Integrand size = 65, antiderivative size = 243 \[ \int \frac {\left (1-2 k^2\right ) x+k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1-d+\left (d-2 k^2\right ) x^2+k^4 x^4\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{d} \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}{2-2 k^2 x^2+\sqrt [3]{d} \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{2 d^{2/3}}+\frac {\log \left (-1+k^2 x^2+\sqrt [3]{d} \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}\right )}{2 d^{2/3}}-\frac {\log \left (1-2 k^2 x^2+k^4 x^4+\left (\sqrt [3]{d}-\sqrt [3]{d} k^2 x^2\right ) \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}+d^{2/3} \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}\right )}{4 d^{2/3}} \]
1/2*3^(1/2)*arctan(3^(1/2)*d^(1/3)*(1+(-k^2-1)*x^2+k^2*x^4)^(1/3)/(2-2*k^2 *x^2+d^(1/3)*(1+(-k^2-1)*x^2+k^2*x^4)^(1/3)))/d^(2/3)+1/2*ln(-1+k^2*x^2+d^ (1/3)*(1+(-k^2-1)*x^2+k^2*x^4)^(1/3))/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)^(1/3)+d^(2/3)*(1+(-k^2-1 )*x^2+k^2*x^4)^(2/3))/d^(2/3)
Time = 19.71 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.84 \[ \int \frac {\left (1-2 k^2\right ) x+k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1-d+\left (d-2 k^2\right ) x^2+k^4 x^4\right )} \, dx=\frac {\sqrt [3]{-1+x^2} \sqrt [3]{-1+k^2 x^2} \left (2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{d} \sqrt [3]{-1+x^2}}{\sqrt [3]{d} \sqrt [3]{-1+x^2}-2 \left (-1+k^2 x^2\right )^{2/3}}\right )+2 \log \left (\sqrt [3]{d} \sqrt [3]{-1+x^2}+\left (-1+k^2 x^2\right )^{2/3}\right )-\log \left (d^{2/3} \left (-1+x^2\right )^{2/3}-\sqrt [3]{d} \sqrt [3]{-1+x^2} \left (-1+k^2 x^2\right )^{2/3}+\left (-1+k^2 x^2\right )^{4/3}\right )\right )}{4 d^{2/3} \sqrt [3]{\left (-1+x^2\right ) \left (-1+k^2 x^2\right )}} \]
Integrate[((1 - 2*k^2)*x + k^2*x^3)/(((1 - x^2)*(1 - k^2*x^2))^(1/3)*(1 - d + (d - 2*k^2)*x^2 + k^4*x^4)),x]
((-1 + x^2)^(1/3)*(-1 + k^2*x^2)^(1/3)*(2*Sqrt[3]*ArcTan[(Sqrt[3]*d^(1/3)* (-1 + x^2)^(1/3))/(d^(1/3)*(-1 + x^2)^(1/3) - 2*(-1 + k^2*x^2)^(2/3))] + 2 *Log[d^(1/3)*(-1 + x^2)^(1/3) + (-1 + k^2*x^2)^(2/3)] - Log[d^(2/3)*(-1 + x^2)^(2/3) - d^(1/3)*(-1 + x^2)^(1/3)*(-1 + k^2*x^2)^(2/3) + (-1 + k^2*x^2 )^(4/3)]))/(4*d^(2/3)*((-1 + x^2)*(-1 + k^2*x^2))^(1/3))
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.
| \(\displaystyle \int \frac {k^2 x^3+\left (1-2 k^2\right ) x}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (x^2 \left (d-2 k^2\right )-d+k^4 x^4+1\right )} \, dx\) |
| \(\Big \downarrow \) 2027 |
| \(\displaystyle \int \frac {x \left (k^2 x^2-2 k^2+1\right )}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (x^2 \left (d-2 k^2\right )-d+k^4 x^4+1\right )}dx\) |
| \(\Big \downarrow \) 2048 |
| \(\displaystyle \int \frac {x \left (k^2 x^2-2 k^2+1\right )}{\sqrt [3]{k^2 x^4+\left (-k^2-1\right ) x^2+1} \left (x^2 \left (d-2 k^2\right )-d+k^4 x^4+1\right )}dx\) |
| \(\Big \downarrow \) 7266 |
| \(\displaystyle \frac {1}{2} \int \frac {x^2 k^2-2 k^2+1}{\sqrt [3]{k^2 x^4-\left (k^2+1\right ) x^2+1} \left (k^4 x^4+\left (d-2 k^2\right ) x^2-d+1\right )}dx^2\) |
| \(\Big \downarrow \) 1375 |
| \(\displaystyle \frac {1}{2} \int \frac {x^2 k^2-2 k^2+1}{\sqrt [3]{k^2 x^4+\left (-k^2-1\right ) x^2+1} \left (k^4 x^4+\left (d-2 k^2\right ) x^2-d+1\right )}dx^2\) |
Int[((1 - 2*k^2)*x + k^2*x^3)/(((1 - x^2)*(1 - k^2*x^2))^(1/3)*(1 - d + (d - 2*k^2)*x^2 + k^4*x^4)),x]
3.27.89.3.1 Defintions of rubi rules used
Int[((g_.) + (h_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_.) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Unintegrable[(g + h*x)*(a + b *x + c*x^2)^p*(d + e*x + f*x^2)^q, x] /; FreeQ[{a, b, c, d, e, f, g, h, p, q}, x]
Int[(Fx_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.))^(p_.), x_Symbol] :> Int[x^ (p*r)*(a + b*x^(s - r))^p*Fx, x] /; FreeQ[{a, b, r, s}, x] && IntegerQ[p] & & PosQ[s - r] && !(EqQ[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]
\[\int \frac {\left (-2 k^{2}+1\right ) x +k^{2} x^{3}}{{\left (\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )\right )}^{\frac {1}{3}} \left (1-d +\left (-2 k^{2}+d \right ) x^{2}+k^{4} x^{4}\right )}d x\]
Timed out. \[ \int \frac {\left (1-2 k^2\right ) x+k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1-d+\left (d-2 k^2\right ) x^2+k^4 x^4\right )} \, dx=\text {Timed out} \]
integrate(((-2*k^2+1)*x+k^2*x^3)/((-x^2+1)*(-k^2*x^2+1))^(1/3)/(1-d+(-2*k^ 2+d)*x^2+k^4*x^4),x, algorithm="fricas")
Timed out. \[ \int \frac {\left (1-2 k^2\right ) x+k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1-d+\left (d-2 k^2\right ) x^2+k^4 x^4\right )} \, dx=\text {Timed out} \]
integrate(((-2*k**2+1)*x+k**2*x**3)/((-x**2+1)*(-k**2*x**2+1))**(1/3)/(1-d +(-2*k**2+d)*x**2+k**4*x**4),x)
\[ \int \frac {\left (1-2 k^2\right ) x+k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1-d+\left (d-2 k^2\right ) x^2+k^4 x^4\right )} \, dx=\int { \frac {k^{2} x^{3} - {\left (2 \, k^{2} - 1\right )} x}{{\left (k^{4} x^{4} - {\left (2 \, k^{2} - d\right )} x^{2} - d + 1\right )} \left ({\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}\right )^{\frac {1}{3}}} \,d x } \]
integrate(((-2*k^2+1)*x+k^2*x^3)/((-x^2+1)*(-k^2*x^2+1))^(1/3)/(1-d+(-2*k^ 2+d)*x^2+k^4*x^4),x, algorithm="maxima")
integrate((k^2*x^3 - (2*k^2 - 1)*x)/((k^4*x^4 - (2*k^2 - d)*x^2 - d + 1)*( (k^2*x^2 - 1)*(x^2 - 1))^(1/3)), x)
\[ \int \frac {\left (1-2 k^2\right ) x+k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1-d+\left (d-2 k^2\right ) x^2+k^4 x^4\right )} \, dx=\int { \frac {k^{2} x^{3} - {\left (2 \, k^{2} - 1\right )} x}{{\left (k^{4} x^{4} - {\left (2 \, k^{2} - d\right )} x^{2} - d + 1\right )} \left ({\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}\right )^{\frac {1}{3}}} \,d x } \]
integrate(((-2*k^2+1)*x+k^2*x^3)/((-x^2+1)*(-k^2*x^2+1))^(1/3)/(1-d+(-2*k^ 2+d)*x^2+k^4*x^4),x, algorithm="giac")
integrate((k^2*x^3 - (2*k^2 - 1)*x)/((k^4*x^4 - (2*k^2 - d)*x^2 - d + 1)*( (k^2*x^2 - 1)*(x^2 - 1))^(1/3)), x)
Timed out. \[ \int \frac {\left (1-2 k^2\right ) x+k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1-d+\left (d-2 k^2\right ) x^2+k^4 x^4\right )} \, dx=-\int -\frac {k^2\,x^3-x\,\left (2\,k^2-1\right )}{{\left (\left (x^2-1\right )\,\left (k^2\,x^2-1\right )\right )}^{1/3}\,\left (k^4\,x^4-d+x^2\,\left (d-2\,k^2\right )+1\right )} \,d x \]
int((k^2*x^3 - x*(2*k^2 - 1))/(((x^2 - 1)*(k^2*x^2 - 1))^(1/3)*(k^4*x^4 - d + x^2*(d - 2*k^2) + 1)),x)