Integrand size = 60, antiderivative size = 255 \[ \int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (b x^4+a \left (q+p x^3\right )^4\right )}{x^7} \, dx=\frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (2 a q^5-a p q^4 x^2+10 a p q^4 x^3+6 b q x^4-3 a p^2 q^3 x^4-3 a p^2 q^3 x^5+20 a p^2 q^3 x^6+6 b p x^7-3 a p^3 q^2 x^7-3 a p^3 q^2 x^8+20 a p^3 q^2 x^9-a p^4 q x^{11}+10 a p^4 q x^{12}+2 a p^5 x^{15}\right )}{12 x^6}+\frac {1}{2} \left (2 b p q+a p^3 q^3\right ) \log (x)+\frac {1}{2} \left (-2 b p q-a p^3 q^3\right ) \log \left (q+p x^3+\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}\right ) \]
1/12*(p^2*x^6+2*p*q*x^3-2*p*q*x^2+q^2)^(1/2)*(2*a*p^5*x^15+10*a*p^4*q*x^12 -a*p^4*q*x^11+20*a*p^3*q^2*x^9-3*a*p^3*q^2*x^8-3*a*p^3*q^2*x^7+20*a*p^2*q^ 3*x^6-3*a*p^2*q^3*x^5-3*a*p^2*q^3*x^4+10*a*p*q^4*x^3+6*b*p*x^7-a*p*q^4*x^2 +2*a*q^5+6*b*q*x^4)/x^6+1/2*(a*p^3*q^3+2*b*p*q)*ln(x)+1/2*(-a*p^3*q^3-2*b* p*q)*ln(q+p*x^3+(p^2*x^6+2*p*q*x^3-2*p*q*x^2+q^2)^(1/2))
Time = 0.34 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.69 \[ \int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (b x^4+a \left (q+p x^3\right )^4\right )}{x^7} \, dx=\frac {1}{12} \left (\frac {\left (q+p x^3\right ) \sqrt {q^2+2 p q (-1+x) x^2+p^2 x^6} \left (6 b x^4+a \left (2 q^4+2 p^4 x^{12}+p q^3 x^2 (-1+8 x)+p^3 q x^8 (-1+8 x)+p^2 q^2 x^4 \left (-3-2 x+12 x^2\right )\right )\right )}{x^6}+6 p q \left (2 b+a p^2 q^2\right ) \log (x)-6 \left (2 b p q+a p^3 q^3\right ) \log \left (q+p x^3+\sqrt {q^2+2 p q (-1+x) x^2+p^2 x^6}\right )\right ) \]
Integrate[((-q + 2*p*x^3)*Sqrt[q^2 - 2*p*q*x^2 + 2*p*q*x^3 + p^2*x^6]*(b*x ^4 + a*(q + p*x^3)^4))/x^7,x]
(((q + p*x^3)*Sqrt[q^2 + 2*p*q*(-1 + x)*x^2 + p^2*x^6]*(6*b*x^4 + a*(2*q^4 + 2*p^4*x^12 + p*q^3*x^2*(-1 + 8*x) + p^3*q*x^8*(-1 + 8*x) + p^2*q^2*x^4* (-3 - 2*x + 12*x^2))))/x^6 + 6*p*q*(2*b + a*p^2*q^2)*Log[x] - 6*(2*b*p*q + a*p^3*q^3)*Log[q + p*x^3 + Sqrt[q^2 + 2*p*q*(-1 + x)*x^2 + p^2*x^6]])/12
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.
| \(\displaystyle \int \frac {\left (2 p x^3-q\right ) \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2} \left (a \left (p x^3+q\right )^4+b x^4\right )}{x^7} \, dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (-\frac {a q^5 \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}}{x^7}-\frac {2 a p q^4 \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}}{x^4}+\frac {2 a p^2 q^3 \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}}{x}+2 a p^5 x^8 \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}+7 a p^4 q x^5 \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}+8 a p^3 q^2 x^2 \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}+2 b p \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}-\frac {b q \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}}{x^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -a q^5 \int \frac {\sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}}{x^7}dx-2 a p q^4 \int \frac {\sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}}{x^4}dx+2 a p^2 q^3 \int \frac {\sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}}{x}dx+2 a p^5 \int x^8 \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}dx+\frac {14}{3} a p^3 q^2 \int x \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}dx+a p^3 q^2 \int x^2 \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}dx+2 b p \int \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}dx-b q \int \frac {\sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}}{x^3}dx+\frac {7}{9} a p^2 q \left (p^2 x^6+2 p q x^3-2 p q x^2+q^2\right )^{3/2}\) |
Int[((-q + 2*p*x^3)*Sqrt[q^2 - 2*p*q*x^2 + 2*p*q*x^3 + p^2*x^6]*(b*x^4 + a *(q + p*x^3)^4))/x^7,x]
3.28.45.3.1 Defintions of rubi rules used
Time = 0.28 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.73
| method | result | size |
| pseudoelliptic | \(\frac {-3 p q \,x^{6} \left (a \,p^{2} q^{2}+2 b \right ) \ln \left (\frac {q +p \,x^{3}+\sqrt {p^{2} x^{6}+2 x^{2} p q \left (-1+x \right )+q^{2}}}{x}\right )+\left (a \,p^{4} x^{12}+4 a \,p^{3} q \,x^{9}-\frac {a \,p^{3} q \,x^{8}}{2}+6 a \,p^{2} q^{2} x^{6}-a \,p^{2} q^{2} x^{5}+3 \left (-\frac {a \,p^{2} q^{2}}{2}+b \right ) x^{4}+4 a p \,q^{3} x^{3}-\frac {a p \,q^{3} x^{2}}{2}+a \,q^{4}\right ) \sqrt {p^{2} x^{6}+2 x^{2} p q \left (-1+x \right )+q^{2}}\, \left (p \,x^{3}+q \right )}{6 x^{6}}\) | \(186\) |
int((2*p*x^3-q)*(p^2*x^6+2*p*q*x^3-2*p*q*x^2+q^2)^(1/2)*(b*x^4+a*(p*x^3+q) ^4)/x^7,x,method=_RETURNVERBOSE)
1/6*(-3*p*q*x^6*(a*p^2*q^2+2*b)*ln((q+p*x^3+(p^2*x^6+2*x^2*p*q*(-1+x)+q^2) ^(1/2))/x)+(a*p^4*x^12+4*a*p^3*q*x^9-1/2*a*p^3*q*x^8+6*a*p^2*q^2*x^6-a*p^2 *q^2*x^5+3*(-1/2*a*p^2*q^2+b)*x^4+4*a*p*q^3*x^3-1/2*a*p*q^3*x^2+a*q^4)*(p^ 2*x^6+2*x^2*p*q*(-1+x)+q^2)^(1/2)*(p*x^3+q))/x^6
Timed out. \[ \int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (b x^4+a \left (q+p x^3\right )^4\right )}{x^7} \, dx=\text {Timed out} \]
integrate((2*p*x^3-q)*(p^2*x^6+2*p*q*x^3-2*p*q*x^2+q^2)^(1/2)*(b*x^4+a*(p* x^3+q)^4)/x^7,x, algorithm="fricas")
\[ \int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (b x^4+a \left (q+p x^3\right )^4\right )}{x^7} \, dx=\int \frac {\left (2 p x^{3} - q\right ) \sqrt {p^{2} x^{6} + 2 p q x^{3} - 2 p q x^{2} + q^{2}} \left (a p^{4} x^{12} + 4 a p^{3} q x^{9} + 6 a p^{2} q^{2} x^{6} + 4 a p q^{3} x^{3} + a q^{4} + b x^{4}\right )}{x^{7}}\, dx \]
integrate((2*p*x**3-q)*(p**2*x**6+2*p*q*x**3-2*p*q*x**2+q**2)**(1/2)*(b*x* *4+a*(p*x**3+q)**4)/x**7,x)
Integral((2*p*x**3 - q)*sqrt(p**2*x**6 + 2*p*q*x**3 - 2*p*q*x**2 + q**2)*( a*p**4*x**12 + 4*a*p**3*q*x**9 + 6*a*p**2*q**2*x**6 + 4*a*p*q**3*x**3 + a* q**4 + b*x**4)/x**7, x)
\[ \int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (b x^4+a \left (q+p x^3\right )^4\right )}{x^7} \, dx=\int { \frac {\sqrt {p^{2} x^{6} + 2 \, p q x^{3} - 2 \, p q x^{2} + q^{2}} {\left ({\left (p x^{3} + q\right )}^{4} a + b x^{4}\right )} {\left (2 \, p x^{3} - q\right )}}{x^{7}} \,d x } \]
integrate((2*p*x^3-q)*(p^2*x^6+2*p*q*x^3-2*p*q*x^2+q^2)^(1/2)*(b*x^4+a*(p* x^3+q)^4)/x^7,x, algorithm="maxima")
integrate(sqrt(p^2*x^6 + 2*p*q*x^3 - 2*p*q*x^2 + q^2)*((p*x^3 + q)^4*a + b *x^4)*(2*p*x^3 - q)/x^7, x)
\[ \int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (b x^4+a \left (q+p x^3\right )^4\right )}{x^7} \, dx=\int { \frac {\sqrt {p^{2} x^{6} + 2 \, p q x^{3} - 2 \, p q x^{2} + q^{2}} {\left ({\left (p x^{3} + q\right )}^{4} a + b x^{4}\right )} {\left (2 \, p x^{3} - q\right )}}{x^{7}} \,d x } \]
integrate((2*p*x^3-q)*(p^2*x^6+2*p*q*x^3-2*p*q*x^2+q^2)^(1/2)*(b*x^4+a*(p* x^3+q)^4)/x^7,x, algorithm="giac")
integrate(sqrt(p^2*x^6 + 2*p*q*x^3 - 2*p*q*x^2 + q^2)*((p*x^3 + q)^4*a + b *x^4)*(2*p*x^3 - q)/x^7, x)
Timed out. \[ \int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (b x^4+a \left (q+p x^3\right )^4\right )}{x^7} \, dx=-\int \frac {\left (q-2\,p\,x^3\right )\,\left (a\,{\left (p\,x^3+q\right )}^4+b\,x^4\right )\,\sqrt {p^2\,x^6+2\,p\,q\,x^3-2\,p\,q\,x^2+q^2}}{x^7} \,d x \]
int(-((q - 2*p*x^3)*(a*(q + p*x^3)^4 + b*x^4)*(p^2*x^6 + q^2 - 2*p*q*x^2 + 2*p*q*x^3)^(1/2))/x^7,x)