Integrand size = 45, antiderivative size = 216 \[ \int \frac {a-2 b+x}{\sqrt [3]{(-a+x) (-b+x)} \left (b+a^2 d-(1+2 a d) x+d x^2\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{a b+(-a-b) x+x^2}}{-2 a \sqrt [3]{d}+2 \sqrt [3]{d} x+\sqrt [3]{a b+(-a-b) x+x^2}}\right )}{\sqrt [3]{d}}+\frac {\log \left (a \sqrt [3]{d}-\sqrt [3]{d} x+\sqrt [3]{a b+(-a-b) x+x^2}\right )}{\sqrt [3]{d}}-\frac {\log \left (a^2 d^{2/3}-2 a d^{2/3} x+d^{2/3} x^2+\left (-a \sqrt [3]{d}+\sqrt [3]{d} x\right ) \sqrt [3]{a b+(-a-b) x+x^2}+\left (a b+(-a-b) x+x^2\right )^{2/3}\right )}{2 \sqrt [3]{d}} \]
3^(1/2)*arctan(3^(1/2)*(a*b+(-a-b)*x+x^2)^(1/3)/(-2*a*d^(1/3)+2*d^(1/3)*x+ (a*b+(-a-b)*x+x^2)^(1/3)))/d^(1/3)+ln(a*d^(1/3)-d^(1/3)*x+(a*b+(-a-b)*x+x^ 2)^(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+(-a-b)*x+x^2)^(1/3)+(a*b+(-a-b)*x+x^2)^(2/3))/d^(1/3)
Time = 0.01 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.81 \[ \int \frac {a-2 b+x}{\sqrt [3]{(-a+x) (-b+x)} \left (b+a^2 d-(1+2 a d) x+d x^2\right )} \, dx=\frac {\sqrt [3]{a-x} \sqrt [3]{b-x} \left (2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b-x}}{-2 \sqrt [3]{d} (a-x)^{2/3}+\sqrt [3]{b-x}}\right )+2 \log \left (\sqrt [3]{d} (a-x)^{2/3}+\sqrt [3]{b-x}\right )-\log \left (d^{2/3} (a-x)^{4/3}-\sqrt [3]{d} (a-x)^{2/3} \sqrt [3]{b-x}+(b-x)^{2/3}\right )\right )}{2 \sqrt [3]{d} \sqrt [3]{(-a+x) (-b+x)}} \]
((a - x)^(1/3)*(b - x)^(1/3)*(2*Sqrt[3]*ArcTan[(Sqrt[3]*(b - x)^(1/3))/(-2 *d^(1/3)*(a - x)^(2/3) + (b - x)^(1/3))] + 2*Log[d^(1/3)*(a - x)^(2/3) + ( b - x)^(1/3)] - Log[d^(2/3)*(a - x)^(4/3) - d^(1/3)*(a - x)^(2/3)*(b - x)^ (1/3) + (b - x)^(2/3)]))/(2*d^(1/3)*((-a + x)*(-b + x))^(1/3))
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.
\(\displaystyle \int \frac {a-2 b+x}{\sqrt [3]{(x-a) (x-b)} \left (a^2 d-x (2 a d+1)+b+d x^2\right )} \, dx\) |
\(\Big \downarrow \) 2048 |
\(\displaystyle \int \frac {a-2 b+x}{\sqrt [3]{x (-a-b)+a b+x^2} \left (a^2 d-x (2 a d+1)+b+d x^2\right )}dx\) |
\(\Big \downarrow \) 1375 |
\(\displaystyle \int \frac {a-2 b+x}{\sqrt [3]{x (-a-b)+a b+x^2} \left (a^2 d+x (-2 a d-1)+b+d x^2\right )}dx\) |
3.26.60.3.1 Defintions of rubi rules used
Int[((g_.) + (h_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_.) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Unintegrable[(g + h*x)*(a + b *x + c*x^2)^p*(d + e*x + f*x^2)^q, x] /; FreeQ[{a, b, c, d, e, f, g, h, p, q}, x]
Int[(u_.)*((e_.)*((a_.) + (b_.)*(x_)^(n_.))*((c_) + (d_.)*(x_)^(n_.)))^(p_) , x_Symbol] :> Int[u*(a*c*e + (b*c + a*d)*e*x^n + b*d*e*x^(2*n))^p, x] /; F reeQ[{a, b, c, d, e, n, p}, x]
\[\int \frac {a -2 b +x}{\left (\left (-a +x \right ) \left (-b +x \right )\right )^{\frac {1}{3}} \left (b +a^{2} d -\left (2 a d +1\right ) x +d \,x^{2}\right )}d x\]
Timed out. \[ \int \frac {a-2 b+x}{\sqrt [3]{(-a+x) (-b+x)} \left (b+a^2 d-(1+2 a d) x+d x^2\right )} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {a-2 b+x}{\sqrt [3]{(-a+x) (-b+x)} \left (b+a^2 d-(1+2 a d) x+d x^2\right )} \, dx=\text {Timed out} \]
\[ \int \frac {a-2 b+x}{\sqrt [3]{(-a+x) (-b+x)} \left (b+a^2 d-(1+2 a d) x+d x^2\right )} \, dx=\int { \frac {a - 2 \, b + x}{{\left (a^{2} d + d x^{2} - {\left (2 \, a d + 1\right )} x + b\right )} \left ({\left (a - x\right )} {\left (b - x\right )}\right )^{\frac {1}{3}}} \,d x } \]
\[ \int \frac {a-2 b+x}{\sqrt [3]{(-a+x) (-b+x)} \left (b+a^2 d-(1+2 a d) x+d x^2\right )} \, dx=\int { \frac {a - 2 \, b + x}{{\left (a^{2} d + d x^{2} - {\left (2 \, a d + 1\right )} x + b\right )} \left ({\left (a - x\right )} {\left (b - x\right )}\right )^{\frac {1}{3}}} \,d x } \]
Timed out. \[ \int \frac {a-2 b+x}{\sqrt [3]{(-a+x) (-b+x)} \left (b+a^2 d-(1+2 a d) x+d x^2\right )} \, dx=\int \frac {a-2\,b+x}{{\left (\left (a-x\right )\,\left (b-x\right )\right )}^{1/3}\,\left (b-x\,\left (2\,a\,d+1\right )+a^2\,d+d\,x^2\right )} \,d x \]