Integrand size = 71, antiderivative size = 242 \[ \int \frac {\left (-q+2 p x^3\right ) \left (a q+b x+a p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^3 \left (c q+d x+c p x^3\right )} \, dx=\frac {\left (a c q+2 b c x-2 a d x+a c p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{2 c^2 x^2}-\frac {2 (-b c+a d) \sqrt {-d^2+2 c^2 p q} \arctan \left (\frac {\sqrt {-d^2+2 c^2 p q} x}{c q+d x+c p x^3+c \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}\right )}{c^3}+\frac {\left (b c d-a d^2+a c^2 p q\right ) \log (x)}{c^3}+\frac {\left (-b c d+a d^2-a c^2 p q\right ) \log \left (q+p x^3+\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}\right )}{c^3} \]
1/2*(a*c*p*x^3+a*c*q-2*a*d*x+2*b*c*x)*(p^2*x^6+2*p*q*x^3-2*p*q*x^2+q^2)^(1 /2)/c^2/x^2-2*(a*d-b*c)*(2*c^2*p*q-d^2)^(1/2)*arctan((2*c^2*p*q-d^2)^(1/2) *x/(c*q+d*x+c*p*x^3+c*(p^2*x^6+2*p*q*x^3-2*p*q*x^2+q^2)^(1/2)))/c^3+(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^3-2*p*q*x^2+q^2)^(1/2))/c^3
Time = 1.15 (sec) , antiderivative size = 390, normalized size of antiderivative = 1.61 \[ \int \frac {\left (-q+2 p x^3\right ) \left (a q+b x+a p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^3 \left (c q+d x+c p x^3\right )} \, dx=\frac {a c^2 q \sqrt {q^2+2 p q (-1+x) x^2+p^2 x^6}+2 b c^2 x \sqrt {q^2+2 p q (-1+x) x^2+p^2 x^6}-2 a c d x \sqrt {q^2+2 p q (-1+x) x^2+p^2 x^6}+a c^2 p x^3 \sqrt {q^2+2 p q (-1+x) x^2+p^2 x^6}-4 (-b c+a d) \sqrt {-d^2+2 c^2 p q} x^2 \arctan \left (\frac {\sqrt {-d^2+2 c^2 p q} x}{d x+c \left (q+p x^3+\sqrt {q^2+2 p q (-1+x) x^2+p^2 x^6}\right )}\right )+2 b c d x^2 \log (x)-2 a d^2 x^2 \log (x)+2 a c^2 p q x^2 \log (x)-2 b c d x^2 \log \left (q+p x^3+\sqrt {q^2+2 p q (-1+x) x^2+p^2 x^6}\right )+2 a d^2 x^2 \log \left (q+p x^3+\sqrt {q^2+2 p q (-1+x) x^2+p^2 x^6}\right )-2 a c^2 p q x^2 \log \left (q+p x^3+\sqrt {q^2+2 p q (-1+x) x^2+p^2 x^6}\right )}{2 c^3 x^2} \]
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^3*(c*q + d*x + c*p*x^3)),x]
(a*c^2*q*Sqrt[q^2 + 2*p*q*(-1 + x)*x^2 + p^2*x^6] + 2*b*c^2*x*Sqrt[q^2 + 2 *p*q*(-1 + x)*x^2 + p^2*x^6] - 2*a*c*d*x*Sqrt[q^2 + 2*p*q*(-1 + x)*x^2 + p ^2*x^6] + a*c^2*p*x^3*Sqrt[q^2 + 2*p*q*(-1 + x)*x^2 + p^2*x^6] - 4*(-(b*c) + a*d)*Sqrt[-d^2 + 2*c^2*p*q]*x^2*ArcTan[(Sqrt[-d^2 + 2*c^2*p*q]*x)/(d*x + c*(q + p*x^3 + Sqrt[q^2 + 2*p*q*(-1 + x)*x^2 + p^2*x^6]))] + 2*b*c*d*x^2 *Log[x] - 2*a*d^2*x^2*Log[x] + 2*a*c^2*p*q*x^2*Log[x] - 2*b*c*d*x^2*Log[q + p*x^3 + Sqrt[q^2 + 2*p*q*(-1 + x)*x^2 + p^2*x^6]] + 2*a*d^2*x^2*Log[q + p*x^3 + Sqrt[q^2 + 2*p*q*(-1 + x)*x^2 + p^2*x^6]] - 2*a*c^2*p*q*x^2*Log[q + p*x^3 + Sqrt[q^2 + 2*p*q*(-1 + x)*x^2 + p^2*x^6]])/(2*c^3*x^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.
| \(\displaystyle \int \frac {\left (2 p x^3-q\right ) \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2} \left (a p x^3+a q+b x\right )}{x^3 \left (c p x^3+c q+d x\right )} \, dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {d \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2} (b c-a d)}{c^3 q x}+\frac {\sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2} (a d-b c)}{c^2 x^2}+\frac {\sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2} (b c-a d) \left (3 c^2 p q x-c d p x^2-d^2\right )}{c^3 q \left (c p x^3+c q+d x\right )}+\frac {2 a p \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}}{c}-\frac {a q \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}}{c x^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {d^2 (b c-a d) \int \frac {\sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}}{c p x^3+d x+c q}dx}{c^3 q}+\frac {d (b c-a d) \int \frac {\sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}}{x}dx}{c^3 q}-\frac {d p (b c-a d) \int \frac {x^2 \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}}{c p x^3+d x+c q}dx}{c^2 q}-\frac {(b c-a d) \int \frac {\sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}}{x^2}dx}{c^2}+\frac {3 p (b c-a d) \int \frac {x \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}}{c p x^3+d x+c q}dx}{c}+\frac {2 a p \int \sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}dx}{c}-\frac {a q \int \frac {\sqrt {p^2 x^6+2 p q x^3-2 p q x^2+q^2}}{x^3}dx}{c}\) |
Int[((-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^3*(c*q + d*x + c*p*x^3)),x]
3.27.78.3.1 Defintions of rubi rules used
Time = 0.32 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.12
| method | result | size |
| pseudoelliptic | \(-\frac {2 \left (-\frac {c^{2} \sqrt {\frac {-2 c^{2} p q +d^{2}}{c^{2}}}\, \left (\left (a p \,x^{3}+a q +2 b x \right ) c -2 a d x \right ) \sqrt {p^{2} x^{6}+2 x^{2} p q \left (-1+x \right )+q^{2}}}{4}+\left (\frac {c \left (a \,c^{2} p q -a \,d^{2}+b c d \right ) \ln \left (\frac {q +p \,x^{3}+\sqrt {p^{2} x^{6}+2 x^{2} p q \left (-1+x \right )+q^{2}}}{x}\right ) \sqrt {\frac {-2 c^{2} p q +d^{2}}{c^{2}}}}{2}+\left (\ln \left (\frac {\sqrt {p^{2} x^{6}+2 x^{2} p q \left (-1+x \right )+q^{2}}\, \sqrt {\frac {-2 c^{2} p q +d^{2}}{c^{2}}}\, c -2 c p q x -\left (p \,x^{3}+q \right ) d}{\left (p \,x^{3}+q \right ) c +d x}\right )+\ln \left (2\right )\right ) \left (c^{2} p q -\frac {d^{2}}{2}\right ) \left (a d -b c \right )\right ) x^{2}\right )}{\sqrt {\frac {-2 c^{2} p q +d^{2}}{c^{2}}}\, c^{4} x^{2}}\) | \(272\) |
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^3/(c*p*x^3+c*q+d*x),x,method=_RETURNVERBOSE)
-2*(-1/4*c^2*((-2*c^2*p*q+d^2)/c^2)^(1/2)*((a*p*x^3+a*q+2*b*x)*c-2*a*d*x)* (p^2*x^6+2*x^2*p*q*(-1+x)+q^2)^(1/2)+(1/2*c*(a*c^2*p*q-a*d^2+b*c*d)*ln((q+ p*x^3+(p^2*x^6+2*x^2*p*q*(-1+x)+q^2)^(1/2))/x)*((-2*c^2*p*q+d^2)/c^2)^(1/2 )+(ln(((p^2*x^6+2*x^2*p*q*(-1+x)+q^2)^(1/2)*((-2*c^2*p*q+d^2)/c^2)^(1/2)*c -2*c*p*q*x-(p*x^3+q)*d)/((p*x^3+q)*c+d*x))+ln(2))*(c^2*p*q-1/2*d^2)*(a*d-b *c))*x^2)/((-2*c^2*p*q+d^2)/c^2)^(1/2)/c^4/x^2
Timed out. \[ \int \frac {\left (-q+2 p x^3\right ) \left (a q+b x+a p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^3 \left (c q+d x+c p x^3\right )} \, dx=\text {Timed out} \]
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^3/(c*p*x^3+c*q+d*x),x, algorithm="fricas")
Timed out. \[ \int \frac {\left (-q+2 p x^3\right ) \left (a q+b x+a p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^3 \left (c q+d x+c p x^3\right )} \, dx=\text {Timed out} \]
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**3/(c*p*x**3+c*q+d*x),x)
\[ \int \frac {\left (-q+2 p x^3\right ) \left (a q+b x+a p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^3 \left (c q+d x+c p x^3\right )} \, dx=\int { \frac {\sqrt {p^{2} x^{6} + 2 \, p q x^{3} - 2 \, p q x^{2} + q^{2}} {\left (a p x^{3} + a q + b x\right )} {\left (2 \, p x^{3} - q\right )}}{{\left (c p x^{3} + c q + d x\right )} x^{3}} \,d x } \]
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^3/(c*p*x^3+c*q+d*x),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)/((c*p*x^3 + c*q + d*x)*x^3), x)
\[ \int \frac {\left (-q+2 p x^3\right ) \left (a q+b x+a p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^3 \left (c q+d x+c p x^3\right )} \, dx=\int { \frac {\sqrt {p^{2} x^{6} + 2 \, p q x^{3} - 2 \, p q x^{2} + q^{2}} {\left (a p x^{3} + a q + b x\right )} {\left (2 \, p x^{3} - q\right )}}{{\left (c p x^{3} + c q + d x\right )} x^{3}} \,d x } \]
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^3/(c*p*x^3+c*q+d*x),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)/((c*p*x^3 + c*q + d*x)*x^3), x)
Timed out. \[ \int \frac {\left (-q+2 p x^3\right ) \left (a q+b x+a p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^3 \left (c q+d x+c p x^3\right )} \, dx=-\int \frac {\left (q-2\,p\,x^3\right )\,\left (a\,p\,x^3+b\,x+a\,q\right )\,\sqrt {p^2\,x^6+2\,p\,q\,x^3-2\,p\,q\,x^2+q^2}}{x^3\,\left (c\,p\,x^3+d\,x+c\,q\right )} \,d x \]
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^3*(c*q + d*x + c*p*x^3)),x)