Integrand size = 29, antiderivative size = 101 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{x^2} \, dx=\frac {\left (4 b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{2 x}+\frac {5}{4} \sqrt [4]{a} b \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b x^2+a x^4}}\right )-\frac {5}{4} \sqrt [4]{a} b \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b x^2+a x^4}}\right ) \]
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Time = 0.26 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.81, number of steps used = 16, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {2077, 2029, 2057, 335, 338, 304, 209, 212, 2045} \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{x^2} \, dx=\frac {5 \sqrt [4]{a} b x^{3/2} \left (a x^2-b\right )^{3/4} \arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2-b}}\right )}{4 \left (a x^4-b x^2\right )^{3/4}}-\frac {5 \sqrt [4]{a} b x^{3/2} \left (a x^2-b\right )^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2-b}}\right )}{4 \left (a x^4-b x^2\right )^{3/4}}+\frac {1}{2} a x \sqrt [4]{a x^4-b x^2}+\frac {2 b \sqrt [4]{a x^4-b x^2}}{x} \]
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Rule 209
Rule 212
Rule 304
Rule 335
Rule 338
Rule 2029
Rule 2045
Rule 2057
Rule 2077
Rubi steps \begin{align*} \text {integral}& = \int \left (a \sqrt [4]{-b x^2+a x^4}-\frac {b \sqrt [4]{-b x^2+a x^4}}{x^2}\right ) \, dx \\ & = a \int \sqrt [4]{-b x^2+a x^4} \, dx-b \int \frac {\sqrt [4]{-b x^2+a x^4}}{x^2} \, dx \\ & = \frac {2 b \sqrt [4]{-b x^2+a x^4}}{x}+\frac {1}{2} a x \sqrt [4]{-b x^2+a x^4}-\frac {1}{4} (a b) \int \frac {x^2}{\left (-b x^2+a x^4\right )^{3/4}} \, dx-(a b) \int \frac {x^2}{\left (-b x^2+a x^4\right )^{3/4}} \, dx \\ & = \frac {2 b \sqrt [4]{-b x^2+a x^4}}{x}+\frac {1}{2} a x \sqrt [4]{-b x^2+a x^4}-\frac {\left (a b x^{3/2} \left (-b+a x^2\right )^{3/4}\right ) \int \frac {\sqrt {x}}{\left (-b+a x^2\right )^{3/4}} \, dx}{4 \left (-b x^2+a x^4\right )^{3/4}}-\frac {\left (a b x^{3/2} \left (-b+a x^2\right )^{3/4}\right ) \int \frac {\sqrt {x}}{\left (-b+a x^2\right )^{3/4}} \, dx}{\left (-b x^2+a x^4\right )^{3/4}} \\ & = \frac {2 b \sqrt [4]{-b x^2+a x^4}}{x}+\frac {1}{2} a x \sqrt [4]{-b x^2+a x^4}-\frac {\left (a b x^{3/2} \left (-b+a x^2\right )^{3/4}\right ) \text {Subst}\left (\int \frac {x^2}{\left (-b+a x^4\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{2 \left (-b x^2+a x^4\right )^{3/4}}-\frac {\left (2 a b x^{3/2} \left (-b+a x^2\right )^{3/4}\right ) \text {Subst}\left (\int \frac {x^2}{\left (-b+a x^4\right )^{3/4}} \, dx,x,\sqrt {x}\right )}{\left (-b x^2+a x^4\right )^{3/4}} \\ & = \frac {2 b \sqrt [4]{-b x^2+a x^4}}{x}+\frac {1}{2} a x \sqrt [4]{-b x^2+a x^4}-\frac {\left (a b x^{3/2} \left (-b+a x^2\right )^{3/4}\right ) \text {Subst}\left (\int \frac {x^2}{1-a x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )}{2 \left (-b x^2+a x^4\right )^{3/4}}-\frac {\left (2 a b x^{3/2} \left (-b+a x^2\right )^{3/4}\right ) \text {Subst}\left (\int \frac {x^2}{1-a x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )}{\left (-b x^2+a x^4\right )^{3/4}} \\ & = \frac {2 b \sqrt [4]{-b x^2+a x^4}}{x}+\frac {1}{2} a x \sqrt [4]{-b x^2+a x^4}-\frac {\left (\sqrt {a} b x^{3/2} \left (-b+a x^2\right )^{3/4}\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )}{4 \left (-b x^2+a x^4\right )^{3/4}}+\frac {\left (\sqrt {a} b x^{3/2} \left (-b+a x^2\right )^{3/4}\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )}{4 \left (-b x^2+a x^4\right )^{3/4}}-\frac {\left (\sqrt {a} b x^{3/2} \left (-b+a x^2\right )^{3/4}\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )}{\left (-b x^2+a x^4\right )^{3/4}}+\frac {\left (\sqrt {a} b x^{3/2} \left (-b+a x^2\right )^{3/4}\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )}{\left (-b x^2+a x^4\right )^{3/4}} \\ & = \frac {2 b \sqrt [4]{-b x^2+a x^4}}{x}+\frac {1}{2} a x \sqrt [4]{-b x^2+a x^4}+\frac {5 \sqrt [4]{a} b x^{3/2} \left (-b+a x^2\right )^{3/4} \arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )}{4 \left (-b x^2+a x^4\right )^{3/4}}-\frac {5 \sqrt [4]{a} b x^{3/2} \left (-b+a x^2\right )^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )}{4 \left (-b x^2+a x^4\right )^{3/4}} \\ \end{align*}
Time = 0.40 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.36 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{x^2} \, dx=\frac {\sqrt [4]{-b x^2+a x^4} \left (2 \sqrt [4]{-b+a x^2} \left (4 b+a x^2\right )+5 \sqrt [4]{a} b \sqrt {x} \arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )-5 \sqrt [4]{a} b \sqrt {x} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )\right )}{4 x \sqrt [4]{-b+a x^2}} \]
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Time = 0.36 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.16
method | result | size |
pseudoelliptic | \(\frac {-5 b x \left (\ln \left (\frac {-a^{\frac {1}{4}} x -\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x -\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )\right ) a^{\frac {1}{4}}+4 \left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}} \left (a \,x^{2}+4 b \right )}{8 x}\) | \(117\) |
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Timed out. \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{x^2} \, dx=\text {Timed out} \]
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\[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{x^2} \, dx=\int \frac {\sqrt [4]{x^{2} \left (a x^{2} - b\right )} \left (a x^{2} - b\right )}{x^{2}}\, dx \]
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\[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{x^2} \, dx=\int { \frac {{\left (a x^{4} - b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{2} - b\right )}}{x^{2}} \,d x } \]
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Leaf count of result is larger than twice the leaf count of optimal. 229 vs. \(2 (81) = 162\).
Time = 0.29 (sec) , antiderivative size = 229, normalized size of antiderivative = 2.27 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{x^2} \, dx=\frac {8 \, {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}} a b x^{2} - 10 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) - 10 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) - 5 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a - \frac {b}{x^{2}}}\right ) + 5 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{2} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a - \frac {b}{x^{2}}}\right ) + 32 \, {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}} b^{2}}{16 \, b} \]
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Time = 6.97 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.90 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{x^2} \, dx=\frac {2\,a\,x\,{\left (a\,x^4-b\,x^2\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {3}{4};\ \frac {7}{4};\ \frac {a\,x^2}{b}\right )}{3\,{\left (1-\frac {a\,x^2}{b}\right )}^{1/4}}+\frac {2\,b\,{\left (a\,x^4-b\,x^2\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},-\frac {1}{4};\ \frac {3}{4};\ \frac {a\,x^2}{b}\right )}{x\,{\left (1-\frac {a\,x^2}{b}\right )}^{1/4}} \]
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