Integrand size = 28, antiderivative size = 165 \[ \int \frac {x^4}{\left (b+a x^4\right )^2 \sqrt [4]{b x^2+a x^4}} \, dx=\frac {\left (-b+a x^2\right ) \left (b x^2+a x^4\right )^{3/4}}{4 a b (a+b) x \left (b+a x^4\right )}+\frac {\text {RootSum}\left [a^2+a b-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {2 a \log (x)-2 a \log \left (\sqrt [4]{b x^2+a x^4}-x \text {$\#$1}\right )-\log (x) \text {$\#$1}^4+\log \left (\sqrt [4]{b x^2+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{a \text {$\#$1}-\text {$\#$1}^5}\&\right ]}{32 a (a+b)} \]
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\[ \int \frac {x^4}{\left (b+a x^4\right )^2 \sqrt [4]{b x^2+a x^4}} \, dx=\int \frac {x^4}{\left (b+a x^4\right )^2 \sqrt [4]{b x^2+a x^4}} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {x} \sqrt [4]{b+a x^2}\right ) \int \frac {x^{7/2}}{\sqrt [4]{b+a x^2} \left (b+a x^4\right )^2} \, dx}{\sqrt [4]{b x^2+a x^4}} \\ & = \frac {\left (2 \sqrt {x} \sqrt [4]{b+a x^2}\right ) \text {Subst}\left (\int \frac {x^8}{\sqrt [4]{b+a x^4} \left (b+a x^8\right )^2} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^4}} \\ \end{align*}
Time = 1.11 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.12 \[ \int \frac {x^4}{\left (b+a x^4\right )^2 \sqrt [4]{b x^2+a x^4}} \, dx=\frac {-8 b^2 x+8 a^2 x^5+\frac {1}{2} b \sqrt {x} \sqrt [4]{b+a x^2} \left (b+a x^4\right ) \text {RootSum}\left [a^2+a b-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {2 a \log (x)-4 a \log \left (\sqrt [4]{b+a x^2}-\sqrt {x} \text {$\#$1}\right )-\log (x) \text {$\#$1}^4+2 \log \left (\sqrt [4]{b+a x^2}-\sqrt {x} \text {$\#$1}\right ) \text {$\#$1}^4}{a \text {$\#$1}-\text {$\#$1}^5}\&\right ]}{32 a b (a+b) \sqrt [4]{x^2 \left (b+a x^2\right )} \left (b+a x^4\right )} \]
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Time = 0.55 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.22
method | result | size |
pseudoelliptic | \(\frac {-\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-2 a \,\textit {\_Z}^{4}+a^{2}+a b \right )}{\sum }\frac {\left (\textit {\_R}^{4}-2 a \right ) \ln \left (\frac {-\textit {\_R} x +\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R} \left (\textit {\_R}^{4}-a \right )}\right ) a b \,x^{5}+8 \left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {3}{4}} a \,x^{2}-\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-2 a \,\textit {\_Z}^{4}+a^{2}+a b \right )}{\sum }\frac {\left (\textit {\_R}^{4}-2 a \right ) \ln \left (\frac {-\textit {\_R} x +\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R} \left (\textit {\_R}^{4}-a \right )}\right ) b^{2} x -8 b \left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {3}{4}}}{32 \left (a \,x^{4}+b \right ) a b \left (a +b \right ) x}\) | \(201\) |
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Timed out. \[ \int \frac {x^4}{\left (b+a x^4\right )^2 \sqrt [4]{b x^2+a x^4}} \, dx=\text {Timed out} \]
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Not integrable
Time = 44.89 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.16 \[ \int \frac {x^4}{\left (b+a x^4\right )^2 \sqrt [4]{b x^2+a x^4}} \, dx=\int \frac {x^{4}}{\sqrt [4]{x^{2} \left (a x^{2} + b\right )} \left (a x^{4} + b\right )^{2}}\, dx \]
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Not integrable
Time = 0.22 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.17 \[ \int \frac {x^4}{\left (b+a x^4\right )^2 \sqrt [4]{b x^2+a x^4}} \, dx=\int { \frac {x^{4}}{{\left (a x^{4} + b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{4} + b\right )}^{2}} \,d x } \]
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Not integrable
Time = 0.87 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.17 \[ \int \frac {x^4}{\left (b+a x^4\right )^2 \sqrt [4]{b x^2+a x^4}} \, dx=\int { \frac {x^{4}}{{\left (a x^{4} + b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{4} + b\right )}^{2}} \,d x } \]
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Not integrable
Time = 6.59 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.17 \[ \int \frac {x^4}{\left (b+a x^4\right )^2 \sqrt [4]{b x^2+a x^4}} \, dx=\int \frac {x^4}{{\left (a\,x^4+b\right )}^2\,{\left (a\,x^4+b\,x^2\right )}^{1/4}} \,d x \]
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