqrt {-\frac {i}{-3 i+\sqrt {15}}} x\right ),\frac {3 i-\sqrt {15}}{3 i+\sqrt {15}}\right )}{60 \sqrt {-\frac {i}{-3 i+\sqrt {15}}} \sqrt {3+3 x^2+2 x^4}} \] Input:

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

Output:

(4*Sqrt[(-I)/(-3*I + Sqrt[15])]*x*(1 - 2*x^2) - (3*I + Sqrt[15])*Sqrt[(-3* 
I + Sqrt[15] - (4*I)*x^2)/(-3*I + Sqrt[15])]*Sqrt[(3*I + Sqrt[15] + (4*I)* 
x^2)/(3*I + Sqrt[15])]*EllipticE[I*ArcSinh[2*Sqrt[(-I)/(-3*I + Sqrt[15])]* 
x], (3*I - Sqrt[15])/(3*I + Sqrt[15])] + (-5*I + Sqrt[15])*Sqrt[(-3*I + Sq 
rt[15] - (4*I)*x^2)/(-3*I + Sqrt[15])]*Sqrt[(3*I + Sqrt[15] + (4*I)*x^2)/( 
3*I + Sqrt[15])]*EllipticF[I*ArcSinh[2*Sqrt[(-I)/(-3*I + Sqrt[15])]*x], (3 
*I - Sqrt[15])/(3*I + Sqrt[15])])/(60*Sqrt[(-I)/(-3*I + Sqrt[15])]*Sqrt[3 
+ 3*x^2 + 2*x^4])
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.02, 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.

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

\(\Big \downarrow \) 1405

\(\displaystyle \frac {1}{45} \int \frac {6 \left (x^2+2\right )}{\sqrt {2 x^4+3 x^2+3}}dx+\frac {x \left (1-2 x^2\right )}{15 \sqrt {2 x^4+3 x^2+3}}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1511

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1416

\(\displaystyle \frac {2}{15} \left (\frac {\left (4+\sqrt {6}\right ) \left (\sqrt {6} x^2+3\right ) \sqrt {\frac {2 x^4+3 x^2+3}{\left (\sqrt {6} x^2+3\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{\frac {2}{3}} x\right ),\frac {1}{8} \left (4-\sqrt {6}\right )\right )}{4 \sqrt [4]{6} \sqrt {2 x^4+3 x^2+3}}-\frac {\int \frac {3-\sqrt {6} x^2}{\sqrt {2 x^4+3 x^2+3}}dx}{\sqrt {6}}\right )+\frac {x \left (1-2 x^2\right )}{15 \sqrt {2 x^4+3 x^2+3}}\)

\(\Big \downarrow \) 1509

\(\displaystyle \frac {2}{15} \left (\frac {\left (4+\sqrt {6}\right ) \left (\sqrt {6} x^2+3\right ) \sqrt {\frac {2 x^4+3 x^2+3}{\left (\sqrt {6} x^2+3\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{\frac {2}{3}} x\right ),\frac {1}{8} \left (4-\sqrt {6}\right )\right )}{4 \sqrt [4]{6} \sqrt {2 x^4+3 x^2+3}}-\frac {\frac {3^{3/4} \left (\sqrt {6} x^2+3\right ) \sqrt {\frac {2 x^4+3 x^2+3}{\left (\sqrt {6} x^2+3\right )^2}} E\left (2 \arctan \left (\sqrt [4]{\frac {2}{3}} x\right )|\frac {1}{8} \left (4-\sqrt {6}\right )\right )}{\sqrt [4]{2} \sqrt {2 x^4+3 x^2+3}}-\frac {3 x \sqrt {2 x^4+3 x^2+3}}{\sqrt {6} x^2+3}}{\sqrt {6}}\right )+\frac {x \left (1-2 x^2\right )}{15 \sqrt {2 x^4+3 x^2+3}}\)

Input:

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

Output:

(x*(1 - 2*x^2))/(15*Sqrt[3 + 3*x^2 + 2*x^4]) + (2*(-(((-3*x*Sqrt[3 + 3*x^2 
 + 2*x^4])/(3 + Sqrt[6]*x^2) + (3^(3/4)*(3 + Sqrt[6]*x^2)*Sqrt[(3 + 3*x^2 
+ 2*x^4)/(3 + Sqrt[6]*x^2)^2]*EllipticE[2*ArcTan[(2/3)^(1/4)*x], (4 - Sqrt 
[6])/8])/(2^(1/4)*Sqrt[3 + 3*x^2 + 2*x^4]))/Sqrt[6]) + ((4 + Sqrt[6])*(3 + 
 Sqrt[6]*x^2)*Sqrt[(3 + 3*x^2 + 2*x^4)/(3 + Sqrt[6]*x^2)^2]*EllipticF[2*Ar 
cTan[(2/3)^(1/4)*x], (4 - Sqrt[6])/8])/(4*6^(1/4)*Sqrt[3 + 3*x^2 + 2*x^4]) 
))/15
 

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 = 1.83 (sec) , antiderivative size = 237, normalized size of antiderivative = 0.91

method	result	size
risch	\(-\frac {x \left (2 x^{2}-1\right )}{15 \sqrt {2 x^{4}+3 x^{2}+3}}+\frac {8 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {15}}{6}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {15}}{6}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-18+6 i \sqrt {15}}}{6}, \frac {\sqrt {-1+i \sqrt {15}}}{2}\right )}{5 \sqrt {-18+6 i \sqrt {15}}\, \sqrt {2 x^{4}+3 x^{2}+3}}-\frac {24 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {15}}{6}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {15}}{6}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {-18+6 i \sqrt {15}}}{6}, \frac {\sqrt {-1+i \sqrt {15}}}{2}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {-18+6 i \sqrt {15}}}{6}, \frac {\sqrt {-1+i \sqrt {15}}}{2}\right )\right )}{5 \sqrt {-18+6 i \sqrt {15}}\, \sqrt {2 x^{4}+3 x^{2}+3}\, \left (3+i \sqrt {15}\right )}\)	\(237\)
default	\(-\frac {4 \left (-\frac {1}{60} x +\frac {1}{30} x^{3}\right )}{\sqrt {2 x^{4}+3 x^{2}+3}}+\frac {8 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {15}}{6}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {15}}{6}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-18+6 i \sqrt {15}}}{6}, \frac {\sqrt {-1+i \sqrt {15}}}{2}\right )}{5 \sqrt {-18+6 i \sqrt {15}}\, \sqrt {2 x^{4}+3 x^{2}+3}}-\frac {24 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {15}}{6}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {15}}{6}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {-18+6 i \sqrt {15}}}{6}, \frac {\sqrt {-1+i \sqrt {15}}}{2}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {-18+6 i \sqrt {15}}}{6}, \frac {\sqrt {-1+i \sqrt {15}}}{2}\right )\right )}{5 \sqrt {-18+6 i \sqrt {15}}\, \sqrt {2 x^{4}+3 x^{2}+3}\, \left (3+i \sqrt {15}\right )}\)	\(238\)
elliptic	\(-\frac {4 \left (-\frac {1}{60} x +\frac {1}{30} x^{3}\right )}{\sqrt {2 x^{4}+3 x^{2}+3}}+\frac {8 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {15}}{6}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {15}}{6}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-18+6 i \sqrt {15}}}{6}, \frac {\sqrt {-1+i \sqrt {15}}}{2}\right )}{5 \sqrt {-18+6 i \sqrt {15}}\, \sqrt {2 x^{4}+3 x^{2}+3}}-\frac {24 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {15}}{6}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {15}}{6}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {-18+6 i \sqrt {15}}}{6}, \frac {\sqrt {-1+i \sqrt {15}}}{2}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {-18+6 i \sqrt {15}}}{6}, \frac {\sqrt {-1+i \sqrt {15}}}{2}\right )\right )}{5 \sqrt {-18+6 i \sqrt {15}}\, \sqrt {2 x^{4}+3 x^{2}+3}\, \left (3+i \sqrt {15}\right )}\)	\(238\)

Input:

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

Output:

-1/15*x*(2*x^2-1)/(2*x^4+3*x^2+3)^(1/2)+8/5/(-18+6*I*15^(1/2))^(1/2)*(1-(- 
1/2+1/6*I*15^(1/2))*x^2)^(1/2)*(1-(-1/2-1/6*I*15^(1/2))*x^2)^(1/2)/(2*x^4+ 
3*x^2+3)^(1/2)*EllipticF(1/6*x*(-18+6*I*15^(1/2))^(1/2),1/2*(-1+I*15^(1/2) 
)^(1/2))-24/5/(-18+6*I*15^(1/2))^(1/2)*(1-(-1/2+1/6*I*15^(1/2))*x^2)^(1/2) 
*(1-(-1/2-1/6*I*15^(1/2))*x^2)^(1/2)/(2*x^4+3*x^2+3)^(1/2)/(3+I*15^(1/2))* 
(EllipticF(1/6*x*(-18+6*I*15^(1/2))^(1/2),1/2*(-1+I*15^(1/2))^(1/2))-Ellip 
ticE(1/6*x*(-18+6*I*15^(1/2))^(1/2),1/2*(-1+I*15^(1/2))^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.64 \[ \int \frac {1}{\left (3+3 x^2+2 x^4\right )^{3/2}} \, dx=\frac {\sqrt {3} {\left (2 \, x^{4} + 3 \, x^{2} - \sqrt {-\frac {5}{3}} {\left (2 \, x^{4} + 3 \, x^{2} + 3\right )} + 3\right )} \sqrt {\frac {1}{2} \, \sqrt {-\frac {5}{3}} - \frac {1}{2}} E(\arcsin \left (x \sqrt {\frac {1}{2} \, \sqrt {-\frac {5}{3}} - \frac {1}{2}}\right )\,|\,\frac {3}{4} \, \sqrt {-\frac {5}{3}} - \frac {1}{4}) - \sqrt {3} {\left (6 \, x^{4} + 9 \, x^{2} + \sqrt {-\frac {5}{3}} {\left (2 \, x^{4} + 3 \, x^{2} + 3\right )} + 9\right )} \sqrt {\frac {1}{2} \, \sqrt {-\frac {5}{3}} - \frac {1}{2}} F(\arcsin \left (x \sqrt {\frac {1}{2} \, \sqrt {-\frac {5}{3}} - \frac {1}{2}}\right )\,|\,\frac {3}{4} \, \sqrt {-\frac {5}{3}} - \frac {1}{4}) - 2 \, \sqrt {2 \, x^{4} + 3 \, x^{2} + 3} {\left (2 \, x^{3} - x\right )}}{30 \, {\left (2 \, x^{4} + 3 \, x^{2} + 3\right )}} \] Input:

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

Output:

1/30*(sqrt(3)*(2*x^4 + 3*x^2 - sqrt(-5/3)*(2*x^4 + 3*x^2 + 3) + 3)*sqrt(1/ 
2*sqrt(-5/3) - 1/2)*elliptic_e(arcsin(x*sqrt(1/2*sqrt(-5/3) - 1/2)), 3/4*s 
qrt(-5/3) - 1/4) - sqrt(3)*(6*x^4 + 9*x^2 + sqrt(-5/3)*(2*x^4 + 3*x^2 + 3) 
 + 9)*sqrt(1/2*sqrt(-5/3) - 1/2)*elliptic_f(arcsin(x*sqrt(1/2*sqrt(-5/3) - 
 1/2)), 3/4*sqrt(-5/3) - 1/4) - 2*sqrt(2*x^4 + 3*x^2 + 3)*(2*x^3 - x))/(2* 
x^4 + 3*x^2 + 3)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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