Integrand size = 23, antiderivative size = 66 \[ \int \frac {1+2 x^2}{1-b x^2+4 x^4} \, dx=-\frac {\arctan \left (\frac {\sqrt {4+b}-4 x}{\sqrt {4-b}}\right )}{\sqrt {4-b}}+\frac {\arctan \left (\frac {\sqrt {4+b}+4 x}{\sqrt {4-b}}\right )}{\sqrt {4-b}} \]
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Time = 0.04 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {1175, 632, 210} \[ \int \frac {1+2 x^2}{1-b x^2+4 x^4} \, dx=\frac {\arctan \left (\frac {\sqrt {b+4}+4 x}{\sqrt {4-b}}\right )}{\sqrt {4-b}}-\frac {\arctan \left (\frac {\sqrt {b+4}-4 x}{\sqrt {4-b}}\right )}{\sqrt {4-b}} \]
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Rule 210
Rule 632
Rule 1175
Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \int \frac {1}{\frac {1}{2}-\frac {1}{2} \sqrt {4+b} x+x^2} \, dx+\frac {1}{4} \int \frac {1}{\frac {1}{2}+\frac {1}{2} \sqrt {4+b} x+x^2} \, dx \\ & = -\left (\frac {1}{2} \text {Subst}\left (\int \frac {1}{\frac {1}{4} (-4+b)-x^2} \, dx,x,-\frac {\sqrt {4+b}}{2}+2 x\right )\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1}{\frac {1}{4} (-4+b)-x^2} \, dx,x,\frac {\sqrt {4+b}}{2}+2 x\right ) \\ & = -\frac {\tan ^{-1}\left (\frac {\sqrt {4+b}-4 x}{\sqrt {4-b}}\right )}{\sqrt {4-b}}+\frac {\tan ^{-1}\left (\frac {\sqrt {4+b}+4 x}{\sqrt {4-b}}\right )}{\sqrt {4-b}} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(134\) vs. \(2(66)=132\).
Time = 0.05 (sec) , antiderivative size = 134, normalized size of antiderivative = 2.03 \[ \int \frac {1+2 x^2}{1-b x^2+4 x^4} \, dx=\frac {\frac {\left (4+b+\sqrt {-16+b^2}\right ) \arctan \left (\frac {2 \sqrt {2} x}{\sqrt {-b-\sqrt {-16+b^2}}}\right )}{\sqrt {-b-\sqrt {-16+b^2}}}+\frac {\left (-4-b+\sqrt {-16+b^2}\right ) \arctan \left (\frac {2 \sqrt {2} x}{\sqrt {-b+\sqrt {-16+b^2}}}\right )}{\sqrt {-b+\sqrt {-16+b^2}}}}{\sqrt {2} \sqrt {-16+b^2}} \]
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Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\frac {\ln \left (2 x^{2} \sqrt {b -4}+\left (4-b \right ) x -\sqrt {b -4}\right )}{2 \sqrt {b -4}}-\frac {\ln \left (2 x^{2} \sqrt {b -4}+x \left (b -4\right )-\sqrt {b -4}\right )}{2 \sqrt {b -4}}\) | \(66\) |
default | \(\frac {\left (4+\sqrt {\left (b -4\right ) \left (4+b \right )}+b \right ) \arctan \left (\frac {4 x}{\sqrt {-2 \sqrt {\left (b -4\right ) \left (4+b \right )}-2 b}}\right )}{\sqrt {\left (b -4\right ) \left (4+b \right )}\, \sqrt {-2 \sqrt {\left (b -4\right ) \left (4+b \right )}-2 b}}+\frac {\left (-4+\sqrt {\left (b -4\right ) \left (4+b \right )}-b \right ) \arctan \left (\frac {4 x}{\sqrt {2 \sqrt {\left (b -4\right ) \left (4+b \right )}-2 b}}\right )}{\sqrt {\left (b -4\right ) \left (4+b \right )}\, \sqrt {2 \sqrt {\left (b -4\right ) \left (4+b \right )}-2 b}}\) | \(124\) |
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Time = 0.25 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.82 \[ \int \frac {1+2 x^2}{1-b x^2+4 x^4} \, dx=\left [\frac {\log \left (\frac {4 \, x^{4} + {\left (b - 8\right )} x^{2} - 2 \, {\left (2 \, x^{3} - x\right )} \sqrt {b - 4} + 1}{4 \, x^{4} - b x^{2} + 1}\right )}{2 \, \sqrt {b - 4}}, \frac {\sqrt {-b + 4} \arctan \left (\frac {{\left (4 \, x^{3} - {\left (b - 2\right )} x\right )} \sqrt {-b + 4}}{b - 4}\right ) + \sqrt {-b + 4} \arctan \left (\frac {2 \, \sqrt {-b + 4} x}{b - 4}\right )}{b - 4}\right ] \]
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Time = 0.22 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.26 \[ \int \frac {1+2 x^2}{1-b x^2+4 x^4} \, dx=\frac {\sqrt {\frac {1}{b - 4}} \log {\left (x^{2} + x \left (- \frac {b \sqrt {\frac {1}{b - 4}}}{2} + 2 \sqrt {\frac {1}{b - 4}}\right ) - \frac {1}{2} \right )}}{2} - \frac {\sqrt {\frac {1}{b - 4}} \log {\left (x^{2} + x \left (\frac {b \sqrt {\frac {1}{b - 4}}}{2} - 2 \sqrt {\frac {1}{b - 4}}\right ) - \frac {1}{2} \right )}}{2} \]
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\[ \int \frac {1+2 x^2}{1-b x^2+4 x^4} \, dx=\int { \frac {2 \, x^{2} + 1}{4 \, x^{4} - b x^{2} + 1} \,d x } \]
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\[ \int \frac {1+2 x^2}{1-b x^2+4 x^4} \, dx=\int { \frac {2 \, x^{2} + 1}{4 \, x^{4} - b x^{2} + 1} \,d x } \]
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Time = 13.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.36 \[ \int \frac {1+2 x^2}{1-b x^2+4 x^4} \, dx=-\frac {\mathrm {atanh}\left (\frac {x\,\sqrt {b-4}}{2\,x^2-1}\right )}{\sqrt {b-4}} \]
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