Integrand size = 32, antiderivative size = 385 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [3]{-x+x^3}}{-d+c x^2} \, dx=\frac {a x \sqrt [3]{-x+x^3}}{2 c}+\frac {(a c+3 b c-3 a d) \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{-x+x^3}}\right )}{2 \sqrt {3} c^2}+\frac {\sqrt {3} \sqrt [3]{c-d} (b c-a d) \arctan \left (\frac {\sqrt {3} \sqrt [3]{c-d} x}{\sqrt [3]{c-d} x-2 \sqrt [3]{d} \sqrt [3]{-x+x^3}}\right )}{2 c^2 \sqrt [3]{d}}+\frac {(a c+3 b c-3 a d) \log \left (-x+\sqrt [3]{-x+x^3}\right )}{6 c^2}+\frac {\sqrt [3]{c-d} (b c-a d) \log \left (\sqrt [3]{c-d} x+\sqrt [3]{d} \sqrt [3]{-x+x^3}\right )}{2 c^2 \sqrt [3]{d}}+\frac {(-a c-3 b c+3 a d) \log \left (x^2+x \sqrt [3]{-x+x^3}+\left (-x+x^3\right )^{2/3}\right )}{12 c^2}-\frac {\sqrt [3]{c-d} (b c-a d) \log \left ((c-d)^{2/3} x^2-\sqrt [3]{c-d} \sqrt [3]{d} x \sqrt [3]{-x+x^3}+d^{2/3} \left (-x+x^3\right )^{2/3}\right )}{4 c^2 \sqrt [3]{d}} \] Output:
1/2*a*x*(x^3-x)^(1/3)/c+1/6*(a*c-3*a*d+3*b*c)*arctan(3^(1/2)*x/(x+2*(x^3-x )^(1/3)))*3^(1/2)/c^2+1/2*3^(1/2)*(c-d)^(1/3)*(-a*d+b*c)*arctan(3^(1/2)*(c -d)^(1/3)*x/((c-d)^(1/3)*x-2*d^(1/3)*(x^3-x)^(1/3)))/c^2/d^(1/3)+1/6*(a*c- 3*a*d+3*b*c)*ln(-x+(x^3-x)^(1/3))/c^2+1/2*(c-d)^(1/3)*(-a*d+b*c)*ln((c-d)^ (1/3)*x+d^(1/3)*(x^3-x)^(1/3))/c^2/d^(1/3)+1/12*(-a*c+3*a*d-3*b*c)*ln(x^2+ x*(x^3-x)^(1/3)+(x^3-x)^(2/3))/c^2-1/4*(c-d)^(1/3)*(-a*d+b*c)*ln((c-d)^(2/ 3)*x^2-(c-d)^(1/3)*d^(1/3)*x*(x^3-x)^(1/3)+d^(2/3)*(x^3-x)^(2/3))/c^2/d^(1 /3)
Time = 9.10 (sec) , antiderivative size = 414, normalized size of antiderivative = 1.08 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [3]{-x+x^3}}{-d+c x^2} \, dx=\frac {x^{2/3} \left (-1+x^2\right )^{2/3} \left (6 a c \sqrt [3]{d} x^{4/3} \sqrt [3]{-1+x^2}+2 \sqrt {3} (3 b c+a (c-3 d)) \sqrt [3]{d} \arctan \left (\frac {\sqrt {3} x^{2/3}}{x^{2/3}+2 \sqrt [3]{-1+x^2}}\right )+6 \sqrt {3} \sqrt [3]{c-d} (b c-a d) \arctan \left (\frac {\sqrt {3} \sqrt [3]{c-d} x^{2/3}}{\sqrt [3]{c-d} x^{2/3}-2 \sqrt [3]{d} \sqrt [3]{-1+x^2}}\right )+2 (3 b c+a (c-3 d)) \sqrt [3]{d} \log \left (-x^{2/3}+\sqrt [3]{-1+x^2}\right )+6 \sqrt [3]{c-d} (b c-a d) \log \left (\sqrt [3]{c-d} x^{2/3}+\sqrt [3]{d} \sqrt [3]{-1+x^2}\right )+\sqrt [3]{d} (-a c-3 b c+3 a d) \log \left (x^{4/3}+x^{2/3} \sqrt [3]{-1+x^2}+\left (-1+x^2\right )^{2/3}\right )-3 \sqrt [3]{c-d} (b c-a d) \log \left ((c-d)^{2/3} x^{4/3}-\sqrt [3]{c-d} \sqrt [3]{d} x^{2/3} \sqrt [3]{-1+x^2}+d^{2/3} \left (-1+x^2\right )^{2/3}\right )\right )}{12 c^2 \sqrt [3]{d} \left (x \left (-1+x^2\right )\right )^{2/3}} \] Input:
Integrate[((-b + a*x^2)*(-x + x^3)^(1/3))/(-d + c*x^2),x]
Output:
(x^(2/3)*(-1 + x^2)^(2/3)*(6*a*c*d^(1/3)*x^(4/3)*(-1 + x^2)^(1/3) + 2*Sqrt [3]*(3*b*c + a*(c - 3*d))*d^(1/3)*ArcTan[(Sqrt[3]*x^(2/3))/(x^(2/3) + 2*(- 1 + x^2)^(1/3))] + 6*Sqrt[3]*(c - d)^(1/3)*(b*c - a*d)*ArcTan[(Sqrt[3]*(c - d)^(1/3)*x^(2/3))/((c - d)^(1/3)*x^(2/3) - 2*d^(1/3)*(-1 + x^2)^(1/3))] + 2*(3*b*c + a*(c - 3*d))*d^(1/3)*Log[-x^(2/3) + (-1 + x^2)^(1/3)] + 6*(c - d)^(1/3)*(b*c - a*d)*Log[(c - d)^(1/3)*x^(2/3) + d^(1/3)*(-1 + x^2)^(1/3 )] + d^(1/3)*(-(a*c) - 3*b*c + 3*a*d)*Log[x^(4/3) + x^(2/3)*(-1 + x^2)^(1/ 3) + (-1 + x^2)^(2/3)] - 3*(c - d)^(1/3)*(b*c - a*d)*Log[(c - d)^(2/3)*x^( 4/3) - (c - d)^(1/3)*d^(1/3)*x^(2/3)*(-1 + x^2)^(1/3) + d^(2/3)*(-1 + x^2) ^(2/3)]))/(12*c^2*d^(1/3)*(x*(-1 + x^2))^(2/3))
Time = 0.90 (sec) , antiderivative size = 315, normalized size of antiderivative = 0.82, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {2467, 443, 27, 446, 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.
| \(\displaystyle \int \frac {\sqrt [3]{x^3-x} \left (a x^2-b\right )}{c x^2-d} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt [3]{x^3-x} \int \frac {\sqrt [3]{x} \sqrt [3]{x^2-1} \left (b-a x^2\right )}{d-c x^2}dx}{\sqrt [3]{x} \sqrt [3]{x^2-1}}\) |
| \(\Big \downarrow \) 443 |
| \(\displaystyle \frac {\sqrt [3]{x^3-x} \left (\frac {a x^{4/3} \sqrt [3]{x^2-1}}{2 c}-\frac {\int \frac {2 \sqrt [3]{x} \left (-\left ((3 b c+a (c-3 d)) x^2\right )+3 b c-2 a d\right )}{3 \left (x^2-1\right )^{2/3} \left (d-c x^2\right )}dx}{2 c}\right )}{\sqrt [3]{x} \sqrt [3]{x^2-1}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt [3]{x^3-x} \left (\frac {a x^{4/3} \sqrt [3]{x^2-1}}{2 c}-\frac {\int \frac {\sqrt [3]{x} \left (-\left ((3 b c+a (c-3 d)) x^2\right )+3 b c-2 a d\right )}{\left (x^2-1\right )^{2/3} \left (d-c x^2\right )}dx}{3 c}\right )}{\sqrt [3]{x} \sqrt [3]{x^2-1}}\) |
| \(\Big \downarrow \) 446 |
| \(\displaystyle \frac {\sqrt [3]{x^3-x} \left (\frac {a x^{4/3} \sqrt [3]{x^2-1}}{2 c}-\frac {\int \left (-\frac {\sqrt [3]{x} (-3 b c-a (c-3 d))}{c \left (x^2-1\right )^{2/3}}-\frac {(-((-3 b c-a (c-3 d)) d)-c (3 b c-2 a d)) \sqrt [3]{x}}{c \left (x^2-1\right )^{2/3} \left (d-c x^2\right )}\right )dx}{3 c}\right )}{\sqrt [3]{x} \sqrt [3]{x^2-1}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt [3]{x^3-x} \left (\frac {a x^{4/3} \sqrt [3]{x^2-1}}{2 c}-\frac {-\frac {\sqrt {3} \arctan \left (\frac {\frac {2 x^{2/3}}{\sqrt [3]{x^2-1}}+1}{\sqrt {3}}\right ) (a (c-3 d)+3 b c)}{2 c}-\frac {3 \sqrt {3} \sqrt [3]{c-d} (b c-a d) \arctan \left (\frac {1-\frac {2 x^{2/3} \sqrt [3]{c-d}}{\sqrt [3]{d} \sqrt [3]{x^2-1}}}{\sqrt {3}}\right )}{2 c \sqrt [3]{d}}+\frac {3 \sqrt [3]{c-d} (b c-a d) \log \left (d-c x^2\right )}{4 c \sqrt [3]{d}}-\frac {3 \log \left (x^{2/3}-\sqrt [3]{x^2-1}\right ) (a (c-3 d)+3 b c)}{4 c}-\frac {9 \sqrt [3]{c-d} (b c-a d) \log \left (x^{2/3} \sqrt [3]{c-d}+\sqrt [3]{d} \sqrt [3]{x^2-1}\right )}{4 c \sqrt [3]{d}}}{3 c}\right )}{\sqrt [3]{x} \sqrt [3]{x^2-1}}\) |
Input:
Int[((-b + a*x^2)*(-x + x^3)^(1/3))/(-d + c*x^2),x]
Output:
((-x + x^3)^(1/3)*((a*x^(4/3)*(-1 + x^2)^(1/3))/(2*c) - (-1/2*(Sqrt[3]*(3* b*c + a*(c - 3*d))*ArcTan[(1 + (2*x^(2/3))/(-1 + x^2)^(1/3))/Sqrt[3]])/c - (3*Sqrt[3]*(c - d)^(1/3)*(b*c - a*d)*ArcTan[(1 - (2*(c - d)^(1/3)*x^(2/3) )/(d^(1/3)*(-1 + x^2)^(1/3)))/Sqrt[3]])/(2*c*d^(1/3)) + (3*(c - d)^(1/3)*( b*c - a*d)*Log[d - c*x^2])/(4*c*d^(1/3)) - (3*(3*b*c + a*(c - 3*d))*Log[x^ (2/3) - (-1 + x^2)^(1/3)])/(4*c) - (9*(c - d)^(1/3)*(b*c - a*d)*Log[(c - d )^(1/3)*x^(2/3) + d^(1/3)*(-1 + x^2)^(1/3)])/(4*c*d^(1/3)))/(3*c)))/(x^(1/ 3)*(-1 + x^2)^(1/3))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q _.)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[f*(g*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(b*g*(m + 2*(p + q + 1) + 1))), x] + Simp[1/(b*(m + 2* (p + q + 1) + 1)) Int[(g*x)^m*(a + b*x^2)^p*(c + d*x^2)^(q - 1)*Simp[c*(( b*e - a*f)*(m + 1) + b*e*2*(p + q + 1)) + (d*(b*e - a*f)*(m + 1) + f*2*q*(b *c - a*d) + b*e*d*2*(p + q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f , g, m, p}, x] && GtQ[q, 0] && !(EqQ[q, 1] && SimplerQ[e + f*x^2, c + d*x^ 2])
Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((e_) + (f_.)*(x_)^2))/( (c_) + (d_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^2)^ p*((e + f*x^2)/(c + d*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x ]
Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p]) Int[x^( p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P olyQ[Px, x] && !IntegerQ[p] && !MonomialQ[Px, x] && !PolyQ[Fx, x]
Time = 0.84 (sec) , antiderivative size = 330, normalized size of antiderivative = 0.86
| method | result | size |
| pseudoelliptic | \(\frac {x \left (\frac {d \left (-3 x a \left (x^{3}-x \right )^{\frac {1}{3}} c +\left (\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x +2 \left (x^{3}-x \right )^{\frac {1}{3}}\right )}{3 x}\right )-\ln \left (\frac {-x +\left (x^{3}-x \right )^{\frac {1}{3}}}{x}\right )+\frac {\ln \left (\frac {x^{2}+x \left (x^{3}-x \right )^{\frac {1}{3}}+\left (x^{3}-x \right )^{\frac {2}{3}}}{x^{2}}\right )}{2}\right ) \left (-3 a d +c \left (a +3 b \right )\right )\right ) \left (\frac {c -d}{d}\right )^{\frac {2}{3}}}{3}+\left (a d -b c \right ) \left (c -d \right ) \left (-\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\left (\frac {c -d}{d}\right )^{\frac {1}{3}} x -2 \left (x^{3}-x \right )^{\frac {1}{3}}\right )}{3 \left (\frac {c -d}{d}\right )^{\frac {1}{3}} x}\right )+\ln \left (\frac {\left (\frac {c -d}{d}\right )^{\frac {1}{3}} x +\left (x^{3}-x \right )^{\frac {1}{3}}}{x}\right )-\frac {\ln \left (\frac {\left (\frac {c -d}{d}\right )^{\frac {2}{3}} x^{2}-\left (\frac {c -d}{d}\right )^{\frac {1}{3}} x \left (x^{3}-x \right )^{\frac {1}{3}}+\left (x^{3}-x \right )^{\frac {2}{3}}}{x^{2}}\right )}{2}\right )\right )}{2 \left (\frac {c -d}{d}\right )^{\frac {2}{3}} c^{2} d \left (\left (x^{3}-x \right )^{\frac {2}{3}}+x \left (\left (x^{3}-x \right )^{\frac {1}{3}}+x \right )\right ) \left (-x +\left (x^{3}-x \right )^{\frac {1}{3}}\right )}\) | \(330\) |
Input:
int((a*x^2-b)*(x^3-x)^(1/3)/(c*x^2-d),x,method=_RETURNVERBOSE)
Output:
1/2/((c-d)/d)^(2/3)*x*(1/3*d*(-3*x*a*(x^3-x)^(1/3)*c+(3^(1/2)*arctan(1/3*3 ^(1/2)/x*(x+2*(x^3-x)^(1/3)))-ln((-x+(x^3-x)^(1/3))/x)+1/2*ln((x^2+x*(x^3- x)^(1/3)+(x^3-x)^(2/3))/x^2))*(-3*a*d+c*(a+3*b)))*((c-d)/d)^(2/3)+(a*d-b*c )*(c-d)*(-3^(1/2)*arctan(1/3*3^(1/2)*(((c-d)/d)^(1/3)*x-2*(x^3-x)^(1/3))/( (c-d)/d)^(1/3)/x)+ln((((c-d)/d)^(1/3)*x+(x^3-x)^(1/3))/x)-1/2*ln((((c-d)/d )^(2/3)*x^2-((c-d)/d)^(1/3)*x*(x^3-x)^(1/3)+(x^3-x)^(2/3))/x^2)))/c^2/d/(( x^3-x)^(2/3)+x*((x^3-x)^(1/3)+x))/(-x+(x^3-x)^(1/3))
Timed out. \[ \int \frac {\left (-b+a x^2\right ) \sqrt [3]{-x+x^3}}{-d+c x^2} \, dx=\text {Timed out} \] Input:
integrate((a*x^2-b)*(x^3-x)^(1/3)/(c*x^2-d),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {\left (-b+a x^2\right ) \sqrt [3]{-x+x^3}}{-d+c x^2} \, dx=\int \frac {\sqrt [3]{x \left (x - 1\right ) \left (x + 1\right )} \left (a x^{2} - b\right )}{c x^{2} - d}\, dx \] Input:
integrate((a*x**2-b)*(x**3-x)**(1/3)/(c*x**2-d),x)
Output:
Integral((x*(x - 1)*(x + 1))**(1/3)*(a*x**2 - b)/(c*x**2 - d), x)
\[ \int \frac {\left (-b+a x^2\right ) \sqrt [3]{-x+x^3}}{-d+c x^2} \, dx=\int { \frac {{\left (a x^{2} - b\right )} {\left (x^{3} - x\right )}^{\frac {1}{3}}}{c x^{2} - d} \,d x } \] Input:
integrate((a*x^2-b)*(x^3-x)^(1/3)/(c*x^2-d),x, algorithm="maxima")
Output:
integrate((a*x^2 - b)*(x^3 - x)^(1/3)/(c*x^2 - d), x)
Time = 0.39 (sec) , antiderivative size = 372, normalized size of antiderivative = 0.97 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [3]{-x+x^3}}{-d+c x^2} \, dx=\frac {a x^{2} {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}}}{2 \, c} - \frac {{\left (b c^{2} - a c d - b c d + a d^{2}\right )} \left (-\frac {c - d}{d}\right )^{\frac {1}{3}} \log \left ({\left | -\left (-\frac {c - d}{d}\right )^{\frac {1}{3}} + {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} \right |}\right )}{2 \, {\left (c^{3} - c^{2} d\right )}} - \frac {\sqrt {3} {\left (a c + 3 \, b c - 3 \, a d\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right )}\right )}{6 \, c^{2}} - \frac {{\left (a c + 3 \, b c - 3 \, a d\right )} \log \left ({\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}} + {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right )}{12 \, c^{2}} + \frac {{\left (a c + 3 \, b c - 3 \, a d\right )} \log \left ({\left | {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right )}{6 \, c^{2}} + \frac {{\left (\sqrt {3} {\left (-c d^{2} + d^{3}\right )}^{\frac {1}{3}} b c - \sqrt {3} {\left (-c d^{2} + d^{3}\right )}^{\frac {1}{3}} a d\right )} \arctan \left (\frac {\sqrt {3} {\left (\left (-\frac {c - d}{d}\right )^{\frac {1}{3}} + 2 \, {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {c - d}{d}\right )^{\frac {1}{3}}}\right )}{2 \, c^{2} d} + \frac {{\left ({\left (-c d^{2} + d^{3}\right )}^{\frac {1}{3}} b c - {\left (-c d^{2} + d^{3}\right )}^{\frac {1}{3}} a d\right )} \log \left (\left (-\frac {c - d}{d}\right )^{\frac {2}{3}} + \left (-\frac {c - d}{d}\right )^{\frac {1}{3}} {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}}\right )}{4 \, c^{2} d} \] Input:
integrate((a*x^2-b)*(x^3-x)^(1/3)/(c*x^2-d),x, algorithm="giac")
Output:
1/2*a*x^2*(-1/x^2 + 1)^(1/3)/c - 1/2*(b*c^2 - a*c*d - b*c*d + a*d^2)*(-(c - d)/d)^(1/3)*log(abs(-(-(c - d)/d)^(1/3) + (-1/x^2 + 1)^(1/3)))/(c^3 - c^ 2*d) - 1/6*sqrt(3)*(a*c + 3*b*c - 3*a*d)*arctan(1/3*sqrt(3)*(2*(-1/x^2 + 1 )^(1/3) + 1))/c^2 - 1/12*(a*c + 3*b*c - 3*a*d)*log((-1/x^2 + 1)^(2/3) + (- 1/x^2 + 1)^(1/3) + 1)/c^2 + 1/6*(a*c + 3*b*c - 3*a*d)*log(abs((-1/x^2 + 1) ^(1/3) - 1))/c^2 + 1/2*(sqrt(3)*(-c*d^2 + d^3)^(1/3)*b*c - sqrt(3)*(-c*d^2 + d^3)^(1/3)*a*d)*arctan(1/3*sqrt(3)*((-(c - d)/d)^(1/3) + 2*(-1/x^2 + 1) ^(1/3))/(-(c - d)/d)^(1/3))/(c^2*d) + 1/4*((-c*d^2 + d^3)^(1/3)*b*c - (-c* d^2 + d^3)^(1/3)*a*d)*log((-(c - d)/d)^(2/3) + (-(c - d)/d)^(1/3)*(-1/x^2 + 1)^(1/3) + (-1/x^2 + 1)^(2/3))/(c^2*d)
Timed out. \[ \int \frac {\left (-b+a x^2\right ) \sqrt [3]{-x+x^3}}{-d+c x^2} \, dx=\int \frac {{\left (x^3-x\right )}^{1/3}\,\left (b-a\,x^2\right )}{d-c\,x^2} \,d x \] Input:
int(((x^3 - x)^(1/3)*(b - a*x^2))/(d - c*x^2),x)
Output:
int(((x^3 - x)^(1/3)*(b - a*x^2))/(d - c*x^2), x)
\[ \int \frac {\left (-b+a x^2\right ) \sqrt [3]{-x+x^3}}{-d+c x^2} \, dx=\frac {3 x^{\frac {4}{3}} \left (x^{2}-1\right )^{\frac {1}{3}} a -2 \left (\int \frac {x^{\frac {7}{3}} \left (x^{2}-1\right )^{\frac {1}{3}}}{c \,x^{4}-c \,x^{2}-d \,x^{2}+d}d x \right ) a c +6 \left (\int \frac {x^{\frac {7}{3}} \left (x^{2}-1\right )^{\frac {1}{3}}}{c \,x^{4}-c \,x^{2}-d \,x^{2}+d}d x \right ) a d -6 \left (\int \frac {x^{\frac {7}{3}} \left (x^{2}-1\right )^{\frac {1}{3}}}{c \,x^{4}-c \,x^{2}-d \,x^{2}+d}d x \right ) b c -4 \left (\int \frac {x^{\frac {1}{3}} \left (x^{2}-1\right )^{\frac {1}{3}}}{c \,x^{4}-c \,x^{2}-d \,x^{2}+d}d x \right ) a d +6 \left (\int \frac {x^{\frac {1}{3}} \left (x^{2}-1\right )^{\frac {1}{3}}}{c \,x^{4}-c \,x^{2}-d \,x^{2}+d}d x \right ) b c}{6 c} \] Input:
int((a*x^2-b)*(x^3-x)^(1/3)/(c*x^2-d),x)
Output:
(3*x**(1/3)*(x**2 - 1)**(1/3)*a*x - 2*int((x**(1/3)*(x**2 - 1)**(1/3)*x**2 )/(c*x**4 - c*x**2 - d*x**2 + d),x)*a*c + 6*int((x**(1/3)*(x**2 - 1)**(1/3 )*x**2)/(c*x**4 - c*x**2 - d*x**2 + d),x)*a*d - 6*int((x**(1/3)*(x**2 - 1) **(1/3)*x**2)/(c*x**4 - c*x**2 - d*x**2 + d),x)*b*c - 4*int((x**(1/3)*(x** 2 - 1)**(1/3))/(c*x**4 - c*x**2 - d*x**2 + d),x)*a*d + 6*int((x**(1/3)*(x* *2 - 1)**(1/3))/(c*x**4 - c*x**2 - d*x**2 + d),x)*b*c)/(6*c)