Integrand size = 64, antiderivative size = 245 \[ \int \frac {a^2 b-2 a^2 x+(2 a-b) x^2}{(x (-a+x) (-b+x))^{2/3} \left (a^2+(-2 a+b d) x+(1-d) x^2\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \left (a b x+(-a-b) x^2+x^3\right )^{2/3}}{-2 b \sqrt [3]{d} x+2 \sqrt [3]{d} x^2+\left (a b x+(-a-b) x^2+x^3\right )^{2/3}}\right )}{\sqrt [3]{d}}+\frac {\log \left (b \sqrt {d} x-\sqrt {d} x^2+\sqrt [6]{d} \left (a b x+(-a-b) x^2+x^3\right )^{2/3}\right )}{\sqrt [3]{d}}-\frac {\log \left (b^2 d x^2-2 b d x^3+d x^4+\left (-b d^{2/3} x+d^{2/3} x^2\right ) \left (a b x+(-a-b) x^2+x^3\right )^{2/3}+\sqrt [3]{d} \left (a b x+(-a-b) x^2+x^3\right )^{4/3}\right )}{2 \sqrt [3]{d}} \]
3^(1/2)*arctan(3^(1/2)*(a*b*x+(-a-b)*x^2+x^3)^(2/3)/(-2*b*d^(1/3)*x+2*d^(1 /3)*x^2+(a*b*x+(-a-b)*x^2+x^3)^(2/3)))/d^(1/3)+ln(b*d^(1/2)*x-d^(1/2)*x^2+ d^(1/6)*(a*b*x+(-a-b)*x^2+x^3)^(2/3))/d^(1/3)-1/2*ln(b^2*d*x^2-2*b*d*x^3+d *x^4+(-b*d^(2/3)*x+d^(2/3)*x^2)*(a*b*x+(-a-b)*x^2+x^3)^(2/3)+d^(1/3)*(a*b* x+(-a-b)*x^2+x^3)^(4/3))/d^(1/3)
\[ \int \frac {a^2 b-2 a^2 x+(2 a-b) x^2}{(x (-a+x) (-b+x))^{2/3} \left (a^2+(-2 a+b d) x+(1-d) x^2\right )} \, dx=\int \frac {a^2 b-2 a^2 x+(2 a-b) x^2}{(x (-a+x) (-b+x))^{2/3} \left (a^2+(-2 a+b d) x+(1-d) x^2\right )} \, dx \]
Integrate[(a^2*b - 2*a^2*x + (2*a - b)*x^2)/((x*(-a + x)*(-b + x))^(2/3)*( a^2 + (-2*a + b*d)*x + (1 - d)*x^2)),x]
Integrate[(a^2*b - 2*a^2*x + (2*a - b)*x^2)/((x*(-a + x)*(-b + x))^(2/3)*( a^2 + (-2*a + b*d)*x + (1 - d)*x^2)), 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.
\(\displaystyle \int \frac {a^2 b-2 a^2 x+x^2 (2 a-b)}{(x (x-a) (x-b))^{2/3} \left (a^2+x (b d-2 a)+(1-d) x^2\right )} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {x^{2/3} \left (-x (a+b)+a b+x^2\right )^{2/3} \int \frac {b a^2-2 x a^2+(2 a-b) x^2}{x^{2/3} \left (x^2-(a+b) x+a b\right )^{2/3} \left (a^2+(1-d) x^2-(2 a-b d) x\right )}dx}{(x (a-x) (b-x))^{2/3}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle \frac {3 x^{2/3} \left (-x (a+b)+a b+x^2\right )^{2/3} \int \frac {b a^2-2 x a^2+(2 a-b) x^2}{\left (x^2-(a+b) x+a b\right )^{2/3} \left (a^2+(1-d) x^2-(2 a-b d) x\right )}d\sqrt [3]{x}}{(x (a-x) (b-x))^{2/3}}\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \frac {3 x^{2/3} \left (-x (a+b)+a b+x^2\right )^{2/3} \int \left (\frac {2 a-b}{(1-d) \left (x^2-(a+b) x+a b\right )^{2/3}}+\frac {a^2 (2 a-b (2-d))-\left (2 (d+1) a^2-2 b (d+1) a+b^2 d\right ) x}{(d-1) \left (x^2-(a+b) x+a b\right )^{2/3} \left (a^2+(1-d) x^2+(b d-2 a) x\right )}\right )d\sqrt [3]{x}}{(x (a-x) (b-x))^{2/3}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3 x^{2/3} \left (-x (a+b)+a b+x^2\right )^{2/3} \left (\frac {\left (\sqrt {d} (2 a-b) \sqrt {4 a^2-4 a b+b^2 d}+2 a^2 (d+1)-2 a b (d+1)+b^2 d\right ) \int \frac {1}{\left (-2 a+b d+2 (1-d) x-\sqrt {d} \sqrt {4 a^2-4 b a+b^2 d}\right ) \left (x^2+(-a-b) x+a b\right )^{2/3}}d\sqrt [3]{x}}{1-d}+\frac {\left (-\sqrt {d} (2 a-b) \sqrt {4 a^2-4 a b+b^2 d}+2 a^2 (d+1)-2 a b (d+1)+b^2 d\right ) \int \frac {1}{\left (-2 a+b d+2 (1-d) x+\sqrt {d} \sqrt {4 a^2-4 b a+b^2 d}\right ) \left (x^2+(-a-b) x+a b\right )^{2/3}}d\sqrt [3]{x}}{1-d}+\frac {\sqrt [3]{x} (2 a-b) \left (1-\frac {x}{a}\right )^{2/3} \left (1-\frac {x}{b}\right )^{2/3} \sqrt [3]{1-\frac {2 x}{-\sqrt {a^2-2 a b+b^2}+a+b}} \left (\frac {1-\frac {2 x}{\sqrt {a^2-2 a b+b^2}+a+b}}{1-\frac {2 x}{-\sqrt {a^2-2 a b+b^2}+a+b}}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {\sqrt {a^2-2 b a+b^2} x}{a b \left (1-\frac {2 x}{a+b-\sqrt {a^2-2 b a+b^2}}\right )}\right )}{(1-d) \left (1-\frac {2 x}{\sqrt {a^2-2 a b+b^2}+a+b}\right )^{2/3} \left (-x (a+b)+a b+x^2\right )^{2/3}}\right )}{(x (a-x) (b-x))^{2/3}}\) |
Int[(a^2*b - 2*a^2*x + (2*a - b)*x^2)/((x*(-a + x)*(-b + x))^(2/3)*(a^2 + (-2*a + b*d)*x + (1 - d)*x^2)),x]
3.27.95.3.1 Defintions of rubi rules used
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]
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]
\[\int \frac {a^{2} b -2 a^{2} x +\left (2 a -b \right ) x^{2}}{\left (x \left (-a +x \right ) \left (-b +x \right )\right )^{\frac {2}{3}} \left (a^{2}+\left (b d -2 a \right ) x +\left (1-d \right ) x^{2}\right )}d x\]
Timed out. \[ \int \frac {a^2 b-2 a^2 x+(2 a-b) x^2}{(x (-a+x) (-b+x))^{2/3} \left (a^2+(-2 a+b d) x+(1-d) x^2\right )} \, dx=\text {Timed out} \]
integrate((a^2*b-2*a^2*x+(2*a-b)*x^2)/(x*(-a+x)*(-b+x))^(2/3)/(a^2+(b*d-2* a)*x+(1-d)*x^2),x, algorithm="fricas")
Timed out. \[ \int \frac {a^2 b-2 a^2 x+(2 a-b) x^2}{(x (-a+x) (-b+x))^{2/3} \left (a^2+(-2 a+b d) x+(1-d) x^2\right )} \, dx=\text {Timed out} \]
integrate((a**2*b-2*a**2*x+(2*a-b)*x**2)/(x*(-a+x)*(-b+x))**(2/3)/(a**2+(b *d-2*a)*x+(1-d)*x**2),x)
\[ \int \frac {a^2 b-2 a^2 x+(2 a-b) x^2}{(x (-a+x) (-b+x))^{2/3} \left (a^2+(-2 a+b d) x+(1-d) x^2\right )} \, dx=\int { -\frac {a^{2} b - 2 \, a^{2} x + {\left (2 \, a - b\right )} x^{2}}{\left ({\left (a - x\right )} {\left (b - x\right )} x\right )^{\frac {2}{3}} {\left ({\left (d - 1\right )} x^{2} - a^{2} - {\left (b d - 2 \, a\right )} x\right )}} \,d x } \]
integrate((a^2*b-2*a^2*x+(2*a-b)*x^2)/(x*(-a+x)*(-b+x))^(2/3)/(a^2+(b*d-2* a)*x+(1-d)*x^2),x, algorithm="maxima")
-integrate((a^2*b - 2*a^2*x + (2*a - b)*x^2)/(((a - x)*(b - x)*x)^(2/3)*(( d - 1)*x^2 - a^2 - (b*d - 2*a)*x)), x)
\[ \int \frac {a^2 b-2 a^2 x+(2 a-b) x^2}{(x (-a+x) (-b+x))^{2/3} \left (a^2+(-2 a+b d) x+(1-d) x^2\right )} \, dx=\int { -\frac {a^{2} b - 2 \, a^{2} x + {\left (2 \, a - b\right )} x^{2}}{\left ({\left (a - x\right )} {\left (b - x\right )} x\right )^{\frac {2}{3}} {\left ({\left (d - 1\right )} x^{2} - a^{2} - {\left (b d - 2 \, a\right )} x\right )}} \,d x } \]
integrate((a^2*b-2*a^2*x+(2*a-b)*x^2)/(x*(-a+x)*(-b+x))^(2/3)/(a^2+(b*d-2* a)*x+(1-d)*x^2),x, algorithm="giac")
integrate(-(a^2*b - 2*a^2*x + (2*a - b)*x^2)/(((a - x)*(b - x)*x)^(2/3)*(( d - 1)*x^2 - a^2 - (b*d - 2*a)*x)), x)
Timed out. \[ \int \frac {a^2 b-2 a^2 x+(2 a-b) x^2}{(x (-a+x) (-b+x))^{2/3} \left (a^2+(-2 a+b d) x+(1-d) x^2\right )} \, dx=\int -\frac {x^2\,\left (2\,a-b\right )+a^2\,b-2\,a^2\,x}{{\left (x\,\left (a-x\right )\,\left (b-x\right )\right )}^{2/3}\,\left (x\,\left (2\,a-b\,d\right )-a^2+x^2\,\left (d-1\right )\right )} \,d x \]
int(-(x^2*(2*a - b) + a^2*b - 2*a^2*x)/((x*(a - x)*(b - x))^(2/3)*(x*(2*a - b*d) - a^2 + x^2*(d - 1))),x)