Integrand size = 69, antiderivative size = 178 \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (a q^2+2 a p q x^3+b x^4+a p^2 x^6\right )}{x^9} \, dx=\frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (a q^3+3 a p q^2 x^3+2 b q x^4-a p q^2 x^4+3 a p^2 q x^6+2 b p x^7-a p^2 q x^7+a p^3 x^9\right )}{4 x^8}+\left (2 b p q+a p^2 q^2\right ) \log (x)+\frac {1}{2} \left (-2 b p q-a p^2 q^2\right ) \log \left (q+p x^3+\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}\right ) \]
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\[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (a q^2+2 a p q x^3+b x^4+a p^2 x^6\right )}{x^9} \, dx=\int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (a q^2+2 a p q x^3+b x^4+a p^2 x^6\right )}{x^9} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (a p^3 \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}-\frac {2 a q^3 \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^9}-\frac {3 a p q^2 \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^6}-\frac {2 b q \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5}+\frac {b p \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^2}\right ) \, dx \\ & = (b p) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^2} \, dx+\left (a p^3\right ) \int \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \, dx-(2 b q) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5} \, dx-\left (3 a p q^2\right ) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^6} \, dx-\left (2 a q^3\right ) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^9} \, dx \\ \end{align*}
Time = 0.38 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.73 \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (a q^2+2 a p q x^3+b x^4+a p^2 x^6\right )}{x^9} \, dx=\frac {\left (q+p x^3\right ) \sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6} \left (2 b x^4+a \left (q^2-p q (-2+x) x^3+p^2 x^6\right )\right )}{4 x^8}+p q (2 b+a p q) \log (x)-\frac {1}{2} p q (2 b+a p q) \log \left (q+p x^3+\sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}\right ) \]
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Time = 0.24 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.75
method | result | size |
pseudoelliptic | \(\frac {-2 p q \,x^{7} \left (a p q +2 b \right ) \ln \left (\frac {p \,x^{3}+\sqrt {\frac {p^{2} x^{6}-2 p q \,x^{3} \left (-1+x \right )+q^{2}}{x^{2}}}\, x +q}{x^{2}}\right )+\left (p \,x^{3}+q \right ) \left (a \,p^{2} x^{6}+\left (-a p q +2 b \right ) x^{4}+2 a p q \,x^{3}+a \,q^{2}\right ) \sqrt {\frac {p^{2} x^{6}-2 p q \,x^{3} \left (-1+x \right )+q^{2}}{x^{2}}}}{4 x^{7}}\) | \(133\) |
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Timed out. \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (a q^2+2 a p q x^3+b x^4+a p^2 x^6\right )}{x^9} \, dx=\text {Timed out} \]
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\[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (a q^2+2 a p q x^3+b x^4+a p^2 x^6\right )}{x^9} \, dx=\int \frac {\left (p x^{3} - 2 q\right ) \sqrt {p^{2} x^{6} - 2 p q x^{4} + 2 p q x^{3} + q^{2}} \left (a p^{2} x^{6} + 2 a p q x^{3} + a q^{2} + b x^{4}\right )}{x^{9}}\, dx \]
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\[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (a q^2+2 a p q x^3+b x^4+a p^2 x^6\right )}{x^9} \, dx=\int { \frac {{\left (a p^{2} x^{6} + 2 \, a p q x^{3} + b x^{4} + a q^{2}\right )} \sqrt {p^{2} x^{6} - 2 \, p q x^{4} + 2 \, p q x^{3} + q^{2}} {\left (p x^{3} - 2 \, q\right )}}{x^{9}} \,d x } \]
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\[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (a q^2+2 a p q x^3+b x^4+a p^2 x^6\right )}{x^9} \, dx=\int { \frac {{\left (a p^{2} x^{6} + 2 \, a p q x^{3} + b x^{4} + a q^{2}\right )} \sqrt {p^{2} x^{6} - 2 \, p q x^{4} + 2 \, p q x^{3} + q^{2}} {\left (p x^{3} - 2 \, q\right )}}{x^{9}} \,d x } \]
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Timed out. \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (a q^2+2 a p q x^3+b x^4+a p^2 x^6\right )}{x^9} \, dx=\int -\frac {\left (2\,q-p\,x^3\right )\,\sqrt {p^2\,x^6-2\,p\,q\,x^4+2\,p\,q\,x^3+q^2}\,\left (a\,p^2\,x^6+2\,a\,p\,q\,x^3+a\,q^2+b\,x^4\right )}{x^9} \,d x \]
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