Integrand size = 43, antiderivative size = 457 \[ \int \frac {x^4 \left (-q+p x^4\right ) \sqrt {q+p x^4}}{b x^8+a \left (q+p x^4\right )^4} \, dx=-\frac {\sqrt {2+\sqrt {2}} \arctan \left (\frac {\left (\sqrt {\frac {2}{2-\sqrt {2}}} \sqrt [8]{a} \sqrt [8]{b}-\frac {2 \sqrt [8]{a} \sqrt [8]{b}}{\sqrt {2-\sqrt {2}}}\right ) x \sqrt {q+p x^4}}{-\sqrt [4]{a} q+\sqrt [4]{b} x^2-\sqrt [4]{a} p x^4}\right )}{8 a^{3/8} b^{5/8}}+\frac {\sqrt {2-\sqrt {2}} \arctan \left (\frac {\sqrt {2+\sqrt {2}} \sqrt [8]{a} \sqrt [8]{b} x \sqrt {q+p x^4}}{\sqrt [4]{a} q-\sqrt [4]{b} x^2+\sqrt [4]{a} p x^4}\right )}{8 a^{3/8} b^{5/8}}-\frac {\sqrt {2+\sqrt {2}} \text {arctanh}\left (\frac {\frac {\sqrt [8]{a} q}{\sqrt {2-\sqrt {2}} \sqrt [8]{b}}+\frac {\sqrt [8]{b} x^2}{\sqrt {2-\sqrt {2}} \sqrt [8]{a}}+\frac {\sqrt [8]{a} p x^4}{\sqrt {2-\sqrt {2}} \sqrt [8]{b}}}{x \sqrt {q+p x^4}}\right )}{8 a^{3/8} b^{5/8}}+\frac {\sqrt {2-\sqrt {2}} \text {arctanh}\left (\frac {\frac {\sqrt [8]{a} q}{\sqrt {2+\sqrt {2}} \sqrt [8]{b}}+\frac {\sqrt [8]{b} x^2}{\sqrt {2+\sqrt {2}} \sqrt [8]{a}}+\frac {\sqrt [8]{a} p x^4}{\sqrt {2+\sqrt {2}} \sqrt [8]{b}}}{x \sqrt {q+p x^4}}\right )}{8 a^{3/8} b^{5/8}} \]
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\[ \int \frac {x^4 \left (-q+p x^4\right ) \sqrt {q+p x^4}}{b x^8+a \left (q+p x^4\right )^4} \, dx=\int \frac {x^4 \left (-q+p x^4\right ) \sqrt {q+p x^4}}{b x^8+a \left (q+p x^4\right )^4} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {q x^4 \sqrt {q+p x^4}}{-a q^4-4 a p q^3 x^4-b \left (1+\frac {6 a p^2 q^2}{b}\right ) x^8-4 a p^3 q x^{12}-a p^4 x^{16}}+\frac {p x^8 \sqrt {q+p x^4}}{a q^4+4 a p q^3 x^4+b \left (1+\frac {6 a p^2 q^2}{b}\right ) x^8+4 a p^3 q x^{12}+a p^4 x^{16}}\right ) \, dx \\ & = p \int \frac {x^8 \sqrt {q+p x^4}}{a q^4+4 a p q^3 x^4+b \left (1+\frac {6 a p^2 q^2}{b}\right ) x^8+4 a p^3 q x^{12}+a p^4 x^{16}} \, dx+q \int \frac {x^4 \sqrt {q+p x^4}}{-a q^4-4 a p q^3 x^4-b \left (1+\frac {6 a p^2 q^2}{b}\right ) x^8-4 a p^3 q x^{12}-a p^4 x^{16}} \, dx \\ & = p \int \frac {x^8 \sqrt {q+p x^4}}{b x^8+a \left (q+p x^4\right )^4} \, dx+q \int \frac {x^4 \sqrt {q+p x^4}}{-b x^8-a \left (q+p x^4\right )^4} \, dx \\ \end{align*}
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 17.59 (sec) , antiderivative size = 15065, normalized size of antiderivative = 32.96 \[ \int \frac {x^4 \left (-q+p x^4\right ) \sqrt {q+p x^4}}{b x^8+a \left (q+p x^4\right )^4} \, dx=\text {Result too large to show} \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.70 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.09
method | result | size |
default | \(\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{8}+b \right )}{\sum }\frac {\ln \left (\frac {-\textit {\_R} x +\sqrt {p \,x^{4}+q}}{x}\right )}{\textit {\_R}^{5}}}{8 a}\) | \(40\) |
pseudoelliptic | \(\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{8}+b \right )}{\sum }\frac {\ln \left (\frac {-\textit {\_R} x +\sqrt {p \,x^{4}+q}}{x}\right )}{\textit {\_R}^{5}}}{8 a}\) | \(40\) |
elliptic | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (16 a \,\textit {\_Z}^{8}+b \right )}{\sum }\frac {\ln \left (\frac {\sqrt {p \,x^{4}+q}\, \sqrt {2}}{2 x}-\textit {\_R} \right )}{\textit {\_R}^{5}}\right ) \sqrt {2}}{64 a}\) | \(47\) |
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Timed out. \[ \int \frac {x^4 \left (-q+p x^4\right ) \sqrt {q+p x^4}}{b x^8+a \left (q+p x^4\right )^4} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {x^4 \left (-q+p x^4\right ) \sqrt {q+p x^4}}{b x^8+a \left (q+p x^4\right )^4} \, dx=\text {Timed out} \]
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\[ \int \frac {x^4 \left (-q+p x^4\right ) \sqrt {q+p x^4}}{b x^8+a \left (q+p x^4\right )^4} \, dx=\int { \frac {\sqrt {p x^{4} + q} {\left (p x^{4} - q\right )} x^{4}}{b x^{8} + {\left (p x^{4} + q\right )}^{4} a} \,d x } \]
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Timed out. \[ \int \frac {x^4 \left (-q+p x^4\right ) \sqrt {q+p x^4}}{b x^8+a \left (q+p x^4\right )^4} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {x^4 \left (-q+p x^4\right ) \sqrt {q+p x^4}}{b x^8+a \left (q+p x^4\right )^4} \, dx=-\int \frac {x^4\,\sqrt {p\,x^4+q}\,\left (q-p\,x^4\right )}{a\,{\left (p\,x^4+q\right )}^4+b\,x^8} \,d x \]
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