Integrand size = 16, antiderivative size = 225 \[ \int \frac {1}{\left (-3+6 x^2+2 x^4\right )^{3/2}} \, dx=-\frac {x \left (4+x^2\right )}{15 \sqrt {-3+6 x^2+2 x^4}}+\frac {\sqrt {\frac {1}{3} \left (3+\sqrt {15}\right )} \sqrt {3-\left (3-\sqrt {15}\right ) x^2} \sqrt {3-\left (3+\sqrt {15}\right ) x^2} E\left (\arcsin \left (\sqrt {\frac {1}{3} \left (3+\sqrt {15}\right )} x\right )|-4+\sqrt {15}\right )}{30 \sqrt {-3+6 x^2+2 x^4}}-\frac {\sqrt {\frac {1}{5} \left (3+\sqrt {15}\right )} \sqrt {3-\left (3-\sqrt {15}\right ) x^2} \sqrt {3-\left (3+\sqrt {15}\right ) x^2} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {1}{3} \left (3+\sqrt {15}\right )} x\right ),-4+\sqrt {15}\right )}{18 \sqrt {-3+6 x^2+2 x^4}} \] Output:
-1/15*x*(x^2+4)/(2*x^4+6*x^2-3)^(1/2)+1/90*(9+3*15^(1/2))^(1/2)*(3-(3-15^( 1/2))*x^2)^(1/2)*(3-(3+15^(1/2))*x^2)^(1/2)*EllipticE(1/3*(9+3*15^(1/2))^( 1/2)*x,1/2*I*10^(1/2)-1/2*I*6^(1/2))/(2*x^4+6*x^2-3)^(1/2)-1/90*(15+5*15^( 1/2))^(1/2)*(3-(3-15^(1/2))*x^2)^(1/2)*(3-(3+15^(1/2))*x^2)^(1/2)*Elliptic F(1/3*(9+3*15^(1/2))^(1/2)*x,1/2*I*10^(1/2)-1/2*I*6^(1/2))/(2*x^4+6*x^2-3) ^(1/2)
Result contains complex when optimal does not.
Time = 6.59 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.70 \[ \int \frac {1}{\left (-3+6 x^2+2 x^4\right )^{3/2}} \, dx=\frac {-4 x \left (4+x^2\right )+2 i \sqrt {-3+\sqrt {15}} \sqrt {3-6 x^2-2 x^4} E\left (i \text {arcsinh}\left (\sqrt {-1+\sqrt {\frac {5}{3}}} x\right )|-4-\sqrt {15}\right )-\frac {2 i \left (-5+\sqrt {15}\right ) \sqrt {3-6 x^2-2 x^4} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-1+\sqrt {\frac {5}{3}}} x\right ),-4-\sqrt {15}\right )}{\sqrt {-3+\sqrt {15}}}}{60 \sqrt {-3+6 x^2+2 x^4}} \] Input:
Integrate[(-3 + 6*x^2 + 2*x^4)^(-3/2),x]
Output:
(-4*x*(4 + x^2) + (2*I)*Sqrt[-3 + Sqrt[15]]*Sqrt[3 - 6*x^2 - 2*x^4]*Ellipt icE[I*ArcSinh[Sqrt[-1 + Sqrt[5/3]]*x], -4 - Sqrt[15]] - ((2*I)*(-5 + Sqrt[ 15])*Sqrt[3 - 6*x^2 - 2*x^4]*EllipticF[I*ArcSinh[Sqrt[-1 + Sqrt[5/3]]*x], -4 - Sqrt[15]])/Sqrt[-3 + Sqrt[15]])/(60*Sqrt[-3 + 6*x^2 + 2*x^4])
Time = 0.75 (sec) , antiderivative size = 373, normalized size of antiderivative = 1.66, 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, 1411, 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.
| \(\displaystyle \int \frac {1}{\left (2 x^4+6 x^2-3\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1405 |
| \(\displaystyle \frac {1}{180} \int -\frac {12 \left (1-x^2\right )}{\sqrt {2 x^4+6 x^2-3}}dx-\frac {x \left (x^2+4\right )}{15 \sqrt {2 x^4+6 x^2-3}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {1}{15} \int \frac {1-x^2}{\sqrt {2 x^4+6 x^2-3}}dx-\frac {x \left (x^2+4\right )}{15 \sqrt {2 x^4+6 x^2-3}}\) |
| \(\Big \downarrow \) 1501 |
| \(\displaystyle \frac {1}{15} \left (\frac {1}{4} \int \frac {2 \left (2 x^2-\sqrt {15}+3\right )}{\sqrt {2 x^4+6 x^2-3}}dx-\frac {1}{2} \left (5-\sqrt {15}\right ) \int \frac {1}{\sqrt {2 x^4+6 x^2-3}}dx\right )-\frac {x \left (x^2+4\right )}{15 \sqrt {2 x^4+6 x^2-3}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{15} \left (\frac {1}{2} \int \frac {2 x^2-\sqrt {15}+3}{\sqrt {2 x^4+6 x^2-3}}dx-\frac {1}{2} \left (5-\sqrt {15}\right ) \int \frac {1}{\sqrt {2 x^4+6 x^2-3}}dx\right )-\frac {x \left (x^2+4\right )}{15 \sqrt {2 x^4+6 x^2-3}}\) |
| \(\Big \downarrow \) 1411 |
| \(\displaystyle \frac {1}{15} \left (\frac {1}{2} \int \frac {2 x^2-\sqrt {15}+3}{\sqrt {2 x^4+6 x^2-3}}dx-\frac {\left (5-\sqrt {15}\right ) \sqrt {\frac {3-\left (3-\sqrt {15}\right ) x^2}{3-\left (3+\sqrt {15}\right ) x^2}} \sqrt {\left (3+\sqrt {15}\right ) x^2-3} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} \sqrt [4]{15} x}{\sqrt {\left (3+\sqrt {15}\right ) x^2-3}}\right ),\frac {1}{10} \left (5+\sqrt {15}\right )\right )}{2 \sqrt {2} 3^{3/4} \sqrt [4]{5} \sqrt {\frac {1}{3-\left (3+\sqrt {15}\right ) x^2}} \sqrt {2 x^4+6 x^2-3}}\right )-\frac {x \left (x^2+4\right )}{15 \sqrt {2 x^4+6 x^2-3}}\) |
| \(\Big \downarrow \) 1498 |
| \(\displaystyle \frac {1}{15} \left (\frac {1}{2} \left (\frac {x \left (2 x^2+\sqrt {15}+3\right )}{\sqrt {2 x^4+6 x^2-3}}-\frac {\sqrt [4]{\frac {5}{3}} \sqrt {2} \sqrt {\frac {3-\left (3-\sqrt {15}\right ) x^2}{3-\left (3+\sqrt {15}\right ) x^2}} \sqrt {\left (3+\sqrt {15}\right ) x^2-3} E\left (\arcsin \left (\frac {\sqrt {2} \sqrt [4]{15} x}{\sqrt {\left (3+\sqrt {15}\right ) x^2-3}}\right )|\frac {1}{10} \left (5+\sqrt {15}\right )\right )}{\sqrt {\frac {1}{3-\left (3+\sqrt {15}\right ) x^2}} \sqrt {2 x^4+6 x^2-3}}\right )-\frac {\left (5-\sqrt {15}\right ) \sqrt {\frac {3-\left (3-\sqrt {15}\right ) x^2}{3-\left (3+\sqrt {15}\right ) x^2}} \sqrt {\left (3+\sqrt {15}\right ) x^2-3} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {2} \sqrt [4]{15} x}{\sqrt {\left (3+\sqrt {15}\right ) x^2-3}}\right ),\frac {1}{10} \left (5+\sqrt {15}\right )\right )}{2 \sqrt {2} 3^{3/4} \sqrt [4]{5} \sqrt {\frac {1}{3-\left (3+\sqrt {15}\right ) x^2}} \sqrt {2 x^4+6 x^2-3}}\right )-\frac {x \left (x^2+4\right )}{15 \sqrt {2 x^4+6 x^2-3}}\) |
Input:
Int[(-3 + 6*x^2 + 2*x^4)^(-3/2),x]
Output:
-1/15*(x*(4 + x^2))/Sqrt[-3 + 6*x^2 + 2*x^4] + (((x*(3 + Sqrt[15] + 2*x^2) )/Sqrt[-3 + 6*x^2 + 2*x^4] - ((5/3)^(1/4)*Sqrt[2]*Sqrt[(3 - (3 - Sqrt[15]) *x^2)/(3 - (3 + Sqrt[15])*x^2)]*Sqrt[-3 + (3 + Sqrt[15])*x^2]*EllipticE[Ar cSin[(Sqrt[2]*15^(1/4)*x)/Sqrt[-3 + (3 + Sqrt[15])*x^2]], (5 + Sqrt[15])/1 0])/(Sqrt[(3 - (3 + Sqrt[15])*x^2)^(-1)]*Sqrt[-3 + 6*x^2 + 2*x^4]))/2 - (( 5 - Sqrt[15])*Sqrt[(3 - (3 - Sqrt[15])*x^2)/(3 - (3 + Sqrt[15])*x^2)]*Sqrt [-3 + (3 + Sqrt[15])*x^2]*EllipticF[ArcSin[(Sqrt[2]*15^(1/4)*x)/Sqrt[-3 + (3 + Sqrt[15])*x^2]], (5 + Sqrt[15])/10])/(2*Sqrt[2]*3^(3/4)*5^(1/4)*Sqrt[ (3 - (3 + Sqrt[15])*x^2)^(-1)]*Sqrt[-3 + 6*x^2 + 2*x^4]))/15
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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)/(2*a + (b + q)*x^2)]*(Sqrt[( 2*a + (b + q)*x^2)/q]/(2*Sqrt[a + b*x^2 + c*x^4]*Sqrt[a/(2*a + (b + q)*x^2) ]))*EllipticF[ArcSin[x/Sqrt[(2*a + (b + q)*x^2)/(2*q)]], (b + q)/(2*q)], x] ] /; FreeQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0] && LtQ[a, 0] && GtQ[c, 0]
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]
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]
Time = 1.79 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.01
| method | result | size |
| risch | \(-\frac {x \left (x^{2}+4\right )}{15 \sqrt {2 x^{4}+6 x^{2}-3}}-\frac {\sqrt {1-\left (1-\frac {\sqrt {15}}{3}\right ) x^{2}}\, \sqrt {1-\left (1+\frac {\sqrt {15}}{3}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {\sqrt {9-3 \sqrt {15}}\, x}{3}, \frac {i \sqrt {6}}{2}+\frac {i \sqrt {10}}{2}\right )}{5 \sqrt {9-3 \sqrt {15}}\, \sqrt {2 x^{4}+6 x^{2}-3}}+\frac {6 \sqrt {1-\left (1-\frac {\sqrt {15}}{3}\right ) x^{2}}\, \sqrt {1-\left (1+\frac {\sqrt {15}}{3}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {\sqrt {9-3 \sqrt {15}}\, x}{3}, \frac {i \sqrt {6}}{2}+\frac {i \sqrt {10}}{2}\right )-\operatorname {EllipticE}\left (\frac {\sqrt {9-3 \sqrt {15}}\, x}{3}, \frac {i \sqrt {6}}{2}+\frac {i \sqrt {10}}{2}\right )\right )}{5 \sqrt {9-3 \sqrt {15}}\, \sqrt {2 x^{4}+6 x^{2}-3}\, \left (6+2 \sqrt {15}\right )}\) | \(228\) |
| default | \(-\frac {4 \left (\frac {1}{15} x +\frac {1}{60} x^{3}\right )}{\sqrt {2 x^{4}+6 x^{2}-3}}-\frac {\sqrt {1-\left (1-\frac {\sqrt {15}}{3}\right ) x^{2}}\, \sqrt {1-\left (1+\frac {\sqrt {15}}{3}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {\sqrt {9-3 \sqrt {15}}\, x}{3}, \frac {i \sqrt {6}}{2}+\frac {i \sqrt {10}}{2}\right )}{5 \sqrt {9-3 \sqrt {15}}\, \sqrt {2 x^{4}+6 x^{2}-3}}+\frac {6 \sqrt {1-\left (1-\frac {\sqrt {15}}{3}\right ) x^{2}}\, \sqrt {1-\left (1+\frac {\sqrt {15}}{3}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {\sqrt {9-3 \sqrt {15}}\, x}{3}, \frac {i \sqrt {6}}{2}+\frac {i \sqrt {10}}{2}\right )-\operatorname {EllipticE}\left (\frac {\sqrt {9-3 \sqrt {15}}\, x}{3}, \frac {i \sqrt {6}}{2}+\frac {i \sqrt {10}}{2}\right )\right )}{5 \sqrt {9-3 \sqrt {15}}\, \sqrt {2 x^{4}+6 x^{2}-3}\, \left (6+2 \sqrt {15}\right )}\) | \(231\) |
| elliptic | \(-\frac {4 \left (\frac {1}{15} x +\frac {1}{60} x^{3}\right )}{\sqrt {2 x^{4}+6 x^{2}-3}}-\frac {\sqrt {1-\left (1-\frac {\sqrt {15}}{3}\right ) x^{2}}\, \sqrt {1-\left (1+\frac {\sqrt {15}}{3}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {\sqrt {9-3 \sqrt {15}}\, x}{3}, \frac {i \sqrt {6}}{2}+\frac {i \sqrt {10}}{2}\right )}{5 \sqrt {9-3 \sqrt {15}}\, \sqrt {2 x^{4}+6 x^{2}-3}}+\frac {6 \sqrt {1-\left (1-\frac {\sqrt {15}}{3}\right ) x^{2}}\, \sqrt {1-\left (1+\frac {\sqrt {15}}{3}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {\sqrt {9-3 \sqrt {15}}\, x}{3}, \frac {i \sqrt {6}}{2}+\frac {i \sqrt {10}}{2}\right )-\operatorname {EllipticE}\left (\frac {\sqrt {9-3 \sqrt {15}}\, x}{3}, \frac {i \sqrt {6}}{2}+\frac {i \sqrt {10}}{2}\right )\right )}{5 \sqrt {9-3 \sqrt {15}}\, \sqrt {2 x^{4}+6 x^{2}-3}\, \left (6+2 \sqrt {15}\right )}\) | \(231\) |
Input:
int(1/(2*x^4+6*x^2-3)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/15*x*(x^2+4)/(2*x^4+6*x^2-3)^(1/2)-1/5/(9-3*15^(1/2))^(1/2)*(1-(1-1/3*1 5^(1/2))*x^2)^(1/2)*(1-(1+1/3*15^(1/2))*x^2)^(1/2)/(2*x^4+6*x^2-3)^(1/2)*E llipticF(1/3*(9-3*15^(1/2))^(1/2)*x,1/2*I*6^(1/2)+1/2*I*10^(1/2))+6/5/(9-3 *15^(1/2))^(1/2)*(1-(1-1/3*15^(1/2))*x^2)^(1/2)*(1-(1+1/3*15^(1/2))*x^2)^( 1/2)/(2*x^4+6*x^2-3)^(1/2)/(6+2*15^(1/2))*(EllipticF(1/3*(9-3*15^(1/2))^(1 /2)*x,1/2*I*6^(1/2)+1/2*I*10^(1/2))-EllipticE(1/3*(9-3*15^(1/2))^(1/2)*x,1 /2*I*6^(1/2)+1/2*I*10^(1/2)))
Time = 0.08 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.66 \[ \int \frac {1}{\left (-3+6 x^2+2 x^4\right )^{3/2}} \, dx=\frac {2 \, \sqrt {\frac {5}{3}} \sqrt {-3} {\left (2 \, x^{4} + 6 \, x^{2} - 3\right )} \sqrt {\sqrt {\frac {5}{3}} + 1} F(\arcsin \left (x \sqrt {\sqrt {\frac {5}{3}} + 1}\right )\,|\,3 \, \sqrt {\frac {5}{3}} - 4) - {\left (\sqrt {\frac {5}{3}} \sqrt {-3} {\left (2 \, x^{4} + 6 \, x^{2} - 3\right )} + \sqrt {-3} {\left (2 \, x^{4} + 6 \, x^{2} - 3\right )}\right )} \sqrt {\sqrt {\frac {5}{3}} + 1} E(\arcsin \left (x \sqrt {\sqrt {\frac {5}{3}} + 1}\right )\,|\,3 \, \sqrt {\frac {5}{3}} - 4) - 2 \, \sqrt {2 \, x^{4} + 6 \, x^{2} - 3} {\left (x^{3} + 4 \, x\right )}}{30 \, {\left (2 \, x^{4} + 6 \, x^{2} - 3\right )}} \] Input:
integrate(1/(2*x^4+6*x^2-3)^(3/2),x, algorithm="fricas")
Output:
1/30*(2*sqrt(5/3)*sqrt(-3)*(2*x^4 + 6*x^2 - 3)*sqrt(sqrt(5/3) + 1)*ellipti c_f(arcsin(x*sqrt(sqrt(5/3) + 1)), 3*sqrt(5/3) - 4) - (sqrt(5/3)*sqrt(-3)* (2*x^4 + 6*x^2 - 3) + sqrt(-3)*(2*x^4 + 6*x^2 - 3))*sqrt(sqrt(5/3) + 1)*el liptic_e(arcsin(x*sqrt(sqrt(5/3) + 1)), 3*sqrt(5/3) - 4) - 2*sqrt(2*x^4 + 6*x^2 - 3)*(x^3 + 4*x))/(2*x^4 + 6*x^2 - 3)
\[ \int \frac {1}{\left (-3+6 x^2+2 x^4\right )^{3/2}} \, dx=\int \frac {1}{\left (2 x^{4} + 6 x^{2} - 3\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/(2*x**4+6*x**2-3)**(3/2),x)
Output:
Integral((2*x**4 + 6*x**2 - 3)**(-3/2), x)
\[ \int \frac {1}{\left (-3+6 x^2+2 x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (2 \, x^{4} + 6 \, x^{2} - 3\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/(2*x^4+6*x^2-3)^(3/2),x, algorithm="maxima")
Output:
integrate((2*x^4 + 6*x^2 - 3)^(-3/2), x)
\[ \int \frac {1}{\left (-3+6 x^2+2 x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (2 \, x^{4} + 6 \, x^{2} - 3\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/(2*x^4+6*x^2-3)^(3/2),x, algorithm="giac")
Output:
integrate((2*x^4 + 6*x^2 - 3)^(-3/2), x)
Timed out. \[ \int \frac {1}{\left (-3+6 x^2+2 x^4\right )^{3/2}} \, dx=\int \frac {1}{{\left (2\,x^4+6\,x^2-3\right )}^{3/2}} \,d x \] Input:
int(1/(6*x^2 + 2*x^4 - 3)^(3/2),x)
Output:
int(1/(6*x^2 + 2*x^4 - 3)^(3/2), x)
\[ \int \frac {1}{\left (-3+6 x^2+2 x^4\right )^{3/2}} \, dx=\int \frac {\sqrt {2 x^{4}+6 x^{2}-3}}{4 x^{8}+24 x^{6}+24 x^{4}-36 x^{2}+9}d x \] Input:
int(1/(2*x^4+6*x^2-3)^(3/2),x)
Output:
int(sqrt(2*x**4 + 6*x**2 - 3)/(4*x**8 + 24*x**6 + 24*x**4 - 36*x**2 + 9),x )