Integrand size = 43, antiderivative size = 399 \[ \int \frac {x^5 \left (7 b+10 a x^3\right )}{\sqrt [4]{b x^3+a x^6} \left (1+b x^7+a x^{10}\right )} \, dx=-\sqrt {2} \arctan \left (\frac {-2^{2/3} x \sqrt [4]{b x^3+a x^6}+x^2 \sqrt [4]{b x^3+a x^6}}{2 \sqrt [6]{2}-\sqrt {2} x+2^{2/3} x \sqrt [4]{b x^3+a x^6}-x^2 \sqrt [4]{b x^3+a x^6}}\right )+\sqrt {2} \arctan \left (\frac {-2^{2/3} x \sqrt [4]{b x^3+a x^6}+x^2 \sqrt [4]{b x^3+a x^6}}{-2 \sqrt [6]{2}+\sqrt {2} x+2^{2/3} x \sqrt [4]{b x^3+a x^6}-x^2 \sqrt [4]{b x^3+a x^6}}\right )-\sqrt {2} \text {arctanh}\left (\frac {-2 2^{5/6} x \sqrt [4]{b x^3+a x^6}+4 \sqrt [6]{2} x^2 \sqrt [4]{b x^3+a x^6}-\sqrt {2} x^3 \sqrt [4]{b x^3+a x^6}}{-2 \sqrt [3]{2}+2\ 2^{2/3} x-x^2-2 \sqrt [3]{2} x^2 \sqrt {b x^3+a x^6}+2\ 2^{2/3} x^3 \sqrt {b x^3+a x^6}-x^4 \sqrt {b x^3+a x^6}}\right ) \]
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\[ \int \frac {x^5 \left (7 b+10 a x^3\right )}{\sqrt [4]{b x^3+a x^6} \left (1+b x^7+a x^{10}\right )} \, dx=\int \frac {x^5 \left (7 b+10 a x^3\right )}{\sqrt [4]{b x^3+a x^6} \left (1+b x^7+a x^{10}\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {\left (x^{3/4} \sqrt [4]{b+a x^3}\right ) \int \frac {x^{17/4} \left (7 b+10 a x^3\right )}{\sqrt [4]{b+a x^3} \left (1+b x^7+a x^{10}\right )} \, dx}{\sqrt [4]{b x^3+a x^6}} \\ & = \frac {\left (4 x^{3/4} \sqrt [4]{b+a x^3}\right ) \text {Subst}\left (\int \frac {x^{20} \left (7 b+10 a x^{12}\right )}{\sqrt [4]{b+a x^{12}} \left (1+b x^{28}+a x^{40}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{b x^3+a x^6}} \\ & = \frac {\left (4 x^{3/4} \sqrt [4]{b+a x^3}\right ) \text {Subst}\left (\int \left (\frac {7 b x^{20}}{\sqrt [4]{b+a x^{12}} \left (1+b x^{28}+a x^{40}\right )}+\frac {10 a x^{32}}{\sqrt [4]{b+a x^{12}} \left (1+b x^{28}+a x^{40}\right )}\right ) \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{b x^3+a x^6}} \\ & = \frac {\left (40 a x^{3/4} \sqrt [4]{b+a x^3}\right ) \text {Subst}\left (\int \frac {x^{32}}{\sqrt [4]{b+a x^{12}} \left (1+b x^{28}+a x^{40}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{b x^3+a x^6}}+\frac {\left (28 b x^{3/4} \sqrt [4]{b+a x^3}\right ) \text {Subst}\left (\int \frac {x^{20}}{\sqrt [4]{b+a x^{12}} \left (1+b x^{28}+a x^{40}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{b x^3+a x^6}} \\ \end{align*}
\[ \int \frac {x^5 \left (7 b+10 a x^3\right )}{\sqrt [4]{b x^3+a x^6} \left (1+b x^7+a x^{10}\right )} \, dx=\int \frac {x^5 \left (7 b+10 a x^3\right )}{\sqrt [4]{b x^3+a x^6} \left (1+b x^7+a x^{10}\right )} \, dx \]
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\[\int \frac {x^{5} \left (10 a \,x^{3}+7 b \right )}{\left (a \,x^{6}+b \,x^{3}\right )^{\frac {1}{4}} \left (a \,x^{10}+b \,x^{7}+1\right )}d x\]
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Timed out. \[ \int \frac {x^5 \left (7 b+10 a x^3\right )}{\sqrt [4]{b x^3+a x^6} \left (1+b x^7+a x^{10}\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {x^5 \left (7 b+10 a x^3\right )}{\sqrt [4]{b x^3+a x^6} \left (1+b x^7+a x^{10}\right )} \, dx=\int \frac {x^{5} \cdot \left (10 a x^{3} + 7 b\right )}{\sqrt [4]{x^{3} \left (a x^{3} + b\right )} \left (a x^{10} + b x^{7} + 1\right )}\, dx \]
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\[ \int \frac {x^5 \left (7 b+10 a x^3\right )}{\sqrt [4]{b x^3+a x^6} \left (1+b x^7+a x^{10}\right )} \, dx=\int { \frac {{\left (10 \, a x^{3} + 7 \, b\right )} x^{5}}{{\left (a x^{10} + b x^{7} + 1\right )} {\left (a x^{6} + b x^{3}\right )}^{\frac {1}{4}}} \,d x } \]
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\[ \int \frac {x^5 \left (7 b+10 a x^3\right )}{\sqrt [4]{b x^3+a x^6} \left (1+b x^7+a x^{10}\right )} \, dx=\int { \frac {{\left (10 \, a x^{3} + 7 \, b\right )} x^{5}}{{\left (a x^{10} + b x^{7} + 1\right )} {\left (a x^{6} + b x^{3}\right )}^{\frac {1}{4}}} \,d x } \]
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Timed out. \[ \int \frac {x^5 \left (7 b+10 a x^3\right )}{\sqrt [4]{b x^3+a x^6} \left (1+b x^7+a x^{10}\right )} \, dx=\int \frac {x^5\,\left (10\,a\,x^3+7\,b\right )}{{\left (a\,x^6+b\,x^3\right )}^{1/4}\,\left (a\,x^{10}+b\,x^7+1\right )} \,d x \]
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