\(\int \frac {1}{(3+5 x^2+2 x^4)^{3/2}} \, dx\) [249]

Optimal result

Integrand size = 16, antiderivative size = 135 \[ \int \frac {1}{\left (3+5 x^2+2 x^4\right )^{3/2}} \, dx=\frac {x}{\sqrt {3+5 x^2+2 x^4}}+\frac {5 \sqrt {2} \sqrt {3+5 x^2+2 x^4} E\left (\arctan \left (\sqrt {\frac {2}{3}} x\right )|-\frac {1}{2}\right )}{3 \sqrt {1+x^2} \sqrt {3+2 x^2}}-\frac {2 \sqrt {2} \sqrt {1+x^2} \sqrt {3+2 x^2} \operatorname {EllipticF}\left (\arctan \left (\sqrt {\frac {2}{3}} x\right ),-\frac {1}{2}\right )}{\sqrt {3+5 x^2+2 x^4}} \] Output:

x/(2*x^4+5*x^2+3)^(1/2)+5/3*2^(1/2)*(2*x^4+5*x^2+3)^(1/2)*EllipticE(x*6^(1 
/2)/(6*x^2+9)^(1/2),1/2*I*2^(1/2))/(x^2+1)^(1/2)/(2*x^2+3)^(1/2)-2*(x^2+1) 
^(1/2)*(2*x^2+3)^(1/2)*InverseJacobiAM(arctan(1/3*x*6^(1/2)),1/2*I*2^(1/2) 
)*2^(1/2)/(2*x^4+5*x^2+3)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 5.84 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\left (3+5 x^2+2 x^4\right )^{3/2}} \, dx=\frac {13 x+10 x^3+5 i \sqrt {2} \sqrt {1+x^2} \sqrt {3+2 x^2} E\left (i \text {arcsinh}\left (\sqrt {\frac {2}{3}} x\right )|\frac {3}{2}\right )+i \sqrt {2} \sqrt {1+x^2} \sqrt {3+2 x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {2}{3}} x\right ),\frac {3}{2}\right )}{3 \sqrt {3+5 x^2+2 x^4}} \] Input:

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

Output:

(13*x + 10*x^3 + (5*I)*Sqrt[2]*Sqrt[1 + x^2]*Sqrt[3 + 2*x^2]*EllipticE[I*A 
rcSinh[Sqrt[2/3]*x], 3/2] + I*Sqrt[2]*Sqrt[1 + x^2]*Sqrt[3 + 2*x^2]*Ellipt 
icF[I*ArcSinh[Sqrt[2/3]*x], 3/2])/(3*Sqrt[3 + 5*x^2 + 2*x^4])
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.28, 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, 1412, 1455}

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.

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

\(\Big \downarrow \) 1405

\(\displaystyle \frac {x \left (10 x^2+13\right )}{3 \sqrt {2 x^4+5 x^2+3}}-\frac {1}{3} \int \frac {2 \left (5 x^2+6\right )}{\sqrt {2 x^4+5 x^2+3}}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x \left (10 x^2+13\right )}{3 \sqrt {2 x^4+5 x^2+3}}-\frac {2}{3} \int \frac {5 x^2+6}{\sqrt {2 x^4+5 x^2+3}}dx\)

\(\Big \downarrow \) 1503

\(\displaystyle \frac {x \left (10 x^2+13\right )}{3 \sqrt {2 x^4+5 x^2+3}}-\frac {2}{3} \left (6 \int \frac {1}{\sqrt {2 x^4+5 x^2+3}}dx+5 \int \frac {x^2}{\sqrt {2 x^4+5 x^2+3}}dx\right )\)

\(\Big \downarrow \) 1412

\(\displaystyle \frac {x \left (10 x^2+13\right )}{3 \sqrt {2 x^4+5 x^2+3}}-\frac {2}{3} \left (5 \int \frac {x^2}{\sqrt {2 x^4+5 x^2+3}}dx+\frac {2 \sqrt {3} \left (x^2+1\right ) \sqrt {\frac {2 x^2+3}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),\frac {1}{3}\right )}{\sqrt {2 x^4+5 x^2+3}}\right )\)

\(\Big \downarrow \) 1455

\(\displaystyle \frac {x \left (10 x^2+13\right )}{3 \sqrt {2 x^4+5 x^2+3}}-\frac {2}{3} \left (\frac {2 \sqrt {3} \left (x^2+1\right ) \sqrt {\frac {2 x^2+3}{x^2+1}} \operatorname {EllipticF}\left (\arctan (x),\frac {1}{3}\right )}{\sqrt {2 x^4+5 x^2+3}}+5 \left (\frac {x \left (2 x^2+3\right )}{2 \sqrt {2 x^4+5 x^2+3}}-\frac {\sqrt {3} \left (x^2+1\right ) \sqrt {\frac {2 x^2+3}{x^2+1}} E\left (\arctan (x)\left |\frac {1}{3}\right .\right )}{2 \sqrt {2 x^4+5 x^2+3}}\right )\right )\)

Input:

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

Output:

(x*(13 + 10*x^2))/(3*Sqrt[3 + 5*x^2 + 2*x^4]) - (2*(5*((x*(3 + 2*x^2))/(2* 
Sqrt[3 + 5*x^2 + 2*x^4]) - (Sqrt[3]*(1 + x^2)*Sqrt[(3 + 2*x^2)/(1 + x^2)]* 
EllipticE[ArcTan[x], 1/3])/(2*Sqrt[3 + 5*x^2 + 2*x^4])) + (2*Sqrt[3]*(1 + 
x^2)*Sqrt[(3 + 2*x^2)/(1 + x^2)]*EllipticF[ArcTan[x], 1/3])/Sqrt[3 + 5*x^2 
 + 2*x^4]))/3
 

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 1412

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] && 
!(PosQ[(b - q)/a] && SimplerSqrtQ[(b - q)/(2*a), (b + q)/(2*a)])] /; FreeQ[ 
{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]
 

rule 1455

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] &&  !(PosQ[ 
(b - q)/a] && SimplerSqrtQ[(b - q)/(2*a), (b + q)/(2*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 = 2.57 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.04

method	result	size
risch	\(\frac {x \left (10 x^{2}+13\right )}{3 \sqrt {2 x^{4}+5 x^{2}+3}}+\frac {2 i \sqrt {6}\, \sqrt {6 x^{2}+9}\, \sqrt {x^{2}+1}\, \operatorname {EllipticF}\left (\frac {i x \sqrt {6}}{3}, \frac {\sqrt {6}}{2}\right )}{3 \sqrt {2 x^{4}+5 x^{2}+3}}-\frac {5 i \sqrt {6}\, \sqrt {6 x^{2}+9}\, \sqrt {x^{2}+1}\, \left (\operatorname {EllipticF}\left (\frac {i x \sqrt {6}}{3}, \frac {\sqrt {6}}{2}\right )-\operatorname {EllipticE}\left (\frac {i x \sqrt {6}}{3}, \frac {\sqrt {6}}{2}\right )\right )}{9 \sqrt {2 x^{4}+5 x^{2}+3}}\)	\(140\)
default	\(-\frac {4 \left (-\frac {5}{6} x^{3}-\frac {13}{12} x \right )}{\sqrt {2 x^{4}+5 x^{2}+3}}+\frac {2 i \sqrt {6}\, \sqrt {6 x^{2}+9}\, \sqrt {x^{2}+1}\, \operatorname {EllipticF}\left (\frac {i x \sqrt {6}}{3}, \frac {\sqrt {6}}{2}\right )}{3 \sqrt {2 x^{4}+5 x^{2}+3}}-\frac {5 i \sqrt {6}\, \sqrt {6 x^{2}+9}\, \sqrt {x^{2}+1}\, \left (\operatorname {EllipticF}\left (\frac {i x \sqrt {6}}{3}, \frac {\sqrt {6}}{2}\right )-\operatorname {EllipticE}\left (\frac {i x \sqrt {6}}{3}, \frac {\sqrt {6}}{2}\right )\right )}{9 \sqrt {2 x^{4}+5 x^{2}+3}}\)	\(141\)
elliptic	\(-\frac {4 \left (-\frac {5}{6} x^{3}-\frac {13}{12} x \right )}{\sqrt {2 x^{4}+5 x^{2}+3}}+\frac {2 i \sqrt {6}\, \sqrt {6 x^{2}+9}\, \sqrt {x^{2}+1}\, \operatorname {EllipticF}\left (\frac {i x \sqrt {6}}{3}, \frac {\sqrt {6}}{2}\right )}{3 \sqrt {2 x^{4}+5 x^{2}+3}}-\frac {5 i \sqrt {6}\, \sqrt {6 x^{2}+9}\, \sqrt {x^{2}+1}\, \left (\operatorname {EllipticF}\left (\frac {i x \sqrt {6}}{3}, \frac {\sqrt {6}}{2}\right )-\operatorname {EllipticE}\left (\frac {i x \sqrt {6}}{3}, \frac {\sqrt {6}}{2}\right )\right )}{9 \sqrt {2 x^{4}+5 x^{2}+3}}\)	\(141\)

Input:

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

Output:

1/3*x*(10*x^2+13)/(2*x^4+5*x^2+3)^(1/2)+2/3*I*6^(1/2)*(6*x^2+9)^(1/2)*(x^2 
+1)^(1/2)/(2*x^4+5*x^2+3)^(1/2)*EllipticF(1/3*I*x*6^(1/2),1/2*6^(1/2))-5/9 
*I*6^(1/2)*(6*x^2+9)^(1/2)*(x^2+1)^(1/2)/(2*x^4+5*x^2+3)^(1/2)*(EllipticF( 
1/3*I*x*6^(1/2),1/2*6^(1/2))-EllipticE(1/3*I*x*6^(1/2),1/2*6^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.73 \[ \int \frac {1}{\left (3+5 x^2+2 x^4\right )^{3/2}} \, dx=-\frac {10 \, \sqrt {3} \sqrt {-\frac {2}{3}} {\left (2 \, x^{4} + 5 \, x^{2} + 3\right )} E(\arcsin \left (\sqrt {-\frac {2}{3}} x\right )\,|\,\frac {3}{2}) - 28 \, \sqrt {3} \sqrt {-\frac {2}{3}} {\left (2 \, x^{4} + 5 \, x^{2} + 3\right )} F(\arcsin \left (\sqrt {-\frac {2}{3}} x\right )\,|\,\frac {3}{2}) - 3 \, \sqrt {2 \, x^{4} + 5 \, x^{2} + 3} {\left (10 \, x^{3} + 13 \, x\right )}}{9 \, {\left (2 \, x^{4} + 5 \, x^{2} + 3\right )}} \] Input:

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

Output:

-1/9*(10*sqrt(3)*sqrt(-2/3)*(2*x^4 + 5*x^2 + 3)*elliptic_e(arcsin(sqrt(-2/ 
3)*x), 3/2) - 28*sqrt(3)*sqrt(-2/3)*(2*x^4 + 5*x^2 + 3)*elliptic_f(arcsin( 
sqrt(-2/3)*x), 3/2) - 3*sqrt(2*x^4 + 5*x^2 + 3)*(10*x^3 + 13*x))/(2*x^4 + 
5*x^2 + 3)
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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