\(\int \frac {1}{(2+5 x^2+3 x^4)^{3/2}} \, dx\) [233]

Optimal result

Integrand size = 16, antiderivative size = 120 \[ \int \frac {1}{\left (2+5 x^2+3 x^4\right )^{3/2}} \, dx=\frac {3 x}{2 \sqrt {2+5 x^2+3 x^4}}+\frac {5 \sqrt {2+5 x^2+3 x^4} E\left (\arctan (x)\left |-\frac {1}{2}\right .\right )}{\sqrt {2} \sqrt {1+x^2} \sqrt {2+3 x^2}}-\frac {3 \sqrt {2} \sqrt {1+x^2} \sqrt {2+3 x^2} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{\sqrt {2+5 x^2+3 x^4}} \] Output:

3/2*x/(3*x^4+5*x^2+2)^(1/2)+5/2*(3*x^4+5*x^2+2)^(1/2)*EllipticE(x/(x^2+1)^ 
(1/2),1/2*I*2^(1/2))*2^(1/2)/(x^2+1)^(1/2)/(3*x^2+2)^(1/2)-3*(x^2+1)^(1/2) 
*(3*x^2+2)^(1/2)*InverseJacobiAM(arctan(x),1/2*I*2^(1/2))*2^(1/2)/(3*x^4+5 
*x^2+2)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 5.77 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.02 \[ \int \frac {1}{\left (2+5 x^2+3 x^4\right )^{3/2}} \, dx=\frac {13 x+15 x^3+5 i \sqrt {3} \sqrt {1+x^2} \sqrt {2+3 x^2} E\left (i \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right )|\frac {2}{3}\right )-i \sqrt {3} \sqrt {1+x^2} \sqrt {2+3 x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {3}{2}} x\right ),\frac {2}{3}\right )}{2 \sqrt {2+5 x^2+3 x^4}} \] Input:

Integrate[(2 + 5*x^2 + 3*x^4)^(-3/2),x]
 

Output:

(13*x + 15*x^3 + (5*I)*Sqrt[3]*Sqrt[1 + x^2]*Sqrt[2 + 3*x^2]*EllipticE[I*A 
rcSinh[Sqrt[3/2]*x], 2/3] - I*Sqrt[3]*Sqrt[1 + x^2]*Sqrt[2 + 3*x^2]*Ellipt 
icF[I*ArcSinh[Sqrt[3/2]*x], 2/3])/(2*Sqrt[2 + 5*x^2 + 3*x^4])
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.44, 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, 1413, 1456}

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.

\(\displaystyle \int \frac {1}{\left (3 x^4+5 x^2+2\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1405

\(\displaystyle \frac {x \left (15 x^2+13\right )}{2 \sqrt {3 x^4+5 x^2+2}}-\frac {1}{2} \int \frac {3 \left (5 x^2+4\right )}{\sqrt {3 x^4+5 x^2+2}}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x \left (15 x^2+13\right )}{2 \sqrt {3 x^4+5 x^2+2}}-\frac {3}{2} \int \frac {5 x^2+4}{\sqrt {3 x^4+5 x^2+2}}dx\)

\(\Big \downarrow \) 1503

\(\displaystyle \frac {x \left (15 x^2+13\right )}{2 \sqrt {3 x^4+5 x^2+2}}-\frac {3}{2} \left (4 \int \frac {1}{\sqrt {3 x^4+5 x^2+2}}dx+5 \int \frac {x^2}{\sqrt {3 x^4+5 x^2+2}}dx\right )\)

\(\Big \downarrow \) 1413

\(\displaystyle \frac {x \left (15 x^2+13\right )}{2 \sqrt {3 x^4+5 x^2+2}}-\frac {3}{2} \left (5 \int \frac {x^2}{\sqrt {3 x^4+5 x^2+2}}dx+\frac {2 \sqrt {2} \left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{\sqrt {3 x^4+5 x^2+2}}\right )\)

\(\Big \downarrow \) 1456

\(\displaystyle \frac {x \left (15 x^2+13\right )}{2 \sqrt {3 x^4+5 x^2+2}}-\frac {3}{2} \left (\frac {2 \sqrt {2} \left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{\sqrt {3 x^4+5 x^2+2}}+5 \left (\frac {x \left (3 x^2+2\right )}{3 \sqrt {3 x^4+5 x^2+2}}-\frac {\sqrt {2} \left (x^2+1\right ) \sqrt {\frac {3 x^2+2}{x^2+1}} E\left (\arctan (x)\left |-\frac {1}{2}\right .\right )}{3 \sqrt {3 x^4+5 x^2+2}}\right )\right )\)

Input:

Int[(2 + 5*x^2 + 3*x^4)^(-3/2),x]
 

Output:

(x*(13 + 15*x^2))/(2*Sqrt[2 + 5*x^2 + 3*x^4]) - (3*(5*((x*(2 + 3*x^2))/(3* 
Sqrt[2 + 5*x^2 + 3*x^4]) - (Sqrt[2]*(1 + x^2)*Sqrt[(2 + 3*x^2)/(1 + x^2)]* 
EllipticE[ArcTan[x], -1/2])/(3*Sqrt[2 + 5*x^2 + 3*x^4])) + (2*Sqrt[2]*(1 + 
 x^2)*Sqrt[(2 + 3*x^2)/(1 + x^2)]*EllipticF[ArcTan[x], -1/2])/Sqrt[2 + 5*x 
^2 + 3*x^4]))/2
 

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 1413

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]] /; 
 FreeQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]
 

rule 1456

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]] /; 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 = 2.65 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.04

method	result	size
risch	\(\frac {x \left (15 x^{2}+13\right )}{2 \sqrt {3 x^{4}+5 x^{2}+2}}+\frac {3 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )}{\sqrt {3 x^{4}+5 x^{2}+2}}-\frac {5 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \left (\operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )-\operatorname {EllipticE}\left (i x , \frac {\sqrt {6}}{2}\right )\right )}{2 \sqrt {3 x^{4}+5 x^{2}+2}}\)	\(125\)
default	\(-\frac {6 \left (-\frac {5}{4} x^{3}-\frac {13}{12} x \right )}{\sqrt {3 x^{4}+5 x^{2}+2}}+\frac {3 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )}{\sqrt {3 x^{4}+5 x^{2}+2}}-\frac {5 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \left (\operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )-\operatorname {EllipticE}\left (i x , \frac {\sqrt {6}}{2}\right )\right )}{2 \sqrt {3 x^{4}+5 x^{2}+2}}\)	\(126\)
elliptic	\(-\frac {6 \left (-\frac {5}{4} x^{3}-\frac {13}{12} x \right )}{\sqrt {3 x^{4}+5 x^{2}+2}}+\frac {3 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )}{\sqrt {3 x^{4}+5 x^{2}+2}}-\frac {5 i \sqrt {x^{2}+1}\, \sqrt {6 x^{2}+4}\, \left (\operatorname {EllipticF}\left (i x , \frac {\sqrt {6}}{2}\right )-\operatorname {EllipticE}\left (i x , \frac {\sqrt {6}}{2}\right )\right )}{2 \sqrt {3 x^{4}+5 x^{2}+2}}\)	\(126\)

Input:

int(1/(3*x^4+5*x^2+2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/2*x*(15*x^2+13)/(3*x^4+5*x^2+2)^(1/2)+3*I*(x^2+1)^(1/2)*(6*x^2+4)^(1/2)/ 
(3*x^4+5*x^2+2)^(1/2)*EllipticF(I*x,1/2*6^(1/2))-5/2*I*(x^2+1)^(1/2)*(6*x^ 
2+4)^(1/2)/(3*x^4+5*x^2+2)^(1/2)*(EllipticF(I*x,1/2*6^(1/2))-EllipticE(I*x 
,1/2*6^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.73 \[ \int \frac {1}{\left (2+5 x^2+3 x^4\right )^{3/2}} \, dx=-\frac {5 \, \sqrt {2} {\left (3 i \, x^{4} + 5 i \, x^{2} + 2 i\right )} E(\arcsin \left (i \, x\right )\,|\,\frac {3}{2}) + 11 \, \sqrt {2} {\left (-3 i \, x^{4} - 5 i \, x^{2} - 2 i\right )} F(\arcsin \left (i \, x\right )\,|\,\frac {3}{2}) - \sqrt {3 \, x^{4} + 5 \, x^{2} + 2} {\left (15 \, x^{3} + 13 \, x\right )}}{2 \, {\left (3 \, x^{4} + 5 \, x^{2} + 2\right )}} \] Input:

integrate(1/(3*x^4+5*x^2+2)^(3/2),x, algorithm="fricas")
 

Output:

-1/2*(5*sqrt(2)*(3*I*x^4 + 5*I*x^2 + 2*I)*elliptic_e(arcsin(I*x), 3/2) + 1 
1*sqrt(2)*(-3*I*x^4 - 5*I*x^2 - 2*I)*elliptic_f(arcsin(I*x), 3/2) - sqrt(3 
*x^4 + 5*x^2 + 2)*(15*x^3 + 13*x))/(3*x^4 + 5*x^2 + 2)
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {1}{\left (2+5 x^2+3 x^4\right )^{3/2}} \, dx=\int \frac {1}{\left (3 x^{4} + 5 x^{2} + 2\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(1/(3*x**4+5*x**2+2)**(3/2),x)
 

Output:

Integral((3*x**4 + 5*x**2 + 2)**(-3/2), x)
 

Maxima [F]

\[ \int \frac {1}{\left (2+5 x^2+3 x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (3 \, x^{4} + 5 \, x^{2} + 2\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/(3*x^4+5*x^2+2)^(3/2),x, algorithm="maxima")
 

Output:

integrate((3*x^4 + 5*x^2 + 2)^(-3/2), x)
 

Giac [F]

\[ \int \frac {1}{\left (2+5 x^2+3 x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (3 \, x^{4} + 5 \, x^{2} + 2\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/(3*x^4+5*x^2+2)^(3/2),x, algorithm="giac")
 

Output:

integrate((3*x^4 + 5*x^2 + 2)^(-3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (2+5 x^2+3 x^4\right )^{3/2}} \, dx=\int \frac {1}{{\left (3\,x^4+5\,x^2+2\right )}^{3/2}} \,d x \] Input:

int(1/(5*x^2 + 3*x^4 + 2)^(3/2),x)
 

Output:

int(1/(5*x^2 + 3*x^4 + 2)^(3/2), x)
 

Reduce [F]

\[ \int \frac {1}{\left (2+5 x^2+3 x^4\right )^{3/2}} \, dx=\int \frac {\sqrt {3 x^{4}+5 x^{2}+2}}{9 x^{8}+30 x^{6}+37 x^{4}+20 x^{2}+4}d x \] Input:

int(1/(3*x^4+5*x^2+2)^(3/2),x)
 

Output:

int(sqrt(3*x**4 + 5*x**2 + 2)/(9*x**8 + 30*x**6 + 37*x**4 + 20*x**2 + 4),x 
)