Integrand size = 26, antiderivative size = 87 \[ \int \frac {x^2}{\left (-b+a x^4\right )^{3/4} \left (b+a x^4\right )} \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2\ 2^{3/4} a^{3/4} b}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2\ 2^{3/4} a^{3/4} b} \]
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Time = 0.06 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {508, 304, 209, 212} \[ \int \frac {x^2}{\left (-b+a x^4\right )^{3/4} \left (b+a x^4\right )} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2\ 2^{3/4} a^{3/4} b}-\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2\ 2^{3/4} a^{3/4} b} \]
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Rule 209
Rule 212
Rule 304
Rule 508
Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {x^2}{b-2 a b x^4} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right ) \\ & = \frac {\text {Subst}\left (\int \frac {1}{1-\sqrt {2} \sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )}{2 \sqrt {2} \sqrt {a} b}-\frac {\text {Subst}\left (\int \frac {1}{1+\sqrt {2} \sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{-b+a x^4}}\right )}{2 \sqrt {2} \sqrt {a} b} \\ & = -\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2\ 2^{3/4} a^{3/4} b}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2\ 2^{3/4} a^{3/4} b} \\ \end{align*}
Time = 0.44 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.83 \[ \int \frac {x^2}{\left (-b+a x^4\right )^{3/4} \left (b+a x^4\right )} \, dx=\frac {-\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )+\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2\ 2^{3/4} a^{3/4} b} \]
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Time = 3.91 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00
method | result | size |
pseudoelliptic | \(\frac {2^{\frac {1}{4}} \left (\ln \left (\frac {-2^{\frac {1}{4}} a^{\frac {1}{4}} x -\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{2^{\frac {1}{4}} a^{\frac {1}{4}} x -\left (a \,x^{4}-b \right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {2^{\frac {3}{4}} \left (a \,x^{4}-b \right )^{\frac {1}{4}}}{2 a^{\frac {1}{4}} x}\right )\right )}{8 a^{\frac {3}{4}} b}\) | \(87\) |
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Timed out. \[ \int \frac {x^2}{\left (-b+a x^4\right )^{3/4} \left (b+a x^4\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {x^2}{\left (-b+a x^4\right )^{3/4} \left (b+a x^4\right )} \, dx=\int \frac {x^{2}}{\left (a x^{4} - b\right )^{\frac {3}{4}} \left (a x^{4} + b\right )}\, dx \]
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\[ \int \frac {x^2}{\left (-b+a x^4\right )^{3/4} \left (b+a x^4\right )} \, dx=\int { \frac {x^{2}}{{\left (a x^{4} + b\right )} {\left (a x^{4} - b\right )}^{\frac {3}{4}}} \,d x } \]
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\[ \int \frac {x^2}{\left (-b+a x^4\right )^{3/4} \left (b+a x^4\right )} \, dx=\int { \frac {x^{2}}{{\left (a x^{4} + b\right )} {\left (a x^{4} - b\right )}^{\frac {3}{4}}} \,d x } \]
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Timed out. \[ \int \frac {x^2}{\left (-b+a x^4\right )^{3/4} \left (b+a x^4\right )} \, dx=\int \frac {x^2}{\left (a\,x^4+b\right )\,{\left (a\,x^4-b\right )}^{3/4}} \,d x \]
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