Integrand size = 39, antiderivative size = 61 \[ \int \frac {-b+a x^4}{\left (b-2 x^2+a x^4\right ) \sqrt [4]{b x^2+a x^6}} \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{b x^2+a x^6}}\right )}{\sqrt [4]{2}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{b x^2+a x^6}}\right )}{\sqrt [4]{2}} \]
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Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 1.56 (sec) , antiderivative size = 444, normalized size of antiderivative = 7.28, number of steps used = 20, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.256, Rules used = {2081, 6847, 6860, 252, 251, 1452, 441, 440, 525, 524} \[ \int \frac {-b+a x^4}{\left (b-2 x^2+a x^4\right ) \sqrt [4]{b x^2+a x^6}} \, dx=-\frac {2 x \sqrt [4]{\frac {a x^4}{b}+1} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {1}{4},\frac {9}{8},\frac {a^2 x^4}{-a b-2 \sqrt {1-a b}+2},-\frac {a x^4}{b}\right )}{\sqrt [4]{a x^6+b x^2}}-\frac {2 x \sqrt [4]{\frac {a x^4}{b}+1} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {1}{4},\frac {9}{8},\frac {a^2 x^4}{-a b+2 \sqrt {1-a b}+2},-\frac {a x^4}{b}\right )}{\sqrt [4]{a x^6+b x^2}}-\frac {2 a x^3 \left (1-\sqrt {1-a b}\right ) \sqrt [4]{\frac {a x^4}{b}+1} \operatorname {AppellF1}\left (\frac {5}{8},1,\frac {1}{4},\frac {13}{8},\frac {a^2 x^4}{-a b-2 \sqrt {1-a b}+2},-\frac {a x^4}{b}\right )}{5 \left (-a b-2 \sqrt {1-a b}+2\right ) \sqrt [4]{a x^6+b x^2}}-\frac {2 a x^3 \left (\sqrt {1-a b}+1\right ) \sqrt [4]{\frac {a x^4}{b}+1} \operatorname {AppellF1}\left (\frac {5}{8},1,\frac {1}{4},\frac {13}{8},\frac {a^2 x^4}{-a b+2 \sqrt {1-a b}+2},-\frac {a x^4}{b}\right )}{5 \left (-a b+2 \sqrt {1-a b}+2\right ) \sqrt [4]{a x^6+b x^2}}+\frac {2 x \sqrt [4]{\frac {a x^4}{b}+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},-\frac {a x^4}{b}\right )}{\sqrt [4]{a x^6+b x^2}} \]
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Rule 251
Rule 252
Rule 440
Rule 441
Rule 524
Rule 525
Rule 1452
Rule 2081
Rule 6847
Rule 6860
Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {x} \sqrt [4]{b+a x^4}\right ) \int \frac {-b+a x^4}{\sqrt {x} \sqrt [4]{b+a x^4} \left (b-2 x^2+a x^4\right )} \, dx}{\sqrt [4]{b x^2+a x^6}} \\ & = \frac {\left (2 \sqrt {x} \sqrt [4]{b+a x^4}\right ) \text {Subst}\left (\int \frac {-b+a x^8}{\sqrt [4]{b+a x^8} \left (b-2 x^4+a x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}} \\ & = \frac {\left (2 \sqrt {x} \sqrt [4]{b+a x^4}\right ) \text {Subst}\left (\int \left (\frac {1}{\sqrt [4]{b+a x^8}}-\frac {2 \left (b-x^4\right )}{\sqrt [4]{b+a x^8} \left (b-2 x^4+a x^8\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}} \\ & = \frac {\left (2 \sqrt {x} \sqrt [4]{b+a x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{b+a x^8}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}}-\frac {\left (4 \sqrt {x} \sqrt [4]{b+a x^4}\right ) \text {Subst}\left (\int \frac {b-x^4}{\sqrt [4]{b+a x^8} \left (b-2 x^4+a x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}} \\ & = -\frac {\left (4 \sqrt {x} \sqrt [4]{b+a x^4}\right ) \text {Subst}\left (\int \left (\frac {-1-\sqrt {1-a b}}{\left (-2-2 \sqrt {1-a b}+2 a x^4\right ) \sqrt [4]{b+a x^8}}+\frac {-1+\sqrt {1-a b}}{\left (-2+2 \sqrt {1-a b}+2 a x^4\right ) \sqrt [4]{b+a x^8}}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}}+\frac {\left (2 \sqrt {x} \sqrt [4]{1+\frac {a x^4}{b}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{1+\frac {a x^8}{b}}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}} \\ & = \frac {2 x \sqrt [4]{1+\frac {a x^4}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},-\frac {a x^4}{b}\right )}{\sqrt [4]{b x^2+a x^6}}-\frac {\left (4 \left (-1-\sqrt {1-a b}\right ) \sqrt {x} \sqrt [4]{b+a x^4}\right ) \text {Subst}\left (\int \frac {1}{\left (-2-2 \sqrt {1-a b}+2 a x^4\right ) \sqrt [4]{b+a x^8}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}}-\frac {\left (4 \left (-1+\sqrt {1-a b}\right ) \sqrt {x} \sqrt [4]{b+a x^4}\right ) \text {Subst}\left (\int \frac {1}{\left (-2+2 \sqrt {1-a b}+2 a x^4\right ) \sqrt [4]{b+a x^8}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}} \\ & = \frac {2 x \sqrt [4]{1+\frac {a x^4}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},-\frac {a x^4}{b}\right )}{\sqrt [4]{b x^2+a x^6}}-\frac {\left (4 \left (-1-\sqrt {1-a b}\right ) \sqrt {x} \sqrt [4]{b+a x^4}\right ) \text {Subst}\left (\int \left (\frac {-1-\sqrt {1-a b}}{2 \sqrt [4]{b+a x^8} \left (2-a b+2 \sqrt {1-a b}-a^2 x^8\right )}-\frac {a x^4}{2 \sqrt [4]{b+a x^8} \left (2-a b+2 \sqrt {1-a b}-a^2 x^8\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}}-\frac {\left (4 \left (-1+\sqrt {1-a b}\right ) \sqrt {x} \sqrt [4]{b+a x^4}\right ) \text {Subst}\left (\int \left (\frac {1-\sqrt {1-a b}}{2 \sqrt [4]{b+a x^8} \left (-2+a b+2 \sqrt {1-a b}+a^2 x^8\right )}+\frac {a x^4}{2 \sqrt [4]{b+a x^8} \left (-2+a b+2 \sqrt {1-a b}+a^2 x^8\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}} \\ & = \frac {2 x \sqrt [4]{1+\frac {a x^4}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},-\frac {a x^4}{b}\right )}{\sqrt [4]{b x^2+a x^6}}+\frac {\left (2 a \left (-1-\sqrt {1-a b}\right ) \sqrt {x} \sqrt [4]{b+a x^4}\right ) \text {Subst}\left (\int \frac {x^4}{\sqrt [4]{b+a x^8} \left (2-a b+2 \sqrt {1-a b}-a^2 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}}-\frac {\left (2 \left (-1-\sqrt {1-a b}\right )^2 \sqrt {x} \sqrt [4]{b+a x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{b+a x^8} \left (2-a b+2 \sqrt {1-a b}-a^2 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}}-\frac {\left (2 a \left (-1+\sqrt {1-a b}\right ) \sqrt {x} \sqrt [4]{b+a x^4}\right ) \text {Subst}\left (\int \frac {x^4}{\sqrt [4]{b+a x^8} \left (-2+a b+2 \sqrt {1-a b}+a^2 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}}-\frac {\left (2 \left (1-\sqrt {1-a b}\right ) \left (-1+\sqrt {1-a b}\right ) \sqrt {x} \sqrt [4]{b+a x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{b+a x^8} \left (-2+a b+2 \sqrt {1-a b}+a^2 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}} \\ & = \frac {2 x \sqrt [4]{1+\frac {a x^4}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},-\frac {a x^4}{b}\right )}{\sqrt [4]{b x^2+a x^6}}+\frac {\left (2 a \left (-1-\sqrt {1-a b}\right ) \sqrt {x} \sqrt [4]{1+\frac {a x^4}{b}}\right ) \text {Subst}\left (\int \frac {x^4}{\left (2-a b+2 \sqrt {1-a b}-a^2 x^8\right ) \sqrt [4]{1+\frac {a x^8}{b}}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}}-\frac {\left (2 \left (-1-\sqrt {1-a b}\right )^2 \sqrt {x} \sqrt [4]{1+\frac {a x^4}{b}}\right ) \text {Subst}\left (\int \frac {1}{\left (2-a b+2 \sqrt {1-a b}-a^2 x^8\right ) \sqrt [4]{1+\frac {a x^8}{b}}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}}-\frac {\left (2 a \left (-1+\sqrt {1-a b}\right ) \sqrt {x} \sqrt [4]{1+\frac {a x^4}{b}}\right ) \text {Subst}\left (\int \frac {x^4}{\left (-2+a b+2 \sqrt {1-a b}+a^2 x^8\right ) \sqrt [4]{1+\frac {a x^8}{b}}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}}-\frac {\left (2 \left (1-\sqrt {1-a b}\right ) \left (-1+\sqrt {1-a b}\right ) \sqrt {x} \sqrt [4]{1+\frac {a x^4}{b}}\right ) \text {Subst}\left (\int \frac {1}{\left (-2+a b+2 \sqrt {1-a b}+a^2 x^8\right ) \sqrt [4]{1+\frac {a x^8}{b}}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{b x^2+a x^6}} \\ & = -\frac {2 x \sqrt [4]{1+\frac {a x^4}{b}} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {1}{4},\frac {9}{8},\frac {a^2 x^4}{2-a b-2 \sqrt {1-a b}},-\frac {a x^4}{b}\right )}{\sqrt [4]{b x^2+a x^6}}-\frac {2 x \sqrt [4]{1+\frac {a x^4}{b}} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {1}{4},\frac {9}{8},\frac {a^2 x^4}{2-a b+2 \sqrt {1-a b}},-\frac {a x^4}{b}\right )}{\sqrt [4]{b x^2+a x^6}}-\frac {2 a \left (1-\sqrt {1-a b}\right ) x^3 \sqrt [4]{1+\frac {a x^4}{b}} \operatorname {AppellF1}\left (\frac {5}{8},1,\frac {1}{4},\frac {13}{8},\frac {a^2 x^4}{2-a b-2 \sqrt {1-a b}},-\frac {a x^4}{b}\right )}{5 \left (2-a b-2 \sqrt {1-a b}\right ) \sqrt [4]{b x^2+a x^6}}-\frac {2 a \left (1+\sqrt {1-a b}\right ) x^3 \sqrt [4]{1+\frac {a x^4}{b}} \operatorname {AppellF1}\left (\frac {5}{8},1,\frac {1}{4},\frac {13}{8},\frac {a^2 x^4}{2-a b+2 \sqrt {1-a b}},-\frac {a x^4}{b}\right )}{5 \left (2-a b+2 \sqrt {1-a b}\right ) \sqrt [4]{b x^2+a x^6}}+\frac {2 x \sqrt [4]{1+\frac {a x^4}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},-\frac {a x^4}{b}\right )}{\sqrt [4]{b x^2+a x^6}} \\ \end{align*}
Time = 13.48 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.89 \[ \int \frac {-b+a x^4}{\left (b-2 x^2+a x^4\right ) \sqrt [4]{b x^2+a x^6}} \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2 \left (b+a x^4\right )}}\right )+\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2 \left (b+a x^4\right )}}\right )}{\sqrt [4]{2}} \]
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Time = 2.87 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.25
method | result | size |
pseudoelliptic | \(\frac {2^{\frac {3}{4}} \left (2 \arctan \left (\frac {2^{\frac {3}{4}} \left (x^{2} \left (a \,x^{4}+b \right )\right )^{\frac {1}{4}}}{2 x}\right )-\ln \left (\frac {2^{\frac {1}{4}} x +\left (x^{2} \left (a \,x^{4}+b \right )\right )^{\frac {1}{4}}}{-2^{\frac {1}{4}} x +\left (x^{2} \left (a \,x^{4}+b \right )\right )^{\frac {1}{4}}}\right )\right )}{4}\) | \(76\) |
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Timed out. \[ \int \frac {-b+a x^4}{\left (b-2 x^2+a x^4\right ) \sqrt [4]{b x^2+a x^6}} \, dx=\text {Timed out} \]
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\[ \int \frac {-b+a x^4}{\left (b-2 x^2+a x^4\right ) \sqrt [4]{b x^2+a x^6}} \, dx=\int \frac {a x^{4} - b}{\sqrt [4]{x^{2} \left (a x^{4} + b\right )} \left (a x^{4} + b - 2 x^{2}\right )}\, dx \]
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\[ \int \frac {-b+a x^4}{\left (b-2 x^2+a x^4\right ) \sqrt [4]{b x^2+a x^6}} \, dx=\int { \frac {a x^{4} - b}{{\left (a x^{6} + b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{4} - 2 \, x^{2} + b\right )}} \,d x } \]
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\[ \int \frac {-b+a x^4}{\left (b-2 x^2+a x^4\right ) \sqrt [4]{b x^2+a x^6}} \, dx=\int { \frac {a x^{4} - b}{{\left (a x^{6} + b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{4} - 2 \, x^{2} + b\right )}} \,d x } \]
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Timed out. \[ \int \frac {-b+a x^4}{\left (b-2 x^2+a x^4\right ) \sqrt [4]{b x^2+a x^6}} \, dx=\int -\frac {b-a\,x^4}{{\left (a\,x^6+b\,x^2\right )}^{1/4}\,\left (a\,x^4-2\,x^2+b\right )} \,d x \]
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