Integrand size = 28, antiderivative size = 16 \[ \int \frac {x \left (2+x^6\right )}{\sqrt {-1+x^6} \left (-1-x^4+x^6\right )} \, dx=-\text {arctanh}\left (\frac {x^2}{\sqrt {-1+x^6}}\right ) \]
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\[ \int \frac {x \left (2+x^6\right )}{\sqrt {-1+x^6} \left (-1-x^4+x^6\right )} \, dx=\int \frac {x \left (2+x^6\right )}{\sqrt {-1+x^6} \left (-1-x^4+x^6\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {2+x^3}{\sqrt {-1+x^3} \left (-1-x^2+x^3\right )} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {1}{\sqrt {-1+x^3}}+\frac {3+x^2}{\sqrt {-1+x^3} \left (-1-x^2+x^3\right )}\right ) \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {1}{\sqrt {-1+x^3}} \, dx,x,x^2\right )+\frac {1}{2} \text {Subst}\left (\int \frac {3+x^2}{\sqrt {-1+x^3} \left (-1-x^2+x^3\right )} \, dx,x,x^2\right ) \\ & = -\frac {\sqrt {2-\sqrt {3}} \left (1-x^2\right ) \sqrt {\frac {1+x^2+x^4}{\left (1-\sqrt {3}-x^2\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1+\sqrt {3}-x^2}{1-\sqrt {3}-x^2}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {-\frac {1-x^2}{\left (1-\sqrt {3}-x^2\right )^2}} \sqrt {-1+x^6}}+\frac {1}{2} \text {Subst}\left (\int \left (\frac {3}{\sqrt {-1+x^3} \left (-1-x^2+x^3\right )}+\frac {x^2}{\sqrt {-1+x^3} \left (-1-x^2+x^3\right )}\right ) \, dx,x,x^2\right ) \\ & = -\frac {\sqrt {2-\sqrt {3}} \left (1-x^2\right ) \sqrt {\frac {1+x^2+x^4}{\left (1-\sqrt {3}-x^2\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1+\sqrt {3}-x^2}{1-\sqrt {3}-x^2}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {-\frac {1-x^2}{\left (1-\sqrt {3}-x^2\right )^2}} \sqrt {-1+x^6}}+\frac {1}{2} \text {Subst}\left (\int \frac {x^2}{\sqrt {-1+x^3} \left (-1-x^2+x^3\right )} \, dx,x,x^2\right )+\frac {3}{2} \text {Subst}\left (\int \frac {1}{\sqrt {-1+x^3} \left (-1-x^2+x^3\right )} \, dx,x,x^2\right ) \\ \end{align*}
Time = 10.80 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {x \left (2+x^6\right )}{\sqrt {-1+x^6} \left (-1-x^4+x^6\right )} \, dx=-\text {arctanh}\left (\frac {x^2}{\sqrt {-1+x^6}}\right ) \]
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Time = 1.46 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94
method | result | size |
pseudoelliptic | \(-\operatorname {arctanh}\left (\frac {\sqrt {x^{6}-1}}{x^{2}}\right )\) | \(15\) |
trager | \(\frac {\ln \left (-\frac {-x^{6}-x^{4}+2 x^{2} \sqrt {x^{6}-1}+1}{x^{6}-x^{4}-1}\right )}{2}\) | \(42\) |
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Leaf count of result is larger than twice the leaf count of optimal. 36 vs. \(2 (14) = 28\).
Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 2.25 \[ \int \frac {x \left (2+x^6\right )}{\sqrt {-1+x^6} \left (-1-x^4+x^6\right )} \, dx=\frac {1}{2} \, \log \left (\frac {x^{6} + x^{4} - 2 \, \sqrt {x^{6} - 1} x^{2} - 1}{x^{6} - x^{4} - 1}\right ) \]
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Timed out. \[ \int \frac {x \left (2+x^6\right )}{\sqrt {-1+x^6} \left (-1-x^4+x^6\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {x \left (2+x^6\right )}{\sqrt {-1+x^6} \left (-1-x^4+x^6\right )} \, dx=\int { \frac {{\left (x^{6} + 2\right )} x}{{\left (x^{6} - x^{4} - 1\right )} \sqrt {x^{6} - 1}} \,d x } \]
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\[ \int \frac {x \left (2+x^6\right )}{\sqrt {-1+x^6} \left (-1-x^4+x^6\right )} \, dx=\int { \frac {{\left (x^{6} + 2\right )} x}{{\left (x^{6} - x^{4} - 1\right )} \sqrt {x^{6} - 1}} \,d x } \]
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Timed out. \[ \int \frac {x \left (2+x^6\right )}{\sqrt {-1+x^6} \left (-1-x^4+x^6\right )} \, dx=\int -\frac {x\,\left (x^6+2\right )}{\sqrt {x^6-1}\,\left (-x^6+x^4+1\right )} \,d x \]
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