Integrand size = 29, antiderivative size = 412 \[ \int \frac {1}{\left (d+e x^2\right ) \sqrt {-a+b x^2-c x^4}} \, dx=\frac {\sqrt {e} \arctan \left (\frac {\sqrt {-c d^2-e (b d+a e)} x}{\sqrt {d} \sqrt {e} \sqrt {-a+b x^2-c x^4}}\right )}{2 \sqrt {d} \sqrt {-c d^2-e (b d+a e)}}+\frac {\sqrt [4]{c} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a-b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (2+\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 \sqrt [4]{a} \left (\sqrt {c} d-\sqrt {a} e\right ) \sqrt {-a+b x^2-c x^4}}-\frac {a^{3/4} \left (\frac {\sqrt {c} d}{\sqrt {a}}+e\right )^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a-b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticPi}\left (-\frac {\left (\sqrt {c} d-\sqrt {a} e\right )^2}{4 \sqrt {a} \sqrt {c} d e},2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (2+\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{4 \sqrt [4]{c} d \left (c d^2-a e^2\right ) \sqrt {-a+b x^2-c x^4}} \]
1/2*arctan(x*(-a*e^2-b*d*e-c*d^2)^(1/2)/d^(1/2)/e^(1/2)/(-c*x^4+b*x^2-a)^( 1/2))*e^(1/2)/d^(1/2)/(-a*e^2-b*d*e-c*d^2)^(1/2)+1/2*c^(1/4)*(cos(2*arctan (c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4)))*EllipticF(s in(2*arctan(c^(1/4)*x/a^(1/4))),1/2*(2+b/a^(1/2)/c^(1/2))^(1/2))*(a^(1/2)+ x^2*c^(1/2))*((c*x^4-b*x^2+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/a^(1/4)/(-e*a ^(1/2)+d*c^(1/2))/(-c*x^4+b*x^2-a)^(1/2)-1/4*a^(3/4)*(cos(2*arctan(c^(1/4) *x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4)))*EllipticPi(sin(2*ar ctan(c^(1/4)*x/a^(1/4))),-1/4*(-e*a^(1/2)+d*c^(1/2))^2/d/e/a^(1/2)/c^(1/2) ,1/2*(2+b/a^(1/2)/c^(1/2))^(1/2))*(a^(1/2)+x^2*c^(1/2))*(e+d*c^(1/2)/a^(1/ 2))^2*((c*x^4-b*x^2+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/c^(1/4)/d/(-a*e^2+c* d^2)/(-c*x^4+b*x^2-a)^(1/2)
Result contains complex when optimal does not.
Time = 10.23 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.50 \[ \int \frac {1}{\left (d+e x^2\right ) \sqrt {-a+b x^2-c x^4}} \, dx=-\frac {i \sqrt {1+\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}} \sqrt {1-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {EllipticPi}\left (-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{2 c d},i \text {arcsinh}\left (\sqrt {2} \sqrt {-\frac {c}{b+\sqrt {b^2-4 a c}}} x\right ),-\frac {b+\sqrt {b^2-4 a c}}{-b+\sqrt {b^2-4 a c}}\right )}{\sqrt {2} \sqrt {-\frac {c}{b+\sqrt {b^2-4 a c}}} d \sqrt {-a+b x^2-c x^4}} \]
((-I)*Sqrt[1 + (2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])]*Sqrt[1 - (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*EllipticPi[-1/2*((b + Sqrt[b^2 - 4*a*c])*e)/(c*d), I* ArcSinh[Sqrt[2]*Sqrt[-(c/(b + Sqrt[b^2 - 4*a*c]))]*x], -((b + Sqrt[b^2 - 4 *a*c])/(-b + Sqrt[b^2 - 4*a*c]))])/(Sqrt[2]*Sqrt[-(c/(b + Sqrt[b^2 - 4*a*c ]))]*d*Sqrt[-a + b*x^2 - c*x^4])
Time = 0.59 (sec) , antiderivative size = 434, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1540, 27, 1416, 2222}
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.
| \(\displaystyle \int \frac {1}{\left (d+e x^2\right ) \sqrt {-a+b x^2-c x^4}} \, dx\) |
| \(\Big \downarrow \) 1540 |
| \(\displaystyle \frac {\sqrt {c} \int \frac {1}{\sqrt {-c x^4+b x^2-a}}dx}{\sqrt {c} d-\sqrt {a} e}-\frac {\sqrt {a} e \int \frac {\sqrt {c} x^2+\sqrt {a}}{\sqrt {a} \left (e x^2+d\right ) \sqrt {-c x^4+b x^2-a}}dx}{\sqrt {c} d-\sqrt {a} e}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {c} \int \frac {1}{\sqrt {-c x^4+b x^2-a}}dx}{\sqrt {c} d-\sqrt {a} e}-\frac {e \int \frac {\sqrt {c} x^2+\sqrt {a}}{\left (e x^2+d\right ) \sqrt {-c x^4+b x^2-a}}dx}{\sqrt {c} d-\sqrt {a} e}\) |
| \(\Big \downarrow \) 1416 |
| \(\displaystyle \frac {\sqrt [4]{c} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a-b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (\frac {b}{\sqrt {a} \sqrt {c}}+2\right )\right )}{2 \sqrt [4]{a} \sqrt {-a+b x^2-c x^4} \left (\sqrt {c} d-\sqrt {a} e\right )}-\frac {e \int \frac {\sqrt {c} x^2+\sqrt {a}}{\left (e x^2+d\right ) \sqrt {-c x^4+b x^2-a}}dx}{\sqrt {c} d-\sqrt {a} e}\) |
| \(\Big \downarrow \) 2222 |
| \(\displaystyle \frac {\sqrt [4]{c} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a-b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (\frac {b}{\sqrt {a} \sqrt {c}}+2\right )\right )}{2 \sqrt [4]{a} \sqrt {-a+b x^2-c x^4} \left (\sqrt {c} d-\sqrt {a} e\right )}-\frac {e \left (\frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a-b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (\sqrt {a} e+\sqrt {c} d\right ) \operatorname {EllipticPi}\left (-\frac {\sqrt {a} \left (\frac {\sqrt {c} d}{\sqrt {a}}-e\right )^2}{4 \sqrt {c} d e},2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (\frac {b}{\sqrt {a} \sqrt {c}}+2\right )\right )}{4 \sqrt [4]{a} \sqrt [4]{c} d e \sqrt {-a+b x^2-c x^4}}-\frac {\left (\sqrt {c} d-\sqrt {a} e\right ) \text {arctanh}\left (\frac {x \sqrt {e (a e+b d)+c d^2}}{\sqrt {d} \sqrt {e} \sqrt {-a+b x^2-c x^4}}\right )}{2 \sqrt {d} \sqrt {e} \sqrt {e (a e+b d)+c d^2}}\right )}{\sqrt {c} d-\sqrt {a} e}\) |
(c^(1/4)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a - b*x^2 + c*x^4)/(Sqrt[a] + Sqrt[ c]*x^2)^2]*EllipticF[2*ArcTan[(c^(1/4)*x)/a^(1/4)], (2 + b/(Sqrt[a]*Sqrt[c ]))/4])/(2*a^(1/4)*(Sqrt[c]*d - Sqrt[a]*e)*Sqrt[-a + b*x^2 - c*x^4]) - (e* (-1/2*((Sqrt[c]*d - Sqrt[a]*e)*ArcTanh[(Sqrt[c*d^2 + e*(b*d + a*e)]*x)/(Sq rt[d]*Sqrt[e]*Sqrt[-a + b*x^2 - c*x^4])])/(Sqrt[d]*Sqrt[e]*Sqrt[c*d^2 + e* (b*d + a*e)]) + ((Sqrt[c]*d + Sqrt[a]*e)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a - b*x^2 + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticPi[-1/4*(Sqrt[a]*((Sqrt [c]*d)/Sqrt[a] - e)^2)/(Sqrt[c]*d*e), 2*ArcTan[(c^(1/4)*x)/a^(1/4)], (2 + b/(Sqrt[a]*Sqrt[c]))/4])/(4*a^(1/4)*c^(1/4)*d*e*Sqrt[-a + b*x^2 - c*x^4])) )/(Sqrt[c]*d - Sqrt[a]*e)
3.4.93.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_S ymbol] :> With[{q = Rt[c/a, 2]}, Simp[(c*d + a*e*q)/(c*d^2 - a*e^2) Int[1 /Sqrt[a + b*x^2 + c*x^4], x], x] - Simp[(a*e*(e + d*q))/(c*d^2 - a*e^2) I nt[(1 + q*x^2)/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a]
Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[B/A, 2]}, Simp[(-(B*d - A*e))*(A rcTanh[Rt[b - c*(d/e) - a*(e/d), 2]*(x/Sqrt[a + b*x^2 + c*x^4])]/(2*d*e*Rt[ b - c*(d/e) - a*(e/d), 2])), x] + Simp[(B*d + A*e)*(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]/(4*d*e*q*Sqrt[a + b*x^2 + c*x^4]))*Ell ipticPi[-(e - d*q^2)^2/(4*d*e*q^2), 2*ArcTan[q*x], 1/2 - b/(4*a*q^2)], x]] /; FreeQ[{a, b, c, d, e, A, B}, x] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a] && EqQ[c*A^2 - a*B^2, 0] && PosQ[B/A] && NegQ[-b + c*(d/e) + a*(e/d)]
Time = 1.01 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.48
| method | result | size |
| default | \(\frac {\sqrt {1-\frac {b \,x^{2}}{2 a}+\frac {x^{2} \sqrt {-4 a c +b^{2}}}{2 a}}\, \sqrt {1-\frac {b \,x^{2}}{2 a}-\frac {x^{2} \sqrt {-4 a c +b^{2}}}{2 a}}\, \Pi \left (\sqrt {-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}}\, x , \frac {2 a e}{\left (-b +\sqrt {-4 a c +b^{2}}\right ) d}, \frac {\sqrt {2}\, \sqrt {\frac {b +\sqrt {-4 a c +b^{2}}}{a}}}{2 \sqrt {-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}}}\right )}{d \sqrt {\frac {b}{2 a}-\frac {\sqrt {-4 a c +b^{2}}}{2 a}}\, \sqrt {-c \,x^{4}+b \,x^{2}-a}}\) | \(199\) |
| elliptic | \(\frac {\sqrt {1-\frac {b \,x^{2}}{2 a}+\frac {x^{2} \sqrt {-4 a c +b^{2}}}{2 a}}\, \sqrt {1-\frac {b \,x^{2}}{2 a}-\frac {x^{2} \sqrt {-4 a c +b^{2}}}{2 a}}\, \Pi \left (\sqrt {-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}}\, x , \frac {2 a e}{\left (-b +\sqrt {-4 a c +b^{2}}\right ) d}, \frac {\sqrt {2}\, \sqrt {\frac {b +\sqrt {-4 a c +b^{2}}}{a}}}{2 \sqrt {-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 a}}}\right )}{d \sqrt {\frac {b}{2 a}-\frac {\sqrt {-4 a c +b^{2}}}{2 a}}\, \sqrt {-c \,x^{4}+b \,x^{2}-a}}\) | \(199\) |
1/d/(1/2*b/a-1/2/a*(-4*a*c+b^2)^(1/2))^(1/2)*(1-1/2*b*x^2/a+1/2*x^2/a*(-4* a*c+b^2)^(1/2))^(1/2)*(1-1/2*b*x^2/a-1/2*x^2/a*(-4*a*c+b^2)^(1/2))^(1/2)/( -c*x^4+b*x^2-a)^(1/2)*EllipticPi((-1/2*(-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*x, 2/(-b+(-4*a*c+b^2)^(1/2))*a*e/d,1/2*2^(1/2)*((b+(-4*a*c+b^2)^(1/2))/a)^(1/ 2)/(-1/2*(-b+(-4*a*c+b^2)^(1/2))/a)^(1/2))
Timed out. \[ \int \frac {1}{\left (d+e x^2\right ) \sqrt {-a+b x^2-c x^4}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{\left (d+e x^2\right ) \sqrt {-a+b x^2-c x^4}} \, dx=\int \frac {1}{\left (d + e x^{2}\right ) \sqrt {- a + b x^{2} - c x^{4}}}\, dx \]
\[ \int \frac {1}{\left (d+e x^2\right ) \sqrt {-a+b x^2-c x^4}} \, dx=\int { \frac {1}{\sqrt {-c x^{4} + b x^{2} - a} {\left (e x^{2} + d\right )}} \,d x } \]
\[ \int \frac {1}{\left (d+e x^2\right ) \sqrt {-a+b x^2-c x^4}} \, dx=\int { \frac {1}{\sqrt {-c x^{4} + b x^{2} - a} {\left (e x^{2} + d\right )}} \,d x } \]
Timed out. \[ \int \frac {1}{\left (d+e x^2\right ) \sqrt {-a+b x^2-c x^4}} \, dx=\int \frac {1}{\left (e\,x^2+d\right )\,\sqrt {-c\,x^4+b\,x^2-a}} \,d x \]