Integrand size = 89, antiderivative size = 227 \[ \int \frac {(-a+x) (-b+x) (-2 a b+(a+b) x)}{\sqrt [3]{x (-a+x) (-b+x)} \left (a^2 b^2 d-2 a b (a+b) d x+\left (a^2+4 a b+b^2\right ) d x^2-2 (a+b) d x^3+(-1+d) x^4\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x-2 \sqrt [6]{d} \sqrt [3]{a b x+(-a-b) x^2+x^3}}\right )}{2 d^{5/6}}-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [6]{d} \sqrt [3]{a b x+(-a-b) x^2+x^3}}\right )}{2 d^{5/6}}-\frac {\text {arctanh}\left (\frac {x}{\sqrt [6]{d} \sqrt [3]{a b x+(-a-b) x^2+x^3}}\right )}{d^{5/6}}-\frac {\text {arctanh}\left (\frac {\frac {x^2}{\sqrt [6]{d}}+\sqrt [6]{d} \left (a b x+(-a-b) x^2+x^3\right )^{2/3}}{x \sqrt [3]{a b x+(-a-b) x^2+x^3}}\right )}{2 d^{5/6}} \]
1/2*3^(1/2)*arctan(3^(1/2)*x/(x-2*d^(1/6)*(a*b*x+(-a-b)*x^2+x^3)^(1/3)))/d ^(5/6)-1/2*3^(1/2)*arctan(3^(1/2)*x/(x+2*d^(1/6)*(a*b*x+(-a-b)*x^2+x^3)^(1 /3)))/d^(5/6)-arctanh(x/d^(1/6)/(a*b*x+(-a-b)*x^2+x^3)^(1/3))/d^(5/6)-1/2* arctanh((x^2/d^(1/6)+d^(1/6)*(a*b*x+(-a-b)*x^2+x^3)^(2/3))/x/(a*b*x+(-a-b) *x^2+x^3)^(1/3))/d^(5/6)
Time = 11.06 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.74 \[ \int \frac {(-a+x) (-b+x) (-2 a b+(a+b) x)}{\sqrt [3]{x (-a+x) (-b+x)} \left (a^2 b^2 d-2 a b (a+b) d x+\left (a^2+4 a b+b^2\right ) d x^2-2 (a+b) d x^3+(-1+d) x^4\right )} \, dx=\frac {\sqrt {3} \left (\arctan \left (\frac {\sqrt {3} x}{x-2 \sqrt [6]{d} \sqrt [3]{x (-a+x) (-b+x)}}\right )-\arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [6]{d} \sqrt [3]{x (-a+x) (-b+x)}}\right )\right )-2 \text {arctanh}\left (\frac {x}{\sqrt [6]{d} \sqrt [3]{x (-a+x) (-b+x)}}\right )-\text {arctanh}\left (\frac {x^2+\sqrt [3]{d} (x (-a+x) (-b+x))^{2/3}}{\sqrt [6]{d} x \sqrt [3]{x (-a+x) (-b+x)}}\right )}{2 d^{5/6}} \]
Integrate[((-a + x)*(-b + x)*(-2*a*b + (a + b)*x))/((x*(-a + x)*(-b + x))^ (1/3)*(a^2*b^2*d - 2*a*b*(a + b)*d*x + (a^2 + 4*a*b + b^2)*d*x^2 - 2*(a + b)*d*x^3 + (-1 + d)*x^4)),x]
(Sqrt[3]*(ArcTan[(Sqrt[3]*x)/(x - 2*d^(1/6)*(x*(-a + x)*(-b + x))^(1/3))] - ArcTan[(Sqrt[3]*x)/(x + 2*d^(1/6)*(x*(-a + x)*(-b + x))^(1/3))]) - 2*Arc Tanh[x/(d^(1/6)*(x*(-a + x)*(-b + x))^(1/3))] - ArcTanh[(x^2 + d^(1/3)*(x* (-a + x)*(-b + x))^(2/3))/(d^(1/6)*x*(x*(-a + x)*(-b + x))^(1/3))])/(2*d^( 5/6))
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.
| \(\displaystyle \int \frac {(x-a) (x-b) (x (a+b)-2 a b)}{\sqrt [3]{x (x-a) (x-b)} \left (d x^2 \left (a^2+4 a b+b^2\right )+a^2 b^2 d-2 d x^3 (a+b)-2 a b d x (a+b)+(d-1) x^4\right )} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt [3]{x} \sqrt [3]{-x (a+b)+a b+x^2} \int -\frac {(a-x) (b-x) (2 a b-(a+b) x)}{\sqrt [3]{x} \sqrt [3]{x^2-(a+b) x+a b} \left (-\left ((1-d) x^4\right )-2 (a+b) d x^3+\left (a^2+4 b a+b^2\right ) d x^2-2 a b (a+b) d x+a^2 b^2 d\right )}dx}{\sqrt [3]{x (a-x) (b-x)}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\sqrt [3]{x} \sqrt [3]{-x (a+b)+a b+x^2} \int \frac {(a-x) (b-x) (2 a b-(a+b) x)}{\sqrt [3]{x} \sqrt [3]{x^2-(a+b) x+a b} \left (-\left ((1-d) x^4\right )-2 (a+b) d x^3+\left (a^2+4 b a+b^2\right ) d x^2-2 a b (a+b) d x+a^2 b^2 d\right )}dx}{\sqrt [3]{x (a-x) (b-x)}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle -\frac {3 \sqrt [3]{x} \sqrt [3]{-x (a+b)+a b+x^2} \int \frac {(a-x) (b-x) \sqrt [3]{x} (2 a b-(a+b) x)}{\sqrt [3]{x^2-(a+b) x+a b} \left (-\left ((1-d) x^4\right )-2 (a+b) d x^3+\left (a^2+4 b a+b^2\right ) d x^2-2 a b (a+b) d x+a^2 b^2 d\right )}d\sqrt [3]{x}}{\sqrt [3]{x (a-x) (b-x)}}\) |
| \(\Big \downarrow \) 1395 |
| \(\displaystyle -\frac {3 \sqrt [3]{x} \sqrt [3]{a-x} \sqrt [3]{b-x} \int \frac {(a-x)^{2/3} (b-x)^{2/3} \sqrt [3]{x} (2 a b-(a+b) x)}{-\left ((1-d) x^4\right )-2 (a+b) d x^3+\left (a^2+4 b a+b^2\right ) d x^2-2 a b (a+b) d x+a^2 b^2 d}d\sqrt [3]{x}}{\sqrt [3]{x (a-x) (b-x)}}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {3 \sqrt [3]{x} \sqrt [3]{a-x} \sqrt [3]{b-x} \int \left (\frac {(-a-b) (a-x)^{2/3} (b-x)^{2/3} x^{4/3}}{-\left ((1-d) x^4\right )-2 a \left (\frac {b}{a}+1\right ) d x^3+a^2 \left (\frac {b (4 a+b)}{a^2}+1\right ) d x^2-2 a^2 b \left (\frac {b}{a}+1\right ) d x+a^2 b^2 d}+\frac {2 a b (a-x)^{2/3} (b-x)^{2/3} \sqrt [3]{x}}{-\left ((1-d) x^4\right )-2 a \left (\frac {b}{a}+1\right ) d x^3+a^2 \left (\frac {b (4 a+b)}{a^2}+1\right ) d x^2-2 a^2 b \left (\frac {b}{a}+1\right ) d x+a^2 b^2 d}\right )d\sqrt [3]{x}}{\sqrt [3]{x (a-x) (b-x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3 \sqrt [3]{x} \sqrt [3]{a-x} \sqrt [3]{b-x} \left (2 a b \int \frac {(a-x)^{2/3} (b-x)^{2/3} \sqrt [3]{x}}{-\left ((1-d) x^4\right )-2 a \left (\frac {b}{a}+1\right ) d x^3+a^2 \left (\frac {b (4 a+b)}{a^2}+1\right ) d x^2-2 a^2 b \left (\frac {b}{a}+1\right ) d x+a^2 b^2 d}d\sqrt [3]{x}-(a+b) \int \frac {(a-x)^{2/3} (b-x)^{2/3} x^{4/3}}{-\left ((1-d) x^4\right )-2 a \left (\frac {b}{a}+1\right ) d x^3+a^2 \left (\frac {b (4 a+b)}{a^2}+1\right ) d x^2-2 a^2 b \left (\frac {b}{a}+1\right ) d x+a^2 b^2 d}d\sqrt [3]{x}\right )}{\sqrt [3]{x (a-x) (b-x)}}\) |
Int[((-a + x)*(-b + x)*(-2*a*b + (a + b)*x))/((x*(-a + x)*(-b + x))^(1/3)* (a^2*b^2*d - 2*a*b*(a + b)*d*x + (a^2 + 4*a*b + b^2)*d*x^2 - 2*(a + b)*d*x ^3 + (-1 + d)*x^4)),x]
3.27.9.3.1 Defintions of rubi rules used
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_)*((d_) + (e_.)*( x_)^(n_))^(q_.), x_Symbol] :> Simp[(a + b*x^n + c*x^(2*n))^FracPart[p]/((d + e*x^n)^FracPart[p]*(a/d + c*(x^n/e))^FracPart[p]) Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && E qQ[n2, 2*n] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && !IntegerQ[p] && !(EqQ[q, 1] && EqQ[n, 2])
Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, 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 = 1.39 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.16
| method | result | size |
| pseudoelliptic | \(\frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x \left (\frac {1}{d}\right )^{\frac {1}{6}}+2 \left (x \left (a -x \right ) \left (b -x \right )\right )^{\frac {1}{3}}\right )}{3 x \left (\frac {1}{d}\right )^{\frac {1}{6}}}\right )-2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x \left (\frac {1}{d}\right )^{\frac {1}{6}}-2 \left (x \left (a -x \right ) \left (b -x \right )\right )^{\frac {1}{3}}\right )}{3 x \left (\frac {1}{d}\right )^{\frac {1}{6}}}\right )-\ln \left (\frac {\left (\frac {1}{d}\right )^{\frac {1}{6}} \left (x \left (a -x \right ) \left (b -x \right )\right )^{\frac {1}{3}} x +\left (\frac {1}{d}\right )^{\frac {1}{3}} x^{2}+\left (x \left (a -x \right ) \left (b -x \right )\right )^{\frac {2}{3}}}{x^{2}}\right )-2 \ln \left (\frac {x \left (\frac {1}{d}\right )^{\frac {1}{6}}+\left (x \left (a -x \right ) \left (b -x \right )\right )^{\frac {1}{3}}}{x}\right )+2 \ln \left (\frac {x \left (\frac {1}{d}\right )^{\frac {1}{6}}-\left (x \left (a -x \right ) \left (b -x \right )\right )^{\frac {1}{3}}}{x}\right )+\ln \left (\frac {\left (\frac {1}{d}\right )^{\frac {1}{6}} \left (x \left (a -x \right ) \left (b -x \right )\right )^{\frac {1}{3}} x -\left (\frac {1}{d}\right )^{\frac {1}{3}} x^{2}-\left (x \left (a -x \right ) \left (b -x \right )\right )^{\frac {2}{3}}}{x^{2}}\right )}{4 \left (\frac {1}{d}\right )^{\frac {1}{6}} d}\) | \(263\) |
int((-a+x)*(-b+x)*(-2*a*b+(a+b)*x)/(x*(-a+x)*(-b+x))^(1/3)/(a^2*b^2*d-2*a* b*(a+b)*d*x+(a^2+4*a*b+b^2)*d*x^2-2*(a+b)*d*x^3+(-1+d)*x^4),x,method=_RETU RNVERBOSE)
1/4*(2*3^(1/2)*arctan(1/3*3^(1/2)*(x*(1/d)^(1/6)+2*(x*(a-x)*(b-x))^(1/3))/ x/(1/d)^(1/6))-2*3^(1/2)*arctan(1/3*3^(1/2)*(x*(1/d)^(1/6)-2*(x*(a-x)*(b-x ))^(1/3))/x/(1/d)^(1/6))-ln(((1/d)^(1/6)*(x*(a-x)*(b-x))^(1/3)*x+(1/d)^(1/ 3)*x^2+(x*(a-x)*(b-x))^(2/3))/x^2)-2*ln((x*(1/d)^(1/6)+(x*(a-x)*(b-x))^(1/ 3))/x)+2*ln((x*(1/d)^(1/6)-(x*(a-x)*(b-x))^(1/3))/x)+ln(((1/d)^(1/6)*(x*(a -x)*(b-x))^(1/3)*x-(1/d)^(1/3)*x^2-(x*(a-x)*(b-x))^(2/3))/x^2))/(1/d)^(1/6 )/d
Timed out. \[ \int \frac {(-a+x) (-b+x) (-2 a b+(a+b) x)}{\sqrt [3]{x (-a+x) (-b+x)} \left (a^2 b^2 d-2 a b (a+b) d x+\left (a^2+4 a b+b^2\right ) d x^2-2 (a+b) d x^3+(-1+d) x^4\right )} \, dx=\text {Timed out} \]
integrate((-a+x)*(-b+x)*(-2*a*b+(a+b)*x)/(x*(-a+x)*(-b+x))^(1/3)/(a^2*b^2* d-2*a*b*(a+b)*d*x+(a^2+4*a*b+b^2)*d*x^2-2*(a+b)*d*x^3+(-1+d)*x^4),x, algor ithm="fricas")
Timed out. \[ \int \frac {(-a+x) (-b+x) (-2 a b+(a+b) x)}{\sqrt [3]{x (-a+x) (-b+x)} \left (a^2 b^2 d-2 a b (a+b) d x+\left (a^2+4 a b+b^2\right ) d x^2-2 (a+b) d x^3+(-1+d) x^4\right )} \, dx=\text {Timed out} \]
integrate((-a+x)*(-b+x)*(-2*a*b+(a+b)*x)/(x*(-a+x)*(-b+x))**(1/3)/(a**2*b* *2*d-2*a*b*(a+b)*d*x+(a**2+4*a*b+b**2)*d*x**2-2*(a+b)*d*x**3+(-1+d)*x**4), x)
\[ \int \frac {(-a+x) (-b+x) (-2 a b+(a+b) x)}{\sqrt [3]{x (-a+x) (-b+x)} \left (a^2 b^2 d-2 a b (a+b) d x+\left (a^2+4 a b+b^2\right ) d x^2-2 (a+b) d x^3+(-1+d) x^4\right )} \, dx=\int { -\frac {{\left (2 \, a b - {\left (a + b\right )} x\right )} {\left (a - x\right )} {\left (b - x\right )}}{{\left (a^{2} b^{2} d - 2 \, {\left (a + b\right )} a b d x - 2 \, {\left (a + b\right )} d x^{3} + {\left (d - 1\right )} x^{4} + {\left (a^{2} + 4 \, a b + b^{2}\right )} d x^{2}\right )} \left ({\left (a - x\right )} {\left (b - x\right )} x\right )^{\frac {1}{3}}} \,d x } \]
integrate((-a+x)*(-b+x)*(-2*a*b+(a+b)*x)/(x*(-a+x)*(-b+x))^(1/3)/(a^2*b^2* d-2*a*b*(a+b)*d*x+(a^2+4*a*b+b^2)*d*x^2-2*(a+b)*d*x^3+(-1+d)*x^4),x, algor ithm="maxima")
-integrate((2*a*b - (a + b)*x)*(a - x)*(b - x)/((a^2*b^2*d - 2*(a + b)*a*b *d*x - 2*(a + b)*d*x^3 + (d - 1)*x^4 + (a^2 + 4*a*b + b^2)*d*x^2)*((a - x) *(b - x)*x)^(1/3)), x)
Time = 0.48 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.39 \[ \int \frac {(-a+x) (-b+x) (-2 a b+(a+b) x)}{\sqrt [3]{x (-a+x) (-b+x)} \left (a^2 b^2 d-2 a b (a+b) d x+\left (a^2+4 a b+b^2\right ) d x^2-2 (a+b) d x^3+(-1+d) x^4\right )} \, dx=-\left (-\frac {1}{d}\right )^{\frac {5}{6}} \arctan \left (\frac {{\left (\frac {a b}{x^{2}} - \frac {a}{x} - \frac {b}{x} + 1\right )}^{\frac {1}{3}}}{\left (-\frac {1}{d}\right )^{\frac {1}{6}}}\right ) + \frac {\sqrt {3} \left (-d^{5}\right )^{\frac {5}{6}} \log \left (\sqrt {3} {\left (\frac {a b}{x^{2}} - \frac {a}{x} - \frac {b}{x} + 1\right )}^{\frac {1}{3}} \left (-\frac {1}{d}\right )^{\frac {1}{6}} + {\left (\frac {a b}{x^{2}} - \frac {a}{x} - \frac {b}{x} + 1\right )}^{\frac {2}{3}} + \left (-\frac {1}{d}\right )^{\frac {1}{3}}\right )}{4 \, d^{5}} - \frac {\sqrt {3} \left (-d^{5}\right )^{\frac {5}{6}} \log \left (-\sqrt {3} {\left (\frac {a b}{x^{2}} - \frac {a}{x} - \frac {b}{x} + 1\right )}^{\frac {1}{3}} \left (-\frac {1}{d}\right )^{\frac {1}{6}} + {\left (\frac {a b}{x^{2}} - \frac {a}{x} - \frac {b}{x} + 1\right )}^{\frac {2}{3}} + \left (-\frac {1}{d}\right )^{\frac {1}{3}}\right )}{4 \, d^{5}} - \frac {\left (-d^{5}\right )^{\frac {5}{6}} \arctan \left (\frac {\sqrt {3} \left (-\frac {1}{d}\right )^{\frac {1}{6}} + 2 \, {\left (\frac {a b}{x^{2}} - \frac {a}{x} - \frac {b}{x} + 1\right )}^{\frac {1}{3}}}{\left (-\frac {1}{d}\right )^{\frac {1}{6}}}\right )}{2 \, d^{5}} - \frac {\left (-d^{5}\right )^{\frac {5}{6}} \arctan \left (-\frac {\sqrt {3} \left (-\frac {1}{d}\right )^{\frac {1}{6}} - 2 \, {\left (\frac {a b}{x^{2}} - \frac {a}{x} - \frac {b}{x} + 1\right )}^{\frac {1}{3}}}{\left (-\frac {1}{d}\right )^{\frac {1}{6}}}\right )}{2 \, d^{5}} \]
integrate((-a+x)*(-b+x)*(-2*a*b+(a+b)*x)/(x*(-a+x)*(-b+x))^(1/3)/(a^2*b^2* d-2*a*b*(a+b)*d*x+(a^2+4*a*b+b^2)*d*x^2-2*(a+b)*d*x^3+(-1+d)*x^4),x, algor ithm="giac")
-(-1/d)^(5/6)*arctan((a*b/x^2 - a/x - b/x + 1)^(1/3)/(-1/d)^(1/6)) + 1/4*s qrt(3)*(-d^5)^(5/6)*log(sqrt(3)*(a*b/x^2 - a/x - b/x + 1)^(1/3)*(-1/d)^(1/ 6) + (a*b/x^2 - a/x - b/x + 1)^(2/3) + (-1/d)^(1/3))/d^5 - 1/4*sqrt(3)*(-d ^5)^(5/6)*log(-sqrt(3)*(a*b/x^2 - a/x - b/x + 1)^(1/3)*(-1/d)^(1/6) + (a*b /x^2 - a/x - b/x + 1)^(2/3) + (-1/d)^(1/3))/d^5 - 1/2*(-d^5)^(5/6)*arctan( (sqrt(3)*(-1/d)^(1/6) + 2*(a*b/x^2 - a/x - b/x + 1)^(1/3))/(-1/d)^(1/6))/d ^5 - 1/2*(-d^5)^(5/6)*arctan(-(sqrt(3)*(-1/d)^(1/6) - 2*(a*b/x^2 - a/x - b /x + 1)^(1/3))/(-1/d)^(1/6))/d^5
Timed out. \[ \int \frac {(-a+x) (-b+x) (-2 a b+(a+b) x)}{\sqrt [3]{x (-a+x) (-b+x)} \left (a^2 b^2 d-2 a b (a+b) d x+\left (a^2+4 a b+b^2\right ) d x^2-2 (a+b) d x^3+(-1+d) x^4\right )} \, dx=-\int \frac {\left (2\,a\,b-x\,\left (a+b\right )\right )\,\left (a-x\right )\,\left (b-x\right )}{{\left (x\,\left (a-x\right )\,\left (b-x\right )\right )}^{1/3}\,\left (x^4\,\left (d-1\right )+a^2\,b^2\,d+d\,x^2\,\left (a^2+4\,a\,b+b^2\right )-2\,d\,x^3\,\left (a+b\right )-2\,a\,b\,d\,x\,\left (a+b\right )\right )} \,d x \]
int(-((2*a*b - x*(a + b))*(a - x)*(b - x))/((x*(a - x)*(b - x))^(1/3)*(x^4 *(d - 1) + a^2*b^2*d + d*x^2*(4*a*b + a^2 + b^2) - 2*d*x^3*(a + b) - 2*a*b *d*x*(a + b))),x)