Integrand size = 60, antiderivative size = 181 \[ \int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (b x^2+a \left (q+p x^3\right )^2\right )}{x^5} \, dx=\frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (a q^3+2 b q x^2-a p q^2 x^2+3 a p q^2 x^3+2 b p x^5-a p^2 q x^5+3 a p^2 q x^6+a p^3 x^9\right )}{4 x^4}+\frac {1}{2} \left (2 b p q+a p^2 q^2\right ) \log (x)+\frac {1}{2} \left (-2 b p q-a p^2 q^2\right ) \log \left (q+p x^3+\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}\right ) \]
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\[ \int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (b x^2+a \left (q+p x^3\right )^2\right )}{x^5} \, dx=\int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (b x^2+a \left (q+p x^3\right )^2\right )}{x^5} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (2 b p \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}-\frac {a q^3 \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^5}-\frac {b q \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^3}+3 a p^2 q x \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}+2 a p^3 x^4 \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}\right ) \, dx \\ & = (2 b p) \int \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \, dx+\left (2 a p^3\right ) \int x^4 \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \, dx-(b q) \int \frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^3} \, dx+\left (3 a p^2 q\right ) \int x \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \, dx-\left (a q^3\right ) \int \frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^5} \, dx \\ \end{align*}
Time = 0.33 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.74 \[ \int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (b x^2+a \left (q+p x^3\right )^2\right )}{x^5} \, dx=\frac {\left (q+p x^3\right ) \sqrt {q^2+2 p q (-1+x) x^2+p^2 x^6} \left (2 b x^2+a \left (q^2+p^2 x^6+p q x^2 (-1+2 x)\right )\right )}{4 x^4}+\frac {1}{2} p q (2 b+a p q) \log (x)-\frac {1}{2} p q (2 b+a p q) \log \left (q+p x^3+\sqrt {q^2+2 p q (-1+x) x^2+p^2 x^6}\right ) \]
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Time = 0.23 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.68
method | result | size |
pseudoelliptic | \(\frac {-2 p q \,x^{4} \left (a p q +2 b \right ) \ln \left (\frac {q +p \,x^{3}+\sqrt {p^{2} x^{6}+2 p q \,x^{2} \left (-1+x \right )+q^{2}}}{x}\right )+\left (p \,x^{3}+q \right ) \left (a \,p^{2} x^{6}+2 a p q \,x^{3}+\left (-a p q +2 b \right ) x^{2}+a \,q^{2}\right ) \sqrt {p^{2} x^{6}+2 p q \,x^{2} \left (-1+x \right )+q^{2}}}{4 x^{4}}\) | \(123\) |
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Timed out. \[ \int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (b x^2+a \left (q+p x^3\right )^2\right )}{x^5} \, dx=\text {Timed out} \]
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\[ \int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (b x^2+a \left (q+p x^3\right )^2\right )}{x^5} \, dx=\int \frac {\left (2 p x^{3} - q\right ) \sqrt {p^{2} x^{6} + 2 p q x^{3} - 2 p q x^{2} + q^{2}} \left (a p^{2} x^{6} + 2 a p q x^{3} + a q^{2} + b x^{2}\right )}{x^{5}}\, dx \]
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\[ \int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (b x^2+a \left (q+p x^3\right )^2\right )}{x^5} \, dx=\int { \frac {\sqrt {p^{2} x^{6} + 2 \, p q x^{3} - 2 \, p q x^{2} + q^{2}} {\left (2 \, p x^{3} - q\right )} {\left ({\left (p x^{3} + q\right )}^{2} a + b x^{2}\right )}}{x^{5}} \,d x } \]
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\[ \int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (b x^2+a \left (q+p x^3\right )^2\right )}{x^5} \, dx=\int { \frac {\sqrt {p^{2} x^{6} + 2 \, p q x^{3} - 2 \, p q x^{2} + q^{2}} {\left (2 \, p x^{3} - q\right )} {\left ({\left (p x^{3} + q\right )}^{2} a + b x^{2}\right )}}{x^{5}} \,d x } \]
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Timed out. \[ \int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (b x^2+a \left (q+p x^3\right )^2\right )}{x^5} \, dx=-\int \frac {\left (q-2\,p\,x^3\right )\,\left (a\,{\left (p\,x^3+q\right )}^2+b\,x^2\right )\,\sqrt {p^2\,x^6+2\,p\,q\,x^3-2\,p\,q\,x^2+q^2}}{x^5} \,d x \]
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