\(\int \frac {(-2 q+p x^3) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} (c x^4+b x^2 (q+p x^3)+a (q+p x^3)^2)}{x^9} \, dx\) [2548]

Optimal result

Integrand size = 71, antiderivative size = 214 \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (c x^4+b x^2 \left (q+p x^3\right )+a \left (q+p x^3\right )^2\right )}{x^9} \, dx=\frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (3 a q^3+4 b q^2 x^2+9 a p q^2 x^3+6 c q x^4-3 a p q^2 x^4+8 b p q x^5-8 b p q x^6+9 a p^2 q x^6+6 c p x^7-3 a p^2 q x^7+4 b p^2 x^8+3 a p^3 x^9\right )}{12 x^8}+\left (2 c p q+a p^2 q^2\right ) \log (x)+\frac {1}{2} \left (-2 c p q-a p^2 q^2\right ) \log \left (q+p x^3+\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}\right ) \] Output:

1/12*(p^2*x^6-2*p*q*x^4+2*p*q*x^3+q^2)^(1/2)*(3*a*p^3*x^9-3*a*p^2*q*x^7+4* 
b*p^2*x^8+9*a*p^2*q*x^6-8*b*p*q*x^6+6*c*p*x^7-3*a*p*q^2*x^4+8*b*p*q*x^5+9* 
a*p*q^2*x^3+6*c*q*x^4+4*b*q^2*x^2+3*a*q^3)/x^8+(a*p^2*q^2+2*c*p*q)*ln(x)+1 
/2*(-a*p^2*q^2-2*c*p*q)*ln(q+p*x^3+(p^2*x^6-2*p*q*x^4+2*p*q*x^3+q^2)^(1/2) 
)
 

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.80 \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (c x^4+b x^2 \left (q+p x^3\right )+a \left (q+p x^3\right )^2\right )}{x^9} \, dx=\frac {\sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6} \left (6 c x^4 \left (q+p x^3\right )+4 b x^2 \left (q^2-2 p q (-1+x) x^3+p^2 x^6\right )+3 a \left (q^3-p q^2 (-3+x) x^3-p^2 q (-3+x) x^6+p^3 x^9\right )\right )}{12 x^8}+p q (2 c+a p q) \log (x)-\frac {1}{2} p q (2 c+a p q) \log \left (q+p x^3+\sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}\right ) \] Input:

Integrate[((-2*q + p*x^3)*Sqrt[q^2 + 2*p*q*x^3 - 2*p*q*x^4 + p^2*x^6]*(c*x 
^4 + b*x^2*(q + p*x^3) + a*(q + p*x^3)^2))/x^9,x]
 

Output:

(Sqrt[q^2 - 2*p*q*(-1 + x)*x^3 + p^2*x^6]*(6*c*x^4*(q + p*x^3) + 4*b*x^2*( 
q^2 - 2*p*q*(-1 + x)*x^3 + p^2*x^6) + 3*a*(q^3 - p*q^2*(-3 + x)*x^3 - p^2* 
q*(-3 + x)*x^6 + p^3*x^9)))/(12*x^8) + p*q*(2*c + a*p*q)*Log[x] - (p*q*(2* 
c + a*p*q)*Log[q + p*x^3 + Sqrt[q^2 - 2*p*q*(-1 + x)*x^3 + p^2*x^6]])/2
 

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.

\(\displaystyle \int \frac {\left (p x^3-2 q\right ) \sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2} \left (a \left (p x^3+q\right )^2+b x^2 \left (p x^3+q\right )+c x^4\right )}{x^9} \, dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (-\frac {3 a p q^2 \sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{x^6}-\frac {2 a q^3 \sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{x^9}+a p^3 \sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}+\frac {b p^2 \sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{x}-\frac {b p q \sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{x^4}-\frac {2 b q^2 \sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{x^7}-\frac {2 c q \sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{x^5}+\frac {c p \sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{x^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -3 a p q^2 \int \frac {\sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{x^6}dx-2 a q^3 \int \frac {\sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{x^9}dx+a p^3 \int \sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}dx+b p^2 \int \frac {\sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{x}dx-b p q \int \frac {\sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{x^4}dx-2 b q^2 \int \frac {\sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{x^7}dx-2 c q \int \frac {\sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{x^5}dx+c p \int \frac {\sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{x^2}dx\)

Input:

Int[((-2*q + p*x^3)*Sqrt[q^2 + 2*p*q*x^3 - 2*p*q*x^4 + p^2*x^6]*(c*x^4 + b 
*x^2*(q + p*x^3) + a*(q + p*x^3)^2))/x^9,x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.71 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.88

method	result	size
pseudoelliptic	\(\frac {\left (3 a \,p^{3} x^{9}+4 b \,p^{2} x^{8}+\left (-3 q a \,p^{2}+6 p c \right ) x^{7}+9 p q \left (a p -\frac {8 b}{9}\right ) x^{6}+8 b p q \,x^{5}+\left (-3 q^{2} a p +6 c q \right ) x^{4}+9 a p \,q^{2} x^{3}+4 b \,q^{2} x^{2}+3 a \,q^{3}\right ) \sqrt {\frac {p^{2} x^{6}-2 p q \,x^{3} \left (-1+x \right )+q^{2}}{x^{2}}}-6 p q \,x^{7} \left (a p q +2 c \right ) \ln \left (\frac {p \,x^{3}+\sqrt {\frac {p^{2} x^{6}-2 p q \,x^{3} \left (-1+x \right )+q^{2}}{x^{2}}}\, x +q}{x^{2}}\right )}{12 x^{7}}\)	\(189\)

Input:

int((p*x^3-2*q)*(p^2*x^6-2*p*q*x^4+2*p*q*x^3+q^2)^(1/2)*(c*x^4+b*x^2*(p*x^ 
3+q)+a*(p*x^3+q)^2)/x^9,x,method=_RETURNVERBOSE)
 

Output:

1/12*((3*a*p^3*x^9+4*b*p^2*x^8+(-3*a*p^2*q+6*c*p)*x^7+9*p*q*(a*p-8/9*b)*x^ 
6+8*b*p*q*x^5+(-3*a*p*q^2+6*c*q)*x^4+9*a*p*q^2*x^3+4*b*q^2*x^2+3*a*q^3)*(( 
p^2*x^6-2*p*q*x^3*(-1+x)+q^2)/x^2)^(1/2)-6*p*q*x^7*(a*p*q+2*c)*ln((p*x^3+( 
(p^2*x^6-2*p*q*x^3*(-1+x)+q^2)/x^2)^(1/2)*x+q)/x^2))/x^7
 

Fricas [A] (verification not implemented)

Time = 78.13 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.91 \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (c x^4+b x^2 \left (q+p x^3\right )+a \left (q+p x^3\right )^2\right )}{x^9} \, dx=-\frac {6 \, {\left (a p^{2} q^{2} + 2 \, c p q\right )} x^{8} \log \left (-\frac {p x^{3} + q + \sqrt {p^{2} x^{6} - 2 \, p q x^{4} + 2 \, p q x^{3} + q^{2}}}{x^{2}}\right ) - {\left (3 \, a p^{3} x^{9} + 4 \, b p^{2} x^{8} + 8 \, b p q x^{5} + {\left (9 \, a p^{2} - 8 \, b p\right )} q x^{6} - 3 \, {\left (a p^{2} q - 2 \, c p\right )} x^{7} + 9 \, a p q^{2} x^{3} + 4 \, b q^{2} x^{2} - 3 \, {\left (a p q^{2} - 2 \, c q\right )} x^{4} + 3 \, a q^{3}\right )} \sqrt {p^{2} x^{6} - 2 \, p q x^{4} + 2 \, p q x^{3} + q^{2}}}{12 \, x^{8}} \] Input:

integrate((p*x^3-2*q)*(p^2*x^6-2*p*q*x^4+2*p*q*x^3+q^2)^(1/2)*(c*x^4+b*x^2 
*(p*x^3+q)+a*(p*x^3+q)^2)/x^9,x, algorithm="fricas")
 

Output:

-1/12*(6*(a*p^2*q^2 + 2*c*p*q)*x^8*log(-(p*x^3 + q + sqrt(p^2*x^6 - 2*p*q* 
x^4 + 2*p*q*x^3 + q^2))/x^2) - (3*a*p^3*x^9 + 4*b*p^2*x^8 + 8*b*p*q*x^5 + 
(9*a*p^2 - 8*b*p)*q*x^6 - 3*(a*p^2*q - 2*c*p)*x^7 + 9*a*p*q^2*x^3 + 4*b*q^ 
2*x^2 - 3*(a*p*q^2 - 2*c*q)*x^4 + 3*a*q^3)*sqrt(p^2*x^6 - 2*p*q*x^4 + 2*p* 
q*x^3 + q^2))/x^8
 

Sympy [F]

\[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (c x^4+b x^2 \left (q+p x^3\right )+a \left (q+p x^3\right )^2\right )}{x^9} \, dx=\int \frac {\left (p x^{3} - 2 q\right ) \sqrt {p^{2} x^{6} - 2 p q x^{4} + 2 p q x^{3} + q^{2}} \left (a p^{2} x^{6} + 2 a p q x^{3} + a q^{2} + b p x^{5} + b q x^{2} + c x^{4}\right )}{x^{9}}\, dx \] Input:

integrate((p*x**3-2*q)*(p**2*x**6-2*p*q*x**4+2*p*q*x**3+q**2)**(1/2)*(c*x* 
*4+b*x**2*(p*x**3+q)+a*(p*x**3+q)**2)/x**9,x)
 

Output:

Integral((p*x**3 - 2*q)*sqrt(p**2*x**6 - 2*p*q*x**4 + 2*p*q*x**3 + q**2)*( 
a*p**2*x**6 + 2*a*p*q*x**3 + a*q**2 + b*p*x**5 + b*q*x**2 + c*x**4)/x**9, 
x)
 

Maxima [F]

\[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (c x^4+b x^2 \left (q+p x^3\right )+a \left (q+p x^3\right )^2\right )}{x^9} \, dx=\int { \frac {\sqrt {p^{2} x^{6} - 2 \, p q x^{4} + 2 \, p q x^{3} + q^{2}} {\left (c x^{4} + {\left (p x^{3} + q\right )} b x^{2} + {\left (p x^{3} + q\right )}^{2} a\right )} {\left (p x^{3} - 2 \, q\right )}}{x^{9}} \,d x } \] Input:

integrate((p*x^3-2*q)*(p^2*x^6-2*p*q*x^4+2*p*q*x^3+q^2)^(1/2)*(c*x^4+b*x^2 
*(p*x^3+q)+a*(p*x^3+q)^2)/x^9,x, algorithm="maxima")
 

Output:

integrate(sqrt(p^2*x^6 - 2*p*q*x^4 + 2*p*q*x^3 + q^2)*(c*x^4 + (p*x^3 + q) 
*b*x^2 + (p*x^3 + q)^2*a)*(p*x^3 - 2*q)/x^9, x)
 

Giac [F]

\[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (c x^4+b x^2 \left (q+p x^3\right )+a \left (q+p x^3\right )^2\right )}{x^9} \, dx=\int { \frac {\sqrt {p^{2} x^{6} - 2 \, p q x^{4} + 2 \, p q x^{3} + q^{2}} {\left (c x^{4} + {\left (p x^{3} + q\right )} b x^{2} + {\left (p x^{3} + q\right )}^{2} a\right )} {\left (p x^{3} - 2 \, q\right )}}{x^{9}} \,d x } \] Input:

integrate((p*x^3-2*q)*(p^2*x^6-2*p*q*x^4+2*p*q*x^3+q^2)^(1/2)*(c*x^4+b*x^2 
*(p*x^3+q)+a*(p*x^3+q)^2)/x^9,x, algorithm="giac")
 

Output:

integrate(sqrt(p^2*x^6 - 2*p*q*x^4 + 2*p*q*x^3 + q^2)*(c*x^4 + (p*x^3 + q) 
*b*x^2 + (p*x^3 + q)^2*a)*(p*x^3 - 2*q)/x^9, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (c x^4+b x^2 \left (q+p x^3\right )+a \left (q+p x^3\right )^2\right )}{x^9} \, dx=\int -\frac {\left (2\,q-p\,x^3\right )\,\left (a\,{\left (p\,x^3+q\right )}^2+c\,x^4+b\,x^2\,\left (p\,x^3+q\right )\right )\,\sqrt {p^2\,x^6-2\,p\,q\,x^4+2\,p\,q\,x^3+q^2}}{x^9} \,d x \] Input:

int(-((2*q - p*x^3)*(a*(q + p*x^3)^2 + c*x^4 + b*x^2*(q + p*x^3))*(p^2*x^6 
 + q^2 + 2*p*q*x^3 - 2*p*q*x^4)^(1/2))/x^9,x)
                                                                                    
                                                                                    
 

Output:

int(-((2*q - p*x^3)*(a*(q + p*x^3)^2 + c*x^4 + b*x^2*(q + p*x^3))*(p^2*x^6 
 + q^2 + 2*p*q*x^3 - 2*p*q*x^4)^(1/2))/x^9, x)
 

Reduce [F]

\[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (c x^4+b x^2 \left (q+p x^3\right )+a \left (q+p x^3\right )^2\right )}{x^9} \, dx=\int \frac {\left (p \,x^{3}-2 q \right ) \sqrt {p^{2} x^{6}-2 p q \,x^{4}+2 p q \,x^{3}+q^{2}}\, \left (c \,x^{4}+b \,x^{2} \left (p \,x^{3}+q \right )+a \left (p \,x^{3}+q \right )^{2}\right )}{x^{9}}d x \] Input:

int((p*x^3-2*q)*(p^2*x^6-2*p*q*x^4+2*p*q*x^3+q^2)^(1/2)*(c*x^4+b*x^2*(p*x^ 
3+q)+a*(p*x^3+q)^2)/x^9,x)
 

Output:

int((p*x^3-2*q)*(p^2*x^6-2*p*q*x^4+2*p*q*x^3+q^2)^(1/2)*(c*x^4+b*x^2*(p*x^ 
3+q)+a*(p*x^3+q)^2)/x^9,x)