Integrand size = 36, antiderivative size = 154 \[ \int \frac {-b+a x^8}{\left (b+a x^8\right ) \sqrt [4]{b-c x^4+a x^8}} \, dx=-\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} x \sqrt [4]{b-c x^4+a x^8}}{-\sqrt {c} x^2+\sqrt {b-c x^4+a x^8}}\right )}{2 \sqrt {2} \sqrt [4]{c}}-\frac {\text {arctanh}\left (\frac {\frac {\sqrt [4]{c} x^2}{\sqrt {2}}+\frac {\sqrt {b-c x^4+a x^8}}{\sqrt {2} \sqrt [4]{c}}}{x \sqrt [4]{b-c x^4+a x^8}}\right )}{2 \sqrt {2} \sqrt [4]{c}} \]
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\[ \int \frac {-b+a x^8}{\left (b+a x^8\right ) \sqrt [4]{b-c x^4+a x^8}} \, dx=\int \frac {-b+a x^8}{\left (b+a x^8\right ) \sqrt [4]{b-c x^4+a x^8}} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{\sqrt [4]{b-c x^4+a x^8}}-\frac {2 b}{\left (b+a x^8\right ) \sqrt [4]{b-c x^4+a x^8}}\right ) \, dx \\ & = -\left ((2 b) \int \frac {1}{\left (b+a x^8\right ) \sqrt [4]{b-c x^4+a x^8}} \, dx\right )+\int \frac {1}{\sqrt [4]{b-c x^4+a x^8}} \, dx \\ & = -\left ((2 b) \int \left (\frac {1}{2 \sqrt {b} \left (\sqrt {b}-\sqrt {-a} x^4\right ) \sqrt [4]{b-c x^4+a x^8}}+\frac {1}{2 \sqrt {b} \left (\sqrt {b}+\sqrt {-a} x^4\right ) \sqrt [4]{b-c x^4+a x^8}}\right ) \, dx\right )+\frac {\left (\sqrt [4]{1+\frac {2 a x^4}{-c-\sqrt {-4 a b+c^2}}} \sqrt [4]{1+\frac {2 a x^4}{-c+\sqrt {-4 a b+c^2}}}\right ) \int \frac {1}{\sqrt [4]{1+\frac {2 a x^4}{-c-\sqrt {-4 a b+c^2}}} \sqrt [4]{1+\frac {2 a x^4}{-c+\sqrt {-4 a b+c^2}}}} \, dx}{\sqrt [4]{b-c x^4+a x^8}} \\ & = \frac {x \sqrt [4]{1-\frac {2 a x^4}{c-\sqrt {-4 a b+c^2}}} \sqrt [4]{1-\frac {2 a x^4}{c+\sqrt {-4 a b+c^2}}} \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{4},\frac {1}{4},\frac {5}{4},\frac {2 a x^4}{c+\sqrt {-4 a b+c^2}},\frac {2 a x^4}{c-\sqrt {-4 a b+c^2}}\right )}{\sqrt [4]{b-c x^4+a x^8}}-\sqrt {b} \int \frac {1}{\left (\sqrt {b}-\sqrt {-a} x^4\right ) \sqrt [4]{b-c x^4+a x^8}} \, dx-\sqrt {b} \int \frac {1}{\left (\sqrt {b}+\sqrt {-a} x^4\right ) \sqrt [4]{b-c x^4+a x^8}} \, dx \\ \end{align*}
Time = 1.78 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.87 \[ \int \frac {-b+a x^8}{\left (b+a x^8\right ) \sqrt [4]{b-c x^4+a x^8}} \, dx=-\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} x \sqrt [4]{b-c x^4+a x^8}}{-\sqrt {c} x^2+\sqrt {b-c x^4+a x^8}}\right )+\text {arctanh}\left (\frac {\sqrt {c} x^2+\sqrt {b-c x^4+a x^8}}{\sqrt {2} \sqrt [4]{c} x \sqrt [4]{b-c x^4+a x^8}}\right )}{2 \sqrt {2} \sqrt [4]{c}} \]
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Time = 2.20 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.16
method | result | size |
pseudoelliptic | \(\frac {\sqrt {2}\, \left (\ln \left (\frac {-\left (a \,x^{8}-c \,x^{4}+b \right )^{\frac {1}{4}} x \,c^{\frac {1}{4}} \sqrt {2}+\sqrt {c}\, x^{2}+\sqrt {a \,x^{8}-c \,x^{4}+b}}{\left (a \,x^{8}-c \,x^{4}+b \right )^{\frac {1}{4}} x \,c^{\frac {1}{4}} \sqrt {2}+\sqrt {c}\, x^{2}+\sqrt {a \,x^{8}-c \,x^{4}+b}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \left (a \,x^{8}-c \,x^{4}+b \right )^{\frac {1}{4}}+c^{\frac {1}{4}} x}{c^{\frac {1}{4}} x}\right )+2 \arctan \left (\frac {\sqrt {2}\, \left (a \,x^{8}-c \,x^{4}+b \right )^{\frac {1}{4}}-c^{\frac {1}{4}} x}{c^{\frac {1}{4}} x}\right )\right )}{8 c^{\frac {1}{4}}}\) | \(178\) |
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Timed out. \[ \int \frac {-b+a x^8}{\left (b+a x^8\right ) \sqrt [4]{b-c x^4+a x^8}} \, dx=\text {Timed out} \]
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\[ \int \frac {-b+a x^8}{\left (b+a x^8\right ) \sqrt [4]{b-c x^4+a x^8}} \, dx=\int \frac {a x^{8} - b}{\left (a x^{8} + b\right ) \sqrt [4]{a x^{8} + b - c x^{4}}}\, dx \]
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\[ \int \frac {-b+a x^8}{\left (b+a x^8\right ) \sqrt [4]{b-c x^4+a x^8}} \, dx=\int { \frac {a x^{8} - b}{{\left (a x^{8} - c x^{4} + b\right )}^{\frac {1}{4}} {\left (a x^{8} + b\right )}} \,d x } \]
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\[ \int \frac {-b+a x^8}{\left (b+a x^8\right ) \sqrt [4]{b-c x^4+a x^8}} \, dx=\int { \frac {a x^{8} - b}{{\left (a x^{8} - c x^{4} + b\right )}^{\frac {1}{4}} {\left (a x^{8} + b\right )}} \,d x } \]
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Timed out. \[ \int \frac {-b+a x^8}{\left (b+a x^8\right ) \sqrt [4]{b-c x^4+a x^8}} \, dx=\int -\frac {b-a\,x^8}{\left (a\,x^8+b\right )\,{\left (a\,x^8-c\,x^4+b\right )}^{1/4}} \,d x \]
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