Integrand size = 24, antiderivative size = 185 \[ \int \frac {1+x^8}{\sqrt [4]{x^2+x^6} \left (-1+x^8\right )} \, dx=-\frac {\left (x^2+x^6\right )^{3/4}}{x \left (1+x^4\right )}-\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2+x^6}}\right )}{4 \sqrt [4]{2}}+\frac {\arctan \left (\frac {2^{3/4} x \sqrt [4]{x^2+x^6}}{\sqrt {2} x^2-\sqrt {x^2+x^6}}\right )}{4\ 2^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2+x^6}}\right )}{4 \sqrt [4]{2}}-\frac {\text {arctanh}\left (\frac {\frac {x^2}{\sqrt [4]{2}}+\frac {\sqrt {x^2+x^6}}{2^{3/4}}}{x \sqrt [4]{x^2+x^6}}\right )}{4\ 2^{3/4}} \]
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Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 0.26 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.44, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2081, 6847, 6857, 251, 1418, 440} \[ \int \frac {1+x^8}{\sqrt [4]{x^2+x^6} \left (-1+x^8\right )} \, dx=\frac {2 x \sqrt [4]{x^4+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},-x^4\right )}{\sqrt [4]{x^6+x^2}}-\frac {4 x \sqrt [4]{x^4+1} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {5}{4},\frac {9}{8},x^4,-x^4\right )}{\sqrt [4]{x^6+x^2}} \]
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Rule 251
Rule 440
Rule 1418
Rule 2081
Rule 6847
Rule 6857
Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {x} \sqrt [4]{1+x^4}\right ) \int \frac {1+x^8}{\sqrt {x} \sqrt [4]{1+x^4} \left (-1+x^8\right )} \, dx}{\sqrt [4]{x^2+x^6}} \\ & = \frac {\left (2 \sqrt {x} \sqrt [4]{1+x^4}\right ) \text {Subst}\left (\int \frac {1+x^{16}}{\sqrt [4]{1+x^8} \left (-1+x^{16}\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^6}} \\ & = \frac {\left (2 \sqrt {x} \sqrt [4]{1+x^4}\right ) \text {Subst}\left (\int \left (\frac {1}{\sqrt [4]{1+x^8}}+\frac {2}{\sqrt [4]{1+x^8} \left (-1+x^{16}\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^6}} \\ & = \frac {\left (2 \sqrt {x} \sqrt [4]{1+x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{1+x^8}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^6}}+\frac {\left (4 \sqrt {x} \sqrt [4]{1+x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{1+x^8} \left (-1+x^{16}\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^6}} \\ & = \frac {2 x \sqrt [4]{1+x^4} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},-x^4\right )}{\sqrt [4]{x^2+x^6}}+\frac {\left (4 \sqrt {x} \sqrt [4]{1+x^4}\right ) \text {Subst}\left (\int \frac {1}{\left (-1+x^8\right ) \left (1+x^8\right )^{5/4}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^6}} \\ & = -\frac {4 x \sqrt [4]{1+x^4} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {5}{4},\frac {9}{8},x^4,-x^4\right )}{\sqrt [4]{x^2+x^6}}+\frac {2 x \sqrt [4]{1+x^4} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},-x^4\right )}{\sqrt [4]{x^2+x^6}} \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.17 \[ \int \frac {1+x^8}{\sqrt [4]{x^2+x^6} \left (-1+x^8\right )} \, dx=-\frac {\sqrt {x} \left (8 \sqrt {x}+2^{3/4} \sqrt [4]{1+x^4} \arctan \left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^4}}\right )-\sqrt [4]{2} \sqrt [4]{1+x^4} \arctan \left (\frac {2^{3/4} \sqrt {x} \sqrt [4]{1+x^4}}{\sqrt {2} x-\sqrt {1+x^4}}\right )+2^{3/4} \sqrt [4]{1+x^4} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^4}}\right )+\sqrt [4]{2} \sqrt [4]{1+x^4} \text {arctanh}\left (\frac {2 \sqrt [4]{2} \sqrt {x} \sqrt [4]{1+x^4}}{2 x+\sqrt {2} \sqrt {1+x^4}}\right )\right )}{8 \sqrt [4]{x^2+x^6}} \]
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Time = 53.67 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.52
method | result | size |
pseudoelliptic | \(\frac {2 \arctan \left (\frac {\left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}} 2^{\frac {3}{4}}}{2 x}\right ) 2^{\frac {3}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}-\ln \left (\frac {-2^{\frac {1}{4}} x -\left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}}{2^{\frac {1}{4}} x -\left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}}\right ) 2^{\frac {3}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}+\ln \left (\frac {-2^{\frac {3}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}} x +\sqrt {2}\, x^{2}+\sqrt {x^{2} \left (x^{4}+1\right )}}{2^{\frac {3}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}} x +\sqrt {2}\, x^{2}+\sqrt {x^{2} \left (x^{4}+1\right )}}\right ) 2^{\frac {1}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}+2 \arctan \left (\frac {2^{\frac {1}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}+x}{x}\right ) 2^{\frac {1}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}+2 \arctan \left (\frac {2^{\frac {1}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}-x}{x}\right ) 2^{\frac {1}{4}} \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}-16 x}{16 \left (x^{2} \left (x^{4}+1\right )\right )^{\frac {1}{4}}}\) | \(281\) |
risch | \(\text {Expression too large to display}\) | \(644\) |
trager | \(\text {Expression too large to display}\) | \(657\) |
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Result contains complex when optimal does not.
Time = 31.70 (sec) , antiderivative size = 733, normalized size of antiderivative = 3.96 \[ \int \frac {1+x^8}{\sqrt [4]{x^2+x^6} \left (-1+x^8\right )} \, dx=\text {Too large to display} \]
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\[ \int \frac {1+x^8}{\sqrt [4]{x^2+x^6} \left (-1+x^8\right )} \, dx=\int \frac {x^{8} + 1}{\sqrt [4]{x^{2} \left (x^{4} + 1\right )} \left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right ) \left (x^{4} + 1\right )}\, dx \]
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\[ \int \frac {1+x^8}{\sqrt [4]{x^2+x^6} \left (-1+x^8\right )} \, dx=\int { \frac {x^{8} + 1}{{\left (x^{8} - 1\right )} {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}}} \,d x } \]
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\[ \int \frac {1+x^8}{\sqrt [4]{x^2+x^6} \left (-1+x^8\right )} \, dx=\int { \frac {x^{8} + 1}{{\left (x^{8} - 1\right )} {\left (x^{6} + x^{2}\right )}^{\frac {1}{4}}} \,d x } \]
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Timed out. \[ \int \frac {1+x^8}{\sqrt [4]{x^2+x^6} \left (-1+x^8\right )} \, dx=\int \frac {x^8+1}{{\left (x^6+x^2\right )}^{1/4}\,\left (x^8-1\right )} \,d x \]
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