Integrand size = 59, antiderivative size = 252 \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (b x^8+a \left (q+p x^3\right )^4\right )}{x^{13}} \, dx=\frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (2 a q^5+10 a p q^4 x^3-a p q^4 x^4+20 a p^2 q^3 x^6-3 a p^2 q^3 x^7+6 b q x^8-3 a p^2 q^3 x^8+20 a p^3 q^2 x^9-3 a p^3 q^2 x^{10}+6 b p x^{11}-3 a p^3 q^2 x^{11}+10 a p^4 q x^{12}-a p^4 q x^{13}+2 a p^5 x^{15}\right )}{12 x^{12}}+\left (2 b p q+a p^3 q^3\right ) \log (x)+\frac {1}{2} \left (-2 b p q-a p^3 q^3\right ) \log \left (q+p x^3+\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}\right ) \]
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\[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (b x^8+a \left (q+p x^3\right )^4\right )}{x^{13}} \, dx=\int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (b x^8+a \left (q+p x^3\right )^4\right )}{x^{13}} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {2 a q^5 \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^{13}}-\frac {7 a p q^4 \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^{10}}-\frac {8 a p^2 q^3 \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^7}-\frac {2 b q \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5}-\frac {2 a p^3 q^2 \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^4}+\frac {b p \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^2}+\frac {2 a p^4 q \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x}+a p^5 x^2 \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}\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^5\right ) \int x^2 \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 (2 a p^4 q\right ) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x} \, dx-\left (2 a p^3 q^2\right ) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^4} \, dx-\left (8 a p^2 q^3\right ) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^7} \, dx-\left (7 a p q^4\right ) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^{10}} \, dx-\left (2 a q^5\right ) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^{13}} \, dx \\ \end{align*}
Time = 0.38 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.69 \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (b x^8+a \left (q+p x^3\right )^4\right )}{x^{13}} \, dx=\frac {\left (q+p x^3\right ) \sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6} \left (6 b x^8+a \left (2 q^4-p q^3 (-8+x) x^3-p^3 q (-8+x) x^9+2 p^4 x^{12}+p^2 q^2 x^6 \left (12-2 x-3 x^2\right )\right )\right )}{12 x^{12}}+\left (2 b p q+a p^3 q^3\right ) \log (x)-\frac {1}{2} p q \left (2 b+a p^2 q^2\right ) \log \left (q+p x^3+\sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}\right ) \]
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Time = 0.31 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.78
method | result | size |
pseudoelliptic | \(\frac {-3 p q \,x^{11} \left (a \,p^{2} q^{2}+2 b \right ) \ln \left (\frac {p \,x^{3}+\sqrt {\frac {p^{2} x^{6}-2 q \,x^{3} \left (-1+x \right ) p +q^{2}}{x^{2}}}\, x +q}{x^{2}}\right )+\left (a \,p^{4} x^{12}-\frac {a \,p^{3} q \,x^{10}}{2}+4 a \,p^{3} q \,x^{9}+3 \left (-\frac {a \,p^{2} q^{2}}{2}+b \right ) x^{8}-a \,p^{2} q^{2} x^{7}+6 a \,p^{2} q^{2} x^{6}-\frac {a p \,q^{3} x^{4}}{2}+4 a p \,q^{3} x^{3}+a \,q^{4}\right ) \sqrt {\frac {p^{2} x^{6}-2 q \,x^{3} \left (-1+x \right ) p +q^{2}}{x^{2}}}\, \left (p \,x^{3}+q \right )}{6 x^{11}}\) | \(196\) |
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Timed out. \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (b x^8+a \left (q+p x^3\right )^4\right )}{x^{13}} \, dx=\text {Timed out} \]
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\[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (b x^8+a \left (q+p x^3\right )^4\right )}{x^{13}} \, dx=\int \frac {\left (p x^{3} - 2 q\right ) \sqrt {p^{2} x^{6} - 2 p q x^{4} + 2 p q x^{3} + q^{2}} \left (a p^{4} x^{12} + 4 a p^{3} q x^{9} + 6 a p^{2} q^{2} x^{6} + 4 a p q^{3} x^{3} + a q^{4} + b x^{8}\right )}{x^{13}}\, dx \]
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\[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (b x^8+a \left (q+p x^3\right )^4\right )}{x^{13}} \, dx=\int { \frac {{\left (b x^{8} + {\left (p x^{3} + q\right )}^{4} a\right )} \sqrt {p^{2} x^{6} - 2 \, p q x^{4} + 2 \, p q x^{3} + q^{2}} {\left (p x^{3} - 2 \, q\right )}}{x^{13}} \,d x } \]
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\[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (b x^8+a \left (q+p x^3\right )^4\right )}{x^{13}} \, dx=\int { \frac {{\left (b x^{8} + {\left (p x^{3} + q\right )}^{4} a\right )} \sqrt {p^{2} x^{6} - 2 \, p q x^{4} + 2 \, p q x^{3} + q^{2}} {\left (p x^{3} - 2 \, q\right )}}{x^{13}} \,d x } \]
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Timed out. \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (b x^8+a \left (q+p x^3\right )^4\right )}{x^{13}} \, dx=\int -\frac {\left (a\,{\left (p\,x^3+q\right )}^4+b\,x^8\right )\,\left (2\,q-p\,x^3\right )\,\sqrt {p^2\,x^6-2\,p\,q\,x^4+2\,p\,q\,x^3+q^2}}{x^{13}} \,d x \]
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