Integrand size = 34, antiderivative size = 40 \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+2 x^2+x^4}}{\left (1+x^2\right ) \left (1+x^4\right )} \, dx=-\frac {\sqrt {\left (1+x^2\right )^2} \text {arctanh}\left (\frac {\sqrt {2} x}{1+x^2}\right )}{\sqrt {2} \left (1+x^2\right )} \]
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Result contains complex when optimal does not.
Time = 0.37 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.85, number of steps used = 12, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.206, Rules used = {6857, 1161, 8, 1162, 396, 212, 209} \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+2 x^2+x^4}}{\left (1+x^2\right ) \left (1+x^4\right )} \, dx=\frac {\sqrt {x^4+2 x^2+1} \text {arctanh}\left ((-1)^{3/4} x\right )}{\sqrt {2} \left (x^2+1\right )}+\frac {i \sqrt {x^4+2 x^2+1} \arctan \left ((-1)^{3/4} x\right )}{\sqrt {2} \left (x^2+1\right )} \]
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Rule 8
Rule 209
Rule 212
Rule 396
Rule 1161
Rule 1162
Rule 6857
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {\sqrt {1+2 x^2+x^4}}{-1-x^2}+\frac {x^2 \sqrt {1+2 x^2+x^4}}{1+x^4}\right ) \, dx \\ & = \int \frac {\sqrt {1+2 x^2+x^4}}{-1-x^2} \, dx+\int \frac {x^2 \sqrt {1+2 x^2+x^4}}{1+x^4} \, dx \\ & = \frac {\sqrt {1+2 x^2+x^4} \int 1 \, dx}{-1-x^2}+\int \left (-\frac {\sqrt {1+2 x^2+x^4}}{2 \left (i-x^2\right )}+\frac {\sqrt {1+2 x^2+x^4}}{2 \left (i+x^2\right )}\right ) \, dx \\ & = -\frac {x \sqrt {1+2 x^2+x^4}}{1+x^2}-\frac {1}{2} \int \frac {\sqrt {1+2 x^2+x^4}}{i-x^2} \, dx+\frac {1}{2} \int \frac {\sqrt {1+2 x^2+x^4}}{i+x^2} \, dx \\ & = -\frac {x \sqrt {1+2 x^2+x^4}}{1+x^2}-\frac {\sqrt {1+2 x^2+x^4} \int \frac {1+x^2}{i-x^2} \, dx}{2 \left (1+x^2\right )}+\frac {\sqrt {1+2 x^2+x^4} \int \frac {1+x^2}{i+x^2} \, dx}{2 \left (1+x^2\right )} \\ & = \frac {\left (\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {1+2 x^2+x^4}\right ) \int \frac {1}{i+x^2} \, dx}{1+x^2}-\frac {\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {1+2 x^2+x^4}\right ) \int \frac {1}{i-x^2} \, dx}{1+x^2} \\ & = \frac {i \sqrt {1+2 x^2+x^4} \arctan \left ((-1)^{3/4} x\right )}{\sqrt {2} \left (1+x^2\right )}+\frac {\sqrt {1+2 x^2+x^4} \text {arctanh}\left ((-1)^{3/4} x\right )}{\sqrt {2} \left (1+x^2\right )} \\ \end{align*}
Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.65 \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+2 x^2+x^4}}{\left (1+x^2\right ) \left (1+x^4\right )} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {\left (1+x^2\right )^2}}\right )}{\sqrt {2}} \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.06 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.75
method | result | size |
pseudoelliptic | \(-\frac {\operatorname {csgn}\left (\frac {x^{2}+1}{x}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (x^{2}+1\right ) \sqrt {2}}{2 x}\right )}{2}\) | \(30\) |
risch | \(\frac {\sqrt {\left (x^{2}+1\right )^{2}}\, \sqrt {2}\, \ln \left (x^{2}-x \sqrt {2}+1\right )}{4 x^{2}+4}-\frac {\sqrt {\left (x^{2}+1\right )^{2}}\, \sqrt {2}\, \ln \left (x^{2}+x \sqrt {2}+1\right )}{4 \left (x^{2}+1\right )}\) | \(67\) |
default | \(-\frac {\sqrt {\left (x^{2}+1\right )^{2}}\, \sqrt {2}\, \left (\ln \left (-\frac {x^{2}+x \sqrt {2}+1}{x \sqrt {2}-x^{2}-1}\right )-\ln \left (-\frac {x \sqrt {2}-x^{2}-1}{x^{2}+x \sqrt {2}+1}\right )\right )}{8 \left (x^{2}+1\right )}\) | \(79\) |
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Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.85 \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+2 x^2+x^4}}{\left (1+x^2\right ) \left (1+x^4\right )} \, dx=\frac {1}{4} \, \sqrt {2} \log \left (\frac {x^{4} + 4 \, x^{2} - 2 \, \sqrt {2} {\left (x^{3} + x\right )} + 1}{x^{4} + 1}\right ) \]
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\[ \int \frac {\left (-1+x^2\right ) \sqrt {1+2 x^2+x^4}}{\left (1+x^2\right ) \left (1+x^4\right )} \, dx=\int \frac {\left (x - 1\right ) \left (x + 1\right ) \sqrt {\left (x^{2} + 1\right )^{2}}}{\left (x^{2} + 1\right ) \left (x^{4} + 1\right )}\, dx \]
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Time = 0.29 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.85 \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+2 x^2+x^4}}{\left (1+x^2\right ) \left (1+x^4\right )} \, dx=-\frac {1}{4} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) \]
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Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.85 \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+2 x^2+x^4}}{\left (1+x^2\right ) \left (1+x^4\right )} \, dx=-\frac {1}{4} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) \]
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Time = 0.10 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.45 \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+2 x^2+x^4}}{\left (1+x^2\right ) \left (1+x^4\right )} \, dx=-\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,x}{x^2+1}\right )}{2} \]
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