Integrand size = 56, antiderivative size = 129 \[ \int \frac {\left (-q+2 p x^3\right ) \left (a q+b x+a p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^4} \, dx=\frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (2 a q^2+3 b q x-4 a p q x^2+4 a p q x^3+3 b p x^4+2 a p^2 x^6\right )}{6 x^3}+b p q \log (x)-b p q \log \left (q+p x^3+\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}\right ) \]
1/6*(p^2*x^6+2*p*q*x^3-2*p*q*x^2+q^2)^(1/2)*(2*a*p^2*x^6+4*a*p*q*x^3+3*b*p *x^4-4*a*p*q*x^2+2*a*q^2+3*b*q*x)/x^3+b*p*q*ln(x)-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 = 0.37 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.88 \[ \int \frac {\left (-q+2 p x^3\right ) \left (a q+b x+a p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^4} \, dx=\frac {\sqrt {q^2+2 p q (-1+x) x^2+p^2 x^6} \left (3 b x \left (q+p x^3\right )+2 a \left (q^2+2 p q (-1+x) x^2+p^2 x^6\right )\right )}{6 x^3}+b p q \log (x)-b p q \log \left (q+p x^3+\sqrt {q^2+2 p q (-1+x) x^2+p^2 x^6}\right ) \]
Integrate[((-q + 2*p*x^3)*(a*q + b*x + a*p*x^3)*Sqrt[q^2 - 2*p*q*x^2 + 2*p *q*x^3 + p^2*x^6])/x^4,x]
(Sqrt[q^2 + 2*p*q*(-1 + x)*x^2 + p^2*x^6]*(3*b*x*(q + p*x^3) + 2*a*(q^2 + 2*p*q*(-1 + x)*x^2 + p^2*x^6)))/(6*x^3) + b*p*q*Log[x] - b*p*q*Log[q + p*x ^3 + Sqrt[q^2 + 2*p*q*(-1 + x)*x^2 + 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.
\(\displaystyle \int \frac {\left (2 p x^3-q\right ) \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2} \left (a p x^3+a q+b x\right )}{x^4} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {a p q \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}}{x}+2 a p^2 x^2 \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}-\frac {a q^2 \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}}{x^4}-\frac {b q \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}}{x^3}+2 b p \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 a p^2 \int x^2 \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}dx+a p q \int \frac {\sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}}{x}dx-a q^2 \int \frac {\sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}}{x^4}dx+2 b p \int \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}dx-b q \int \frac {\sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}}{x^3}dx\) |
3.19.75.3.1 Defintions of rubi rules used
Time = 0.21 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.91
method | result | size |
pseudoelliptic | \(\frac {\sqrt {p^{2} x^{6}+2 p q \,x^{2} \left (-1+x \right )+q^{2}}\, \left (2 a \,p^{2} x^{6}+4 a p q \,x^{3}+3 b p \,x^{4}-4 a p q \,x^{2}+2 a \,q^{2}+3 b q x \right )-6 b p q \ln \left (\frac {q +p \,x^{3}+\sqrt {p^{2} x^{6}+2 p q \,x^{2} \left (-1+x \right )+q^{2}}}{x}\right ) x^{3}}{6 x^{3}}\) | \(118\) |
int((2*p*x^3-q)*(a*p*x^3+a*q+b*x)*(p^2*x^6+2*p*q*x^3-2*p*q*x^2+q^2)^(1/2)/ x^4,x,method=_RETURNVERBOSE)
1/6*((p^2*x^6+2*p*q*x^2*(-1+x)+q^2)^(1/2)*(2*a*p^2*x^6+4*a*p*q*x^3+3*b*p*x ^4-4*a*p*q*x^2+2*a*q^2+3*b*q*x)-6*b*p*q*ln((q+p*x^3+(p^2*x^6+2*p*q*x^2*(-1 +x)+q^2)^(1/2))/x)*x^3)/x^3
Timed out. \[ \int \frac {\left (-q+2 p x^3\right ) \left (a q+b x+a p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^4} \, dx=\text {Timed out} \]
integrate((2*p*x^3-q)*(a*p*x^3+a*q+b*x)*(p^2*x^6+2*p*q*x^3-2*p*q*x^2+q^2)^ (1/2)/x^4,x, algorithm="fricas")
\[ \int \frac {\left (-q+2 p x^3\right ) \left (a q+b x+a p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^4} \, dx=\int \frac {\left (2 p x^{3} - q\right ) \left (a p x^{3} + a q + b x\right ) \sqrt {p^{2} x^{6} + 2 p q x^{3} - 2 p q x^{2} + q^{2}}}{x^{4}}\, dx \]
integrate((2*p*x**3-q)*(a*p*x**3+a*q+b*x)*(p**2*x**6+2*p*q*x**3-2*p*q*x**2 +q**2)**(1/2)/x**4,x)
Integral((2*p*x**3 - q)*(a*p*x**3 + a*q + b*x)*sqrt(p**2*x**6 + 2*p*q*x**3 - 2*p*q*x**2 + q**2)/x**4, x)
\[ \int \frac {\left (-q+2 p x^3\right ) \left (a q+b x+a p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^4} \, dx=\int { \frac {\sqrt {p^{2} x^{6} + 2 \, p q x^{3} - 2 \, p q x^{2} + q^{2}} {\left (a p x^{3} + a q + b x\right )} {\left (2 \, p x^{3} - q\right )}}{x^{4}} \,d x } \]
integrate((2*p*x^3-q)*(a*p*x^3+a*q+b*x)*(p^2*x^6+2*p*q*x^3-2*p*q*x^2+q^2)^ (1/2)/x^4,x, algorithm="maxima")
integrate(sqrt(p^2*x^6 + 2*p*q*x^3 - 2*p*q*x^2 + q^2)*(a*p*x^3 + a*q + b*x )*(2*p*x^3 - q)/x^4, x)
\[ \int \frac {\left (-q+2 p x^3\right ) \left (a q+b x+a p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^4} \, dx=\int { \frac {\sqrt {p^{2} x^{6} + 2 \, p q x^{3} - 2 \, p q x^{2} + q^{2}} {\left (a p x^{3} + a q + b x\right )} {\left (2 \, p x^{3} - q\right )}}{x^{4}} \,d x } \]
integrate((2*p*x^3-q)*(a*p*x^3+a*q+b*x)*(p^2*x^6+2*p*q*x^3-2*p*q*x^2+q^2)^ (1/2)/x^4,x, algorithm="giac")
integrate(sqrt(p^2*x^6 + 2*p*q*x^3 - 2*p*q*x^2 + q^2)*(a*p*x^3 + a*q + b*x )*(2*p*x^3 - q)/x^4, x)
Timed out. \[ \int \frac {\left (-q+2 p x^3\right ) \left (a q+b x+a p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^4} \, dx=-\int \frac {\left (q-2\,p\,x^3\right )\,\left (a\,p\,x^3+b\,x+a\,q\right )\,\sqrt {p^2\,x^6+2\,p\,q\,x^3-2\,p\,q\,x^2+q^2}}{x^4} \,d x \]
int(-((q - 2*p*x^3)*(a*q + b*x + a*p*x^3)*(p^2*x^6 + q^2 - 2*p*q*x^2 + 2*p *q*x^3)^(1/2))/x^4,x)