Integrand size = 34, antiderivative size = 16 \[ \int \frac {-2 b+3 a x^5}{\sqrt {b+a x^5} \left (b+x^2+a x^5\right )} \, dx=-2 \arctan \left (\frac {x}{\sqrt {b+a x^5}}\right ) \]
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\[ \int \frac {-2 b+3 a x^5}{\sqrt {b+a x^5} \left (b+x^2+a x^5\right )} \, dx=\int \frac {-2 b+3 a x^5}{\sqrt {b+a x^5} \left (b+x^2+a x^5\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {3}{\sqrt {b+a x^5}}-\frac {5 b+3 x^2}{\sqrt {b+a x^5} \left (b+x^2+a x^5\right )}\right ) \, dx \\ & = 3 \int \frac {1}{\sqrt {b+a x^5}} \, dx-\int \frac {5 b+3 x^2}{\sqrt {b+a x^5} \left (b+x^2+a x^5\right )} \, dx \\ & = \frac {\left (3 \sqrt {1+\frac {a x^5}{b}}\right ) \int \frac {1}{\sqrt {1+\frac {a x^5}{b}}} \, dx}{\sqrt {b+a x^5}}-\int \left (\frac {5 b}{\sqrt {b+a x^5} \left (b+x^2+a x^5\right )}+\frac {3 x^2}{\sqrt {b+a x^5} \left (b+x^2+a x^5\right )}\right ) \, dx \\ & = \frac {3 x \sqrt {1+\frac {a x^5}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{5},\frac {1}{2},\frac {6}{5},-\frac {a x^5}{b}\right )}{\sqrt {b+a x^5}}-3 \int \frac {x^2}{\sqrt {b+a x^5} \left (b+x^2+a x^5\right )} \, dx-(5 b) \int \frac {1}{\sqrt {b+a x^5} \left (b+x^2+a x^5\right )} \, dx \\ \end{align*}
Time = 2.75 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {-2 b+3 a x^5}{\sqrt {b+a x^5} \left (b+x^2+a x^5\right )} \, dx=-2 \arctan \left (\frac {x}{\sqrt {b+a x^5}}\right ) \]
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Time = 1.72 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06
method | result | size |
pseudoelliptic | \(2 \arctan \left (\frac {\sqrt {a \,x^{5}+b}}{x}\right )\) | \(17\) |
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Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (14) = 28\).
Time = 0.29 (sec) , antiderivative size = 35, normalized size of antiderivative = 2.19 \[ \int \frac {-2 b+3 a x^5}{\sqrt {b+a x^5} \left (b+x^2+a x^5\right )} \, dx=\arctan \left (\frac {{\left (a x^{5} - x^{2} + b\right )} \sqrt {a x^{5} + b}}{2 \, {\left (a x^{6} + b x\right )}}\right ) \]
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\[ \int \frac {-2 b+3 a x^5}{\sqrt {b+a x^5} \left (b+x^2+a x^5\right )} \, dx=\int \frac {3 a x^{5} - 2 b}{\sqrt {a x^{5} + b} \left (a x^{5} + b + x^{2}\right )}\, dx \]
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\[ \int \frac {-2 b+3 a x^5}{\sqrt {b+a x^5} \left (b+x^2+a x^5\right )} \, dx=\int { \frac {3 \, a x^{5} - 2 \, b}{{\left (a x^{5} + x^{2} + b\right )} \sqrt {a x^{5} + b}} \,d x } \]
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\[ \int \frac {-2 b+3 a x^5}{\sqrt {b+a x^5} \left (b+x^2+a x^5\right )} \, dx=\int { \frac {3 \, a x^{5} - 2 \, b}{{\left (a x^{5} + x^{2} + b\right )} \sqrt {a x^{5} + b}} \,d x } \]
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Time = 7.13 (sec) , antiderivative size = 42, normalized size of antiderivative = 2.62 \[ \int \frac {-2 b+3 a x^5}{\sqrt {b+a x^5} \left (b+x^2+a x^5\right )} \, dx=\ln \left (\frac {b+a\,x^5-x^2+x\,\sqrt {a\,x^5+b}\,2{}\mathrm {i}}{a\,x^5+x^2+b}\right )\,1{}\mathrm {i} \]
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