Integrand size = 33, antiderivative size = 94 \[ \int \frac {4 b+a x^3}{\sqrt [4]{b+a x^3} \left (b+a x^3+x^4\right )} \, dx=-\sqrt {2} \arctan \left (\frac {-\frac {x^2}{\sqrt {2}}+\frac {\sqrt {b+a x^3}}{\sqrt {2}}}{x \sqrt [4]{b+a x^3}}\right )+\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} x \sqrt [4]{b+a x^3}}{x^2+\sqrt {b+a x^3}}\right ) \]
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\[ \int \frac {4 b+a x^3}{\sqrt [4]{b+a x^3} \left (b+a x^3+x^4\right )} \, dx=\int \frac {4 b+a x^3}{\sqrt [4]{b+a x^3} \left (b+a x^3+x^4\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \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 \\ & = a \int \frac {x^3}{\sqrt [4]{b+a x^3} \left (b+a x^3+x^4\right )} \, dx+(4 b) \int \frac {1}{\sqrt [4]{b+a x^3} \left (b+a x^3+x^4\right )} \, dx \\ \end{align*}
Time = 0.94 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.88 \[ \int \frac {4 b+a x^3}{\sqrt [4]{b+a x^3} \left (b+a x^3+x^4\right )} \, dx=\sqrt {2} \left (-\arctan \left (\frac {-x^2+\sqrt {b+a x^3}}{\sqrt {2} x \sqrt [4]{b+a x^3}}\right )+\text {arctanh}\left (\frac {\sqrt {2} x \sqrt [4]{b+a x^3}}{x^2+\sqrt {b+a x^3}}\right )\right ) \]
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\[\int \frac {a \,x^{3}+4 b}{\left (a \,x^{3}+b \right )^{\frac {1}{4}} \left (a \,x^{3}+x^{4}+b \right )}d x\]
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Timed out. \[ \int \frac {4 b+a x^3}{\sqrt [4]{b+a x^3} \left (b+a x^3+x^4\right )} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {4 b+a x^3}{\sqrt [4]{b+a x^3} \left (b+a x^3+x^4\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {4 b+a x^3}{\sqrt [4]{b+a x^3} \left (b+a x^3+x^4\right )} \, dx=\int { \frac {a x^{3} + 4 \, b}{{\left (a x^{3} + x^{4} + b\right )} {\left (a x^{3} + b\right )}^{\frac {1}{4}}} \,d x } \]
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\[ \int \frac {4 b+a x^3}{\sqrt [4]{b+a x^3} \left (b+a x^3+x^4\right )} \, dx=\int { \frac {a x^{3} + 4 \, b}{{\left (a x^{3} + x^{4} + b\right )} {\left (a x^{3} + b\right )}^{\frac {1}{4}}} \,d x } \]
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Timed out. \[ \int \frac {4 b+a x^3}{\sqrt [4]{b+a x^3} \left (b+a x^3+x^4\right )} \, dx=\int \frac {a\,x^3+4\,b}{{\left (a\,x^3+b\right )}^{1/4}\,\left (x^4+a\,x^3+b\right )} \,d x \]
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