Integrand size = 42, antiderivative size = 56 \[ \int \frac {\sqrt {q+p x^5} \left (-2 q+3 p x^5\right )}{x^2 \left (a q+b x^2+a p x^5\right )} \, dx=\frac {2 \sqrt {q+p x^5}}{a x}+\frac {2 \sqrt {b} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a} \sqrt {q+p x^5}}\right )}{a^{3/2}} \]
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\[ \int \frac {\sqrt {q+p x^5} \left (-2 q+3 p x^5\right )}{x^2 \left (a q+b x^2+a p x^5\right )} \, dx=\int \frac {\sqrt {q+p x^5} \left (-2 q+3 p x^5\right )}{x^2 \left (a q+b x^2+a p x^5\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {2 \sqrt {q+p x^5}}{a x^2}+\frac {\left (2 b+5 a p x^3\right ) \sqrt {q+p x^5}}{a \left (a q+b x^2+a p x^5\right )}\right ) \, dx \\ & = \frac {\int \frac {\left (2 b+5 a p x^3\right ) \sqrt {q+p x^5}}{a q+b x^2+a p x^5} \, dx}{a}-\frac {2 \int \frac {\sqrt {q+p x^5}}{x^2} \, dx}{a} \\ & = \frac {2 \sqrt {q+p x^5}}{a x}+\frac {\int \left (\frac {2 b \sqrt {q+p x^5}}{a q+b x^2+a p x^5}+\frac {5 a p x^3 \sqrt {q+p x^5}}{a q+b x^2+a p x^5}\right ) \, dx}{a}-\frac {(5 p) \int \frac {x^3}{\sqrt {q+p x^5}} \, dx}{a} \\ & = \frac {2 \sqrt {q+p x^5}}{a x}+\frac {(2 b) \int \frac {\sqrt {q+p x^5}}{a q+b x^2+a p x^5} \, dx}{a}+(5 p) \int \frac {x^3 \sqrt {q+p x^5}}{a q+b x^2+a p x^5} \, dx-\frac {\left (5 p \sqrt {1+\frac {p x^5}{q}}\right ) \int \frac {x^3}{\sqrt {1+\frac {p x^5}{q}}} \, dx}{a \sqrt {q+p x^5}} \\ & = \frac {2 \sqrt {q+p x^5}}{a x}-\frac {5 p x^4 \sqrt {1+\frac {p x^5}{q}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4}{5},\frac {9}{5},-\frac {p x^5}{q}\right )}{4 a \sqrt {q+p x^5}}+\frac {(2 b) \int \frac {\sqrt {q+p x^5}}{a q+b x^2+a p x^5} \, dx}{a}+(5 p) \int \frac {x^3 \sqrt {q+p x^5}}{a q+b x^2+a p x^5} \, dx \\ \end{align*}
Time = 0.81 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {q+p x^5} \left (-2 q+3 p x^5\right )}{x^2 \left (a q+b x^2+a p x^5\right )} \, dx=\frac {2 \sqrt {q+p x^5}}{a x}+\frac {2 \sqrt {b} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a} \sqrt {q+p x^5}}\right )}{a^{3/2}} \]
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Time = 2.70 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.96
method | result | size |
pseudoelliptic | \(-\frac {2 \left (b \arctan \left (\frac {\sqrt {p \,x^{5}+q}\, a}{x \sqrt {a b}}\right ) x -\sqrt {p \,x^{5}+q}\, \sqrt {a b}\right )}{a x \sqrt {a b}}\) | \(54\) |
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Timed out. \[ \int \frac {\sqrt {q+p x^5} \left (-2 q+3 p x^5\right )}{x^2 \left (a q+b x^2+a p x^5\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {\sqrt {q+p x^5} \left (-2 q+3 p x^5\right )}{x^2 \left (a q+b x^2+a p x^5\right )} \, dx=\int \frac {\sqrt {p x^{5} + q} \left (3 p x^{5} - 2 q\right )}{x^{2} \left (a p x^{5} + a q + b x^{2}\right )}\, dx \]
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\[ \int \frac {\sqrt {q+p x^5} \left (-2 q+3 p x^5\right )}{x^2 \left (a q+b x^2+a p x^5\right )} \, dx=\int { \frac {{\left (3 \, p x^{5} - 2 \, q\right )} \sqrt {p x^{5} + q}}{{\left (a p x^{5} + b x^{2} + a q\right )} x^{2}} \,d x } \]
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\[ \int \frac {\sqrt {q+p x^5} \left (-2 q+3 p x^5\right )}{x^2 \left (a q+b x^2+a p x^5\right )} \, dx=\int { \frac {{\left (3 \, p x^{5} - 2 \, q\right )} \sqrt {p x^{5} + q}}{{\left (a p x^{5} + b x^{2} + a q\right )} x^{2}} \,d x } \]
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Time = 12.58 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.82 \[ \int \frac {\sqrt {q+p x^5} \left (-2 q+3 p x^5\right )}{x^2 \left (a q+b x^2+a p x^5\right )} \, dx=\frac {2\,\sqrt {p\,x^5+q}}{a\,x}+\frac {\sqrt {b}\,\ln \left (\frac {a^5\,b\,p^4\,x^2-a^6\,p^4\,\left (p\,x^5+q\right )+a^{11/2}\,\sqrt {b}\,p^4\,x\,\sqrt {p\,x^5+q}\,2{}\mathrm {i}}{4\,b^2\,q\,x^2+4\,a\,b\,q\,\left (p\,x^5+q\right )}\right )\,1{}\mathrm {i}}{a^{3/2}} \]
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