Integrand size = 36, antiderivative size = 301 \[ \int \frac {-b+x}{\left ((-a+x) (-b+x)^2\right )^{2/3} (a-b d+(-1+d) x)} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} b \sqrt [3]{d}-\sqrt {3} \sqrt [3]{d} x}{b \sqrt [3]{d}-\sqrt [3]{d} x-2 \sqrt [3]{-a b^2+\left (2 a b+b^2\right ) x+(-a-2 b) x^2+x^3}}\right )}{(a-b) d^{2/3}}-\frac {\log \left (b \sqrt [3]{d}-\sqrt [3]{d} x+\sqrt [3]{-a b^2+\left (2 a b+b^2\right ) x+(-a-2 b) x^2+x^3}\right )}{(a-b) d^{2/3}}+\frac {\log \left (b^2 d^{2/3}-2 b d^{2/3} x+d^{2/3} x^2+\left (-b \sqrt [3]{d}+\sqrt [3]{d} x\right ) \sqrt [3]{-a b^2+\left (2 a b+b^2\right ) x+(-a-2 b) x^2+x^3}+\left (-a b^2+\left (2 a b+b^2\right ) x+(-a-2 b) x^2+x^3\right )^{2/3}\right )}{2 (a-b) d^{2/3}} \]
-3^(1/2)*arctan((3^(1/2)*b*d^(1/3)-3^(1/2)*d^(1/3)*x)/(b*d^(1/3)-d^(1/3)*x -2*(-a*b^2+(2*a*b+b^2)*x+(-a-2*b)*x^2+x^3)^(1/3)))/(a-b)/d^(2/3)-ln(b*d^(1 /3)-d^(1/3)*x+(-a*b^2+(2*a*b+b^2)*x+(-a-2*b)*x^2+x^3)^(1/3))/(a-b)/d^(2/3) +1/2*ln(b^2*d^(2/3)-2*b*d^(2/3)*x+d^(2/3)*x^2+(-b*d^(1/3)+d^(1/3)*x)*(-a*b ^2+(2*a*b+b^2)*x+(-a-2*b)*x^2+x^3)^(1/3)+(-a*b^2+(2*a*b+b^2)*x+(-a-2*b)*x^ 2+x^3)^(2/3))/(a-b)/d^(2/3)
Time = 0.21 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.57 \[ \int \frac {-b+x}{\left ((-a+x) (-b+x)^2\right )^{2/3} (a-b d+(-1+d) x)} \, dx=\frac {(b-x)^{4/3} (-a+x)^{2/3} \left (2 \sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{-a+x}}{\sqrt [3]{d} \sqrt [3]{b-x}}}{\sqrt {3}}\right )-2 \log \left (\sqrt [3]{d}+\frac {\sqrt [3]{-a+x}}{\sqrt [3]{b-x}}\right )+\log \left (d^{2/3}-\frac {\sqrt [3]{d} \sqrt [3]{-a+x}}{\sqrt [3]{b-x}}+\frac {(-a+x)^{2/3}}{(b-x)^{2/3}}\right )\right )}{2 (a-b) d^{2/3} \left ((b-x)^2 (-a+x)\right )^{2/3}} \]
((b - x)^(4/3)*(-a + x)^(2/3)*(2*Sqrt[3]*ArcTan[(1 - (2*(-a + x)^(1/3))/(d ^(1/3)*(b - x)^(1/3)))/Sqrt[3]] - 2*Log[d^(1/3) + (-a + x)^(1/3)/(b - x)^( 1/3)] + Log[d^(2/3) - (d^(1/3)*(-a + x)^(1/3))/(b - x)^(1/3) + (-a + x)^(2 /3)/(b - x)^(2/3)]))/(2*(a - b)*d^(2/3)*((b - x)^2*(-a + x))^(2/3))
Time = 0.66 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.56, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {7270, 102}
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.
| \(\displaystyle \int \frac {x-b}{\left ((x-a) (x-b)^2\right )^{2/3} (a-b d+(d-1) x)} \, dx\) |
| \(\Big \downarrow \) 7270 |
| \(\displaystyle \frac {(x-a)^{2/3} (x-b)^{4/3} \int \frac {1}{(x-a)^{2/3} \sqrt [3]{x-b} (a-b d-(1-d) x)}dx}{\left (-\left ((a-x) (b-x)^2\right )\right )^{2/3}}\) |
| \(\Big \downarrow \) 102 |
| \(\displaystyle \frac {(x-a)^{2/3} (x-b)^{4/3} \left (-\frac {\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{d} \sqrt [3]{x-b}}{\sqrt {3} \sqrt [3]{x-a}}+\frac {1}{\sqrt {3}}\right )}{d^{2/3} (a-b)}+\frac {\log (a-b d-(1-d) x)}{2 d^{2/3} (a-b)}-\frac {3 \log \left (\sqrt [3]{d} \sqrt [3]{x-b}-\sqrt [3]{x-a}\right )}{2 d^{2/3} (a-b)}\right )}{\left (-\left ((a-x) (b-x)^2\right )\right )^{2/3}}\) |
((-a + x)^(2/3)*(-b + x)^(4/3)*(-((Sqrt[3]*ArcTan[1/Sqrt[3] + (2*d^(1/3)*( -b + x)^(1/3))/(Sqrt[3]*(-a + x)^(1/3))])/((a - b)*d^(2/3))) + Log[a - b*d - (1 - d)*x]/(2*(a - b)*d^(2/3)) - (3*Log[-(-a + x)^(1/3) + d^(1/3)*(-b + x)^(1/3)])/(2*(a - b)*d^(2/3))))/(-((a - x)*(b - x)^2))^(2/3)
3.29.62.3.1 Defintions of rubi rules used
Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)*((e_.) + (f_.) *(x_))), x_] :> With[{q = Rt[(d*e - c*f)/(b*e - a*f), 3]}, Simp[(-Sqrt[3])* q*(ArcTan[1/Sqrt[3] + 2*q*((a + b*x)^(1/3)/(Sqrt[3]*(c + d*x)^(1/3)))]/(d*e - c*f)), x] + (Simp[q*(Log[e + f*x]/(2*(d*e - c*f))), x] - Simp[3*q*(Log[q *(a + b*x)^(1/3) - (c + d*x)^(1/3)]/(2*(d*e - c*f))), x])] /; FreeQ[{a, b, c, d, e, f}, x]
Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Simp[a^IntPart[p ]*((a*v^m*w^n)^FracPart[p]/(v^(m*FracPart[p])*w^(n*FracPart[p]))) Int[u*v ^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] && !IntegerQ[p] && !Free Q[v, x] && !FreeQ[w, x]
\[\int \frac {-b +x}{\left (\left (-a +x \right ) \left (-b +x \right )^{2}\right )^{\frac {2}{3}} \left (a -b d +\left (-1+d \right ) x \right )}d x\]
Time = 0.29 (sec) , antiderivative size = 290, normalized size of antiderivative = 0.96 \[ \int \frac {-b+x}{\left ((-a+x) (-b+x)^2\right )^{2/3} (a-b d+(-1+d) x)} \, dx=\frac {2 \, \sqrt {3} {\left (d^{2}\right )}^{\frac {1}{6}} d \arctan \left (\frac {\sqrt {3} {\left (d^{2}\right )}^{\frac {1}{6}} {\left ({\left (b d - d x\right )} {\left (d^{2}\right )}^{\frac {1}{3}} - 2 \, {\left (-a b^{2} - {\left (a + 2 \, b\right )} x^{2} + x^{3} + {\left (2 \, a b + b^{2}\right )} x\right )}^{\frac {1}{3}} {\left (d^{2}\right )}^{\frac {2}{3}}\right )}}{3 \, {\left (b d^{2} - d^{2} x\right )}}\right ) + {\left (d^{2}\right )}^{\frac {2}{3}} \log \left (-\frac {{\left (-a b^{2} - {\left (a + 2 \, b\right )} x^{2} + x^{3} + {\left (2 \, a b + b^{2}\right )} x\right )}^{\frac {1}{3}} {\left (d^{2}\right )}^{\frac {2}{3}} {\left (b - x\right )} - {\left (-a b^{2} - {\left (a + 2 \, b\right )} x^{2} + x^{3} + {\left (2 \, a b + b^{2}\right )} x\right )}^{\frac {2}{3}} d - {\left (b^{2} d - 2 \, b d x + d x^{2}\right )} {\left (d^{2}\right )}^{\frac {1}{3}}}{b^{2} - 2 \, b x + x^{2}}\right ) - 2 \, {\left (d^{2}\right )}^{\frac {2}{3}} \log \left (-\frac {{\left (d^{2}\right )}^{\frac {2}{3}} {\left (b - x\right )} + {\left (-a b^{2} - {\left (a + 2 \, b\right )} x^{2} + x^{3} + {\left (2 \, a b + b^{2}\right )} x\right )}^{\frac {1}{3}} d}{b - x}\right )}{2 \, {\left (a - b\right )} d^{2}} \]
1/2*(2*sqrt(3)*(d^2)^(1/6)*d*arctan(1/3*sqrt(3)*(d^2)^(1/6)*((b*d - d*x)*( d^2)^(1/3) - 2*(-a*b^2 - (a + 2*b)*x^2 + x^3 + (2*a*b + b^2)*x)^(1/3)*(d^2 )^(2/3))/(b*d^2 - d^2*x)) + (d^2)^(2/3)*log(-((-a*b^2 - (a + 2*b)*x^2 + x^ 3 + (2*a*b + b^2)*x)^(1/3)*(d^2)^(2/3)*(b - x) - (-a*b^2 - (a + 2*b)*x^2 + x^3 + (2*a*b + b^2)*x)^(2/3)*d - (b^2*d - 2*b*d*x + d*x^2)*(d^2)^(1/3))/( b^2 - 2*b*x + x^2)) - 2*(d^2)^(2/3)*log(-((d^2)^(2/3)*(b - x) + (-a*b^2 - (a + 2*b)*x^2 + x^3 + (2*a*b + b^2)*x)^(1/3)*d)/(b - x)))/((a - b)*d^2)
\[ \int \frac {-b+x}{\left ((-a+x) (-b+x)^2\right )^{2/3} (a-b d+(-1+d) x)} \, dx=\int \frac {- b + x}{\left (\left (- a + x\right ) \left (- b + x\right )^{2}\right )^{\frac {2}{3}} \left (a - b d + d x - x\right )}\, dx \]
\[ \int \frac {-b+x}{\left ((-a+x) (-b+x)^2\right )^{2/3} (a-b d+(-1+d) x)} \, dx=\int { \frac {b - x}{\left (-{\left (a - x\right )} {\left (b - x\right )}^{2}\right )^{\frac {2}{3}} {\left (b d - {\left (d - 1\right )} x - a\right )}} \,d x } \]
\[ \int \frac {-b+x}{\left ((-a+x) (-b+x)^2\right )^{2/3} (a-b d+(-1+d) x)} \, dx=\int { \frac {b - x}{\left (-{\left (a - x\right )} {\left (b - x\right )}^{2}\right )^{\frac {2}{3}} {\left (b d - {\left (d - 1\right )} x - a\right )}} \,d x } \]
Timed out. \[ \int \frac {-b+x}{\left ((-a+x) (-b+x)^2\right )^{2/3} (a-b d+(-1+d) x)} \, dx=\int -\frac {b-x}{{\left (-\left (a-x\right )\,{\left (b-x\right )}^2\right )}^{2/3}\,\left (a-b\,d+x\,\left (d-1\right )\right )} \,d x \]