\(\int \frac {\sqrt {q+p x^5} (-2 q+3 p x^5)}{x^2 (a q+b x^2+a p x^5)} \, dx\) [727]

Optimal result

Integrand size = 42, antiderivative size = 56 \[ \int \frac {\sqrt {q+p x^5} \left (-2 q+3 p x^5\right )}{x^2 \left (a q+b x^2+a p x^5\right )} \, dx=\frac {2 \sqrt {q+p x^5}}{a x}+\frac {2 \sqrt {b} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a} \sqrt {q+p x^5}}\right )}{a^{3/2}} \]

[Out]

2*(p*x^5+q)^(1/2)/a/x+2*b^(1/2)*arctan(b^(1/2)*x/a^(1/2)/(p*x^5+q)^(1/2))/a^(3/2)

Rubi [F]

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

[In]

Int[(Sqrt[q + p*x^5]*(-2*q + 3*p*x^5))/(x^2*(a*q + b*x^2 + a*p*x^5)),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.81 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {q+p x^5} \left (-2 q+3 p x^5\right )}{x^2 \left (a q+b x^2+a p x^5\right )} \, dx=\frac {2 \sqrt {q+p x^5}}{a x}+\frac {2 \sqrt {b} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a} \sqrt {q+p x^5}}\right )}{a^{3/2}} \]

[In]

Integrate[(Sqrt[q + p*x^5]*(-2*q + 3*p*x^5))/(x^2*(a*q + b*x^2 + a*p*x^5)),x]

[Out]

(2*Sqrt[q + p*x^5])/(a*x) + (2*Sqrt[b]*ArcTan[(Sqrt[b]*x)/(Sqrt[a]*Sqrt[q + p*x^5])])/a^(3/2)

Maple [A] (verified)

Time = 2.70 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.96

method	result	size
pseudoelliptic	\(-\frac {2 \left (b \arctan \left (\frac {\sqrt {p \,x^{5}+q}\, a}{x \sqrt {a b}}\right ) x -\sqrt {p \,x^{5}+q}\, \sqrt {a b}\right )}{a x \sqrt {a b}}\)	\(54\)

[In]

int((p*x^5+q)^(1/2)*(3*p*x^5-2*q)/x^2/(a*p*x^5+b*x^2+a*q),x,method=_RETURNVERBOSE)

[Out]

-2*(b*arctan((p*x^5+q)^(1/2)/x*a/(a*b)^(1/2))*x-(p*x^5+q)^(1/2)*(a*b)^(1/2))/a/x/(a*b)^(1/2)

Fricas [F(-1)]

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

[In]

integrate((p*x^5+q)^(1/2)*(3*p*x^5-2*q)/x^2/(a*p*x^5+b*x^2+a*q),x, algorithm="fricas")

[Out]

Timed out

Sympy [F]

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

[In]

integrate((p*x**5+q)**(1/2)*(3*p*x**5-2*q)/x**2/(a*p*x**5+b*x**2+a*q),x)

[Out]

Integral(sqrt(p*x**5 + q)*(3*p*x**5 - 2*q)/(x**2*(a*p*x**5 + a*q + b*x**2)), x)

Maxima [F]

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

[In]

integrate((p*x^5+q)^(1/2)*(3*p*x^5-2*q)/x^2/(a*p*x^5+b*x^2+a*q),x, algorithm="maxima")

[Out]

integrate((3*p*x^5 - 2*q)*sqrt(p*x^5 + q)/((a*p*x^5 + b*x^2 + a*q)*x^2), x)

Giac [F]

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

[In]

integrate((p*x^5+q)^(1/2)*(3*p*x^5-2*q)/x^2/(a*p*x^5+b*x^2+a*q),x, algorithm="giac")

[Out]

integrate((3*p*x^5 - 2*q)*sqrt(p*x^5 + q)/((a*p*x^5 + b*x^2 + a*q)*x^2), x)

Mupad [B] (verification not implemented)

Time = 12.58 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.82 \[ \int \frac {\sqrt {q+p x^5} \left (-2 q+3 p x^5\right )}{x^2 \left (a q+b x^2+a p x^5\right )} \, dx=\frac {2\,\sqrt {p\,x^5+q}}{a\,x}+\frac {\sqrt {b}\,\ln \left (\frac {a^5\,b\,p^4\,x^2-a^6\,p^4\,\left (p\,x^5+q\right )+a^{11/2}\,\sqrt {b}\,p^4\,x\,\sqrt {p\,x^5+q}\,2{}\mathrm {i}}{4\,b^2\,q\,x^2+4\,a\,b\,q\,\left (p\,x^5+q\right )}\right )\,1{}\mathrm {i}}{a^{3/2}} \]

[In]

int(-((q + p*x^5)^(1/2)*(2*q - 3*p*x^5))/(x^2*(a*q + b*x^2 + a*p*x^5)),x)

[Out]

(2*(q + p*x^5)^(1/2))/(a*x) + (b^(1/2)*log((a^5*b*p^4*x^2 - a^6*p^4*(q + p*x^5) + a^(11/2)*b^(1/2)*p^4*x*(q +
p*x^5)^(1/2)*2i)/(4*b^2*q*x^2 + 4*a*b*q*(q + p*x^5)))*1i)/a^(3/2)