Integrand size = 52, antiderivative size = 146 \[ \int \frac {x^4 \left (-4 b+a x^3\right )}{\sqrt [4]{-b+a x^3} \left (-b^2+2 a b x^3-a^2 x^6+x^8\right )} \, dx=\arctan \left (\frac {\sqrt [4]{-b+a x^3}}{x}\right )-\frac {\arctan \left (\frac {-\frac {x^2}{\sqrt {2}}+\frac {\sqrt {-b+a x^3}}{\sqrt {2}}}{x \sqrt [4]{-b+a x^3}}\right )}{\sqrt {2}}+\text {arctanh}\left (\frac {x \left (-b+a x^3\right )^{3/4}}{b-a x^3}\right )+\frac {\text {arctanh}\left (\frac {\sqrt {2} x \sqrt [4]{-b+a x^3}}{x^2+\sqrt {-b+a x^3}}\right )}{\sqrt {2}} \]
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\[ \int \frac {x^4 \left (-4 b+a x^3\right )}{\sqrt [4]{-b+a x^3} \left (-b^2+2 a b x^3-a^2 x^6+x^8\right )} \, dx=\int \frac {x^4 \left (-4 b+a x^3\right )}{\sqrt [4]{-b+a x^3} \left (-b^2+2 a b x^3-a^2 x^6+x^8\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {4 b-a x^3}{2 \sqrt [4]{-b+a x^3} \left (b-a x^3-x^4\right )}+\frac {-4 b+a x^3}{2 \sqrt [4]{-b+a x^3} \left (b-a x^3+x^4\right )}\right ) \, dx \\ & = \frac {1}{2} \int \frac {4 b-a x^3}{\sqrt [4]{-b+a x^3} \left (b-a x^3-x^4\right )} \, dx+\frac {1}{2} \int \frac {-4 b+a x^3}{\sqrt [4]{-b+a x^3} \left (b-a x^3+x^4\right )} \, dx \\ & = \frac {1}{2} \int \left (-\frac {a x^3}{\sqrt [4]{-b+a x^3} \left (-b+a x^3-x^4\right )}-\frac {4 b}{\sqrt [4]{-b+a x^3} \left (b-a x^3+x^4\right )}\right ) \, dx+\frac {1}{2} \int \left (\frac {4 b}{\sqrt [4]{-b+a x^3} \left (b-a x^3-x^4\right )}+\frac {a x^3}{\sqrt [4]{-b+a x^3} \left (-b+a x^3+x^4\right )}\right ) \, dx \\ & = -\left (\frac {1}{2} a \int \frac {x^3}{\sqrt [4]{-b+a x^3} \left (-b+a x^3-x^4\right )} \, dx\right )+\frac {1}{2} a \int \frac {x^3}{\sqrt [4]{-b+a x^3} \left (-b+a x^3+x^4\right )} \, dx+(2 b) \int \frac {1}{\sqrt [4]{-b+a x^3} \left (b-a x^3-x^4\right )} \, dx-(2 b) \int \frac {1}{\sqrt [4]{-b+a x^3} \left (b-a x^3+x^4\right )} \, dx \\ \end{align*}
Time = 1.57 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.90 \[ \int \frac {x^4 \left (-4 b+a x^3\right )}{\sqrt [4]{-b+a x^3} \left (-b^2+2 a b x^3-a^2 x^6+x^8\right )} \, dx=\arctan \left (\frac {\sqrt [4]{-b+a x^3}}{x}\right )-\frac {\arctan \left (\frac {-x^2+\sqrt {-b+a x^3}}{\sqrt {2} x \sqrt [4]{-b+a x^3}}\right )}{\sqrt {2}}-\text {arctanh}\left (\frac {x}{\sqrt [4]{-b+a x^3}}\right )+\frac {\text {arctanh}\left (\frac {\sqrt {2} x \sqrt [4]{-b+a x^3}}{x^2+\sqrt {-b+a x^3}}\right )}{\sqrt {2}} \]
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\[\int \frac {x^{4} \left (a \,x^{3}-4 b \right )}{\left (a \,x^{3}-b \right )^{\frac {1}{4}} \left (-a^{2} x^{6}+x^{8}+2 a b \,x^{3}-b^{2}\right )}d x\]
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Timed out. \[ \int \frac {x^4 \left (-4 b+a x^3\right )}{\sqrt [4]{-b+a x^3} \left (-b^2+2 a b x^3-a^2 x^6+x^8\right )} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {x^4 \left (-4 b+a x^3\right )}{\sqrt [4]{-b+a x^3} \left (-b^2+2 a b x^3-a^2 x^6+x^8\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {x^4 \left (-4 b+a x^3\right )}{\sqrt [4]{-b+a x^3} \left (-b^2+2 a b x^3-a^2 x^6+x^8\right )} \, dx=\int { -\frac {{\left (a x^{3} - 4 \, b\right )} x^{4}}{{\left (a^{2} x^{6} - x^{8} - 2 \, a b x^{3} + b^{2}\right )} {\left (a x^{3} - b\right )}^{\frac {1}{4}}} \,d x } \]
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\[ \int \frac {x^4 \left (-4 b+a x^3\right )}{\sqrt [4]{-b+a x^3} \left (-b^2+2 a b x^3-a^2 x^6+x^8\right )} \, dx=\int { -\frac {{\left (a x^{3} - 4 \, b\right )} x^{4}}{{\left (a^{2} x^{6} - x^{8} - 2 \, a b x^{3} + b^{2}\right )} {\left (a x^{3} - b\right )}^{\frac {1}{4}}} \,d x } \]
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Timed out. \[ \int \frac {x^4 \left (-4 b+a x^3\right )}{\sqrt [4]{-b+a x^3} \left (-b^2+2 a b x^3-a^2 x^6+x^8\right )} \, dx=-\int -\frac {x^4\,\left (4\,b-a\,x^3\right )}{{\left (a\,x^3-b\right )}^{1/4}\,\left (a^2\,x^6-2\,a\,b\,x^3+b^2-x^8\right )} \,d x \]
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