Integrand size = 57, antiderivative size = 132 \[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^7} \, dx=\frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (2 a q^2+3 b q x^2+4 a p q x^3-4 a p q x^4+3 b p x^5+2 a p^2 x^6\right )}{6 x^6}+2 b p q \log (x)-b p q \log \left (q+p x^3+\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}\right ) \]
1/6*(p^2*x^6-2*p*q*x^4+2*p*q*x^3+q^2)^(1/2)*(2*a*p^2*x^6-4*a*p*q*x^4+3*b*p *x^5+4*a*p*q*x^3+3*b*q*x^2+2*a*q^2)/x^6+2*b*p*q*ln(x)-b*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))
Time = 0.42 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.88 \[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^7} \, dx=\frac {\sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6} \left (3 b x^2 \left (q+p x^3\right )+2 a \left (q^2-2 p q (-1+x) x^3+p^2 x^6\right )\right )}{6 x^6}+2 b p q \log (x)-b p q \log \left (q+p x^3+\sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}\right ) \]
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^7,x]
(Sqrt[q^2 - 2*p*q*(-1 + x)*x^3 + p^2*x^6]*(3*b*x^2*(q + p*x^3) + 2*a*(q^2 - 2*p*q*(-1 + x)*x^3 + p^2*x^6)))/(6*x^6) + 2*b*p*q*Log[x] - b*p*q*Log[q + p*x^3 + Sqrt[q^2 - 2*p*q*(-1 + x)*x^3 + p^2*x^6]]
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.
\(\displaystyle \int \frac {\left (p x^3-2 q\right ) \sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2} \left (a p x^3+a q+b x^2\right )}{x^7} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {a p^2 \sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{x}-\frac {a p q \sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{x^4}-\frac {2 a q^2 \sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{x^7}-\frac {2 b q \sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{x^5}+\frac {b p \sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle a p^2 \int \frac {\sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{x}dx-a p q \int \frac {\sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{x^4}dx-2 a q^2 \int \frac {\sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{x^7}dx-2 b q \int \frac {\sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{x^5}dx+b p \int \frac {\sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{x^2}dx\) |
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^7,x]
3.20.2.3.1 Defintions of rubi rules used
Time = 0.28 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.98
method | result | size |
pseudoelliptic | \(\frac {\sqrt {\frac {p^{2} x^{6}-2 p q \,x^{3} \left (-1+x \right )+q^{2}}{x^{2}}}\, \left (2 a \,p^{2} x^{6}-4 a p q \,x^{4}+3 b p \,x^{5}+4 a p q \,x^{3}+3 b q \,x^{2}+2 a \,q^{2}\right )-6 b p q \ln \left (\frac {p \,x^{3}+\sqrt {\frac {p^{2} x^{6}-2 p q \,x^{3} \left (-1+x \right )+q^{2}}{x^{2}}}\, x +q}{x^{2}}\right ) x^{5}}{6 x^{5}}\) | \(130\) |
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^7,x,method=_RETURNVERBOSE)
1/6*(((p^2*x^6-2*p*q*x^3*(-1+x)+q^2)/x^2)^(1/2)*(2*a*p^2*x^6-4*a*p*q*x^4+3 *b*p*x^5+4*a*p*q*x^3+3*b*q*x^2+2*a*q^2)-6*b*p*q*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^5)/x^5
Timed out. \[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^7} \, dx=\text {Timed out} \]
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^7,x, algorithm="fricas")
\[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^7} \, dx=\int \frac {\left (p x^{3} - 2 q\right ) \left (a p x^{3} + a q + b x^{2}\right ) \sqrt {p^{2} x^{6} - 2 p q x^{4} + 2 p q x^{3} + q^{2}}}{x^{7}}\, dx \]
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**7,x)
Integral((p*x**3 - 2*q)*(a*p*x**3 + a*q + b*x**2)*sqrt(p**2*x**6 - 2*p*q*x **4 + 2*p*q*x**3 + q**2)/x**7, x)
\[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^7} \, dx=\int { \frac {\sqrt {p^{2} x^{6} - 2 \, p q x^{4} + 2 \, p q x^{3} + q^{2}} {\left (a p x^{3} + b x^{2} + a q\right )} {\left (p x^{3} - 2 \, q\right )}}{x^{7}} \,d x } \]
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^7,x, algorithm="maxima")
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)/x^7, x)
\[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^7} \, dx=\int { \frac {\sqrt {p^{2} x^{6} - 2 \, p q x^{4} + 2 \, p q x^{3} + q^{2}} {\left (a p x^{3} + b x^{2} + a q\right )} {\left (p x^{3} - 2 \, q\right )}}{x^{7}} \,d x } \]
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^7,x, algorithm="giac")
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)/x^7, x)
Timed out. \[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^7} \, dx=\int -\frac {\left (2\,q-p\,x^3\right )\,\left (a\,p\,x^3+b\,x^2+a\,q\right )\,\sqrt {p^2\,x^6-2\,p\,q\,x^4+2\,p\,q\,x^3+q^2}}{x^7} \,d x \]
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^7,x)