Integrand size = 57, antiderivative size = 96 \[ \int \frac {\sqrt {q+p x^5} \left (-2 q+3 p x^5\right ) \left (a q+b x^2+a p x^5\right )}{x^4 \left (c q+d x^2+c p x^5\right )} \, dx=\frac {2 \sqrt {q+p x^5} \left (a c q+3 b c x^2-3 a d x^2+a c p x^5\right )}{3 c^2 x^3}+\frac {2 \left (b c \sqrt {d}-a d^{3/2}\right ) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c} \sqrt {q+p x^5}}\right )}{c^{5/2}} \]
2/3*(p*x^5+q)^(1/2)*(a*c*p*x^5-3*a*d*x^2+3*b*c*x^2+a*c*q)/c^2/x^3+2*(b*c*d ^(1/2)-a*d^(3/2))*arctan(d^(1/2)*x/c^(1/2)/(p*x^5+q)^(1/2))/c^(5/2)
Time = 6.89 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.96 \[ \int \frac {\sqrt {q+p x^5} \left (-2 q+3 p x^5\right ) \left (a q+b x^2+a p x^5\right )}{x^4 \left (c q+d x^2+c p x^5\right )} \, dx=\frac {2 \sqrt {q+p x^5} \left (a c q+3 b c x^2-3 a d x^2+a c p x^5\right )}{3 c^2 x^3}+\frac {2 \sqrt {d} (b c-a d) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c} \sqrt {q+p x^5}}\right )}{c^{5/2}} \]
Integrate[(Sqrt[q + p*x^5]*(-2*q + 3*p*x^5)*(a*q + b*x^2 + a*p*x^5))/(x^4* (c*q + d*x^2 + c*p*x^5)),x]
(2*Sqrt[q + p*x^5]*(a*c*q + 3*b*c*x^2 - 3*a*d*x^2 + a*c*p*x^5))/(3*c^2*x^3 ) + (2*Sqrt[d]*(b*c - a*d)*ArcTan[(Sqrt[d]*x)/(Sqrt[c]*Sqrt[q + p*x^5])])/ c^(5/2)
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sqrt {p x^5+q} \left (3 p x^5-2 q\right ) \left (a p x^5+a q+b x^2\right )}{x^4 \left (c p x^5+c q+d x^2\right )} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {2 \sqrt {p x^5+q} (b c-a d)}{c^2 x^2}+\frac {\sqrt {p x^5+q} (b c-a d) \left (5 c p x^3+2 d\right )}{c^2 \left (c p x^5+c q+d x^2\right )}+\frac {3 a p x \sqrt {p x^5+q}}{c}-\frac {2 a q \sqrt {p x^5+q}}{c x^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 d (b c-a d) \int \frac {\sqrt {p x^5+q}}{c p x^5+d x^2+c q}dx}{c^2}+\frac {5 p (b c-a d) \int \frac {x^3 \sqrt {p x^5+q}}{c p x^5+d x^2+c q}dx}{c}-\frac {5 p x^4 \sqrt {\frac {p x^5}{q}+1} (b c-a d) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4}{5},\frac {9}{5},-\frac {p x^5}{q}\right )}{4 c^2 \sqrt {p x^5+q}}+\frac {2 \sqrt {p x^5+q} (b c-a d)}{c^2 x}+\frac {2 a q \sqrt {p x^5+q}}{3 c x^3}+\frac {2 a p x^2 \sqrt {p x^5+q}}{3 c}\) |
Int[(Sqrt[q + p*x^5]*(-2*q + 3*p*x^5)*(a*q + b*x^2 + a*p*x^5))/(x^4*(c*q + d*x^2 + c*p*x^5)),x]
3.14.32.3.1 Defintions of rubi rules used
Time = 3.25 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.93
method | result | size |
pseudoelliptic | \(\frac {2 d \,x^{3} \left (a d -b c \right ) \arctan \left (\frac {\sqrt {p \,x^{5}+q}\, c}{x \sqrt {c d}}\right )+\frac {2 \left (a c p \,x^{5}+\left (-3 a d +3 b c \right ) x^{2}+a c q \right ) \sqrt {p \,x^{5}+q}\, \sqrt {c d}}{3}}{c^{2} x^{3} \sqrt {c d}}\) | \(89\) |
int((p*x^5+q)^(1/2)*(3*p*x^5-2*q)*(a*p*x^5+b*x^2+a*q)/x^4/(c*p*x^5+d*x^2+c *q),x,method=_RETURNVERBOSE)
2/3*(3*d*x^3*(a*d-b*c)*arctan((p*x^5+q)^(1/2)/x*c/(c*d)^(1/2))+(a*c*p*x^5+ (-3*a*d+3*b*c)*x^2+a*c*q)*(p*x^5+q)^(1/2)*(c*d)^(1/2))/(c*d)^(1/2)/c^2/x^3
Timed out. \[ \int \frac {\sqrt {q+p x^5} \left (-2 q+3 p x^5\right ) \left (a q+b x^2+a p x^5\right )}{x^4 \left (c q+d x^2+c p x^5\right )} \, dx=\text {Timed out} \]
integrate((p*x^5+q)^(1/2)*(3*p*x^5-2*q)*(a*p*x^5+b*x^2+a*q)/x^4/(c*p*x^5+d *x^2+c*q),x, algorithm="fricas")
\[ \int \frac {\sqrt {q+p x^5} \left (-2 q+3 p x^5\right ) \left (a q+b x^2+a p x^5\right )}{x^4 \left (c q+d x^2+c p x^5\right )} \, dx=\int \frac {\sqrt {p x^{5} + q} \left (3 p x^{5} - 2 q\right ) \left (a p x^{5} + a q + b x^{2}\right )}{x^{4} \left (c p x^{5} + c q + d x^{2}\right )}\, dx \]
integrate((p*x**5+q)**(1/2)*(3*p*x**5-2*q)*(a*p*x**5+b*x**2+a*q)/x**4/(c*p *x**5+d*x**2+c*q),x)
Integral(sqrt(p*x**5 + q)*(3*p*x**5 - 2*q)*(a*p*x**5 + a*q + b*x**2)/(x**4 *(c*p*x**5 + c*q + d*x**2)), x)
\[ \int \frac {\sqrt {q+p x^5} \left (-2 q+3 p x^5\right ) \left (a q+b x^2+a p x^5\right )}{x^4 \left (c q+d x^2+c p x^5\right )} \, dx=\int { \frac {{\left (a p x^{5} + b x^{2} + a q\right )} {\left (3 \, p x^{5} - 2 \, q\right )} \sqrt {p x^{5} + q}}{{\left (c p x^{5} + d x^{2} + c q\right )} x^{4}} \,d x } \]
integrate((p*x^5+q)^(1/2)*(3*p*x^5-2*q)*(a*p*x^5+b*x^2+a*q)/x^4/(c*p*x^5+d *x^2+c*q),x, algorithm="maxima")
integrate((a*p*x^5 + b*x^2 + a*q)*(3*p*x^5 - 2*q)*sqrt(p*x^5 + q)/((c*p*x^ 5 + d*x^2 + c*q)*x^4), x)
\[ \int \frac {\sqrt {q+p x^5} \left (-2 q+3 p x^5\right ) \left (a q+b x^2+a p x^5\right )}{x^4 \left (c q+d x^2+c p x^5\right )} \, dx=\int { \frac {{\left (a p x^{5} + b x^{2} + a q\right )} {\left (3 \, p x^{5} - 2 \, q\right )} \sqrt {p x^{5} + q}}{{\left (c p x^{5} + d x^{2} + c q\right )} x^{4}} \,d x } \]
integrate((p*x^5+q)^(1/2)*(3*p*x^5-2*q)*(a*p*x^5+b*x^2+a*q)/x^4/(c*p*x^5+d *x^2+c*q),x, algorithm="giac")
integrate((a*p*x^5 + b*x^2 + a*q)*(3*p*x^5 - 2*q)*sqrt(p*x^5 + q)/((c*p*x^ 5 + d*x^2 + c*q)*x^4), x)
Time = 15.12 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.92 \[ \int \frac {\sqrt {q+p x^5} \left (-2 q+3 p x^5\right ) \left (a q+b x^2+a p x^5\right )}{x^4 \left (c q+d x^2+c p x^5\right )} \, dx=\frac {2\,a\,{\left (p\,x^5+q\right )}^{3/2}}{3\,c\,x^3}+\frac {2\,b\,\sqrt {p\,x^5+q}}{c\,x}-\frac {2\,a\,d\,\sqrt {p\,x^5+q}}{c^2\,x}+\frac {a\,d^{3/2}\,\ln \left (\frac {c\,q-d\,x^2+c\,p\,x^5+\sqrt {c}\,\sqrt {d}\,x\,\sqrt {p\,x^5+q}\,2{}\mathrm {i}}{c\,p\,x^5+d\,x^2+c\,q}\right )\,1{}\mathrm {i}}{c^{5/2}}-\frac {b\,\sqrt {d}\,\ln \left (\frac {c\,q-d\,x^2+c\,p\,x^5+\sqrt {c}\,\sqrt {d}\,x\,\sqrt {p\,x^5+q}\,2{}\mathrm {i}}{c\,p\,x^5+d\,x^2+c\,q}\right )\,1{}\mathrm {i}}{c^{3/2}} \]
int(-((q + p*x^5)^(1/2)*(2*q - 3*p*x^5)*(a*q + b*x^2 + a*p*x^5))/(x^4*(c*q + d*x^2 + c*p*x^5)),x)
(a*d^(3/2)*log((c*q - d*x^2 + c*p*x^5 + c^(1/2)*d^(1/2)*x*(q + p*x^5)^(1/2 )*2i)/(c*q + d*x^2 + c*p*x^5))*1i)/c^(5/2) - (b*d^(1/2)*log((c*q - d*x^2 + c*p*x^5 + c^(1/2)*d^(1/2)*x*(q + p*x^5)^(1/2)*2i)/(c*q + d*x^2 + c*p*x^5) )*1i)/c^(3/2) + (2*a*(q + p*x^5)^(3/2))/(3*c*x^3) + (2*b*(q + p*x^5)^(1/2) )/(c*x) - (2*a*d*(q + p*x^5)^(1/2))/(c^2*x)