Integrand size = 28, antiderivative size = 24 \[ \int \frac {-1+4 x^5}{\left (1-a x+x^5\right ) \sqrt {x+x^6}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {x+x^6}}\right )}{\sqrt {a}} \]
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\[ \int \frac {-1+4 x^5}{\left (1-a x+x^5\right ) \sqrt {x+x^6}} \, dx=\int \frac {-1+4 x^5}{\left (1-a x+x^5\right ) \sqrt {x+x^6}} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {x} \sqrt {1+x^5}\right ) \int \frac {-1+4 x^5}{\sqrt {x} \sqrt {1+x^5} \left (1-a x+x^5\right )} \, dx}{\sqrt {x+x^6}} \\ & = \frac {\left (2 \sqrt {x} \sqrt {1+x^5}\right ) \text {Subst}\left (\int \frac {-1+4 x^{10}}{\sqrt {1+x^{10}} \left (1-a x^2+x^{10}\right )} \, dx,x,\sqrt {x}\right )}{\sqrt {x+x^6}} \\ & = \frac {\left (2 \sqrt {x} \sqrt {1+x^5}\right ) \text {Subst}\left (\int \left (\frac {4}{\sqrt {1+x^{10}}}-\frac {5-4 a x^2}{\sqrt {1+x^{10}} \left (1-a x^2+x^{10}\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt {x+x^6}} \\ & = -\frac {\left (2 \sqrt {x} \sqrt {1+x^5}\right ) \text {Subst}\left (\int \frac {5-4 a x^2}{\sqrt {1+x^{10}} \left (1-a x^2+x^{10}\right )} \, dx,x,\sqrt {x}\right )}{\sqrt {x+x^6}}+\frac {\left (8 \sqrt {x} \sqrt {1+x^5}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+x^{10}}} \, dx,x,\sqrt {x}\right )}{\sqrt {x+x^6}} \\ & = \frac {8 x \sqrt {1+x^5} \operatorname {Hypergeometric2F1}\left (\frac {1}{10},\frac {1}{2},\frac {11}{10},-x^5\right )}{\sqrt {x+x^6}}-\frac {\left (2 \sqrt {x} \sqrt {1+x^5}\right ) \text {Subst}\left (\int \left (-\frac {5}{\left (-1+a x^2-x^{10}\right ) \sqrt {1+x^{10}}}+\frac {4 a x^2}{\left (-1+a x^2-x^{10}\right ) \sqrt {1+x^{10}}}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt {x+x^6}} \\ & = \frac {8 x \sqrt {1+x^5} \operatorname {Hypergeometric2F1}\left (\frac {1}{10},\frac {1}{2},\frac {11}{10},-x^5\right )}{\sqrt {x+x^6}}+\frac {\left (10 \sqrt {x} \sqrt {1+x^5}\right ) \text {Subst}\left (\int \frac {1}{\left (-1+a x^2-x^{10}\right ) \sqrt {1+x^{10}}} \, dx,x,\sqrt {x}\right )}{\sqrt {x+x^6}}-\frac {\left (8 a \sqrt {x} \sqrt {1+x^5}\right ) \text {Subst}\left (\int \frac {x^2}{\left (-1+a x^2-x^{10}\right ) \sqrt {1+x^{10}}} \, dx,x,\sqrt {x}\right )}{\sqrt {x+x^6}} \\ \end{align*}
\[ \int \frac {-1+4 x^5}{\left (1-a x+x^5\right ) \sqrt {x+x^6}} \, dx=\int \frac {-1+4 x^5}{\left (1-a x+x^5\right ) \sqrt {x+x^6}} \, dx \]
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Time = 1.63 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88
method | result | size |
pseudoelliptic | \(-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {x^{6}+x}}{x \sqrt {a}}\right )}{\sqrt {a}}\) | \(21\) |
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Time = 0.35 (sec) , antiderivative size = 119, normalized size of antiderivative = 4.96 \[ \int \frac {-1+4 x^5}{\left (1-a x+x^5\right ) \sqrt {x+x^6}} \, dx=\left [\frac {\log \left (-\frac {x^{10} + 6 \, a x^{6} + 2 \, x^{5} + a^{2} x^{2} - 4 \, \sqrt {x^{6} + x} {\left (x^{5} + a x + 1\right )} \sqrt {a} + 6 \, a x + 1}{x^{10} - 2 \, a x^{6} + 2 \, x^{5} + a^{2} x^{2} - 2 \, a x + 1}\right )}{2 \, \sqrt {a}}, \frac {\sqrt {-a} \arctan \left (\frac {2 \, \sqrt {x^{6} + x} \sqrt {-a}}{x^{5} + a x + 1}\right )}{a}\right ] \]
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\[ \int \frac {-1+4 x^5}{\left (1-a x+x^5\right ) \sqrt {x+x^6}} \, dx=\int \frac {4 x^{5} - 1}{\sqrt {x \left (x + 1\right ) \left (x^{4} - x^{3} + x^{2} - x + 1\right )} \left (- a x + x^{5} + 1\right )}\, dx \]
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\[ \int \frac {-1+4 x^5}{\left (1-a x+x^5\right ) \sqrt {x+x^6}} \, dx=\int { \frac {4 \, x^{5} - 1}{\sqrt {x^{6} + x} {\left (x^{5} - a x + 1\right )}} \,d x } \]
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\[ \int \frac {-1+4 x^5}{\left (1-a x+x^5\right ) \sqrt {x+x^6}} \, dx=\int { \frac {4 \, x^{5} - 1}{\sqrt {x^{6} + x} {\left (x^{5} - a x + 1\right )}} \,d x } \]
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Timed out. \[ \int \frac {-1+4 x^5}{\left (1-a x+x^5\right ) \sqrt {x+x^6}} \, dx=\int \frac {4\,x^5-1}{\sqrt {x^6+x}\,\left (x^5-a\,x+1\right )} \,d x \]
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