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

3.8.27.1 Optimal result

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

output

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)
 

3.8.27.2 Mathematica [A] (verified)

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

input

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

output

(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)
 

3.8.27.3 Rubi [F]

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.

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 b \int \frac {\sqrt {p x^5+q}}{a p x^5+b x^2+a q}dx}{a}+5 p \int \frac {x^3 \sqrt {p x^5+q}}{a p x^5+b x^2+a q}dx-\frac {5 p x^4 \sqrt {\frac {p x^5}{q}+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4}{5},\frac {9}{5},-\frac {p x^5}{q}\right )}{4 a \sqrt {p x^5+q}}+\frac {2 \sqrt {p x^5+q}}{a x}\)

input

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

output

$Aborted
 

3.8.27.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

3.8.27.4 Maple [A] (verified)

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

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

input

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=_RETURN 
VERBOSE)
 

output

-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)
 

3.8.27.5 Fricas [F(-1)]

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

input

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, algorit 
hm="fricas")
 

output

Timed out
 

3.8.27.6 Sympy [F]

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

input

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)
 

output

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

3.8.27.7 Maxima [F]

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

input

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, algorit 
hm="maxima")
 

output

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

3.8.27.8 Giac [F]

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

input

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, algorit 
hm="giac")
 

output

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

3.8.27.9 Mupad [B] (verification not implemented)

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

input

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)
 

output

(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)