Integrand size = 43, antiderivative size = 497 \[ \int \frac {-1+k^2 x^4}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1+k^2 x^4\right )} \, dx=-\frac {i \left (-i \sqrt {1-2 k+k^2-2 i \sqrt {2} \sqrt {k} \sqrt {1+k^2}}-i k^2 \sqrt {1-2 k+k^2-2 i \sqrt {2} \sqrt {k} \sqrt {1+k^2}}+\sqrt {2} \sqrt {k} \sqrt {1+k^2} \sqrt {1-2 k+k^2-2 i \sqrt {2} \sqrt {k} \sqrt {1+k^2}}\right ) \arctan \left (\frac {\sqrt {1-2 k+k^2-2 i \sqrt {2} \sqrt {k} \sqrt {1+k^2}} x}{1+k x^2+\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{\left (-i+\sqrt {k}\right )^2 \left (i+\sqrt {k}\right )^2 (-i+k) (i+k)}+\frac {i \left (i \sqrt {1-2 k+k^2+2 i \sqrt {2} \sqrt {k} \sqrt {1+k^2}}+i k^2 \sqrt {1-2 k+k^2+2 i \sqrt {2} \sqrt {k} \sqrt {1+k^2}}+\sqrt {2} \sqrt {k} \sqrt {1+k^2} \sqrt {1-2 k+k^2+2 i \sqrt {2} \sqrt {k} \sqrt {1+k^2}}\right ) \arctan \left (\frac {\sqrt {1-2 k+k^2+2 i \sqrt {2} \sqrt {k} \sqrt {1+k^2}} x}{1+k x^2+\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{\left (-i+\sqrt {k}\right )^2 \left (i+\sqrt {k}\right )^2 (-i+k) (i+k)} \]
-I*(-I*(1-2*k+k^2-2*I*2^(1/2)*k^(1/2)*(k^2+1)^(1/2))^(1/2)-I*k^2*(1-2*k+k^ 2-2*I*2^(1/2)*k^(1/2)*(k^2+1)^(1/2))^(1/2)+2^(1/2)*k^(1/2)*(k^2+1)^(1/2)*( 1-2*k+k^2-2*I*2^(1/2)*k^(1/2)*(k^2+1)^(1/2))^(1/2))*arctan((1-2*k+k^2-2*I* 2^(1/2)*k^(1/2)*(k^2+1)^(1/2))^(1/2)*x/(1+k*x^2+(1+(-k^2-1)*x^2+k^2*x^4)^( 1/2)))/(-I+k^(1/2))^2/(I+k^(1/2))^2/(-I+k)/(I+k)+I*(I*(1-2*k+k^2+2*I*2^(1/ 2)*k^(1/2)*(k^2+1)^(1/2))^(1/2)+I*k^2*(1-2*k+k^2+2*I*2^(1/2)*k^(1/2)*(k^2+ 1)^(1/2))^(1/2)+2^(1/2)*k^(1/2)*(k^2+1)^(1/2)*(1-2*k+k^2+2*I*2^(1/2)*k^(1/ 2)*(k^2+1)^(1/2))^(1/2))*arctan((1-2*k+k^2+2*I*2^(1/2)*k^(1/2)*(k^2+1)^(1/ 2))^(1/2)*x/(1+k*x^2+(1+(-k^2-1)*x^2+k^2*x^4)^(1/2)))/(-I+k^(1/2))^2/(I+k^ (1/2))^2/(-I+k)/(I+k)
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 9.54 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.16 \[ \int \frac {-1+k^2 x^4}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1+k^2 x^4\right )} \, dx=\frac {\sqrt {1-x^2} \sqrt {1-k^2 x^2} \left (\operatorname {EllipticF}\left (\arcsin (x),k^2\right )-\operatorname {EllipticPi}\left (-i k,\arcsin (x),k^2\right )-\operatorname {EllipticPi}\left (i k,\arcsin (x),k^2\right )\right )}{\sqrt {\left (-1+x^2\right ) \left (-1+k^2 x^2\right )}} \]
(Sqrt[1 - x^2]*Sqrt[1 - k^2*x^2]*(EllipticF[ArcSin[x], k^2] - EllipticPi[( -I)*k, ArcSin[x], k^2] - EllipticPi[I*k, ArcSin[x], k^2]))/Sqrt[(-1 + x^2) *(-1 + k^2*x^2)]
Time = 0.36 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.09, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {2048, 2537, 216}
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.
\(\displaystyle \int \frac {k^2 x^4-1}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (k^2 x^4+1\right )} \, dx\) |
\(\Big \downarrow \) 2048 |
\(\displaystyle \int \frac {k^2 x^4-1}{\left (k^2 x^4+1\right ) \sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}}dx\) |
\(\Big \downarrow \) 2537 |
\(\displaystyle -\int \frac {1}{\frac {\left (k^2+1\right ) x^2}{k^2 x^4-\left (k^2+1\right ) x^2+1}+1}d\frac {x}{\sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {\arctan \left (\frac {\sqrt {k^2+1} x}{\sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\right )}{\sqrt {k^2+1}}\) |
3.31.71.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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]
Int[((u_)*((A_) + (B_.)*(x_)^4))/Sqrt[v_], x_Symbol] :> With[{a = Coeff[v, x, 0], b = Coeff[v, x, 2], c = Coeff[v, x, 4], d = Coeff[1/u, x, 0], e = Co eff[1/u, x, 2], f = Coeff[1/u, x, 4]}, Simp[A Subst[Int[1/(d - (b*d - a*e )*x^2), x], x, x/Sqrt[v]], x] /; EqQ[a*B + A*c, 0] && EqQ[c*d - a*f, 0]] /; FreeQ[{A, B}, x] && PolyQ[v, x^2, 2] && PolyQ[1/u, x^2, 2]
Time = 2.00 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.10
method | result | size |
elliptic | \(\frac {\arctan \left (\frac {\sqrt {\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )}\, \sqrt {2}}{x \sqrt {2 k^{2}+2}}\right ) \sqrt {2}}{\sqrt {2 k^{2}+2}}\) | \(51\) |
default | \(-\frac {2 \ln \left (2\right )+\ln \left (\frac {\sqrt {-k^{2}-1}\, \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}-\sqrt {2}\, k^{\frac {3}{2}} x^{2}-\sqrt {2}\, \sqrt {k}-x \left (1+k \right )^{2}}{\sqrt {2}\, \sqrt {k}\, x +k \,x^{2}+1}\right )+\ln \left (\frac {\sqrt {-k^{2}-1}\, \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}+\sqrt {2}\, k^{\frac {3}{2}} x^{2}+\sqrt {2}\, \sqrt {k}-x \left (1+k \right )^{2}}{-\sqrt {2}\, \sqrt {k}\, x +k \,x^{2}+1}\right )}{2 \sqrt {-k^{2}-1}}\) | \(164\) |
pseudoelliptic | \(-\frac {2 \ln \left (2\right )+\ln \left (\frac {\sqrt {-k^{2}-1}\, \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}-\sqrt {2}\, k^{\frac {3}{2}} x^{2}-\sqrt {2}\, \sqrt {k}-x \left (1+k \right )^{2}}{\sqrt {2}\, \sqrt {k}\, x +k \,x^{2}+1}\right )+\ln \left (\frac {\sqrt {-k^{2}-1}\, \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}+\sqrt {2}\, k^{\frac {3}{2}} x^{2}+\sqrt {2}\, \sqrt {k}-x \left (1+k \right )^{2}}{-\sqrt {2}\, \sqrt {k}\, x +k \,x^{2}+1}\right )}{2 \sqrt {-k^{2}-1}}\) | \(164\) |
Time = 0.27 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.12 \[ \int \frac {-1+k^2 x^4}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1+k^2 x^4\right )} \, dx=-\frac {\arctan \left (\frac {2 \, \sqrt {k^{2} x^{4} - {\left (k^{2} + 1\right )} x^{2} + 1} \sqrt {k^{2} + 1} x}{k^{2} x^{4} - 2 \, {\left (k^{2} + 1\right )} x^{2} + 1}\right )}{2 \, \sqrt {k^{2} + 1}} \]
-1/2*arctan(2*sqrt(k^2*x^4 - (k^2 + 1)*x^2 + 1)*sqrt(k^2 + 1)*x/(k^2*x^4 - 2*(k^2 + 1)*x^2 + 1))/sqrt(k^2 + 1)
\[ \int \frac {-1+k^2 x^4}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1+k^2 x^4\right )} \, dx=\int \frac {\left (k x^{2} - 1\right ) \left (k x^{2} + 1\right )}{\sqrt {\left (x - 1\right ) \left (x + 1\right ) \left (k x - 1\right ) \left (k x + 1\right )} \left (k^{2} x^{4} + 1\right )}\, dx \]
Integral((k*x**2 - 1)*(k*x**2 + 1)/(sqrt((x - 1)*(x + 1)*(k*x - 1)*(k*x + 1))*(k**2*x**4 + 1)), x)
\[ \int \frac {-1+k^2 x^4}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1+k^2 x^4\right )} \, dx=\int { \frac {k^{2} x^{4} - 1}{{\left (k^{2} x^{4} + 1\right )} \sqrt {{\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}}} \,d x } \]
\[ \int \frac {-1+k^2 x^4}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1+k^2 x^4\right )} \, dx=\int { \frac {k^{2} x^{4} - 1}{{\left (k^{2} x^{4} + 1\right )} \sqrt {{\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}}} \,d x } \]
Timed out. \[ \int \frac {-1+k^2 x^4}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1+k^2 x^4\right )} \, dx=\int \frac {k^2\,x^4-1}{\left (k^2\,x^4+1\right )\,\sqrt {\left (x^2-1\right )\,\left (k^2\,x^2-1\right )}} \,d x \]