3.28.31 \(\int \frac {(-2 q+p x^3) (a q+b x^2+a p x^3) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5 (c q+d x^2+c p x^3)} \, dx\) [2731]

3.28.31.1 Optimal result

Integrand size = 74, antiderivative size = 251 \[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5 \left (c q+d x^2+c p x^3\right )} \, dx=\frac {\left (a c q+2 b c x^2-2 a d x^2+a c p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{2 c^2 x^4}-\frac {2 (-b c+a d) \sqrt {-d^2+2 c^2 p q} \arctan \left (\frac {\sqrt {-d^2+2 c^2 p q} x^2}{c q+d x^2+c p x^3+c \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}\right )}{c^3}+\frac {2 \left (b c d-a d^2+a c^2 p q\right ) \log (x)}{c^3}+\frac {\left (-b c d+a d^2-a c^2 p q\right ) \log \left (q+p x^3+\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}\right )}{c^3} \]

output

1/2*(a*c*p*x^3-2*a*d*x^2+2*b*c*x^2+a*c*q)*(p^2*x^6-2*p*q*x^4+2*p*q*x^3+q^2 
)^(1/2)/c^2/x^4-2*(a*d-b*c)*(2*c^2*p*q-d^2)^(1/2)*arctan((2*c^2*p*q-d^2)^( 
1/2)*x^2/(c*q+d*x^2+c*p*x^3+c*(p^2*x^6-2*p*q*x^4+2*p*q*x^3+q^2)^(1/2)))/c^ 
3+2*(a*c^2*p*q-a*d^2+b*c*d)*ln(x)/c^3+(-a*c^2*p*q+a*d^2-b*c*d)*ln(q+p*x^3+ 
(p^2*x^6-2*p*q*x^4+2*p*q*x^3+q^2)^(1/2))/c^3
 

3.28.31.2 Mathematica [A] (verified)

Time = 1.16 (sec) , antiderivative size = 398, normalized size of antiderivative = 1.59 \[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5 \left (c q+d x^2+c p x^3\right )} \, dx=\frac {a c^2 q \sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}+2 b c^2 x^2 \sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}-2 a c d x^2 \sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}+a c^2 p x^3 \sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}-4 (-b c+a d) \sqrt {-d^2+2 c^2 p q} x^4 \arctan \left (\frac {\sqrt {-d^2+2 c^2 p q} x^2}{d x^2+c \left (q+p x^3+\sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}\right )}\right )+4 b c d x^4 \log (x)-4 a d^2 x^4 \log (x)+4 a c^2 p q x^4 \log (x)-2 b c d x^4 \log \left (q+p x^3+\sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}\right )+2 a d^2 x^4 \log \left (q+p x^3+\sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}\right )-2 a c^2 p q x^4 \log \left (q+p x^3+\sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}\right )}{2 c^3 x^4} \]

input

Integrate[((-2*q + p*x^3)*(a*q + b*x^2 + a*p*x^3)*Sqrt[q^2 + 2*p*q*x^3 - 2 
*p*q*x^4 + p^2*x^6])/(x^5*(c*q + d*x^2 + c*p*x^3)),x]
 

output

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

3.28.31.3 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.

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

input

Int[((-2*q + p*x^3)*(a*q + b*x^2 + a*p*x^3)*Sqrt[q^2 + 2*p*q*x^3 - 2*p*q*x 
^4 + p^2*x^6])/(x^5*(c*q + d*x^2 + c*p*x^3)),x]
 

output

$Aborted
 

3.28.31.3.1 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] 
]
 

3.28.31.4 Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.18

method	result	size
pseudoelliptic	\(-\frac {2 \left (-\frac {c^{2} \left (\left (a p \,x^{3}+2 b \,x^{2}+a q \right ) c -2 a d \,x^{2}\right ) \sqrt {\frac {-2 c^{2} p q +d^{2}}{c^{2}}}\, \sqrt {\frac {p^{2} x^{6}-2 q \,x^{3} \left (-1+x \right ) p +q^{2}}{x^{2}}}}{4}+\left (\frac {\left (a \,c^{2} p q -a \,d^{2}+b c d \right ) c \ln \left (\frac {p \,x^{3}+\sqrt {\frac {p^{2} x^{6}-2 q \,x^{3} \left (-1+x \right ) p +q^{2}}{x^{2}}}\, x +q}{x^{2}}\right ) \sqrt {\frac {-2 c^{2} p q +d^{2}}{c^{2}}}}{2}+\left (c^{2} p q -\frac {d^{2}}{2}\right ) \left (a d -b c \right ) \left (\ln \left (\frac {-2 c p q \,x^{2}-d p \,x^{3}+\sqrt {\frac {-2 c^{2} p q +d^{2}}{c^{2}}}\, \sqrt {\frac {p^{2} x^{6}-2 q \,x^{3} \left (-1+x \right ) p +q^{2}}{x^{2}}}\, c x -d q}{c p \,x^{3}+d \,x^{2}+c q}\right )+\ln \left (2\right )\right )\right ) x^{3}\right )}{\sqrt {\frac {-2 c^{2} p q +d^{2}}{c^{2}}}\, c^{4} x^{3}}\)	\(296\)

input

int((p*x^3-2*q)*(a*p*x^3+b*x^2+a*q)*(p^2*x^6-2*p*q*x^4+2*p*q*x^3+q^2)^(1/2 
)/x^5/(c*p*x^3+d*x^2+c*q),x,method=_RETURNVERBOSE)
 

output

-2*(-1/4*c^2*((a*p*x^3+2*b*x^2+a*q)*c-2*a*d*x^2)*((-2*c^2*p*q+d^2)/c^2)^(1 
/2)*((p^2*x^6-2*q*x^3*(-1+x)*p+q^2)/x^2)^(1/2)+(1/2*(a*c^2*p*q-a*d^2+b*c*d 
)*c*ln((p*x^3+((p^2*x^6-2*q*x^3*(-1+x)*p+q^2)/x^2)^(1/2)*x+q)/x^2)*((-2*c^ 
2*p*q+d^2)/c^2)^(1/2)+(c^2*p*q-1/2*d^2)*(a*d-b*c)*(ln((-2*c*p*q*x^2-d*p*x^ 
3+((-2*c^2*p*q+d^2)/c^2)^(1/2)*((p^2*x^6-2*q*x^3*(-1+x)*p+q^2)/x^2)^(1/2)* 
c*x-d*q)/(c*p*x^3+d*x^2+c*q))+ln(2)))*x^3)/((-2*c^2*p*q+d^2)/c^2)^(1/2)/c^ 
4/x^3
 

3.28.31.5 Fricas [F(-1)]

Timed out. \[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5 \left (c q+d x^2+c p x^3\right )} \, dx=\text {Timed out} \]

input

integrate((p*x^3-2*q)*(a*p*x^3+b*x^2+a*q)*(p^2*x^6-2*p*q*x^4+2*p*q*x^3+q^2 
)^(1/2)/x^5/(c*p*x^3+d*x^2+c*q),x, algorithm="fricas")
 

output

Timed out
 

3.28.31.6 Sympy [F(-1)]

Timed out. \[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5 \left (c q+d x^2+c p x^3\right )} \, dx=\text {Timed out} \]

input

integrate((p*x**3-2*q)*(a*p*x**3+b*x**2+a*q)*(p**2*x**6-2*p*q*x**4+2*p*q*x 
**3+q**2)**(1/2)/x**5/(c*p*x**3+d*x**2+c*q),x)
 

output

Timed out
 

3.28.31.7 Maxima [F]

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

input

integrate((p*x^3-2*q)*(a*p*x^3+b*x^2+a*q)*(p^2*x^6-2*p*q*x^4+2*p*q*x^3+q^2 
)^(1/2)/x^5/(c*p*x^3+d*x^2+c*q),x, algorithm="maxima")
 

output

integrate(sqrt(p^2*x^6 - 2*p*q*x^4 + 2*p*q*x^3 + q^2)*(a*p*x^3 + b*x^2 + a 
*q)*(p*x^3 - 2*q)/((c*p*x^3 + d*x^2 + c*q)*x^5), x)
 

3.28.31.8 Giac [F]

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

input

integrate((p*x^3-2*q)*(a*p*x^3+b*x^2+a*q)*(p^2*x^6-2*p*q*x^4+2*p*q*x^3+q^2 
)^(1/2)/x^5/(c*p*x^3+d*x^2+c*q),x, algorithm="giac")
 

output

integrate(sqrt(p^2*x^6 - 2*p*q*x^4 + 2*p*q*x^3 + q^2)*(a*p*x^3 + b*x^2 + a 
*q)*(p*x^3 - 2*q)/((c*p*x^3 + d*x^2 + c*q)*x^5), x)
 

3.28.31.9 Mupad [F(-1)]

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

input

int(-((2*q - p*x^3)*(a*q + b*x^2 + a*p*x^3)*(p^2*x^6 + q^2 + 2*p*q*x^3 - 2 
*p*q*x^4)^(1/2))/(x^5*(c*q + d*x^2 + c*p*x^3)),x)
 

output

int(-((2*q - p*x^3)*(a*q + b*x^2 + a*p*x^3)*(p^2*x^6 + q^2 + 2*p*q*x^3 - 2 
*p*q*x^4)^(1/2))/(x^5*(c*q + d*x^2 + c*p*x^3)), x)