Integrand size = 17, antiderivative size = 115 \[ \int \frac {1}{\sqrt {a-b x^2+c x^4}} \, dx=\frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a-b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (2+\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a-b x^2+c x^4}} \] Output:
1/2*(a^(1/2)+c^(1/2)*x^2)*((c*x^4-b*x^2+a)/(a^(1/2)+c^(1/2)*x^2)^2)^(1/2)* InverseJacobiAM(2*arctan(c^(1/4)*x/a^(1/4)),1/2*(2+b/a^(1/2)/c^(1/2))^(1/2 ))/a^(1/4)/c^(1/4)/(c*x^4-b*x^2+a)^(1/2)
Result contains complex when optimal does not.
Time = 10.13 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.57 \[ \int \frac {1}{\sqrt {a-b x^2+c x^4}} \, dx=-\frac {i \sqrt {1-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \sqrt {1-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {-\frac {c}{b-\sqrt {b^2-4 a c}}} x\right ),\frac {b-\sqrt {b^2-4 a c}}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {2} \sqrt {-\frac {c}{b-\sqrt {b^2-4 a c}}} \sqrt {a-b x^2+c x^4}} \] Input:
Integrate[1/Sqrt[a - b*x^2 + c*x^4],x]
Output:
((-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])]*EllipticF[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]*Sq rt[-(c/(b - Sqrt[b^2 - 4*a*c]))]*Sqrt[a - b*x^2 + c*x^4])
Time = 0.33 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {1416}
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.
\(\displaystyle \int \frac {1}{\sqrt {a-b x^2+c x^4}} \, dx\) |
\(\Big \downarrow \) 1416 |
\(\displaystyle \frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a-b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (\frac {b}{\sqrt {a} \sqrt {c}}+2\right )\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a-b x^2+c x^4}}\) |
Input:
Int[1/Sqrt[a - b*x^2 + c*x^4],x]
Output:
((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)*c^(1/4)*Sqrt[a - b*x^2 + c*x^4])
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]
Time = 0.52 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.24
method | result | size |
default | \(\frac {\sqrt {2}\, \sqrt {4-\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4-\frac {2 b \left (-b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )}{4 \sqrt {\frac {b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}-b \,x^{2}+a}}\) | \(143\) |
elliptic | \(\frac {\sqrt {2}\, \sqrt {4-\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4-\frac {2 b \left (-b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )}{4 \sqrt {\frac {b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}-b \,x^{2}+a}}\) | \(143\) |
Input:
int(1/(c*x^4-b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/4*2^(1/2)/((b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*(4-2*(b+(-4*a*c+b^2)^(1/2))/a *x^2)^(1/2)*(4+2*(-b+(-4*a*c+b^2)^(1/2))/a*x^2)^(1/2)/(c*x^4-b*x^2+a)^(1/2 )*EllipticF(1/2*x*2^(1/2)*((b+(-4*a*c+b^2)^(1/2))/a)^(1/2),1/2*(-4-2*b*(-b +(-4*a*c+b^2)^(1/2))/a/c)^(1/2))
Time = 0.08 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.05 \[ \int \frac {1}{\sqrt {a-b x^2+c x^4}} \, dx=-\frac {\sqrt {\frac {1}{2}} {\left (a \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} - b\right )} \sqrt {\frac {a \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} + b}{a}} F(\arcsin \left (\sqrt {\frac {1}{2}} x \sqrt {\frac {a \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} + b}{a}}\right )\,|\,-\frac {a b \sqrt {\frac {b^{2} - 4 \, a c}{a^{2}}} - b^{2} + 2 \, a c}{2 \, a c})}{2 \, \sqrt {a} c} \] Input:
integrate(1/(c*x^4-b*x^2+a)^(1/2),x, algorithm="fricas")
Output:
-1/2*sqrt(1/2)*(a*sqrt((b^2 - 4*a*c)/a^2) - b)*sqrt((a*sqrt((b^2 - 4*a*c)/ a^2) + b)/a)*elliptic_f(arcsin(sqrt(1/2)*x*sqrt((a*sqrt((b^2 - 4*a*c)/a^2) + b)/a)), -1/2*(a*b*sqrt((b^2 - 4*a*c)/a^2) - b^2 + 2*a*c)/(a*c))/(sqrt(a )*c)
\[ \int \frac {1}{\sqrt {a-b x^2+c x^4}} \, dx=\int \frac {1}{\sqrt {a - b x^{2} + c x^{4}}}\, dx \] Input:
integrate(1/(c*x**4-b*x**2+a)**(1/2),x)
Output:
Integral(1/sqrt(a - b*x**2 + c*x**4), x)
\[ \int \frac {1}{\sqrt {a-b x^2+c x^4}} \, dx=\int { \frac {1}{\sqrt {c x^{4} - b x^{2} + a}} \,d x } \] Input:
integrate(1/(c*x^4-b*x^2+a)^(1/2),x, algorithm="maxima")
Output:
integrate(1/sqrt(c*x^4 - b*x^2 + a), x)
\[ \int \frac {1}{\sqrt {a-b x^2+c x^4}} \, dx=\int { \frac {1}{\sqrt {c x^{4} - b x^{2} + a}} \,d x } \] Input:
integrate(1/(c*x^4-b*x^2+a)^(1/2),x, algorithm="giac")
Output:
integrate(1/sqrt(c*x^4 - b*x^2 + a), x)
Timed out. \[ \int \frac {1}{\sqrt {a-b x^2+c x^4}} \, dx=\int \frac {1}{\sqrt {c\,x^4-b\,x^2+a}} \,d x \] Input:
int(1/(a - b*x^2 + c*x^4)^(1/2),x)
Output:
int(1/(a - b*x^2 + c*x^4)^(1/2), x)
\[ \int \frac {1}{\sqrt {a-b x^2+c x^4}} \, dx=\int \frac {\sqrt {c \,x^{4}-b \,x^{2}+a}}{c \,x^{4}-b \,x^{2}+a}d x \] Input:
int(1/(c*x^4-b*x^2+a)^(1/2),x)
Output:
int(sqrt(a - b*x**2 + c*x**4)/(a - b*x**2 + c*x**4),x)