Integrand size = 22, antiderivative size = 89 \[ \int \frac {1-b x^2}{\sqrt {-1+b^2 x^4}} \, dx=-\frac {\sqrt {1-b^2 x^4} E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{\sqrt {b} \sqrt {-1+b^2 x^4}}+\frac {2 \sqrt {1-b^2 x^4} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {b} x\right ),-1\right )}{\sqrt {b} \sqrt {-1+b^2 x^4}} \]
-EllipticE(x*b^(1/2),I)*(-b^2*x^4+1)^(1/2)/b^(1/2)/(b^2*x^4-1)^(1/2)+2*Ell ipticF(x*b^(1/2),I)*(-b^2*x^4+1)^(1/2)/b^(1/2)/(b^2*x^4-1)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.83 \[ \int \frac {1-b x^2}{\sqrt {-1+b^2 x^4}} \, dx=-\frac {\sqrt {1-b^2 x^4} \left (-3 x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},b^2 x^4\right )+b x^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},b^2 x^4\right )\right )}{3 \sqrt {-1+b^2 x^4}} \]
-1/3*(Sqrt[1 - b^2*x^4]*(-3*x*Hypergeometric2F1[1/4, 1/2, 5/4, b^2*x^4] + b*x^3*Hypergeometric2F1[1/2, 3/4, 7/4, b^2*x^4]))/Sqrt[-1 + b^2*x^4]
Time = 0.21 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.71, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1390, 1388, 326, 284, 327, 762}
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.
\(\displaystyle \int \frac {1-b x^2}{\sqrt {b^2 x^4-1}} \, dx\) |
\(\Big \downarrow \) 1390 |
\(\displaystyle \frac {\sqrt {1-b^2 x^4} \int \frac {1-b x^2}{\sqrt {1-b^2 x^4}}dx}{\sqrt {b^2 x^4-1}}\) |
\(\Big \downarrow \) 1388 |
\(\displaystyle \frac {\sqrt {1-b^2 x^4} \int \frac {\sqrt {1-b x^2}}{\sqrt {b x^2+1}}dx}{\sqrt {b^2 x^4-1}}\) |
\(\Big \downarrow \) 326 |
\(\displaystyle \frac {\sqrt {1-b^2 x^4} \left (2 \int \frac {1}{\sqrt {1-b x^2} \sqrt {b x^2+1}}dx-\int \frac {\sqrt {b x^2+1}}{\sqrt {1-b x^2}}dx\right )}{\sqrt {b^2 x^4-1}}\) |
\(\Big \downarrow \) 284 |
\(\displaystyle \frac {\sqrt {1-b^2 x^4} \left (2 \int \frac {1}{\sqrt {1-b^2 x^4}}dx-\int \frac {\sqrt {b x^2+1}}{\sqrt {1-b x^2}}dx\right )}{\sqrt {b^2 x^4-1}}\) |
\(\Big \downarrow \) 327 |
\(\displaystyle \frac {\sqrt {1-b^2 x^4} \left (2 \int \frac {1}{\sqrt {1-b^2 x^4}}dx-\frac {E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{\sqrt {b}}\right )}{\sqrt {b^2 x^4-1}}\) |
\(\Big \downarrow \) 762 |
\(\displaystyle \frac {\sqrt {1-b^2 x^4} \left (\frac {2 \operatorname {EllipticF}\left (\arcsin \left (\sqrt {b} x\right ),-1\right )}{\sqrt {b}}-\frac {E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{\sqrt {b}}\right )}{\sqrt {b^2 x^4-1}}\) |
(Sqrt[1 - b^2*x^4]*(-(EllipticE[ArcSin[Sqrt[b]*x], -1]/Sqrt[b]) + (2*Ellip ticF[ArcSin[Sqrt[b]*x], -1])/Sqrt[b]))/Sqrt[-1 + b^2*x^4]
3.1.17.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> I nt[(a*c + b*d*x^4)^p, x] /; FreeQ[{a, b, c, d, p}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0]))
Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ b/d Int[Sqrt[c + d*x^2]/Sqrt[a + b*x^2], x], x] - Simp[(b*c - a*d)/d In t[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d}, x] && PosQ[d/c] && NegQ[b/a]
Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ (Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*EllipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d) )], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0]
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) )*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x] /; FreeQ[{a, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && (Integer Q[p] || (GtQ[a, 0] && GtQ[d, 0]))
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Simp[Sqrt [1 + c*(x^4/a)]/Sqrt[a + c*x^4] Int[(d + e*x^2)/Sqrt[1 + c*(x^4/a)], x], x] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && NegQ[c/a] && !GtQ [a, 0] && !(LtQ[a, 0] && GtQ[c, 0])
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 1.14 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.99
method | result | size |
meijerg | \(-\frac {b \sqrt {-\operatorname {signum}\left (b^{2} x^{4}-1\right )}\, x^{3} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};b^{2} x^{4}\right )}{3 \sqrt {\operatorname {signum}\left (b^{2} x^{4}-1\right )}}+\frac {\sqrt {-\operatorname {signum}\left (b^{2} x^{4}-1\right )}\, x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};b^{2} x^{4}\right )}{\sqrt {\operatorname {signum}\left (b^{2} x^{4}-1\right )}}\) | \(88\) |
default | \(\frac {\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, F\left (x \sqrt {-b}, i\right )}{\sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}-\frac {\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \left (F\left (x \sqrt {-b}, i\right )-E\left (x \sqrt {-b}, i\right )\right )}{\sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}\) | \(108\) |
elliptic | \(\frac {\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, F\left (x \sqrt {-b}, i\right )}{\sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}-\frac {\sqrt {b \,x^{2}+1}\, \sqrt {-b \,x^{2}+1}\, \left (F\left (x \sqrt {-b}, i\right )-E\left (x \sqrt {-b}, i\right )\right )}{\sqrt {-b}\, \sqrt {b^{2} x^{4}-1}}\) | \(108\) |
-1/3*b/signum(b^2*x^4-1)^(1/2)*(-signum(b^2*x^4-1))^(1/2)*x^3*hypergeom([1 /2,3/4],[7/4],b^2*x^4)+1/signum(b^2*x^4-1)^(1/2)*(-signum(b^2*x^4-1))^(1/2 )*x*hypergeom([1/4,1/2],[5/4],b^2*x^4)
Time = 0.07 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.60 \[ \int \frac {1-b x^2}{\sqrt {-1+b^2 x^4}} \, dx=-\frac {\frac {{\left (b - 1\right )} x F(\arcsin \left (\frac {1}{\sqrt {b} x}\right )\,|\,-1)}{\sqrt {b}} + \frac {x E(\arcsin \left (\frac {1}{\sqrt {b} x}\right )\,|\,-1)}{\sqrt {b}} + \sqrt {b^{2} x^{4} - 1}}{b x} \]
-((b - 1)*x*elliptic_f(arcsin(1/(sqrt(b)*x)), -1)/sqrt(b) + x*elliptic_e(a rcsin(1/(sqrt(b)*x)), -1)/sqrt(b) + sqrt(b^2*x^4 - 1))/(b*x)
Time = 0.82 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.67 \[ \int \frac {1-b x^2}{\sqrt {-1+b^2 x^4}} \, dx=\frac {i b x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {b^{2} x^{4}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} - \frac {i x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {b^{2} x^{4}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \]
I*b*x**3*gamma(3/4)*hyper((1/2, 3/4), (7/4,), b**2*x**4)/(4*gamma(7/4)) - I*x*gamma(1/4)*hyper((1/4, 1/2), (5/4,), b**2*x**4)/(4*gamma(5/4))
\[ \int \frac {1-b x^2}{\sqrt {-1+b^2 x^4}} \, dx=\int { -\frac {b x^{2} - 1}{\sqrt {b^{2} x^{4} - 1}} \,d x } \]
\[ \int \frac {1-b x^2}{\sqrt {-1+b^2 x^4}} \, dx=\int { -\frac {b x^{2} - 1}{\sqrt {b^{2} x^{4} - 1}} \,d x } \]
Timed out. \[ \int \frac {1-b x^2}{\sqrt {-1+b^2 x^4}} \, dx=-\int \frac {b\,x^2-1}{\sqrt {b^2\,x^4-1}} \,d x \]