Integrand size = 60, antiderivative size = 255 \[ \int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (b x^4+a \left (q+p x^3\right )^4\right )}{x^7} \, dx=\frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (2 a q^5-a p q^4 x^2+10 a p q^4 x^3+6 b q x^4-3 a p^2 q^3 x^4-3 a p^2 q^3 x^5+20 a p^2 q^3 x^6+6 b p x^7-3 a p^3 q^2 x^7-3 a p^3 q^2 x^8+20 a p^3 q^2 x^9-a p^4 q x^{11}+10 a p^4 q x^{12}+2 a p^5 x^{15}\right )}{12 x^6}+\frac {1}{2} \left (2 b p q+a p^3 q^3\right ) \log (x)+\frac {1}{2} \left (-2 b p q-a p^3 q^3\right ) \log \left (q+p x^3+\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}\right ) \]
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\[ \int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (b x^4+a \left (q+p x^3\right )^4\right )}{x^7} \, dx=\int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (b x^4+a \left (q+p x^3\right )^4\right )}{x^7} \, dx \]
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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^5 \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^7}-\frac {2 a p q^4 \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^4}-\frac {b q \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^3}+\frac {2 a p^2 q^3 \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x}+8 a p^3 q^2 x^2 \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}+7 a p^4 q x^5 \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}+2 a p^5 x^8 \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^5\right ) \int x^8 \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 (7 a p^4 q\right ) \int x^5 \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \, dx+\left (8 a p^3 q^2\right ) \int x^2 \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \, dx+\left (2 a p^2 q^3\right ) \int \frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x} \, dx-\left (2 a p q^4\right ) \int \frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^4} \, dx-\left (a q^5\right ) \int \frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^7} \, dx \\ & = \frac {7}{9} a p^2 q \left (q^2-2 p q x^2+2 p q x^3+p^2 x^6\right )^{3/2}+(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^5\right ) \int x^8 \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+\frac {1}{6} \left (7 a p^2 q\right ) \int \left (4 p q x-6 p q x^2\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \, dx+\left (8 a p^3 q^2\right ) \int x^2 \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \, dx+\left (2 a p^2 q^3\right ) \int \frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x} \, dx-\left (2 a p q^4\right ) \int \frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^4} \, dx-\left (a q^5\right ) \int \frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^7} \, dx \\ & = \frac {7}{9} a p^2 q \left (q^2-2 p q x^2+2 p q x^3+p^2 x^6\right )^{3/2}+(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^5\right ) \int x^8 \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+\frac {1}{6} \left (7 a p^2 q\right ) \int x (4 p q-6 p q x) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \, dx+\left (8 a p^3 q^2\right ) \int x^2 \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \, dx+\left (2 a p^2 q^3\right ) \int \frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x} \, dx-\left (2 a p q^4\right ) \int \frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^4} \, dx-\left (a q^5\right ) \int \frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^7} \, dx \\ & = \frac {7}{9} a p^2 q \left (q^2-2 p q x^2+2 p q x^3+p^2 x^6\right )^{3/2}+(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^5\right ) \int x^8 \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+\frac {1}{6} \left (7 a p^2 q\right ) \int \left (4 p q x \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}-6 p q x^2 \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}\right ) \, dx+\left (8 a p^3 q^2\right ) \int x^2 \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \, dx+\left (2 a p^2 q^3\right ) \int \frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x} \, dx-\left (2 a p q^4\right ) \int \frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^4} \, dx-\left (a q^5\right ) \int \frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^7} \, dx \\ & = \frac {7}{9} a p^2 q \left (q^2-2 p q x^2+2 p q x^3+p^2 x^6\right )^{3/2}+(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^5\right ) \int x^8 \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+\frac {1}{3} \left (14 a p^3 q^2\right ) \int x \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \, dx-\left (7 a p^3 q^2\right ) \int x^2 \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \, dx+\left (8 a p^3 q^2\right ) \int x^2 \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \, dx+\left (2 a p^2 q^3\right ) \int \frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x} \, dx-\left (2 a p q^4\right ) \int \frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^4} \, dx-\left (a q^5\right ) \int \frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^7} \, dx \\ \end{align*}
Time = 0.34 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.69 \[ \int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (b x^4+a \left (q+p x^3\right )^4\right )}{x^7} \, dx=\frac {1}{12} \left (\frac {\left (q+p x^3\right ) \sqrt {q^2+2 p q (-1+x) x^2+p^2 x^6} \left (6 b x^4+a \left (2 q^4+2 p^4 x^{12}+p q^3 x^2 (-1+8 x)+p^3 q x^8 (-1+8 x)+p^2 q^2 x^4 \left (-3-2 x+12 x^2\right )\right )\right )}{x^6}+6 p q \left (2 b+a p^2 q^2\right ) \log (x)-6 \left (2 b p q+a p^3 q^3\right ) \log \left (q+p x^3+\sqrt {q^2+2 p q (-1+x) x^2+p^2 x^6}\right )\right ) \]
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Time = 0.28 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.73
method | result | size |
pseudoelliptic | \(\frac {-3 p q \,x^{6} \left (a \,p^{2} q^{2}+2 b \right ) \ln \left (\frac {q +p \,x^{3}+\sqrt {p^{2} x^{6}+2 x^{2} p q \left (-1+x \right )+q^{2}}}{x}\right )+\left (a \,p^{4} x^{12}+4 a \,p^{3} q \,x^{9}-\frac {a \,p^{3} q \,x^{8}}{2}+6 a \,p^{2} q^{2} x^{6}-a \,p^{2} q^{2} x^{5}+3 \left (-\frac {a \,p^{2} q^{2}}{2}+b \right ) x^{4}+4 a p \,q^{3} x^{3}-\frac {a p \,q^{3} x^{2}}{2}+a \,q^{4}\right ) \sqrt {p^{2} x^{6}+2 x^{2} p q \left (-1+x \right )+q^{2}}\, \left (p \,x^{3}+q \right )}{6 x^{6}}\) | \(186\) |
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Timed out. \[ \int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (b x^4+a \left (q+p x^3\right )^4\right )}{x^7} \, dx=\text {Timed out} \]
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\[ \int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (b x^4+a \left (q+p x^3\right )^4\right )}{x^7} \, dx=\int \frac {\left (2 p x^{3} - q\right ) \sqrt {p^{2} x^{6} + 2 p q x^{3} - 2 p q x^{2} + q^{2}} \left (a p^{4} x^{12} + 4 a p^{3} q x^{9} + 6 a p^{2} q^{2} x^{6} + 4 a p q^{3} x^{3} + a q^{4} + b x^{4}\right )}{x^{7}}\, dx \]
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\[ \int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (b x^4+a \left (q+p x^3\right )^4\right )}{x^7} \, dx=\int { \frac {\sqrt {p^{2} x^{6} + 2 \, p q x^{3} - 2 \, p q x^{2} + q^{2}} {\left ({\left (p x^{3} + q\right )}^{4} a + b x^{4}\right )} {\left (2 \, p x^{3} - q\right )}}{x^{7}} \,d x } \]
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\[ \int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (b x^4+a \left (q+p x^3\right )^4\right )}{x^7} \, dx=\int { \frac {\sqrt {p^{2} x^{6} + 2 \, p q x^{3} - 2 \, p q x^{2} + q^{2}} {\left ({\left (p x^{3} + q\right )}^{4} a + b x^{4}\right )} {\left (2 \, p x^{3} - q\right )}}{x^{7}} \,d x } \]
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Timed out. \[ \int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (b x^4+a \left (q+p x^3\right )^4\right )}{x^7} \, dx=-\int \frac {\left (q-2\,p\,x^3\right )\,\left (a\,{\left (p\,x^3+q\right )}^4+b\,x^4\right )\,\sqrt {p^2\,x^6+2\,p\,q\,x^3-2\,p\,q\,x^2+q^2}}{x^7} \,d x \]
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