Integrand size = 19, antiderivative size = 94 \[ \int \frac {1}{\sqrt {\left (a+b x^2\right ) \left (c+d x^2\right )}} \, dx=\frac {\left (a+b x^2\right ) \sqrt {\frac {a \left (c+d x^2\right )}{c \left (a+b x^2\right )}} \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),1-\frac {a d}{b c}\right )}{\sqrt {a} \sqrt {b} \sqrt {a c+(b c+a d) x^2+b d x^4}} \] Output:
(b*x^2+a)*(a*(d*x^2+c)/c/(b*x^2+a))^(1/2)*InverseJacobiAM(arctan(b^(1/2)*x /a^(1/2)),(1-a*d/b/c)^(1/2))/a^(1/2)/b^(1/2)/(a*c+(a*d+b*c)*x^2+b*d*x^4)^( 1/2)
Time = 0.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\sqrt {\left (a+b x^2\right ) \left (c+d x^2\right )}} \, dx=\frac {\sqrt {\frac {a+b x^2}{a}} \sqrt {\frac {c+d x^2}{c}} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {-\frac {b}{a}} x\right ),\frac {a d}{b c}\right )}{\sqrt {-\frac {b}{a}} \sqrt {\left (a+b x^2\right ) \left (c+d x^2\right )}} \] Input:
Integrate[1/Sqrt[(a + b*x^2)*(c + d*x^2)],x]
Output:
(Sqrt[(a + b*x^2)/a]*Sqrt[(c + d*x^2)/c]*EllipticF[ArcSin[Sqrt[-(b/a)]*x], (a*d)/(b*c)])/(Sqrt[-(b/a)]*Sqrt[(a + b*x^2)*(c + d*x^2)])
Leaf count is larger than twice the leaf count of optimal. \(190\) vs. \(2(94)=188\).
Time = 0.41 (sec) , antiderivative size = 190, normalized size of antiderivative = 2.02, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2048, 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.
\(\displaystyle \int \frac {1}{\sqrt {\left (a+b x^2\right ) \left (c+d x^2\right )}} \, dx\) |
\(\Big \downarrow \) 2048 |
\(\displaystyle \int \frac {1}{\sqrt {x^2 (a d+b c)+a c+b d x^4}}dx\) |
\(\Big \downarrow \) 1416 |
\(\displaystyle \frac {\left (\sqrt {a} \sqrt {c}+\sqrt {b} \sqrt {d} x^2\right ) \sqrt {\frac {x^2 (a d+b c)+a c+b d x^4}{\left (\sqrt {a} \sqrt {c}+\sqrt {b} \sqrt {d} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt [4]{d} x}{\sqrt [4]{a} \sqrt [4]{c}}\right ),\frac {1}{4} \left (2-\frac {b c+a d}{\sqrt {a} \sqrt {b} \sqrt {c} \sqrt {d}}\right )\right )}{2 \sqrt [4]{a} \sqrt [4]{b} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x^2 (a d+b c)+a c+b d x^4}}\) |
Input:
Int[1/Sqrt[(a + b*x^2)*(c + d*x^2)],x]
Output:
((Sqrt[a]*Sqrt[c] + Sqrt[b]*Sqrt[d]*x^2)*Sqrt[(a*c + (b*c + a*d)*x^2 + b*d *x^4)/(Sqrt[a]*Sqrt[c] + Sqrt[b]*Sqrt[d]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/ 4)*d^(1/4)*x)/(a^(1/4)*c^(1/4))], (2 - (b*c + a*d)/(Sqrt[a]*Sqrt[b]*Sqrt[c ]*Sqrt[d]))/4])/(2*a^(1/4)*b^(1/4)*c^(1/4)*d^(1/4)*Sqrt[a*c + (b*c + a*d)* x^2 + b*d*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]
Int[(u_.)*((e_.)*((a_.) + (b_.)*(x_)^(n_.))*((c_) + (d_.)*(x_)^(n_.)))^(p_) , x_Symbol] :> Int[u*(a*c*e + (b*c + a*d)*e*x^n + b*d*e*x^(2*n))^p, x] /; F reeQ[{a, b, c, d, e, n, p}, x]
Time = 0.75 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.93
method | result | size |
default | \(\frac {\sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {d \,x^{2}}{c}}\, \operatorname {EllipticF}\left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )}{\sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+x^{2} d a +b c \,x^{2}+a c}}\) | \(87\) |
elliptic | \(\frac {\sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {d \,x^{2}}{c}}\, \operatorname {EllipticF}\left (x \sqrt {-\frac {b}{a}}, \sqrt {-1+\frac {a d +b c}{c b}}\right )}{\sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+x^{2} d a +b c \,x^{2}+a c}}\) | \(87\) |
Input:
int(1/((b*x^2+a)*(d*x^2+c))^(1/2),x,method=_RETURNVERBOSE)
Output:
1/(-b/a)^(1/2)*(1+b*x^2/a)^(1/2)*(1+d*x^2/c)^(1/2)/(b*d*x^4+a*d*x^2+b*c*x^ 2+a*c)^(1/2)*EllipticF(x*(-b/a)^(1/2),(-1+(a*d+b*c)/c/b)^(1/2))
Time = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.45 \[ \int \frac {1}{\sqrt {\left (a+b x^2\right ) \left (c+d x^2\right )}} \, dx=-\frac {\sqrt {a c} \sqrt {-\frac {b}{a}} F(\arcsin \left (x \sqrt {-\frac {b}{a}}\right )\,|\,\frac {a d}{b c})}{b c} \] Input:
integrate(1/((b*x^2+a)*(d*x^2+c))^(1/2),x, algorithm="fricas")
Output:
-sqrt(a*c)*sqrt(-b/a)*elliptic_f(arcsin(x*sqrt(-b/a)), a*d/(b*c))/(b*c)
\[ \int \frac {1}{\sqrt {\left (a+b x^2\right ) \left (c+d x^2\right )}} \, dx=\int \frac {1}{\sqrt {\left (a + b x^{2}\right ) \left (c + d x^{2}\right )}}\, dx \] Input:
integrate(1/((b*x**2+a)*(d*x**2+c))**(1/2),x)
Output:
Integral(1/sqrt((a + b*x**2)*(c + d*x**2)), x)
\[ \int \frac {1}{\sqrt {\left (a+b x^2\right ) \left (c+d x^2\right )}} \, dx=\int { \frac {1}{\sqrt {{\left (b x^{2} + a\right )} {\left (d x^{2} + c\right )}}} \,d x } \] Input:
integrate(1/((b*x^2+a)*(d*x^2+c))^(1/2),x, algorithm="maxima")
Output:
integrate(1/sqrt((b*x^2 + a)*(d*x^2 + c)), x)
\[ \int \frac {1}{\sqrt {\left (a+b x^2\right ) \left (c+d x^2\right )}} \, dx=\int { \frac {1}{\sqrt {{\left (b x^{2} + a\right )} {\left (d x^{2} + c\right )}}} \,d x } \] Input:
integrate(1/((b*x^2+a)*(d*x^2+c))^(1/2),x, algorithm="giac")
Output:
integrate(1/sqrt((b*x^2 + a)*(d*x^2 + c)), x)
Timed out. \[ \int \frac {1}{\sqrt {\left (a+b x^2\right ) \left (c+d x^2\right )}} \, dx=\int \frac {1}{\sqrt {\left (b\,x^2+a\right )\,\left (d\,x^2+c\right )}} \,d x \] Input:
int(1/((a + b*x^2)*(c + d*x^2))^(1/2),x)
Output:
int(1/((a + b*x^2)*(c + d*x^2))^(1/2), x)
\[ \int \frac {1}{\sqrt {\left (a+b x^2\right ) \left (c+d x^2\right )}} \, dx=\int \frac {\sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}}{b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}d x \] Input:
int(1/((b*x^2+a)*(d*x^2+c))^(1/2),x)
Output:
int((sqrt(c + d*x**2)*sqrt(a + b*x**2))/(a*c + a*d*x**2 + b*c*x**2 + b*d*x **4),x)