Integrand size = 51, antiderivative size = 233 \[ \int \frac {a b+(-2 a+b) x}{\sqrt [3]{x (-a+x) (-b+x)} \left (a^2 d+(b-2 a d) x+(-1+d) x^2\right )} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{a b x+(-a-b) x^2+x^3}}{-2 a \sqrt [3]{d}+2 \sqrt [3]{d} x+\sqrt [3]{a b x+(-a-b) x^2+x^3}}\right )}{\sqrt [3]{d}}-\frac {\log \left (a \sqrt [3]{d}-\sqrt [3]{d} x+\sqrt [3]{a b x+(-a-b) x^2+x^3}\right )}{\sqrt [3]{d}}+\frac {\log \left (a^2 d^{2/3}-2 a d^{2/3} x+d^{2/3} x^2+\left (-a \sqrt [3]{d}+\sqrt [3]{d} x\right ) \sqrt [3]{a b x+(-a-b) x^2+x^3}+\left (a b x+(-a-b) x^2+x^3\right )^{2/3}\right )}{2 \sqrt [3]{d}} \]
-3^(1/2)*arctan(3^(1/2)*(a*b*x+(-a-b)*x^2+x^3)^(1/3)/(-2*a*d^(1/3)+2*d^(1/ 3)*x+(a*b*x+(-a-b)*x^2+x^3)^(1/3)))/d^(1/3)-ln(a*d^(1/3)-d^(1/3)*x+(a*b*x+ (-a-b)*x^2+x^3)^(1/3))/d^(1/3)+1/2*ln(a^2*d^(2/3)-2*a*d^(2/3)*x+d^(2/3)*x^ 2+(-a*d^(1/3)+d^(1/3)*x)*(a*b*x+(-a-b)*x^2+x^3)^(1/3)+(a*b*x+(-a-b)*x^2+x^ 3)^(2/3))/d^(1/3)
\[ \int \frac {a b+(-2 a+b) x}{\sqrt [3]{x (-a+x) (-b+x)} \left (a^2 d+(b-2 a d) x+(-1+d) x^2\right )} \, dx=\int \frac {a b+(-2 a+b) x}{\sqrt [3]{x (-a+x) (-b+x)} \left (a^2 d+(b-2 a d) x+(-1+d) x^2\right )} \, dx \]
Integrate[(a*b + (-2*a + b)*x)/((x*(-a + x)*(-b + x))^(1/3)*(a^2*d + (b - 2*a*d)*x + (-1 + d)*x^2)),x]
Integrate[(a*b + (-2*a + b)*x)/((x*(-a + x)*(-b + x))^(1/3)*(a^2*d + (b - 2*a*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.
| \(\displaystyle \int \frac {x (b-2 a)+a b}{\sqrt [3]{x (x-a) (x-b)} \left (a^2 d+x (b-2 a d)+(d-1) x^2\right )} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt [3]{x} \sqrt [3]{-x (a+b)+a b+x^2} \int \frac {a b-(2 a-b) x}{\sqrt [3]{x} \sqrt [3]{x^2-(a+b) x+a b} \left (d a^2-(1-d) x^2+(b-2 a d) x\right )}dx}{\sqrt [3]{x (a-x) (b-x)}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle \frac {3 \sqrt [3]{x} \sqrt [3]{-x (a+b)+a b+x^2} \int \frac {\sqrt [3]{x} (a b-(2 a-b) x)}{\sqrt [3]{x^2-(a+b) x+a b} \left (d a^2-(1-d) x^2+(b-2 a d) x\right )}d\sqrt [3]{x}}{\sqrt [3]{x (a-x) (b-x)}}\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \frac {3 \sqrt [3]{x} \sqrt [3]{-x (a+b)+a b+x^2} \int \left (\frac {(b-2 a) x^{4/3}}{\sqrt [3]{x^2-(a+b) x+a b} \left (d a^2-(1-d) x^2+(b-2 a d) x\right )}+\frac {a b \sqrt [3]{x}}{\sqrt [3]{x^2-(a+b) x+a b} \left (d a^2-(1-d) x^2+(b-2 a d) x\right )}\right )d\sqrt [3]{x}}{\sqrt [3]{x (a-x) (b-x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {3 \sqrt [3]{x} \sqrt [3]{-x (a+b)+a b+x^2} \left (\frac {(2 a-b) \left (-\sqrt {4 a^2 d-4 a b d+b^2}-2 a d+b\right ) \int \frac {\sqrt [3]{x}}{\left (b-2 a d-2 (1-d) x-\sqrt {4 d a^2-4 b d a+b^2}\right ) \sqrt [3]{x^2+(-a-b) x+a b}}d\sqrt [3]{x}}{\sqrt {4 a^2 d-4 a b d+b^2}}-\frac {2 a b (1-d) \int \frac {\sqrt [3]{x}}{\left (b-2 a d-2 (1-d) x-\sqrt {4 d a^2-4 b d a+b^2}\right ) \sqrt [3]{x^2+(-a-b) x+a b}}d\sqrt [3]{x}}{\sqrt {4 a^2 d-4 a b d+b^2}}-\frac {(2 a-b) \left (\sqrt {4 a^2 d-4 a b d+b^2}-2 a d+b\right ) \int \frac {\sqrt [3]{x}}{\left (b-2 a d-2 (1-d) x+\sqrt {4 d a^2-4 b d a+b^2}\right ) \sqrt [3]{x^2+(-a-b) x+a b}}d\sqrt [3]{x}}{\sqrt {4 a^2 d-4 a b d+b^2}}-\frac {2 a b (1-d) \int \frac {\sqrt [3]{x}}{\left (-b+2 a d+2 (1-d) x-\sqrt {4 d a^2-4 b d a+b^2}\right ) \sqrt [3]{x^2+(-a-b) x+a b}}d\sqrt [3]{x}}{\sqrt {4 a^2 d-4 a b d+b^2}}\right )}{\sqrt [3]{x (a-x) (b-x)}}\) |
Int[(a*b + (-2*a + b)*x)/((x*(-a + x)*(-b + x))^(1/3)*(a^2*d + (b - 2*a*d) *x + (-1 + d)*x^2)),x]
3.27.36.3.1 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]
\[\int \frac {a b +\left (-2 a +b \right ) x}{\left (x \left (-a +x \right ) \left (-b +x \right )\right )^{\frac {1}{3}} \left (a^{2} d +\left (-2 a d +b \right ) x +\left (-1+d \right ) x^{2}\right )}d x\]
Timed out. \[ \int \frac {a b+(-2 a+b) x}{\sqrt [3]{x (-a+x) (-b+x)} \left (a^2 d+(b-2 a d) x+(-1+d) x^2\right )} \, dx=\text {Timed out} \]
integrate((a*b+(-2*a+b)*x)/(x*(-a+x)*(-b+x))^(1/3)/(a^2*d+(-2*a*d+b)*x+(-1 +d)*x^2),x, algorithm="fricas")
Timed out. \[ \int \frac {a b+(-2 a+b) x}{\sqrt [3]{x (-a+x) (-b+x)} \left (a^2 d+(b-2 a d) x+(-1+d) x^2\right )} \, dx=\text {Timed out} \]
\[ \int \frac {a b+(-2 a+b) x}{\sqrt [3]{x (-a+x) (-b+x)} \left (a^2 d+(b-2 a d) x+(-1+d) x^2\right )} \, dx=\int { \frac {a b - {\left (2 \, a - b\right )} x}{{\left (a^{2} d + {\left (d - 1\right )} x^{2} - {\left (2 \, a d - b\right )} x\right )} \left ({\left (a - x\right )} {\left (b - x\right )} x\right )^{\frac {1}{3}}} \,d x } \]
integrate((a*b+(-2*a+b)*x)/(x*(-a+x)*(-b+x))^(1/3)/(a^2*d+(-2*a*d+b)*x+(-1 +d)*x^2),x, algorithm="maxima")
integrate((a*b - (2*a - b)*x)/((a^2*d + (d - 1)*x^2 - (2*a*d - b)*x)*((a - x)*(b - x)*x)^(1/3)), x)
\[ \int \frac {a b+(-2 a+b) x}{\sqrt [3]{x (-a+x) (-b+x)} \left (a^2 d+(b-2 a d) x+(-1+d) x^2\right )} \, dx=\int { \frac {a b - {\left (2 \, a - b\right )} x}{{\left (a^{2} d + {\left (d - 1\right )} x^{2} - {\left (2 \, a d - b\right )} x\right )} \left ({\left (a - x\right )} {\left (b - x\right )} x\right )^{\frac {1}{3}}} \,d x } \]
integrate((a*b+(-2*a+b)*x)/(x*(-a+x)*(-b+x))^(1/3)/(a^2*d+(-2*a*d+b)*x+(-1 +d)*x^2),x, algorithm="giac")
integrate((a*b - (2*a - b)*x)/((a^2*d + (d - 1)*x^2 - (2*a*d - b)*x)*((a - x)*(b - x)*x)^(1/3)), x)
Timed out. \[ \int \frac {a b+(-2 a+b) x}{\sqrt [3]{x (-a+x) (-b+x)} \left (a^2 d+(b-2 a d) x+(-1+d) x^2\right )} \, dx=\int \frac {a\,b-x\,\left (2\,a-b\right )}{{\left (x\,\left (a-x\right )\,\left (b-x\right )\right )}^{1/3}\,\left (a^2\,d+x\,\left (b-2\,a\,d\right )+x^2\,\left (d-1\right )\right )} \,d x \]