Integrand size = 74, antiderivative size = 251 \[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5 \left (c q+d x^2+c p x^3\right )} \, dx=\frac {\left (a c q+2 b c x^2-2 a d x^2+a c p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{2 c^2 x^4}-\frac {2 (-b c+a d) \sqrt {-d^2+2 c^2 p q} \arctan \left (\frac {\sqrt {-d^2+2 c^2 p q} x^2}{c q+d x^2+c p x^3+c \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}\right )}{c^3}+\frac {2 \left (b c d-a d^2+a c^2 p q\right ) \log (x)}{c^3}+\frac {\left (-b c d+a d^2-a c^2 p q\right ) \log \left (q+p x^3+\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}\right )}{c^3} \] Output:
1/2*(a*c*p*x^3-2*a*d*x^2+2*b*c*x^2+a*c*q)*(p^2*x^6-2*p*q*x^4+2*p*q*x^3+q^2 )^(1/2)/c^2/x^4-2*(a*d-b*c)*(2*c^2*p*q-d^2)^(1/2)*arctan((2*c^2*p*q-d^2)^( 1/2)*x^2/(c*q+d*x^2+c*p*x^3+c*(p^2*x^6-2*p*q*x^4+2*p*q*x^3+q^2)^(1/2)))/c^ 3+2*(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^4+2*p*q*x^3+q^2)^(1/2))/c^3
Time = 1.47 (sec) , antiderivative size = 398, normalized size of antiderivative = 1.59 \[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5 \left (c q+d x^2+c p x^3\right )} \, dx=\frac {a c^2 q \sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}+2 b c^2 x^2 \sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}-2 a c d x^2 \sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}+a c^2 p x^3 \sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}-4 (-b c+a d) \sqrt {-d^2+2 c^2 p q} x^4 \arctan \left (\frac {\sqrt {-d^2+2 c^2 p q} x^2}{d x^2+c \left (q+p x^3+\sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}\right )}\right )+4 b c d x^4 \log (x)-4 a d^2 x^4 \log (x)+4 a c^2 p q x^4 \log (x)-2 b c d x^4 \log \left (q+p x^3+\sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}\right )+2 a d^2 x^4 \log \left (q+p x^3+\sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}\right )-2 a c^2 p q x^4 \log \left (q+p x^3+\sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}\right )}{2 c^3 x^4} \] Input:
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^5*(c*q + d*x^2 + c*p*x^3)),x]
Output:
(a*c^2*q*Sqrt[q^2 - 2*p*q*(-1 + x)*x^3 + p^2*x^6] + 2*b*c^2*x^2*Sqrt[q^2 - 2*p*q*(-1 + x)*x^3 + p^2*x^6] - 2*a*c*d*x^2*Sqrt[q^2 - 2*p*q*(-1 + x)*x^3 + p^2*x^6] + a*c^2*p*x^3*Sqrt[q^2 - 2*p*q*(-1 + x)*x^3 + p^2*x^6] - 4*(-( b*c) + a*d)*Sqrt[-d^2 + 2*c^2*p*q]*x^4*ArcTan[(Sqrt[-d^2 + 2*c^2*p*q]*x^2) /(d*x^2 + c*(q + p*x^3 + Sqrt[q^2 - 2*p*q*(-1 + x)*x^3 + p^2*x^6]))] + 4*b *c*d*x^4*Log[x] - 4*a*d^2*x^4*Log[x] + 4*a*c^2*p*q*x^4*Log[x] - 2*b*c*d*x^ 4*Log[q + p*x^3 + Sqrt[q^2 - 2*p*q*(-1 + x)*x^3 + p^2*x^6]] + 2*a*d^2*x^4* Log[q + p*x^3 + Sqrt[q^2 - 2*p*q*(-1 + x)*x^3 + p^2*x^6]] - 2*a*c^2*p*q*x^ 4*Log[q + p*x^3 + Sqrt[q^2 - 2*p*q*(-1 + x)*x^3 + p^2*x^6]])/(2*c^3*x^4)
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.
\(\displaystyle \int \frac {\left (p x^3-2 q\right ) \sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2} \left (a p x^3+a q+b x^2\right )}{x^5 \left (c p x^3+c q+d x^2\right )} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 d \sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2} (b c-a d)}{c^3 q x}-\frac {2 \sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2} (b c-a d)}{c^2 x^3}+\frac {\sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2} (b c-a d) \left (3 c^2 p q-2 c d p x^2-2 d^2 x\right )}{c^3 q \left (c p x^3+c q+d x^2\right )}-\frac {2 a q \sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{c x^5}+\frac {a p \sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{c x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 d^2 (b c-a d) \int \frac {x \sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{c p x^3+d x^2+c q}dx}{c^3 q}+\frac {2 d (b c-a d) \int \frac {\sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{x}dx}{c^3 q}-\frac {2 (b c-a d) \int \frac {\sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{x^3}dx}{c^2}-\frac {2 d p (b c-a d) \int \frac {x^2 \sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{c p x^3+d x^2+c q}dx}{c^2 q}+\frac {3 p (b c-a d) \int \frac {\sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{c p x^3+d x^2+c q}dx}{c}-\frac {2 a q \int \frac {\sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{x^5}dx}{c}+\frac {a p \int \frac {\sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{x^2}dx}{c}\) |
Input:
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^5*(c*q + d*x^2 + c*p*x^3)),x]
Output:
$Aborted
Time = 0.52 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.18
method | result | size |
pseudoelliptic | \(-\frac {2 \left (-\frac {c^{2} \sqrt {\frac {-2 c^{2} p q +d^{2}}{c^{2}}}\, \left (\left (a p \,x^{3}+2 b \,x^{2}+a q \right ) c -2 a d \,x^{2}\right ) \sqrt {\frac {p^{2} x^{6}-2 q \,x^{3} \left (-1+x \right ) p +q^{2}}{x^{2}}}}{4}+x^{3} \left (\frac {c \ln \left (\frac {p \,x^{3}+\sqrt {\frac {p^{2} x^{6}-2 q \,x^{3} \left (-1+x \right ) p +q^{2}}{x^{2}}}\, x +q}{x^{2}}\right ) \left (a \,c^{2} p q -a \,d^{2}+b c d \right ) \sqrt {\frac {-2 c^{2} p q +d^{2}}{c^{2}}}}{2}+\left (a d -b c \right ) \left (\ln \left (\frac {-2 c p q \,x^{2}-d p \,x^{3}+\sqrt {\frac {-2 c^{2} p q +d^{2}}{c^{2}}}\, \sqrt {\frac {p^{2} x^{6}-2 q \,x^{3} \left (-1+x \right ) p +q^{2}}{x^{2}}}\, c x -d q}{c p \,x^{3}+d \,x^{2}+c q}\right )+\ln \left (2\right )\right ) \left (c^{2} p q -\frac {d^{2}}{2}\right )\right )\right )}{\sqrt {\frac {-2 c^{2} p q +d^{2}}{c^{2}}}\, c^{4} x^{3}}\) | \(296\) |
Input:
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^5/(c*p*x^3+d*x^2+c*q),x,method=_RETURNVERBOSE)
Output:
-2/((-2*c^2*p*q+d^2)/c^2)^(1/2)*(-1/4*c^2*((-2*c^2*p*q+d^2)/c^2)^(1/2)*((a *p*x^3+2*b*x^2+a*q)*c-2*a*d*x^2)*((p^2*x^6-2*q*x^3*(-1+x)*p+q^2)/x^2)^(1/2 )+x^3*(1/2*c*ln((p*x^3+((p^2*x^6-2*q*x^3*(-1+x)*p+q^2)/x^2)^(1/2)*x+q)/x^2 )*(a*c^2*p*q-a*d^2+b*c*d)*((-2*c^2*p*q+d^2)/c^2)^(1/2)+(a*d-b*c)*(ln((-2*c *p*q*x^2-d*p*x^3+((-2*c^2*p*q+d^2)/c^2)^(1/2)*((p^2*x^6-2*q*x^3*(-1+x)*p+q ^2)/x^2)^(1/2)*c*x-d*q)/(c*p*x^3+d*x^2+c*q))+ln(2))*(c^2*p*q-1/2*d^2)))/c^ 4/x^3
Timed out. \[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5 \left (c q+d x^2+c p x^3\right )} \, dx=\text {Timed out} \] Input:
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^5/(c*p*x^3+d*x^2+c*q),x, algorithm="fricas")
Output:
Timed out
Timed out. \[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5 \left (c q+d x^2+c p x^3\right )} \, dx=\text {Timed out} \] Input:
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**5/(c*p*x**3+d*x**2+c*q),x)
Output:
Timed out
\[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5 \left (c q+d x^2+c p x^3\right )} \, dx=\int { \frac {\sqrt {p^{2} x^{6} - 2 \, p q x^{4} + 2 \, p q x^{3} + q^{2}} {\left (a p x^{3} + b x^{2} + a q\right )} {\left (p x^{3} - 2 \, q\right )}}{{\left (c p x^{3} + d x^{2} + c q\right )} x^{5}} \,d x } \] Input:
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^5/(c*p*x^3+d*x^2+c*q),x, algorithm="maxima")
Output:
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)/((c*p*x^3 + d*x^2 + c*q)*x^5), x)
\[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5 \left (c q+d x^2+c p x^3\right )} \, dx=\int { \frac {\sqrt {p^{2} x^{6} - 2 \, p q x^{4} + 2 \, p q x^{3} + q^{2}} {\left (a p x^{3} + b x^{2} + a q\right )} {\left (p x^{3} - 2 \, q\right )}}{{\left (c p x^{3} + d x^{2} + c q\right )} x^{5}} \,d x } \] Input:
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^5/(c*p*x^3+d*x^2+c*q),x, algorithm="giac")
Output:
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)/((c*p*x^3 + d*x^2 + c*q)*x^5), x)
Timed out. \[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5 \left (c q+d x^2+c p x^3\right )} \, dx=\int -\frac {\left (2\,q-p\,x^3\right )\,\left (a\,p\,x^3+b\,x^2+a\,q\right )\,\sqrt {p^2\,x^6-2\,p\,q\,x^4+2\,p\,q\,x^3+q^2}}{x^5\,\left (c\,p\,x^3+d\,x^2+c\,q\right )} \,d x \] Input:
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^5*(c*q + d*x^2 + c*p*x^3)),x)
Output:
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^5*(c*q + d*x^2 + c*p*x^3)), x)
\[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5 \left (c q+d x^2+c p x^3\right )} \, dx=\int \frac {\left (p \,x^{3}-2 q \right ) \left (a p \,x^{3}+b \,x^{2}+a q \right ) \sqrt {p^{2} x^{6}-2 p q \,x^{4}+2 p q \,x^{3}+q^{2}}}{x^{5} \left (c p \,x^{3}+d \,x^{2}+c q \right )}d x \] Input:
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^5/(c*p*x^3+d*x^2+c*q),x)
Output:
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^5/(c*p*x^3+d*x^2+c*q),x)