Integrand size = 66, antiderivative size = 243 \[ \int \frac {\left (2-k^2\right ) x-2 x^3+k^2 x^5}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (-1+d+\left (-2 d+k^2\right ) x^2+d x^4\right )} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{d} \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}}{2-2 k^2 x^2+\sqrt [3]{d} \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}}\right )}{2 d^{2/3}}-\frac {\log \left (-1+k^2 x^2+\sqrt [3]{d} \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}\right )}{2 d^{2/3}}+\frac {\log \left (1-2 k^2 x^2+k^4 x^4+\left (\sqrt [3]{d}-\sqrt [3]{d} k^2 x^2\right ) \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}+d^{2/3} \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{4/3}\right )}{4 d^{2/3}} \]
-1/2*3^(1/2)*arctan(3^(1/2)*d^(1/3)*(1+(-k^2-1)*x^2+k^2*x^4)^(2/3)/(2-2*k^ 2*x^2+d^(1/3)*(1+(-k^2-1)*x^2+k^2*x^4)^(2/3)))/d^(2/3)-1/2*ln(-1+k^2*x^2+d ^(1/3)*(1+(-k^2-1)*x^2+k^2*x^4)^(2/3))/d^(2/3)+1/4*ln(1-2*k^2*x^2+k^4*x^4+ (d^(1/3)-d^(1/3)*k^2*x^2)*(1+(-k^2-1)*x^2+k^2*x^4)^(2/3)+d^(2/3)*(1+(-k^2- 1)*x^2+k^2*x^4)^(4/3))/d^(2/3)
\[ \int \frac {\left (2-k^2\right ) x-2 x^3+k^2 x^5}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (-1+d+\left (-2 d+k^2\right ) x^2+d x^4\right )} \, dx=\int \frac {\left (2-k^2\right ) x-2 x^3+k^2 x^5}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (-1+d+\left (-2 d+k^2\right ) x^2+d x^4\right )} \, dx \]
Integrate[((2 - k^2)*x - 2*x^3 + k^2*x^5)/(((1 - x^2)*(1 - k^2*x^2))^(2/3) *(-1 + d + (-2*d + k^2)*x^2 + d*x^4)),x]
Integrate[((2 - k^2)*x - 2*x^3 + k^2*x^5)/(((1 - x^2)*(1 - k^2*x^2))^(2/3) *(-1 + d + (-2*d + k^2)*x^2 + d*x^4)), x]
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.
| \(\displaystyle \int \frac {k^2 x^5+\left (2-k^2\right ) x-2 x^3}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (x^2 \left (k^2-2 d\right )+d x^4+d-1\right )} \, dx\) |
| \(\Big \downarrow \) 2028 |
| \(\displaystyle \int \frac {x \left (k^2 x^4-k^2-2 x^2+2\right )}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (x^2 \left (k^2-2 d\right )+d x^4+d-1\right )}dx\) |
| \(\Big \downarrow \) 2048 |
| \(\displaystyle \int \frac {x \left (k^2 x^4-k^2-2 x^2+2\right )}{\left (k^2 x^4+\left (-k^2-1\right ) x^2+1\right )^{2/3} \left (x^2 \left (k^2-2 d\right )+d x^4+d-1\right )}dx\) |
| \(\Big \downarrow \) 7266 |
| \(\displaystyle \frac {1}{2} \int -\frac {k^2 x^4-2 x^2-k^2+2}{\left (-d x^4+\left (2 d-k^2\right ) x^2-d+1\right ) \left (k^2 x^4-\left (k^2+1\right ) x^2+1\right )^{2/3}}dx^2\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {1}{2} \int \frac {k^2 x^4-2 x^2-k^2+2}{\left (-d x^4+\left (2 d-k^2\right ) x^2-d+1\right ) \left (k^2 x^4-\left (k^2+1\right ) x^2+1\right )^{2/3}}dx^2\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle -\frac {1}{2} \int \left (\frac {k^2-\left (k^4+2 d \left (1-k^2\right )\right ) x^2+2 d \left (1-k^2\right )}{d \left (-d x^4+\left (2 d-k^2\right ) x^2-d+1\right ) \left (k^2 x^4-\left (k^2+1\right ) x^2+1\right )^{2/3}}-\frac {k^2}{d \left (k^2 x^4-\left (k^2+1\right ) x^2+1\right )^{2/3}}\right )dx^2\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} \left (-\frac {\int \frac {k^2+\left (-k^4-2 d \left (1-k^2\right )\right ) x^2+2 d \left (1-k^2\right )}{\left (-d x^4+\left (2 d-k^2\right ) x^2-d+1\right ) \left (k^2 x^4+\left (-k^2-1\right ) x^2+1\right )^{2/3}}dx^2}{d}-\frac {\sqrt [3]{2} 3^{3/4} \sqrt {2+\sqrt {3}} k^{4/3} \sqrt {\left (2 k^2 x^2-k^2-1\right )^2} \left (\left (k^2-1\right )^{2/3}+2^{2/3} k^{2/3} \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}\right ) \sqrt {\frac {\left (k^2-1\right )^{4/3}+2 \sqrt [3]{2} k^{4/3} \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}-2^{2/3} \left (k^2-1\right )^{2/3} k^{2/3} \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}{\left (\left (1+\sqrt {3}\right ) \left (k^2-1\right )^{2/3}+2^{2/3} k^{2/3} \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {2^{2/3} \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} k^{2/3}+\left (1-\sqrt {3}\right ) \left (k^2-1\right )^{2/3}}{2^{2/3} \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} k^{2/3}+\left (1+\sqrt {3}\right ) \left (k^2-1\right )^{2/3}}\right ),-7-4 \sqrt {3}\right )}{d \left (-2 k^2 x^2+k^2+1\right ) \sqrt {\left (-\left (k^2 \left (1-2 x^2\right )\right )-1\right )^2} \sqrt {\frac {\left (k^2-1\right )^{2/3} \left (\left (k^2-1\right )^{2/3}+2^{2/3} k^{2/3} \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}\right )}{\left (\left (1+\sqrt {3}\right ) \left (k^2-1\right )^{2/3}+2^{2/3} k^{2/3} \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}\right )^2}}}\right )\) |
Int[((2 - k^2)*x - 2*x^3 + k^2*x^5)/(((1 - x^2)*(1 - k^2*x^2))^(2/3)*(-1 + d + (-2*d + k^2)*x^2 + d*x^4)),x]
3.27.90.3.1 Defintions of rubi rules used
Int[(Fx_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.) + (c_.)*(x_)^(t_.))^(p_.), x_Symbol] :> Int[x^(p*r)*(a + b*x^(s - r) + c*x^(t - r))^p*Fx, x] /; FreeQ[ {a, b, c, r, s, t}, x] && IntegerQ[p] && PosQ[s - r] && PosQ[t - r] && !(E qQ[p, 1] && EqQ[u, 1])
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_)*(x_)^(m_.), x_Symbol] :> Simp[1/(m + 1) Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /; FreeQ[m, x] && NeQ[m, -1] && Function OfQ[x^(m + 1), u, x]
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
\[\int \frac {\left (-k^{2}+2\right ) x -2 x^{3}+k^{2} x^{5}}{{\left (\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )\right )}^{\frac {2}{3}} \left (-1+d +\left (k^{2}-2 d \right ) x^{2}+d \,x^{4}\right )}d x\]
Timed out. \[ \int \frac {\left (2-k^2\right ) x-2 x^3+k^2 x^5}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (-1+d+\left (-2 d+k^2\right ) x^2+d x^4\right )} \, dx=\text {Timed out} \]
integrate(((-k^2+2)*x-2*x^3+k^2*x^5)/((-x^2+1)*(-k^2*x^2+1))^(2/3)/(-1+d+( k^2-2*d)*x^2+d*x^4),x, algorithm="fricas")
Timed out. \[ \int \frac {\left (2-k^2\right ) x-2 x^3+k^2 x^5}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (-1+d+\left (-2 d+k^2\right ) x^2+d x^4\right )} \, dx=\text {Timed out} \]
integrate(((-k**2+2)*x-2*x**3+k**2*x**5)/((-x**2+1)*(-k**2*x**2+1))**(2/3) /(-1+d+(k**2-2*d)*x**2+d*x**4),x)
\[ \int \frac {\left (2-k^2\right ) x-2 x^3+k^2 x^5}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (-1+d+\left (-2 d+k^2\right ) x^2+d x^4\right )} \, dx=\int { \frac {k^{2} x^{5} - 2 \, x^{3} - {\left (k^{2} - 2\right )} x}{{\left (d x^{4} + {\left (k^{2} - 2 \, d\right )} x^{2} + d - 1\right )} \left ({\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}\right )^{\frac {2}{3}}} \,d x } \]
integrate(((-k^2+2)*x-2*x^3+k^2*x^5)/((-x^2+1)*(-k^2*x^2+1))^(2/3)/(-1+d+( k^2-2*d)*x^2+d*x^4),x, algorithm="maxima")
integrate((k^2*x^5 - 2*x^3 - (k^2 - 2)*x)/((d*x^4 + (k^2 - 2*d)*x^2 + d - 1)*((k^2*x^2 - 1)*(x^2 - 1))^(2/3)), x)
\[ \int \frac {\left (2-k^2\right ) x-2 x^3+k^2 x^5}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (-1+d+\left (-2 d+k^2\right ) x^2+d x^4\right )} \, dx=\int { \frac {k^{2} x^{5} - 2 \, x^{3} - {\left (k^{2} - 2\right )} x}{{\left (d x^{4} + {\left (k^{2} - 2 \, d\right )} x^{2} + d - 1\right )} \left ({\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}\right )^{\frac {2}{3}}} \,d x } \]
integrate(((-k^2+2)*x-2*x^3+k^2*x^5)/((-x^2+1)*(-k^2*x^2+1))^(2/3)/(-1+d+( k^2-2*d)*x^2+d*x^4),x, algorithm="giac")
integrate((k^2*x^5 - 2*x^3 - (k^2 - 2)*x)/((d*x^4 + (k^2 - 2*d)*x^2 + d - 1)*((k^2*x^2 - 1)*(x^2 - 1))^(2/3)), x)
Timed out. \[ \int \frac {\left (2-k^2\right ) x-2 x^3+k^2 x^5}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (-1+d+\left (-2 d+k^2\right ) x^2+d x^4\right )} \, dx=-\int \frac {x\,\left (k^2-2\right )-k^2\,x^5+2\,x^3}{{\left (\left (x^2-1\right )\,\left (k^2\,x^2-1\right )\right )}^{2/3}\,\left (d\,x^4+\left (k^2-2\,d\right )\,x^2+d-1\right )} \,d x \]
int(-(x*(k^2 - 2) - k^2*x^5 + 2*x^3)/(((x^2 - 1)*(k^2*x^2 - 1))^(2/3)*(d - x^2*(2*d - k^2) + d*x^4 - 1)),x)