Integrand size = 60, antiderivative size = 181 \[ \int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (b x^2+a \left (q+p x^3\right )^2\right )}{x^5} \, dx=\frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (a q^3+2 b q x^2-a p q^2 x^2+3 a p q^2 x^3+2 b p x^5-a p^2 q x^5+3 a p^2 q x^6+a p^3 x^9\right )}{4 x^4}+\frac {1}{2} \left (2 b p q+a p^2 q^2\right ) \log (x)+\frac {1}{2} \left (-2 b p q-a p^2 q^2\right ) \log \left (q+p x^3+\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}\right ) \]
1/4*(p^2*x^6+2*p*q*x^3-2*p*q*x^2+q^2)^(1/2)*(a*p^3*x^9+3*a*p^2*q*x^6-a*p^2 *q*x^5+3*a*p*q^2*x^3+2*b*p*x^5-a*p*q^2*x^2+a*q^3+2*b*q*x^2)/x^4+1/2*(a*p^2 *q^2+2*b*p*q)*ln(x)+1/2*(-a*p^2*q^2-2*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 = 0.33 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.74 \[ \int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (b x^2+a \left (q+p x^3\right )^2\right )}{x^5} \, dx=\frac {\left (q+p x^3\right ) \sqrt {q^2+2 p q (-1+x) x^2+p^2 x^6} \left (2 b x^2+a \left (q^2+p^2 x^6+p q x^2 (-1+2 x)\right )\right )}{4 x^4}+\frac {1}{2} p q (2 b+a p q) \log (x)-\frac {1}{2} p q (2 b+a p q) \log \left (q+p x^3+\sqrt {q^2+2 p q (-1+x) x^2+p^2 x^6}\right ) \]
Integrate[((-q + 2*p*x^3)*Sqrt[q^2 - 2*p*q*x^2 + 2*p*q*x^3 + p^2*x^6]*(b*x ^2 + a*(q + p*x^3)^2))/x^5,x]
((q + p*x^3)*Sqrt[q^2 + 2*p*q*(-1 + x)*x^2 + p^2*x^6]*(2*b*x^2 + a*(q^2 + p^2*x^6 + p*q*x^2*(-1 + 2*x))))/(4*x^4) + (p*q*(2*b + a*p*q)*Log[x])/2 - ( p*q*(2*b + a*p*q)*Log[q + p*x^3 + Sqrt[q^2 + 2*p*q*(-1 + x)*x^2 + p^2*x^6] ])/2
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.
| \(\displaystyle \int \frac {\left (2 p x^3-q\right ) \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2} \left (a \left (p x^3+q\right )^2+b x^2\right )}{x^5} \, dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (3 a p^2 q x \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}-\frac {a q^3 \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}}{x^5}+2 a p^3 x^4 \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}+2 b p \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}-\frac {b q \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}}{x^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 3 a p^2 q \int x \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}dx-a q^3 \int \frac {\sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}}{x^5}dx+2 a p^3 \int x^4 \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}dx+2 b p \int \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}dx-b q \int \frac {\sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}}{x^3}dx\) |
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^2 + a *(q + p*x^3)^2))/x^5,x]
3.24.28.3.1 Defintions of rubi rules used
Time = 0.23 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.68
| method | result | size |
| pseudoelliptic | \(\frac {-2 p q \,x^{4} \left (a p q +2 b \right ) \ln \left (\frac {q +p \,x^{3}+\sqrt {p^{2} x^{6}+2 p q \,x^{2} \left (-1+x \right )+q^{2}}}{x}\right )+\left (p \,x^{3}+q \right ) \left (a \,p^{2} x^{6}+2 a p q \,x^{3}+\left (-a p q +2 b \right ) x^{2}+a \,q^{2}\right ) \sqrt {p^{2} x^{6}+2 p q \,x^{2} \left (-1+x \right )+q^{2}}}{4 x^{4}}\) | \(123\) |
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^2+a*(p*x^3+q) ^2)/x^5,x,method=_RETURNVERBOSE)
1/4*(-2*p*q*x^4*(a*p*q+2*b)*ln((q+p*x^3+(p^2*x^6+2*p*q*x^2*(-1+x)+q^2)^(1/ 2))/x)+(p*x^3+q)*(a*p^2*x^6+2*a*p*q*x^3+(-a*p*q+2*b)*x^2+a*q^2)*(p^2*x^6+2 *p*q*x^2*(-1+x)+q^2)^(1/2))/x^4
Timed out. \[ \int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (b x^2+a \left (q+p x^3\right )^2\right )}{x^5} \, dx=\text {Timed out} \]
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^2+a*(p* x^3+q)^2)/x^5,x, algorithm="fricas")
\[ \int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (b x^2+a \left (q+p x^3\right )^2\right )}{x^5} \, dx=\int \frac {\left (2 p x^{3} - q\right ) \sqrt {p^{2} x^{6} + 2 p q x^{3} - 2 p q x^{2} + q^{2}} \left (a p^{2} x^{6} + 2 a p q x^{3} + a q^{2} + b x^{2}\right )}{x^{5}}\, dx \]
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* *2+a*(p*x**3+q)**2)/x**5,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**2*x**6 + 2*a*p*q*x**3 + a*q**2 + b*x**2)/x**5, x)
\[ \int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (b x^2+a \left (q+p x^3\right )^2\right )}{x^5} \, dx=\int { \frac {\sqrt {p^{2} x^{6} + 2 \, p q x^{3} - 2 \, p q x^{2} + q^{2}} {\left (2 \, p x^{3} - q\right )} {\left ({\left (p x^{3} + q\right )}^{2} a + b x^{2}\right )}}{x^{5}} \,d x } \]
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^2+a*(p* x^3+q)^2)/x^5,x, algorithm="maxima")
integrate(sqrt(p^2*x^6 + 2*p*q*x^3 - 2*p*q*x^2 + q^2)*(2*p*x^3 - q)*((p*x^ 3 + q)^2*a + b*x^2)/x^5, x)
\[ \int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (b x^2+a \left (q+p x^3\right )^2\right )}{x^5} \, dx=\int { \frac {\sqrt {p^{2} x^{6} + 2 \, p q x^{3} - 2 \, p q x^{2} + q^{2}} {\left (2 \, p x^{3} - q\right )} {\left ({\left (p x^{3} + q\right )}^{2} a + b x^{2}\right )}}{x^{5}} \,d x } \]
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^2+a*(p* x^3+q)^2)/x^5,x, algorithm="giac")
integrate(sqrt(p^2*x^6 + 2*p*q*x^3 - 2*p*q*x^2 + q^2)*(2*p*x^3 - q)*((p*x^ 3 + q)^2*a + b*x^2)/x^5, x)
Timed out. \[ \int \frac {\left (-q+2 p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (b x^2+a \left (q+p x^3\right )^2\right )}{x^5} \, dx=-\int \frac {\left (q-2\,p\,x^3\right )\,\left (a\,{\left (p\,x^3+q\right )}^2+b\,x^2\right )\,\sqrt {p^2\,x^6+2\,p\,q\,x^3-2\,p\,q\,x^2+q^2}}{x^5} \,d x \]
int(-((q - 2*p*x^3)*(a*(q + p*x^3)^2 + b*x^2)*(p^2*x^6 + q^2 - 2*p*q*x^2 + 2*p*q*x^3)^(1/2))/x^5,x)