\(\int \frac {1}{(-3+5 x^2+2 x^4)^{3/2}} \, dx\) [219]

Optimal result

Integrand size = 16, antiderivative size = 147 \[ \int \frac {1}{\left (-3+5 x^2+2 x^4\right )^{3/2}} \, dx=-\frac {x \left (37+10 x^2\right )}{147 \sqrt {-3+5 x^2+2 x^4}}+\frac {5 \sqrt {\frac {2}{3}} \sqrt {1-2 x^2} \sqrt {3+x^2} E\left (\arcsin \left (\sqrt {2} x\right )|-\frac {1}{6}\right )}{49 \sqrt {-3+5 x^2+2 x^4}}-\frac {\sqrt {\frac {2}{3}} \sqrt {1-2 x^2} \sqrt {3+x^2} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {2} x\right ),-\frac {1}{6}\right )}{7 \sqrt {-3+5 x^2+2 x^4}} \] Output:

-1/147*x*(10*x^2+37)/(2*x^4+5*x^2-3)^(1/2)+5/147*6^(1/2)*(-2*x^2+1)^(1/2)* 
(x^2+3)^(1/2)*EllipticE(x*2^(1/2),1/6*I*6^(1/2))/(2*x^4+5*x^2-3)^(1/2)-1/2 
1*(-2*x^2+1)^(1/2)*(x^2+3)^(1/2)*EllipticF(x*2^(1/2),1/6*I*6^(1/2))*6^(1/2 
)/(2*x^4+5*x^2-3)^(1/2)
 

Mathematica [A] (verified)

Time = 6.51 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.66 \[ \int \frac {1}{\left (-3+5 x^2+2 x^4\right )^{3/2}} \, dx=\frac {-37 x-10 x^3+5 \sqrt {6-12 x^2} \sqrt {3+x^2} E\left (\arcsin \left (\sqrt {2} x\right )|-\frac {1}{6}\right )-7 \sqrt {6-12 x^2} \sqrt {3+x^2} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {2} x\right ),-\frac {1}{6}\right )}{147 \sqrt {-3+5 x^2+2 x^4}} \] Input:

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

Output:

(-37*x - 10*x^3 + 5*Sqrt[6 - 12*x^2]*Sqrt[3 + x^2]*EllipticE[ArcSin[Sqrt[2 
]*x], -1/6] - 7*Sqrt[6 - 12*x^2]*Sqrt[3 + x^2]*EllipticF[ArcSin[Sqrt[2]*x] 
, -1/6])/(147*Sqrt[-3 + 5*x^2 + 2*x^4])
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.52, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {1405, 27, 1501, 27, 1410, 1498}

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

\(\displaystyle \frac {1}{147} \int -\frac {2 \left (6-5 x^2\right )}{\sqrt {2 x^4+5 x^2-3}}dx-\frac {x \left (10 x^2+37\right )}{147 \sqrt {2 x^4+5 x^2-3}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {2}{147} \int \frac {6-5 x^2}{\sqrt {2 x^4+5 x^2-3}}dx-\frac {x \left (10 x^2+37\right )}{147 \sqrt {2 x^4+5 x^2-3}}\)

\(\Big \downarrow \) 1501

\(\displaystyle -\frac {2}{147} \left (\frac {7}{2} \int \frac {1}{\sqrt {2 x^4+5 x^2-3}}dx-\frac {5}{4} \int -\frac {2 \left (1-2 x^2\right )}{\sqrt {2 x^4+5 x^2-3}}dx\right )-\frac {x \left (10 x^2+37\right )}{147 \sqrt {2 x^4+5 x^2-3}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {2}{147} \left (\frac {7}{2} \int \frac {1}{\sqrt {2 x^4+5 x^2-3}}dx+\frac {5}{2} \int \frac {1-2 x^2}{\sqrt {2 x^4+5 x^2-3}}dx\right )-\frac {x \left (10 x^2+37\right )}{147 \sqrt {2 x^4+5 x^2-3}}\)

\(\Big \downarrow \) 1410

\(\displaystyle -\frac {2}{147} \left (\frac {5}{2} \int \frac {1-2 x^2}{\sqrt {2 x^4+5 x^2-3}}dx+\frac {\sqrt {7} \sqrt {x^2+3} \sqrt {2 x^2-1} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {7}{3}} x}{\sqrt {2 x^2-1}}\right ),\frac {6}{7}\right )}{2 \sqrt {2 x^4+5 x^2-3}}\right )-\frac {x \left (10 x^2+37\right )}{147 \sqrt {2 x^4+5 x^2-3}}\)

\(\Big \downarrow \) 1498

\(\displaystyle -\frac {2}{147} \left (\frac {\sqrt {7} \sqrt {x^2+3} \sqrt {2 x^2-1} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {\frac {7}{3}} x}{\sqrt {2 x^2-1}}\right ),\frac {6}{7}\right )}{2 \sqrt {2 x^4+5 x^2-3}}+\frac {5}{2} \left (\frac {\sqrt {7} \sqrt {\frac {x^2+3}{1-2 x^2}} \sqrt {2 x^2-1} E\left (\arcsin \left (\frac {\sqrt {\frac {7}{3}} x}{\sqrt {2 x^2-1}}\right )|\frac {6}{7}\right )}{\sqrt {\frac {1}{1-2 x^2}} \sqrt {2 x^4+5 x^2-3}}-\frac {2 x \left (x^2+3\right )}{\sqrt {2 x^4+5 x^2-3}}\right )\right )-\frac {x \left (10 x^2+37\right )}{147 \sqrt {2 x^4+5 x^2-3}}\)

Input:

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

Output:

-1/147*(x*(37 + 10*x^2))/Sqrt[-3 + 5*x^2 + 2*x^4] - (2*((5*((-2*x*(3 + x^2 
))/Sqrt[-3 + 5*x^2 + 2*x^4] + (Sqrt[7]*Sqrt[(3 + x^2)/(1 - 2*x^2)]*Sqrt[-1 
 + 2*x^2]*EllipticE[ArcSin[(Sqrt[7/3]*x)/Sqrt[-1 + 2*x^2]], 6/7])/(Sqrt[(1 
 - 2*x^2)^(-1)]*Sqrt[-3 + 5*x^2 + 2*x^4])))/2 + (Sqrt[7]*Sqrt[3 + x^2]*Sqr 
t[-1 + 2*x^2]*EllipticF[ArcSin[(Sqrt[7/3]*x)/Sqrt[-1 + 2*x^2]], 6/7])/(2*S 
qrt[-3 + 5*x^2 + 2*x^4])))/147
 

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 1410

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b 
^2 - 4*a*c, 2]}, Simp[Sqrt[-2*a - (b - q)*x^2]*(Sqrt[(2*a + (b + q)*x^2)/q] 
/(2*Sqrt[-a]*Sqrt[a + b*x^2 + c*x^4]))*EllipticF[ArcSin[x/Sqrt[(2*a + (b + 
q)*x^2)/(2*q)]], (b + q)/(2*q)], x] /; IntegerQ[q]] /; FreeQ[{a, b, c}, x] 
&& GtQ[b^2 - 4*a*c, 0] && LtQ[a, 0] && GtQ[c, 0]
 

rule 1498

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[e*x*((b + q + 2*c*x^2)/(2*c*Sqrt[ 
a + b*x^2 + c*x^4])), x] - Simp[e*q*Sqrt[(2*a + (b - q)*x^2)/(2*a + (b + q) 
*x^2)]*(Sqrt[(2*a + (b + q)*x^2)/q]/(2*c*Sqrt[a + b*x^2 + c*x^4]*Sqrt[a/(2* 
a + (b + q)*x^2)]))*EllipticE[ArcSin[x/Sqrt[(2*a + (b + q)*x^2)/(2*q)]], (b 
 + q)/(2*q)], x] /; EqQ[2*c*d - e*(b - q), 0]] /; FreeQ[{a, b, c, d, e}, x] 
 && GtQ[b^2 - 4*a*c, 0] && LtQ[a, 0] && GtQ[c, 0]
 

rule 1501

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[(2*c*d - e*(b - q))/(2*c)   Int[1 
/Sqrt[a + b*x^2 + c*x^4], x], x] + Simp[e/(2*c)   Int[(b - q + 2*c*x^2)/Sqr 
t[a + b*x^2 + c*x^4], x], x] /; NeQ[2*c*d - e*(b - q), 0]] /; FreeQ[{a, b, 
c, d, e}, x] && GtQ[b^2 - 4*a*c, 0] && LtQ[a, 0] && GtQ[c, 0]
 

Maple [A] (verified)

Time = 2.29 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00

method	result	size
risch	\(-\frac {x \left (10 x^{2}+37\right )}{147 \sqrt {2 x^{4}+5 x^{2}-3}}+\frac {4 i \sqrt {3}\, \sqrt {3 x^{2}+9}\, \sqrt {-2 x^{2}+1}\, \operatorname {EllipticF}\left (\frac {i x \sqrt {3}}{3}, i \sqrt {6}\right )}{147 \sqrt {2 x^{4}+5 x^{2}-3}}-\frac {5 i \sqrt {3}\, \sqrt {3 x^{2}+9}\, \sqrt {-2 x^{2}+1}\, \left (\operatorname {EllipticF}\left (\frac {i x \sqrt {3}}{3}, i \sqrt {6}\right )-\operatorname {EllipticE}\left (\frac {i x \sqrt {3}}{3}, i \sqrt {6}\right )\right )}{441 \sqrt {2 x^{4}+5 x^{2}-3}}\)	\(147\)
default	\(-\frac {4 \left (\frac {37}{588} x +\frac {5}{294} x^{3}\right )}{\sqrt {2 x^{4}+5 x^{2}-3}}+\frac {4 i \sqrt {3}\, \sqrt {3 x^{2}+9}\, \sqrt {-2 x^{2}+1}\, \operatorname {EllipticF}\left (\frac {i x \sqrt {3}}{3}, i \sqrt {6}\right )}{147 \sqrt {2 x^{4}+5 x^{2}-3}}-\frac {5 i \sqrt {3}\, \sqrt {3 x^{2}+9}\, \sqrt {-2 x^{2}+1}\, \left (\operatorname {EllipticF}\left (\frac {i x \sqrt {3}}{3}, i \sqrt {6}\right )-\operatorname {EllipticE}\left (\frac {i x \sqrt {3}}{3}, i \sqrt {6}\right )\right )}{441 \sqrt {2 x^{4}+5 x^{2}-3}}\)	\(148\)
elliptic	\(-\frac {4 \left (\frac {37}{588} x +\frac {5}{294} x^{3}\right )}{\sqrt {2 x^{4}+5 x^{2}-3}}+\frac {4 i \sqrt {3}\, \sqrt {3 x^{2}+9}\, \sqrt {-2 x^{2}+1}\, \operatorname {EllipticF}\left (\frac {i x \sqrt {3}}{3}, i \sqrt {6}\right )}{147 \sqrt {2 x^{4}+5 x^{2}-3}}-\frac {5 i \sqrt {3}\, \sqrt {3 x^{2}+9}\, \sqrt {-2 x^{2}+1}\, \left (\operatorname {EllipticF}\left (\frac {i x \sqrt {3}}{3}, i \sqrt {6}\right )-\operatorname {EllipticE}\left (\frac {i x \sqrt {3}}{3}, i \sqrt {6}\right )\right )}{441 \sqrt {2 x^{4}+5 x^{2}-3}}\)	\(148\)

Input:

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

Output:

-1/147*x*(10*x^2+37)/(2*x^4+5*x^2-3)^(1/2)+4/147*I*3^(1/2)*(3*x^2+9)^(1/2) 
*(-2*x^2+1)^(1/2)/(2*x^4+5*x^2-3)^(1/2)*EllipticF(1/3*I*x*3^(1/2),I*6^(1/2 
))-5/441*I*3^(1/2)*(3*x^2+9)^(1/2)*(-2*x^2+1)^(1/2)/(2*x^4+5*x^2-3)^(1/2)* 
(EllipticF(1/3*I*x*3^(1/2),I*6^(1/2))-EllipticE(1/3*I*x*3^(1/2),I*6^(1/2)) 
)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.66 \[ \int \frac {1}{\left (-3+5 x^2+2 x^4\right )^{3/2}} \, dx=-\frac {10 \, \sqrt {2} \sqrt {-3} {\left (2 \, x^{4} + 5 \, x^{2} - 3\right )} E(\arcsin \left (\sqrt {2} x\right )\,|\,-\frac {1}{6}) - 12 \, \sqrt {2} \sqrt {-3} {\left (2 \, x^{4} + 5 \, x^{2} - 3\right )} F(\arcsin \left (\sqrt {2} x\right )\,|\,-\frac {1}{6}) + \sqrt {2 \, x^{4} + 5 \, x^{2} - 3} {\left (10 \, x^{3} + 37 \, x\right )}}{147 \, {\left (2 \, x^{4} + 5 \, x^{2} - 3\right )}} \] Input:

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

Output:

-1/147*(10*sqrt(2)*sqrt(-3)*(2*x^4 + 5*x^2 - 3)*elliptic_e(arcsin(sqrt(2)* 
x), -1/6) - 12*sqrt(2)*sqrt(-3)*(2*x^4 + 5*x^2 - 3)*elliptic_f(arcsin(sqrt 
(2)*x), -1/6) + sqrt(2*x^4 + 5*x^2 - 3)*(10*x^3 + 37*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]

\[ \int \frac {1}{\left (-3+5 x^2+2 x^4\right )^{3/2}} \, dx=\int \frac {\sqrt {2 x^{4}+5 x^{2}-3}}{4 x^{8}+20 x^{6}+13 x^{4}-30 x^{2}+9}d x \] 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 + 13*x**4 - 30*x**2 + 9),x 
)