\(\int \frac {\sqrt {q+p x^5} (-2 q+3 p x^5) (a q+b x^2+a p x^5)}{x^4 (c q+d x^2+c p x^5)} \, dx\) [1332]

Optimal result

Integrand size = 57, antiderivative size = 96 \[ \int \frac {\sqrt {q+p x^5} \left (-2 q+3 p x^5\right ) \left (a q+b x^2+a p x^5\right )}{x^4 \left (c q+d x^2+c p x^5\right )} \, dx=\frac {2 \sqrt {q+p x^5} \left (a c q+3 b c x^2-3 a d x^2+a c p x^5\right )}{3 c^2 x^3}+\frac {2 \left (b c \sqrt {d}-a d^{3/2}\right ) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c} \sqrt {q+p x^5}}\right )}{c^{5/2}} \]

[Out]

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)

Rubi [F]

\[ \int \frac {\sqrt {q+p x^5} \left (-2 q+3 p x^5\right ) \left (a q+b x^2+a p x^5\right )}{x^4 \left (c q+d x^2+c p x^5\right )} \, dx=\int \frac {\sqrt {q+p x^5} \left (-2 q+3 p x^5\right ) \left (a q+b x^2+a p x^5\right )}{x^4 \left (c q+d x^2+c p x^5\right )} \, dx \]

[In]

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]

[Out]

(2*a*q*Sqrt[q + p*x^5])/(3*c*x^3) + (2*(b*c - a*d)*Sqrt[q + p*x^5])/(c^2*x) + (2*a*p*x^2*Sqrt[q + p*x^5])/(3*c
) - (5*(b*c - a*d)*p*x^4*Sqrt[1 + (p*x^5)/q]*Hypergeometric2F1[1/2, 4/5, 9/5, -((p*x^5)/q)])/(4*c^2*Sqrt[q + p
*x^5]) + (2*d*(b*c - a*d)*Defer[Int][Sqrt[q + p*x^5]/(c*q + d*x^2 + c*p*x^5), x])/c^2 + (5*(b*c - a*d)*p*Defer
[Int][(x^3*Sqrt[q + p*x^5])/(c*q + d*x^2 + c*p*x^5), x])/c

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {2 a q \sqrt {q+p x^5}}{c x^4}-\frac {2 (b c-a d) \sqrt {q+p x^5}}{c^2 x^2}+\frac {3 a p x \sqrt {q+p x^5}}{c}+\frac {(b c-a d) \left (2 d+5 c p x^3\right ) \sqrt {q+p x^5}}{c^2 \left (c q+d x^2+c p x^5\right )}\right ) \, dx \\ & = \frac {(b c-a d) \int \frac {\left (2 d+5 c p x^3\right ) \sqrt {q+p x^5}}{c q+d x^2+c p x^5} \, dx}{c^2}-\frac {(2 (b c-a d)) \int \frac {\sqrt {q+p x^5}}{x^2} \, dx}{c^2}+\frac {(3 a p) \int x \sqrt {q+p x^5} \, dx}{c}-\frac {(2 a q) \int \frac {\sqrt {q+p x^5}}{x^4} \, dx}{c} \\ & = \frac {2 a q \sqrt {q+p x^5}}{3 c x^3}+\frac {2 (b c-a d) \sqrt {q+p x^5}}{c^2 x}+\frac {2 a p x^2 \sqrt {q+p x^5}}{3 c}+\frac {(b c-a d) \int \left (\frac {2 d \sqrt {q+p x^5}}{c q+d x^2+c p x^5}+\frac {5 c p x^3 \sqrt {q+p x^5}}{c q+d x^2+c p x^5}\right ) \, dx}{c^2}-\frac {(5 (b c-a d) p) \int \frac {x^3}{\sqrt {q+p x^5}} \, dx}{c^2} \\ & = \frac {2 a q \sqrt {q+p x^5}}{3 c x^3}+\frac {2 (b c-a d) \sqrt {q+p x^5}}{c^2 x}+\frac {2 a p x^2 \sqrt {q+p x^5}}{3 c}+\frac {(2 d (b c-a d)) \int \frac {\sqrt {q+p x^5}}{c q+d x^2+c p x^5} \, dx}{c^2}+\frac {(5 (b c-a d) p) \int \frac {x^3 \sqrt {q+p x^5}}{c q+d x^2+c p x^5} \, dx}{c}-\frac {\left (5 (b c-a d) p \sqrt {1+\frac {p x^5}{q}}\right ) \int \frac {x^3}{\sqrt {1+\frac {p x^5}{q}}} \, dx}{c^2 \sqrt {q+p x^5}} \\ & = \frac {2 a q \sqrt {q+p x^5}}{3 c x^3}+\frac {2 (b c-a d) \sqrt {q+p x^5}}{c^2 x}+\frac {2 a p x^2 \sqrt {q+p x^5}}{3 c}-\frac {5 (b c-a d) p x^4 \sqrt {1+\frac {p x^5}{q}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4}{5},\frac {9}{5},-\frac {p x^5}{q}\right )}{4 c^2 \sqrt {q+p x^5}}+\frac {(2 d (b c-a d)) \int \frac {\sqrt {q+p x^5}}{c q+d x^2+c p x^5} \, dx}{c^2}+\frac {(5 (b c-a d) p) \int \frac {x^3 \sqrt {q+p x^5}}{c q+d x^2+c p x^5} \, dx}{c} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.89 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.96 \[ \int \frac {\sqrt {q+p x^5} \left (-2 q+3 p x^5\right ) \left (a q+b x^2+a p x^5\right )}{x^4 \left (c q+d x^2+c p x^5\right )} \, dx=\frac {2 \sqrt {q+p x^5} \left (a c q+3 b c x^2-3 a d x^2+a c p x^5\right )}{3 c^2 x^3}+\frac {2 \sqrt {d} (b c-a d) \arctan \left (\frac {\sqrt {d} x}{\sqrt {c} \sqrt {q+p x^5}}\right )}{c^{5/2}} \]

[In]

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]

[Out]

(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[(S
qrt[d]*x)/(Sqrt[c]*Sqrt[q + p*x^5])])/c^(5/2)

Maple [A] (verified)

Time = 3.25 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.93

method	result	size
pseudoelliptic	\(\frac {2 d \,x^{3} \left (a d -b c \right ) \arctan \left (\frac {\sqrt {p \,x^{5}+q}\, c}{x \sqrt {c d}}\right )+\frac {2 \left (a c p \,x^{5}+\left (-3 a d +3 b c \right ) x^{2}+a c q \right ) \sqrt {p \,x^{5}+q}\, \sqrt {c d}}{3}}{c^{2} x^{3} \sqrt {c d}}\)	\(89\)

[In]

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)

[Out]

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

Fricas [F(-1)]

Timed out. \[ \int \frac {\sqrt {q+p x^5} \left (-2 q+3 p x^5\right ) \left (a q+b x^2+a p x^5\right )}{x^4 \left (c q+d x^2+c p x^5\right )} \, dx=\text {Timed out} \]

[In]

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")

[Out]

Timed out

Sympy [F]

\[ \int \frac {\sqrt {q+p x^5} \left (-2 q+3 p x^5\right ) \left (a q+b x^2+a p x^5\right )}{x^4 \left (c q+d x^2+c p x^5\right )} \, dx=\int \frac {\sqrt {p x^{5} + q} \left (3 p x^{5} - 2 q\right ) \left (a p x^{5} + a q + b x^{2}\right )}{x^{4} \left (c p x^{5} + c q + d x^{2}\right )}\, dx \]

[In]

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)

[Out]

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)

Maxima [F]

\[ \int \frac {\sqrt {q+p x^5} \left (-2 q+3 p x^5\right ) \left (a q+b x^2+a p x^5\right )}{x^4 \left (c q+d x^2+c p x^5\right )} \, dx=\int { \frac {{\left (a p x^{5} + b x^{2} + a q\right )} {\left (3 \, p x^{5} - 2 \, q\right )} \sqrt {p x^{5} + q}}{{\left (c p x^{5} + d x^{2} + c q\right )} x^{4}} \,d x } \]

[In]

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")

[Out]

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)

Giac [F]

\[ \int \frac {\sqrt {q+p x^5} \left (-2 q+3 p x^5\right ) \left (a q+b x^2+a p x^5\right )}{x^4 \left (c q+d x^2+c p x^5\right )} \, dx=\int { \frac {{\left (a p x^{5} + b x^{2} + a q\right )} {\left (3 \, p x^{5} - 2 \, q\right )} \sqrt {p x^{5} + q}}{{\left (c p x^{5} + d x^{2} + c q\right )} x^{4}} \,d x } \]

[In]

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")

[Out]

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)

Mupad [B] (verification not implemented)

Time = 15.12 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.92 \[ \int \frac {\sqrt {q+p x^5} \left (-2 q+3 p x^5\right ) \left (a q+b x^2+a p x^5\right )}{x^4 \left (c q+d x^2+c p x^5\right )} \, dx=\frac {2\,a\,{\left (p\,x^5+q\right )}^{3/2}}{3\,c\,x^3}+\frac {2\,b\,\sqrt {p\,x^5+q}}{c\,x}-\frac {2\,a\,d\,\sqrt {p\,x^5+q}}{c^2\,x}+\frac {a\,d^{3/2}\,\ln \left (\frac {c\,q-d\,x^2+c\,p\,x^5+\sqrt {c}\,\sqrt {d}\,x\,\sqrt {p\,x^5+q}\,2{}\mathrm {i}}{c\,p\,x^5+d\,x^2+c\,q}\right )\,1{}\mathrm {i}}{c^{5/2}}-\frac {b\,\sqrt {d}\,\ln \left (\frac {c\,q-d\,x^2+c\,p\,x^5+\sqrt {c}\,\sqrt {d}\,x\,\sqrt {p\,x^5+q}\,2{}\mathrm {i}}{c\,p\,x^5+d\,x^2+c\,q}\right )\,1{}\mathrm {i}}{c^{3/2}} \]

[In]

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)

[Out]

(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)