Integrand size = 74, antiderivative size = 251 \[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5 \left (c q+d x^2+c p x^3\right )} \, dx=\frac {\left (a c q+2 b c x^2-2 a d x^2+a c p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{2 c^2 x^4}-\frac {2 (-b c+a d) \sqrt {-d^2+2 c^2 p q} \arctan \left (\frac {\sqrt {-d^2+2 c^2 p q} x^2}{c q+d x^2+c p x^3+c \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}\right )}{c^3}+\frac {2 \left (b c d-a d^2+a c^2 p q\right ) \log (x)}{c^3}+\frac {\left (-b c d+a d^2-a c^2 p q\right ) \log \left (q+p x^3+\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}\right )}{c^3} \]
[Out]
\[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5 \left (c q+d x^2+c p x^3\right )} \, dx=\int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5 \left (c q+d x^2+c p x^3\right )} \, dx \]
[In]
[Out]
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {2 a q \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{c x^5}-\frac {2 (b c-a d) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{c^2 x^3}+\frac {a p \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{c x^2}+\frac {2 d (b c-a d) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{c^3 q x}+\frac {(b c-a d) \left (3 c^2 p q-2 d^2 x-2 c d p x^2\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{c^3 q \left (c q+d x^2+c p x^3\right )}\right ) \, dx \\ & = -\frac {(2 (b c-a d)) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^3} \, dx}{c^2}+\frac {(a p) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^2} \, dx}{c}+\frac {(b c-a d) \int \frac {\left (3 c^2 p q-2 d^2 x-2 c d p x^2\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{c q+d x^2+c p x^3} \, dx}{c^3 q}+\frac {(2 d (b c-a d)) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x} \, dx}{c^3 q}-\frac {(2 a q) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5} \, dx}{c} \\ & = -\frac {(2 (b c-a d)) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^3} \, dx}{c^2}+\frac {(a p) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^2} \, dx}{c}+\frac {(b c-a d) \int \left (\frac {3 c^2 p q \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{c q+d x^2+c p x^3}-\frac {2 d^2 x \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{c q+d x^2+c p x^3}-\frac {2 c d p x^2 \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{c q+d x^2+c p x^3}\right ) \, dx}{c^3 q}+\frac {(2 d (b c-a d)) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x} \, dx}{c^3 q}-\frac {(2 a q) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5} \, dx}{c} \\ & = -\frac {(2 (b c-a d)) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^3} \, dx}{c^2}+\frac {(a p) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^2} \, dx}{c}+\frac {(3 (b c-a d) p) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{c q+d x^2+c p x^3} \, dx}{c}+\frac {(2 d (b c-a d)) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x} \, dx}{c^3 q}-\frac {\left (2 d^2 (b c-a d)\right ) \int \frac {x \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{c q+d x^2+c p x^3} \, dx}{c^3 q}-\frac {(2 d (b c-a d) p) \int \frac {x^2 \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{c q+d x^2+c p x^3} \, dx}{c^2 q}-\frac {(2 a q) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5} \, dx}{c} \\ \end{align*}
Time = 1.16 (sec) , antiderivative size = 398, normalized size of antiderivative = 1.59 \[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5 \left (c q+d x^2+c p x^3\right )} \, dx=\frac {a c^2 q \sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}+2 b c^2 x^2 \sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}-2 a c d x^2 \sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}+a c^2 p x^3 \sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}-4 (-b c+a d) \sqrt {-d^2+2 c^2 p q} x^4 \arctan \left (\frac {\sqrt {-d^2+2 c^2 p q} x^2}{d x^2+c \left (q+p x^3+\sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}\right )}\right )+4 b c d x^4 \log (x)-4 a d^2 x^4 \log (x)+4 a c^2 p q x^4 \log (x)-2 b c d x^4 \log \left (q+p x^3+\sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}\right )+2 a d^2 x^4 \log \left (q+p x^3+\sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}\right )-2 a c^2 p q x^4 \log \left (q+p x^3+\sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}\right )}{2 c^3 x^4} \]
[In]
[Out]
Time = 0.34 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.18
method | result | size |
pseudoelliptic | \(-\frac {2 \left (-\frac {c^{2} \left (\left (a p \,x^{3}+2 b \,x^{2}+a q \right ) c -2 a d \,x^{2}\right ) \sqrt {\frac {-2 c^{2} p q +d^{2}}{c^{2}}}\, \sqrt {\frac {p^{2} x^{6}-2 q \,x^{3} \left (-1+x \right ) p +q^{2}}{x^{2}}}}{4}+\left (\frac {\left (a \,c^{2} p q -a \,d^{2}+b c d \right ) c \ln \left (\frac {p \,x^{3}+\sqrt {\frac {p^{2} x^{6}-2 q \,x^{3} \left (-1+x \right ) p +q^{2}}{x^{2}}}\, x +q}{x^{2}}\right ) \sqrt {\frac {-2 c^{2} p q +d^{2}}{c^{2}}}}{2}+\left (c^{2} p q -\frac {d^{2}}{2}\right ) \left (a d -b c \right ) \left (\ln \left (\frac {-2 c p q \,x^{2}-d p \,x^{3}+\sqrt {\frac {-2 c^{2} p q +d^{2}}{c^{2}}}\, \sqrt {\frac {p^{2} x^{6}-2 q \,x^{3} \left (-1+x \right ) p +q^{2}}{x^{2}}}\, c x -d q}{c p \,x^{3}+d \,x^{2}+c q}\right )+\ln \left (2\right )\right )\right ) x^{3}\right )}{\sqrt {\frac {-2 c^{2} p q +d^{2}}{c^{2}}}\, c^{4} x^{3}}\) | \(296\) |
[In]
[Out]
Timed out. \[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5 \left (c q+d x^2+c p x^3\right )} \, dx=\text {Timed out} \]
[In]
[Out]
Timed out. \[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5 \left (c q+d x^2+c p x^3\right )} \, dx=\text {Timed out} \]
[In]
[Out]
\[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5 \left (c q+d x^2+c p x^3\right )} \, dx=\int { \frac {\sqrt {p^{2} x^{6} - 2 \, p q x^{4} + 2 \, p q x^{3} + q^{2}} {\left (a p x^{3} + b x^{2} + a q\right )} {\left (p x^{3} - 2 \, q\right )}}{{\left (c p x^{3} + d x^{2} + c q\right )} x^{5}} \,d x } \]
[In]
[Out]
\[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5 \left (c q+d x^2+c p x^3\right )} \, dx=\int { \frac {\sqrt {p^{2} x^{6} - 2 \, p q x^{4} + 2 \, p q x^{3} + q^{2}} {\left (a p x^{3} + b x^{2} + a q\right )} {\left (p x^{3} - 2 \, q\right )}}{{\left (c p x^{3} + d x^{2} + c q\right )} x^{5}} \,d x } \]
[In]
[Out]
Timed out. \[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5 \left (c q+d x^2+c p x^3\right )} \, dx=\int -\frac {\left (2\,q-p\,x^3\right )\,\left (a\,p\,x^3+b\,x^2+a\,q\right )\,\sqrt {p^2\,x^6-2\,p\,q\,x^4+2\,p\,q\,x^3+q^2}}{x^5\,\left (c\,p\,x^3+d\,x^2+c\,q\right )} \,d x \]
[In]
[Out]