Integrand size = 58, antiderivative size = 204 \[ \int \frac {x \left (4 a b-3 (a+b) x+2 x^2\right )}{\sqrt [3]{x^2 (-a+x) (-b+x)} \left (-a b+(a+b) x-x^2+d x^4\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{a b x^2+(-a-b) x^3+x^4}}{2 \sqrt [3]{d} x^2+\sqrt [3]{a b x^2+(-a-b) x^3+x^4}}\right )}{\sqrt [3]{d}}+\frac {\log \left (-\sqrt [3]{d} x^2+\sqrt [3]{a b x^2+(-a-b) x^3+x^4}\right )}{\sqrt [3]{d}}-\frac {\log \left (d^{2/3} x^4+\sqrt [3]{d} x^2 \sqrt [3]{a b x^2+(-a-b) x^3+x^4}+\left (a b x^2+(-a-b) x^3+x^4\right )^{2/3}\right )}{2 \sqrt [3]{d}} \]
[Out]
\[ \int \frac {x \left (4 a b-3 (a+b) x+2 x^2\right )}{\sqrt [3]{x^2 (-a+x) (-b+x)} \left (-a b+(a+b) x-x^2+d x^4\right )} \, dx=\int \frac {x \left (4 a b-3 (a+b) x+2 x^2\right )}{\sqrt [3]{x^2 (-a+x) (-b+x)} \left (-a b+(a+b) x-x^2+d x^4\right )} \, dx \]
[In]
[Out]
Rubi steps \begin{align*} \text {integral}& = \frac {\left (x^{2/3} \sqrt [3]{-a+x} \sqrt [3]{-b+x}\right ) \int \frac {\sqrt [3]{x} \left (4 a b-3 (a+b) x+2 x^2\right )}{\sqrt [3]{-a+x} \sqrt [3]{-b+x} \left (-a b+(a+b) x-x^2+d x^4\right )} \, dx}{\sqrt [3]{x^2 (-a+x) (-b+x)}} \\ & = \frac {\left (3 x^{2/3} \sqrt [3]{-a+x} \sqrt [3]{-b+x}\right ) \text {Subst}\left (\int \frac {x^3 \left (4 a b-3 (a+b) x^3+2 x^6\right )}{\sqrt [3]{-a+x^3} \sqrt [3]{-b+x^3} \left (-a b+(a+b) x^3-x^6+d x^{12}\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{x^2 (-a+x) (-b+x)}} \\ & = \frac {\left (3 x^{2/3} \sqrt [3]{-a+x} \sqrt [3]{-b+x}\right ) \text {Subst}\left (\int \left (\frac {3 (a+b) x^6}{\sqrt [3]{-a+x^3} \sqrt [3]{-b+x^3} \left (a b-a \left (1+\frac {b}{a}\right ) x^3+x^6-d x^{12}\right )}+\frac {4 a b x^3}{\sqrt [3]{-a+x^3} \sqrt [3]{-b+x^3} \left (-a b+a \left (1+\frac {b}{a}\right ) x^3-x^6+d x^{12}\right )}+\frac {2 x^9}{\sqrt [3]{-a+x^3} \sqrt [3]{-b+x^3} \left (-a b+a \left (1+\frac {b}{a}\right ) x^3-x^6+d x^{12}\right )}\right ) \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{x^2 (-a+x) (-b+x)}} \\ & = \frac {\left (6 x^{2/3} \sqrt [3]{-a+x} \sqrt [3]{-b+x}\right ) \text {Subst}\left (\int \frac {x^9}{\sqrt [3]{-a+x^3} \sqrt [3]{-b+x^3} \left (-a b+a \left (1+\frac {b}{a}\right ) x^3-x^6+d x^{12}\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{x^2 (-a+x) (-b+x)}}+\frac {\left (12 a b x^{2/3} \sqrt [3]{-a+x} \sqrt [3]{-b+x}\right ) \text {Subst}\left (\int \frac {x^3}{\sqrt [3]{-a+x^3} \sqrt [3]{-b+x^3} \left (-a b+a \left (1+\frac {b}{a}\right ) x^3-x^6+d x^{12}\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{x^2 (-a+x) (-b+x)}}+\frac {\left (9 (a+b) x^{2/3} \sqrt [3]{-a+x} \sqrt [3]{-b+x}\right ) \text {Subst}\left (\int \frac {x^6}{\sqrt [3]{-a+x^3} \sqrt [3]{-b+x^3} \left (a b-a \left (1+\frac {b}{a}\right ) x^3+x^6-d x^{12}\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{x^2 (-a+x) (-b+x)}} \\ \end{align*}
Time = 12.10 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.80 \[ \int \frac {x \left (4 a b-3 (a+b) x+2 x^2\right )}{\sqrt [3]{x^2 (-a+x) (-b+x)} \left (-a b+(a+b) x-x^2+d x^4\right )} \, dx=\frac {2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{x^2 (-a+x) (-b+x)}}{2 \sqrt [3]{d} x^2+\sqrt [3]{x^2 (-a+x) (-b+x)}}\right )+2 \log \left (-\sqrt [3]{d} x^2+\sqrt [3]{x^2 (-a+x) (-b+x)}\right )-\log \left (d^{2/3} x^4+\sqrt [3]{d} x^2 \sqrt [3]{x^2 (-a+x) (-b+x)}+\left (x^2 (-a+x) (-b+x)\right )^{2/3}\right )}{2 \sqrt [3]{d}} \]
[In]
[Out]
\[\int \frac {x \left (4 a b -3 \left (a +b \right ) x +2 x^{2}\right )}{\left (x^{2} \left (-a +x \right ) \left (-b +x \right )\right )^{\frac {1}{3}} \left (-a b +\left (a +b \right ) x -x^{2}+d \,x^{4}\right )}d x\]
[In]
[Out]
Timed out. \[ \int \frac {x \left (4 a b-3 (a+b) x+2 x^2\right )}{\sqrt [3]{x^2 (-a+x) (-b+x)} \left (-a b+(a+b) x-x^2+d x^4\right )} \, dx=\text {Timed out} \]
[In]
[Out]
Timed out. \[ \int \frac {x \left (4 a b-3 (a+b) x+2 x^2\right )}{\sqrt [3]{x^2 (-a+x) (-b+x)} \left (-a b+(a+b) x-x^2+d x^4\right )} \, dx=\text {Timed out} \]
[In]
[Out]
\[ \int \frac {x \left (4 a b-3 (a+b) x+2 x^2\right )}{\sqrt [3]{x^2 (-a+x) (-b+x)} \left (-a b+(a+b) x-x^2+d x^4\right )} \, dx=\int { \frac {{\left (4 \, a b - 3 \, {\left (a + b\right )} x + 2 \, x^{2}\right )} x}{{\left (d x^{4} - a b + {\left (a + b\right )} x - x^{2}\right )} \left ({\left (a - x\right )} {\left (b - x\right )} x^{2}\right )^{\frac {1}{3}}} \,d x } \]
[In]
[Out]
\[ \int \frac {x \left (4 a b-3 (a+b) x+2 x^2\right )}{\sqrt [3]{x^2 (-a+x) (-b+x)} \left (-a b+(a+b) x-x^2+d x^4\right )} \, dx=\int { \frac {{\left (4 \, a b - 3 \, {\left (a + b\right )} x + 2 \, x^{2}\right )} x}{{\left (d x^{4} - a b + {\left (a + b\right )} x - x^{2}\right )} \left ({\left (a - x\right )} {\left (b - x\right )} x^{2}\right )^{\frac {1}{3}}} \,d x } \]
[In]
[Out]
Timed out. \[ \int \frac {x \left (4 a b-3 (a+b) x+2 x^2\right )}{\sqrt [3]{x^2 (-a+x) (-b+x)} \left (-a b+(a+b) x-x^2+d x^4\right )} \, dx=-\int \frac {x\,\left (4\,a\,b+2\,x^2-3\,x\,\left (a+b\right )\right )}{{\left (x^2\,\left (a-x\right )\,\left (b-x\right )\right )}^{1/3}\,\left (-d\,x^4+x^2+\left (-a-b\right )\,x+a\,b\right )} \,d x \]
[In]
[Out]