Integrand size = 36, antiderivative size = 54 \[ \int \frac {2 b+a x^2}{\sqrt [4]{b+a x^2} \left (-2 b-2 a x^2+x^4\right )} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{b+a x^2}}{x}\right )}{2^{3/4}}-\frac {\text {arctanh}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{b+a x^2}}\right )}{2^{3/4}} \]
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Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.40 (sec) , antiderivative size = 443, normalized size of antiderivative = 8.20, number of steps used = 10, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1706, 408, 504, 1232} \[ \int \frac {2 b+a x^2}{\sqrt [4]{b+a x^2} \left (-2 b-2 a x^2+x^4\right )} \, dx=\frac {\sqrt [4]{b} \left (a-\sqrt {a^2+2 b}\right ) \sqrt {-\frac {a x^2}{b}} \operatorname {EllipticPi}\left (-\frac {\sqrt {b}}{\sqrt {a^2-\sqrt {a^2+2 b} a+b}},\arcsin \left (\frac {\sqrt [4]{a x^2+b}}{\sqrt [4]{b}}\right ),-1\right )}{2 x \sqrt {-a \sqrt {a^2+2 b}+a^2+b}}-\frac {\sqrt [4]{b} \left (a-\sqrt {a^2+2 b}\right ) \sqrt {-\frac {a x^2}{b}} \operatorname {EllipticPi}\left (\frac {\sqrt {b}}{\sqrt {a^2-\sqrt {a^2+2 b} a+b}},\arcsin \left (\frac {\sqrt [4]{a x^2+b}}{\sqrt [4]{b}}\right ),-1\right )}{2 x \sqrt {-a \sqrt {a^2+2 b}+a^2+b}}+\frac {\sqrt [4]{b} \left (\sqrt {a^2+2 b}+a\right ) \sqrt {-\frac {a x^2}{b}} \operatorname {EllipticPi}\left (-\frac {\sqrt {b}}{\sqrt {a^2+\sqrt {a^2+2 b} a+b}},\arcsin \left (\frac {\sqrt [4]{a x^2+b}}{\sqrt [4]{b}}\right ),-1\right )}{2 x \sqrt {a \sqrt {a^2+2 b}+a^2+b}}-\frac {\sqrt [4]{b} \left (\sqrt {a^2+2 b}+a\right ) \sqrt {-\frac {a x^2}{b}} \operatorname {EllipticPi}\left (\frac {\sqrt {b}}{\sqrt {a^2+\sqrt {a^2+2 b} a+b}},\arcsin \left (\frac {\sqrt [4]{a x^2+b}}{\sqrt [4]{b}}\right ),-1\right )}{2 x \sqrt {a \sqrt {a^2+2 b}+a^2+b}} \]
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Rule 408
Rule 504
Rule 1232
Rule 1706
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {a+\sqrt {a^2+2 b}}{\left (-2 a-2 \sqrt {a^2+2 b}+2 x^2\right ) \sqrt [4]{b+a x^2}}+\frac {a-\sqrt {a^2+2 b}}{\left (-2 a+2 \sqrt {a^2+2 b}+2 x^2\right ) \sqrt [4]{b+a x^2}}\right ) \, dx \\ & = \left (a-\sqrt {a^2+2 b}\right ) \int \frac {1}{\left (-2 a+2 \sqrt {a^2+2 b}+2 x^2\right ) \sqrt [4]{b+a x^2}} \, dx+\left (a+\sqrt {a^2+2 b}\right ) \int \frac {1}{\left (-2 a-2 \sqrt {a^2+2 b}+2 x^2\right ) \sqrt [4]{b+a x^2}} \, dx \\ & = \frac {\left (2 \left (a-\sqrt {a^2+2 b}\right ) \sqrt {-\frac {a x^2}{b}}\right ) \text {Subst}\left (\int \frac {x^2}{\left (-2 b+a \left (-2 a+2 \sqrt {a^2+2 b}\right )+2 x^4\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x^2}\right )}{x}+\frac {\left (2 \left (a+\sqrt {a^2+2 b}\right ) \sqrt {-\frac {a x^2}{b}}\right ) \text {Subst}\left (\int \frac {x^2}{\left (-2 b+a \left (-2 a-2 \sqrt {a^2+2 b}\right )+2 x^4\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x^2}\right )}{x} \\ & = -\frac {\left (\left (a-\sqrt {a^2+2 b}\right ) \sqrt {-\frac {a x^2}{b}}\right ) \text {Subst}\left (\int \frac {1}{\left (\sqrt {a^2+b-a \sqrt {a^2+2 b}}-x^2\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x^2}\right )}{2 x}+\frac {\left (\left (a-\sqrt {a^2+2 b}\right ) \sqrt {-\frac {a x^2}{b}}\right ) \text {Subst}\left (\int \frac {1}{\left (\sqrt {a^2+b-a \sqrt {a^2+2 b}}+x^2\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x^2}\right )}{2 x}-\frac {\left (\left (a+\sqrt {a^2+2 b}\right ) \sqrt {-\frac {a x^2}{b}}\right ) \text {Subst}\left (\int \frac {1}{\left (\sqrt {a^2+b+a \sqrt {a^2+2 b}}-x^2\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x^2}\right )}{2 x}+\frac {\left (\left (a+\sqrt {a^2+2 b}\right ) \sqrt {-\frac {a x^2}{b}}\right ) \text {Subst}\left (\int \frac {1}{\left (\sqrt {a^2+b+a \sqrt {a^2+2 b}}+x^2\right ) \sqrt {1-\frac {x^4}{b}}} \, dx,x,\sqrt [4]{b+a x^2}\right )}{2 x} \\ & = \frac {\sqrt [4]{b} \left (a-\sqrt {a^2+2 b}\right ) \sqrt {-\frac {a x^2}{b}} \operatorname {EllipticPi}\left (-\frac {\sqrt {b}}{\sqrt {a^2+b-a \sqrt {a^2+2 b}}},\arcsin \left (\frac {\sqrt [4]{b+a x^2}}{\sqrt [4]{b}}\right ),-1\right )}{2 \sqrt {a^2+b-a \sqrt {a^2+2 b}} x}-\frac {\sqrt [4]{b} \left (a-\sqrt {a^2+2 b}\right ) \sqrt {-\frac {a x^2}{b}} \operatorname {EllipticPi}\left (\frac {\sqrt {b}}{\sqrt {a^2+b-a \sqrt {a^2+2 b}}},\arcsin \left (\frac {\sqrt [4]{b+a x^2}}{\sqrt [4]{b}}\right ),-1\right )}{2 \sqrt {a^2+b-a \sqrt {a^2+2 b}} x}+\frac {\sqrt [4]{b} \left (a+\sqrt {a^2+2 b}\right ) \sqrt {-\frac {a x^2}{b}} \operatorname {EllipticPi}\left (-\frac {\sqrt {b}}{\sqrt {a^2+b+a \sqrt {a^2+2 b}}},\arcsin \left (\frac {\sqrt [4]{b+a x^2}}{\sqrt [4]{b}}\right ),-1\right )}{2 \sqrt {a^2+b+a \sqrt {a^2+2 b}} x}-\frac {\sqrt [4]{b} \left (a+\sqrt {a^2+2 b}\right ) \sqrt {-\frac {a x^2}{b}} \operatorname {EllipticPi}\left (\frac {\sqrt {b}}{\sqrt {a^2+b+a \sqrt {a^2+2 b}}},\arcsin \left (\frac {\sqrt [4]{b+a x^2}}{\sqrt [4]{b}}\right ),-1\right )}{2 \sqrt {a^2+b+a \sqrt {a^2+2 b}} x} \\ \end{align*}
Time = 0.18 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.91 \[ \int \frac {2 b+a x^2}{\sqrt [4]{b+a x^2} \left (-2 b-2 a x^2+x^4\right )} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{b+a x^2}}{x}\right )-\text {arctanh}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{b+a x^2}}\right )}{2^{3/4}} \]
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\[\int \frac {a \,x^{2}+2 b}{\left (a \,x^{2}+b \right )^{\frac {1}{4}} \left (x^{4}-2 a \,x^{2}-2 b \right )}d x\]
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Timed out. \[ \int \frac {2 b+a x^2}{\sqrt [4]{b+a x^2} \left (-2 b-2 a x^2+x^4\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {2 b+a x^2}{\sqrt [4]{b+a x^2} \left (-2 b-2 a x^2+x^4\right )} \, dx=\int \frac {a x^{2} + 2 b}{\sqrt [4]{a x^{2} + b} \left (- 2 a x^{2} - 2 b + x^{4}\right )}\, dx \]
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\[ \int \frac {2 b+a x^2}{\sqrt [4]{b+a x^2} \left (-2 b-2 a x^2+x^4\right )} \, dx=\int { \frac {a x^{2} + 2 \, b}{{\left (x^{4} - 2 \, a x^{2} - 2 \, b\right )} {\left (a x^{2} + b\right )}^{\frac {1}{4}}} \,d x } \]
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\[ \int \frac {2 b+a x^2}{\sqrt [4]{b+a x^2} \left (-2 b-2 a x^2+x^4\right )} \, dx=\int { \frac {a x^{2} + 2 \, b}{{\left (x^{4} - 2 \, a x^{2} - 2 \, b\right )} {\left (a x^{2} + b\right )}^{\frac {1}{4}}} \,d x } \]
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Timed out. \[ \int \frac {2 b+a x^2}{\sqrt [4]{b+a x^2} \left (-2 b-2 a x^2+x^4\right )} \, dx=\int -\frac {a\,x^2+2\,b}{{\left (a\,x^2+b\right )}^{1/4}\,\left (-x^4+2\,a\,x^2+2\,b\right )} \,d x \]
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