Integrand size = 41, antiderivative size = 56 \[ \int \frac {x^3 \left (5 b+8 a x^3\right )}{\sqrt [4]{b x+a x^4} \left (-2+b x^5+a x^8\right )} \, dx=2^{3/4} \arctan \left (\frac {x \sqrt [4]{b x+a x^4}}{\sqrt [4]{2}}\right )-2^{3/4} \text {arctanh}\left (\frac {x \sqrt [4]{b x+a x^4}}{\sqrt [4]{2}}\right ) \]
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\[ \int \frac {x^3 \left (5 b+8 a x^3\right )}{\sqrt [4]{b x+a x^4} \left (-2+b x^5+a x^8\right )} \, dx=\int \frac {x^3 \left (5 b+8 a x^3\right )}{\sqrt [4]{b x+a x^4} \left (-2+b x^5+a x^8\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt [4]{x} \sqrt [4]{b+a x^3}\right ) \int \frac {x^{11/4} \left (5 b+8 a x^3\right )}{\sqrt [4]{b+a x^3} \left (-2+b x^5+a x^8\right )} \, dx}{\sqrt [4]{b x+a x^4}} \\ & = \frac {\left (4 \sqrt [4]{x} \sqrt [4]{b+a x^3}\right ) \text {Subst}\left (\int \frac {x^{14} \left (5 b+8 a x^{12}\right )}{\sqrt [4]{b+a x^{12}} \left (-2+b x^{20}+a x^{32}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{b x+a x^4}} \\ & = \frac {\left (4 \sqrt [4]{x} \sqrt [4]{b+a x^3}\right ) \text {Subst}\left (\int \left (\frac {5 b x^{14}}{\sqrt [4]{b+a x^{12}} \left (-2+b x^{20}+a x^{32}\right )}+\frac {8 a x^{26}}{\sqrt [4]{b+a x^{12}} \left (-2+b x^{20}+a x^{32}\right )}\right ) \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{b x+a x^4}} \\ & = \frac {\left (32 a \sqrt [4]{x} \sqrt [4]{b+a x^3}\right ) \text {Subst}\left (\int \frac {x^{26}}{\sqrt [4]{b+a x^{12}} \left (-2+b x^{20}+a x^{32}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{b x+a x^4}}+\frac {\left (20 b \sqrt [4]{x} \sqrt [4]{b+a x^3}\right ) \text {Subst}\left (\int \frac {x^{14}}{\sqrt [4]{b+a x^{12}} \left (-2+b x^{20}+a x^{32}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{b x+a x^4}} \\ \end{align*}
\[ \int \frac {x^3 \left (5 b+8 a x^3\right )}{\sqrt [4]{b x+a x^4} \left (-2+b x^5+a x^8\right )} \, dx=\int \frac {x^3 \left (5 b+8 a x^3\right )}{\sqrt [4]{b x+a x^4} \left (-2+b x^5+a x^8\right )} \, dx \]
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\[\int \frac {x^{3} \left (8 a \,x^{3}+5 b \right )}{\left (a \,x^{4}+b x \right )^{\frac {1}{4}} \left (a \,x^{8}+b \,x^{5}-2\right )}d x\]
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Timed out. \[ \int \frac {x^3 \left (5 b+8 a x^3\right )}{\sqrt [4]{b x+a x^4} \left (-2+b x^5+a x^8\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {x^3 \left (5 b+8 a x^3\right )}{\sqrt [4]{b x+a x^4} \left (-2+b x^5+a x^8\right )} \, dx=\int \frac {x^{3} \cdot \left (8 a x^{3} + 5 b\right )}{\sqrt [4]{x \left (a x^{3} + b\right )} \left (a x^{8} + b x^{5} - 2\right )}\, dx \]
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\[ \int \frac {x^3 \left (5 b+8 a x^3\right )}{\sqrt [4]{b x+a x^4} \left (-2+b x^5+a x^8\right )} \, dx=\int { \frac {{\left (8 \, a x^{3} + 5 \, b\right )} x^{3}}{{\left (a x^{8} + b x^{5} - 2\right )} {\left (a x^{4} + b x\right )}^{\frac {1}{4}}} \,d x } \]
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\[ \int \frac {x^3 \left (5 b+8 a x^3\right )}{\sqrt [4]{b x+a x^4} \left (-2+b x^5+a x^8\right )} \, dx=\int { \frac {{\left (8 \, a x^{3} + 5 \, b\right )} x^{3}}{{\left (a x^{8} + b x^{5} - 2\right )} {\left (a x^{4} + b x\right )}^{\frac {1}{4}}} \,d x } \]
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Timed out. \[ \int \frac {x^3 \left (5 b+8 a x^3\right )}{\sqrt [4]{b x+a x^4} \left (-2+b x^5+a x^8\right )} \, dx=\int \frac {x^3\,\left (8\,a\,x^3+5\,b\right )}{{\left (a\,x^4+b\,x\right )}^{1/4}\,\left (a\,x^8+b\,x^5-2\right )} \,d x \]
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