Integrand size = 43, antiderivative size = 87 \[ \int \frac {1+k^2 x^4}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+k^2 x^4\right )} \, dx=-\frac {\arctan \left (\frac {(-1+k) x}{\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{2 (-1+k)}+\frac {\arctan \left (\frac {(1+k) x}{1+k x^2+\sqrt {1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{-1-k} \]
-arctan((-1+k)*x/(1+(-k^2-1)*x^2+k^2*x^4)^(1/2))/(-2+2*k)+arctan((1+k)*x/( 1+k*x^2+(1+(-k^2-1)*x^2+k^2*x^4)^(1/2)))/(-1-k)
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 11.35 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.83 \[ \int \frac {1+k^2 x^4}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+k^2 x^4\right )} \, dx=\frac {\sqrt {1-x^2} \sqrt {1-k^2 x^2} \left (\operatorname {EllipticF}\left (\arcsin (x),k^2\right )-\operatorname {EllipticPi}\left (-k,\arcsin (x),k^2\right )-\operatorname {EllipticPi}\left (k,\arcsin (x),k^2\right )\right )}{\sqrt {\left (-1+x^2\right ) \left (-1+k^2 x^2\right )}} \]
(Sqrt[1 - x^2]*Sqrt[1 - k^2*x^2]*(EllipticF[ArcSin[x], k^2] - EllipticPi[- k, ArcSin[x], k^2] - EllipticPi[k, ArcSin[x], k^2]))/Sqrt[(-1 + x^2)*(-1 + k^2*x^2)]
Time = 0.96 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.93, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {2048, 7276, 2009}
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 |
\(\displaystyle \int \frac {k^2 x^4+1}{\left (k^2 x^4-1\right ) \sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}}dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (\frac {2}{\left (k^2 x^4-1\right ) \sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}}+\frac {1}{\sqrt {k^2 x^4+\left (-k^2-1\right ) x^2+1}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\arctan \left (\frac {(1-k) x}{\sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\right )}{2 (1-k)}-\frac {\arctan \left (\frac {(k+1) x}{\sqrt {k^2 x^4-\left (k^2+1\right ) x^2+1}}\right )}{2 (k+1)}\) |
-1/2*ArcTan[((1 - k)*x)/Sqrt[1 - (1 + k^2)*x^2 + k^2*x^4]]/(1 - k) - ArcTa n[((1 + k)*x)/Sqrt[1 - (1 + k^2)*x^2 + k^2*x^4]]/(2*(1 + k))
3.12.87.3.1 Defintions of rubi rules used
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_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 5.67 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00
method | result | size |
elliptic | \(\frac {\left (\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )}}{x \left (-1+k \right )}\right )}{-2+2 k}+\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )}}{x \left (1+k \right )}\right )}{2+2 k}\right ) \sqrt {2}}{2}\) | \(87\) |
default | \(-\frac {\frac {\sqrt {-\left (1+k \right )^{2}}\, \ln \left (\frac {\sqrt {-\left (-1+k \right )^{2}}\, \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}-2 k^{\frac {3}{2}} x^{2}-2 \sqrt {k}+\left (-k^{2}-2 k -1\right ) x}{1+2 \sqrt {k}\, x +k \,x^{2}}\right )}{2}+\frac {\sqrt {-\left (1+k \right )^{2}}\, \ln \left (\frac {\sqrt {-\left (-1+k \right )^{2}}\, \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}+2 k^{\frac {3}{2}} x^{2}+2 \sqrt {k}+\left (-k^{2}-2 k -1\right ) x}{1-2 \sqrt {k}\, x +k \,x^{2}}\right )}{2}+\sqrt {-\left (-1+k \right )^{2}}\, \ln \left (\frac {\sqrt {-\left (1+k \right )^{2}}\, \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}-x \left (1+k \right )^{2}}{k \,x^{2}+1}\right )+\ln \left (2\right ) \left (\sqrt {-\left (-1+k \right )^{2}}+\sqrt {-\left (1+k \right )^{2}}\right )}{2 \sqrt {-\left (1+k \right )^{2}}\, \sqrt {-\left (-1+k \right )^{2}}}\) | \(263\) |
pseudoelliptic | \(-\frac {\frac {\sqrt {-\left (1+k \right )^{2}}\, \ln \left (\frac {\sqrt {-\left (-1+k \right )^{2}}\, \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}-2 k^{\frac {3}{2}} x^{2}-2 \sqrt {k}+\left (-k^{2}-2 k -1\right ) x}{1+2 \sqrt {k}\, x +k \,x^{2}}\right )}{2}+\frac {\sqrt {-\left (1+k \right )^{2}}\, \ln \left (\frac {\sqrt {-\left (-1+k \right )^{2}}\, \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}+2 k^{\frac {3}{2}} x^{2}+2 \sqrt {k}+\left (-k^{2}-2 k -1\right ) x}{1-2 \sqrt {k}\, x +k \,x^{2}}\right )}{2}+\sqrt {-\left (-1+k \right )^{2}}\, \ln \left (\frac {\sqrt {-\left (1+k \right )^{2}}\, \sqrt {\left (x^{2}-1\right ) \left (k^{2} x^{2}-1\right )}-x \left (1+k \right )^{2}}{k \,x^{2}+1}\right )+\ln \left (2\right ) \left (\sqrt {-\left (-1+k \right )^{2}}+\sqrt {-\left (1+k \right )^{2}}\right )}{2 \sqrt {-\left (1+k \right )^{2}}\, \sqrt {-\left (-1+k \right )^{2}}}\) | \(263\) |
1/2*(1/2*2^(1/2)/(-1+k)*arctan(((-x^2+1)*(-k^2*x^2+1))^(1/2)/x/(-1+k))+1/2 *2^(1/2)/(1+k)*arctan(((-x^2+1)*(-k^2*x^2+1))^(1/2)/x/(1+k)))*2^(1/2)
Time = 0.33 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.92 \[ \int \frac {1+k^2 x^4}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+k^2 x^4\right )} \, dx=\frac {{\left (k - 1\right )} \arctan \left (\frac {\sqrt {k^{2} x^{4} - {\left (k^{2} + 1\right )} x^{2} + 1}}{{\left (k + 1\right )} x}\right ) + {\left (k + 1\right )} \arctan \left (\frac {\sqrt {k^{2} x^{4} - {\left (k^{2} + 1\right )} x^{2} + 1}}{{\left (k - 1\right )} x}\right )}{2 \, {\left (k^{2} - 1\right )}} \]
1/2*((k - 1)*arctan(sqrt(k^2*x^4 - (k^2 + 1)*x^2 + 1)/((k + 1)*x)) + (k + 1)*arctan(sqrt(k^2*x^4 - (k^2 + 1)*x^2 + 1)/((k - 1)*x)))/(k^2 - 1)
\[ \int \frac {1+k^2 x^4}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+k^2 x^4\right )} \, dx=\int \frac {k^{2} x^{4} + 1}{\sqrt {\left (x - 1\right ) \left (x + 1\right ) \left (k x - 1\right ) \left (k x + 1\right )} \left (k x^{2} - 1\right ) \left (k x^{2} + 1\right )}\, dx \]
Integral((k**2*x**4 + 1)/(sqrt((x - 1)*(x + 1)*(k*x - 1)*(k*x + 1))*(k*x** 2 - 1)*(k*x**2 + 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. \[ \int \frac {1+k^2 x^4}{\sqrt {\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+k^2 x^4\right )} \, dx=\int \frac {k^2\,x^4+1}{\left (k^2\,x^4-1\right )\,\sqrt {\left (x^2-1\right )\,\left (k^2\,x^2-1\right )}} \,d x \]