Integrand size = 36, antiderivative size = 59 \[ \int \frac {b+a x^8}{\sqrt [4]{b-a x^8} \left (-b+c x^4+a x^8\right )} \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{b-a x^8}}\right )}{2 \sqrt [4]{c}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{b-a x^8}}\right )}{2 \sqrt [4]{c}} \]
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Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 1.04 (sec) , antiderivative size = 460, normalized size of antiderivative = 7.80, number of steps used = 18, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6860, 252, 251, 1452, 441, 440, 525, 524} \[ \int \frac {b+a x^8}{\sqrt [4]{b-a x^8} \left (-b+c x^4+a x^8\right )} \, dx=-\frac {x \sqrt [4]{1-\frac {a x^8}{b}} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {1}{4},\frac {9}{8},\frac {2 a^2 x^8}{c^2-\sqrt {c^2+4 a b} c+2 a b},\frac {a x^8}{b}\right )}{\sqrt [4]{b-a x^8}}-\frac {x \sqrt [4]{1-\frac {a x^8}{b}} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {1}{4},\frac {9}{8},\frac {2 a^2 x^8}{2 a b+c \left (c+\sqrt {c^2+4 a b}\right )},\frac {a x^8}{b}\right )}{\sqrt [4]{b-a x^8}}+\frac {a x^5 \left (c-\sqrt {4 a b+c^2}\right ) \sqrt [4]{1-\frac {a x^8}{b}} \operatorname {AppellF1}\left (\frac {5}{8},1,\frac {1}{4},\frac {13}{8},\frac {2 a^2 x^8}{c^2-\sqrt {c^2+4 a b} c+2 a b},\frac {a x^8}{b}\right )}{5 \left (c \left (c-\sqrt {4 a b+c^2}\right )+2 a b\right ) \sqrt [4]{b-a x^8}}+\frac {a x^5 \left (\sqrt {4 a b+c^2}+c\right ) \sqrt [4]{1-\frac {a x^8}{b}} \operatorname {AppellF1}\left (\frac {5}{8},1,\frac {1}{4},\frac {13}{8},\frac {2 a^2 x^8}{2 a b+c \left (c+\sqrt {c^2+4 a b}\right )},\frac {a x^8}{b}\right )}{5 \left (c \left (\sqrt {4 a b+c^2}+c\right )+2 a b\right ) \sqrt [4]{b-a x^8}}+\frac {x \sqrt [4]{1-\frac {a x^8}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},\frac {a x^8}{b}\right )}{\sqrt [4]{b-a x^8}} \]
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Rule 251
Rule 252
Rule 440
Rule 441
Rule 524
Rule 525
Rule 1452
Rule 6860
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{\sqrt [4]{b-a x^8}}+\frac {2 b-c x^4}{\sqrt [4]{b-a x^8} \left (-b+c x^4+a x^8\right )}\right ) \, dx \\ & = \int \frac {1}{\sqrt [4]{b-a x^8}} \, dx+\int \frac {2 b-c x^4}{\sqrt [4]{b-a x^8} \left (-b+c x^4+a x^8\right )} \, dx \\ & = \frac {\sqrt [4]{1-\frac {a x^8}{b}} \int \frac {1}{\sqrt [4]{1-\frac {a x^8}{b}}} \, dx}{\sqrt [4]{b-a x^8}}+\int \left (\frac {-c+\sqrt {4 a b+c^2}}{\left (c-\sqrt {4 a b+c^2}+2 a x^4\right ) \sqrt [4]{b-a x^8}}+\frac {-c-\sqrt {4 a b+c^2}}{\left (c+\sqrt {4 a b+c^2}+2 a x^4\right ) \sqrt [4]{b-a x^8}}\right ) \, dx \\ & = \frac {x \sqrt [4]{1-\frac {a x^8}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},\frac {a x^8}{b}\right )}{\sqrt [4]{b-a x^8}}+\left (-c-\sqrt {4 a b+c^2}\right ) \int \frac {1}{\left (c+\sqrt {4 a b+c^2}+2 a x^4\right ) \sqrt [4]{b-a x^8}} \, dx+\left (-c+\sqrt {4 a b+c^2}\right ) \int \frac {1}{\left (c-\sqrt {4 a b+c^2}+2 a x^4\right ) \sqrt [4]{b-a x^8}} \, dx \\ & = \frac {x \sqrt [4]{1-\frac {a x^8}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},\frac {a x^8}{b}\right )}{\sqrt [4]{b-a x^8}}+\left (-c-\sqrt {4 a b+c^2}\right ) \int \left (\frac {c+\sqrt {4 a b+c^2}}{2 \sqrt [4]{b-a x^8} \left (2 a b+c^2+c \sqrt {4 a b+c^2}-2 a^2 x^8\right )}+\frac {a x^4}{\sqrt [4]{b-a x^8} \left (-2 a b-c^2-c \sqrt {4 a b+c^2}+2 a^2 x^8\right )}\right ) \, dx+\left (-c+\sqrt {4 a b+c^2}\right ) \int \left (\frac {-c+\sqrt {4 a b+c^2}}{2 \sqrt [4]{b-a x^8} \left (-2 a b-c^2+c \sqrt {4 a b+c^2}+2 a^2 x^8\right )}+\frac {a x^4}{\sqrt [4]{b-a x^8} \left (-2 a b-c^2+c \sqrt {4 a b+c^2}+2 a^2 x^8\right )}\right ) \, dx \\ & = \frac {x \sqrt [4]{1-\frac {a x^8}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},\frac {a x^8}{b}\right )}{\sqrt [4]{b-a x^8}}-\left (a \left (c-\sqrt {4 a b+c^2}\right )\right ) \int \frac {x^4}{\sqrt [4]{b-a x^8} \left (-2 a b-c^2+c \sqrt {4 a b+c^2}+2 a^2 x^8\right )} \, dx+\frac {1}{2} \left (c-\sqrt {4 a b+c^2}\right )^2 \int \frac {1}{\sqrt [4]{b-a x^8} \left (-2 a b-c^2+c \sqrt {4 a b+c^2}+2 a^2 x^8\right )} \, dx-\left (a \left (c+\sqrt {4 a b+c^2}\right )\right ) \int \frac {x^4}{\sqrt [4]{b-a x^8} \left (-2 a b-c^2-c \sqrt {4 a b+c^2}+2 a^2 x^8\right )} \, dx-\frac {1}{2} \left (c+\sqrt {4 a b+c^2}\right )^2 \int \frac {1}{\sqrt [4]{b-a x^8} \left (2 a b+c^2+c \sqrt {4 a b+c^2}-2 a^2 x^8\right )} \, dx \\ & = \frac {x \sqrt [4]{1-\frac {a x^8}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},\frac {a x^8}{b}\right )}{\sqrt [4]{b-a x^8}}-\frac {\left (a \left (c-\sqrt {4 a b+c^2}\right ) \sqrt [4]{1-\frac {a x^8}{b}}\right ) \int \frac {x^4}{\left (-2 a b-c^2+c \sqrt {4 a b+c^2}+2 a^2 x^8\right ) \sqrt [4]{1-\frac {a x^8}{b}}} \, dx}{\sqrt [4]{b-a x^8}}+\frac {\left (\left (c-\sqrt {4 a b+c^2}\right )^2 \sqrt [4]{1-\frac {a x^8}{b}}\right ) \int \frac {1}{\left (-2 a b-c^2+c \sqrt {4 a b+c^2}+2 a^2 x^8\right ) \sqrt [4]{1-\frac {a x^8}{b}}} \, dx}{2 \sqrt [4]{b-a x^8}}-\frac {\left (a \left (c+\sqrt {4 a b+c^2}\right ) \sqrt [4]{1-\frac {a x^8}{b}}\right ) \int \frac {x^4}{\left (-2 a b-c^2-c \sqrt {4 a b+c^2}+2 a^2 x^8\right ) \sqrt [4]{1-\frac {a x^8}{b}}} \, dx}{\sqrt [4]{b-a x^8}}-\frac {\left (\left (c+\sqrt {4 a b+c^2}\right )^2 \sqrt [4]{1-\frac {a x^8}{b}}\right ) \int \frac {1}{\left (2 a b+c^2+c \sqrt {4 a b+c^2}-2 a^2 x^8\right ) \sqrt [4]{1-\frac {a x^8}{b}}} \, dx}{2 \sqrt [4]{b-a x^8}} \\ & = -\frac {x \sqrt [4]{1-\frac {a x^8}{b}} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {1}{4},\frac {9}{8},\frac {2 a^2 x^8}{2 a b+c^2-c \sqrt {4 a b+c^2}},\frac {a x^8}{b}\right )}{\sqrt [4]{b-a x^8}}-\frac {x \sqrt [4]{1-\frac {a x^8}{b}} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {1}{4},\frac {9}{8},\frac {2 a^2 x^8}{2 a b+c \left (c+\sqrt {4 a b+c^2}\right )},\frac {a x^8}{b}\right )}{\sqrt [4]{b-a x^8}}+\frac {a \left (c-\sqrt {4 a b+c^2}\right ) x^5 \sqrt [4]{1-\frac {a x^8}{b}} \operatorname {AppellF1}\left (\frac {5}{8},1,\frac {1}{4},\frac {13}{8},\frac {2 a^2 x^8}{2 a b+c^2-c \sqrt {4 a b+c^2}},\frac {a x^8}{b}\right )}{5 \left (2 a b+c \left (c-\sqrt {4 a b+c^2}\right )\right ) \sqrt [4]{b-a x^8}}+\frac {a \left (c+\sqrt {4 a b+c^2}\right ) x^5 \sqrt [4]{1-\frac {a x^8}{b}} \operatorname {AppellF1}\left (\frac {5}{8},1,\frac {1}{4},\frac {13}{8},\frac {2 a^2 x^8}{2 a b+c \left (c+\sqrt {4 a b+c^2}\right )},\frac {a x^8}{b}\right )}{5 \left (2 a b+c \left (c+\sqrt {4 a b+c^2}\right )\right ) \sqrt [4]{b-a x^8}}+\frac {x \sqrt [4]{1-\frac {a x^8}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},\frac {a x^8}{b}\right )}{\sqrt [4]{b-a x^8}} \\ \end{align*}
Time = 8.85 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.85 \[ \int \frac {b+a x^8}{\sqrt [4]{b-a x^8} \left (-b+c x^4+a x^8\right )} \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{b-a x^8}}\right )+\text {arctanh}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{b-a x^8}}\right )}{2 \sqrt [4]{c}} \]
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Time = 2.64 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.19
method | result | size |
pseudoelliptic | \(\frac {2 \arctan \left (\frac {\left (-a \,x^{8}+b \right )^{\frac {1}{4}}}{c^{\frac {1}{4}} x}\right )-\ln \left (\frac {-c^{\frac {1}{4}} x -\left (-a \,x^{8}+b \right )^{\frac {1}{4}}}{c^{\frac {1}{4}} x -\left (-a \,x^{8}+b \right )^{\frac {1}{4}}}\right )}{4 c^{\frac {1}{4}}}\) | \(70\) |
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Timed out. \[ \int \frac {b+a x^8}{\sqrt [4]{b-a x^8} \left (-b+c x^4+a x^8\right )} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {b+a x^8}{\sqrt [4]{b-a x^8} \left (-b+c x^4+a x^8\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {b+a x^8}{\sqrt [4]{b-a x^8} \left (-b+c x^4+a x^8\right )} \, dx=\int { \frac {a x^{8} + b}{{\left (a x^{8} + c x^{4} - b\right )} {\left (-a x^{8} + b\right )}^{\frac {1}{4}}} \,d x } \]
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\[ \int \frac {b+a x^8}{\sqrt [4]{b-a x^8} \left (-b+c x^4+a x^8\right )} \, dx=\int { \frac {a x^{8} + b}{{\left (a x^{8} + c x^{4} - b\right )} {\left (-a x^{8} + b\right )}^{\frac {1}{4}}} \,d x } \]
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Timed out. \[ \int \frac {b+a x^8}{\sqrt [4]{b-a x^8} \left (-b+c x^4+a x^8\right )} \, dx=\int \frac {a\,x^8+b}{{\left (b-a\,x^8\right )}^{1/4}\,\left (a\,x^8+c\,x^4-b\right )} \,d x \]
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