Integrand size = 39, antiderivative size = 211 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b-a x^2+x^4} \, dx=-a^{5/4} \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b x^2+a x^4}}\right )+a^{5/4} \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b x^2+a x^4}}\right )-\frac {1}{2} \text {RootSum}\left [b-a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {a b \log (x)-a b \log \left (\sqrt [4]{-b x^2+a x^4}-x \text {$\#$1}\right )-a^2 \log (x) \text {$\#$1}^4+b \log (x) \text {$\#$1}^4+a^2 \log \left (\sqrt [4]{-b x^2+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4-b \log \left (\sqrt [4]{-b x^2+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{a \text {$\#$1}^3-2 \text {$\#$1}^7}\&\right ] \]
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Time = 0.33 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.95, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.128, Rules used = {2081, 1283, 1542, 525, 524} \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b-a x^2+x^4} \, dx=\frac {4 b x \sqrt [4]{a x^4-b x^2} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {5}{4},\frac {7}{4},\frac {2 x^2}{a-\sqrt {a^2-4 b}},\frac {a x^2}{b}\right )}{3 \left (-a \sqrt {a^2-4 b}+a^2-4 b\right ) \sqrt [4]{1-\frac {a x^2}{b}}}+\frac {4 b x \sqrt [4]{a x^4-b x^2} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {5}{4},\frac {7}{4},\frac {2 x^2}{a+\sqrt {a^2-4 b}},\frac {a x^2}{b}\right )}{3 \left (a \sqrt {a^2-4 b}+a^2-4 b\right ) \sqrt [4]{1-\frac {a x^2}{b}}} \]
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Rule 524
Rule 525
Rule 1283
Rule 1542
Rule 2081
Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt [4]{-b x^2+a x^4} \int \frac {\sqrt {x} \left (-b+a x^2\right )^{5/4}}{b-a x^2+x^4} \, dx}{\sqrt {x} \sqrt [4]{-b+a x^2}} \\ & = \frac {\left (2 \sqrt [4]{-b x^2+a x^4}\right ) \text {Subst}\left (\int \frac {x^2 \left (-b+a x^4\right )^{5/4}}{b-a x^4+x^8} \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{-b+a x^2}} \\ & = \frac {\left (2 \sqrt [4]{-b x^2+a x^4}\right ) \text {Subst}\left (\int \left (-\frac {2 x^2 \left (-b+a x^4\right )^{5/4}}{\sqrt {a^2-4 b} \left (a+\sqrt {a^2-4 b}-2 x^4\right )}-\frac {2 x^2 \left (-b+a x^4\right )^{5/4}}{\sqrt {a^2-4 b} \left (-a+\sqrt {a^2-4 b}+2 x^4\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt {x} \sqrt [4]{-b+a x^2}} \\ & = -\frac {\left (4 \sqrt [4]{-b x^2+a x^4}\right ) \text {Subst}\left (\int \frac {x^2 \left (-b+a x^4\right )^{5/4}}{a+\sqrt {a^2-4 b}-2 x^4} \, dx,x,\sqrt {x}\right )}{\sqrt {a^2-4 b} \sqrt {x} \sqrt [4]{-b+a x^2}}-\frac {\left (4 \sqrt [4]{-b x^2+a x^4}\right ) \text {Subst}\left (\int \frac {x^2 \left (-b+a x^4\right )^{5/4}}{-a+\sqrt {a^2-4 b}+2 x^4} \, dx,x,\sqrt {x}\right )}{\sqrt {a^2-4 b} \sqrt {x} \sqrt [4]{-b+a x^2}} \\ & = \frac {\left (4 b \sqrt [4]{-b x^2+a x^4}\right ) \text {Subst}\left (\int \frac {x^2 \left (1-\frac {a x^4}{b}\right )^{5/4}}{a+\sqrt {a^2-4 b}-2 x^4} \, dx,x,\sqrt {x}\right )}{\sqrt {a^2-4 b} \sqrt {x} \sqrt [4]{1-\frac {a x^2}{b}}}+\frac {\left (4 b \sqrt [4]{-b x^2+a x^4}\right ) \text {Subst}\left (\int \frac {x^2 \left (1-\frac {a x^4}{b}\right )^{5/4}}{-a+\sqrt {a^2-4 b}+2 x^4} \, dx,x,\sqrt {x}\right )}{\sqrt {a^2-4 b} \sqrt {x} \sqrt [4]{1-\frac {a x^2}{b}}} \\ & = \frac {4 b x \sqrt [4]{-b x^2+a x^4} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {5}{4},\frac {7}{4},\frac {2 x^2}{a-\sqrt {a^2-4 b}},\frac {a x^2}{b}\right )}{3 \left (a^2-a \sqrt {a^2-4 b}-4 b\right ) \sqrt [4]{1-\frac {a x^2}{b}}}+\frac {4 b x \sqrt [4]{-b x^2+a x^4} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {5}{4},\frac {7}{4},\frac {2 x^2}{a+\sqrt {a^2-4 b}},\frac {a x^2}{b}\right )}{3 \left (a^2+a \sqrt {a^2-4 b}-4 b\right ) \sqrt [4]{1-\frac {a x^2}{b}}} \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.18 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b-a x^2+x^4} \, dx=\frac {\sqrt [4]{-b x^2+a x^4} \left (4 a^{5/4} \left (-\arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )+\text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{-b+a x^2}}\right )\right )+\text {RootSum}\left [b-a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-a b \log (x)+2 a b \log \left (\sqrt [4]{-b+a x^2}-\sqrt {x} \text {$\#$1}\right )+a^2 \log (x) \text {$\#$1}^4-b \log (x) \text {$\#$1}^4-2 a^2 \log \left (\sqrt [4]{-b+a x^2}-\sqrt {x} \text {$\#$1}\right ) \text {$\#$1}^4+2 b \log \left (\sqrt [4]{-b+a x^2}-\sqrt {x} \text {$\#$1}\right ) \text {$\#$1}^4}{a \text {$\#$1}^3-2 \text {$\#$1}^7}\&\right ]\right )}{4 \sqrt {x} \sqrt [4]{-b+a x^2}} \]
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Time = 0.26 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.73
method | result | size |
pseudoelliptic | \(a^{\frac {5}{4}} \arctan \left (\frac {\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )+\frac {a^{\frac {5}{4}} \ln \left (\frac {x \,a^{\frac {1}{4}}+\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}{-x \,a^{\frac {1}{4}}+\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}\right )}{2}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-a \,\textit {\_Z}^{4}+b \right )}{\sum }\left (-\frac {\left (\left (a^{2}-b \right ) \textit {\_R}^{4}-a b \right ) \ln \left (\frac {-\textit {\_R} x +\left (x^{2} \left (a \,x^{2}-b \right )\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}^{3} \left (-2 \textit {\_R}^{4}+a \right )}\right )\right )}{2}\) | \(153\) |
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Timed out. \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b-a x^2+x^4} \, dx=\text {Timed out} \]
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Not integrable
Time = 13.39 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.15 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b-a x^2+x^4} \, dx=\int \frac {\sqrt [4]{x^{2} \left (a x^{2} - b\right )} \left (a x^{2} - b\right )}{- a x^{2} + b + x^{4}}\, dx \]
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Not integrable
Time = 0.23 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.18 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b-a x^2+x^4} \, dx=\int { \frac {{\left (a x^{4} - b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{2} - b\right )}}{x^{4} - a x^{2} + b} \,d x } \]
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Result contains higher order function than in optimal. Order 3 vs. order 1.
Time = 7.97 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.87 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b-a x^2+x^4} \, dx=\frac {1}{2} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} a \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) + \frac {1}{2} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} a \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) + \frac {1}{4} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} a \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a - \frac {b}{x^{2}}}\right ) - \frac {1}{4} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} a \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a - \frac {b}{x^{2}}}\right ) \]
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Not integrable
Time = 0.00 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.18 \[ \int \frac {\left (-b+a x^2\right ) \sqrt [4]{-b x^2+a x^4}}{b-a x^2+x^4} \, dx=\int -\frac {\left (b-a\,x^2\right )\,{\left (a\,x^4-b\,x^2\right )}^{1/4}}{x^4-a\,x^2+b} \,d x \]
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