\(\int \frac {(-2 q+p x^3) (a q+b x^2+a p x^3) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5 (c q+d x^2+c p x^3)} \, dx\) [2731]

   Optimal result
   Rubi [F]
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [F(-1)]
   Sympy [F(-1)]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 74, antiderivative size = 251 \[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5 \left (c q+d x^2+c p x^3\right )} \, dx=\frac {\left (a c q+2 b c x^2-2 a d x^2+a c p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{2 c^2 x^4}-\frac {2 (-b c+a d) \sqrt {-d^2+2 c^2 p q} \arctan \left (\frac {\sqrt {-d^2+2 c^2 p q} x^2}{c q+d x^2+c p x^3+c \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}\right )}{c^3}+\frac {2 \left (b c d-a d^2+a c^2 p q\right ) \log (x)}{c^3}+\frac {\left (-b c d+a d^2-a c^2 p q\right ) \log \left (q+p x^3+\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}\right )}{c^3} \]

[Out]

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

Rubi [F]

\[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5 \left (c q+d x^2+c p x^3\right )} \, dx=\int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5 \left (c q+d x^2+c p x^3\right )} \, dx \]

[In]

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]

[Out]

(-2*a*q*Defer[Int][Sqrt[q^2 + 2*p*q*x^3 - 2*p*q*x^4 + p^2*x^6]/x^5, x])/c - (2*(b*c - a*d)*Defer[Int][Sqrt[q^2
 + 2*p*q*x^3 - 2*p*q*x^4 + p^2*x^6]/x^3, x])/c^2 + (a*p*Defer[Int][Sqrt[q^2 + 2*p*q*x^3 - 2*p*q*x^4 + p^2*x^6]
/x^2, x])/c + (2*d*(b*c - a*d)*Defer[Int][Sqrt[q^2 + 2*p*q*x^3 - 2*p*q*x^4 + p^2*x^6]/x, x])/(c^3*q) + (3*(b*c
 - a*d)*p*Defer[Int][Sqrt[q^2 + 2*p*q*x^3 - 2*p*q*x^4 + p^2*x^6]/(c*q + d*x^2 + c*p*x^3), x])/c - (2*d^2*(b*c
- a*d)*Defer[Int][(x*Sqrt[q^2 + 2*p*q*x^3 - 2*p*q*x^4 + p^2*x^6])/(c*q + d*x^2 + c*p*x^3), x])/(c^3*q) - (2*d*
(b*c - a*d)*p*Defer[Int][(x^2*Sqrt[q^2 + 2*p*q*x^3 - 2*p*q*x^4 + p^2*x^6])/(c*q + d*x^2 + c*p*x^3), x])/(c^2*q
)

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {2 a q \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{c x^5}-\frac {2 (b c-a d) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{c^2 x^3}+\frac {a p \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{c x^2}+\frac {2 d (b c-a d) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{c^3 q x}+\frac {(b c-a d) \left (3 c^2 p q-2 d^2 x-2 c d p x^2\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{c^3 q \left (c q+d x^2+c p x^3\right )}\right ) \, dx \\ & = -\frac {(2 (b c-a d)) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^3} \, dx}{c^2}+\frac {(a p) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^2} \, dx}{c}+\frac {(b c-a d) \int \frac {\left (3 c^2 p q-2 d^2 x-2 c d p x^2\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{c q+d x^2+c p x^3} \, dx}{c^3 q}+\frac {(2 d (b c-a d)) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x} \, dx}{c^3 q}-\frac {(2 a q) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5} \, dx}{c} \\ & = -\frac {(2 (b c-a d)) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^3} \, dx}{c^2}+\frac {(a p) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^2} \, dx}{c}+\frac {(b c-a d) \int \left (\frac {3 c^2 p q \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{c q+d x^2+c p x^3}-\frac {2 d^2 x \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{c q+d x^2+c p x^3}-\frac {2 c d p x^2 \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{c q+d x^2+c p x^3}\right ) \, dx}{c^3 q}+\frac {(2 d (b c-a d)) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x} \, dx}{c^3 q}-\frac {(2 a q) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5} \, dx}{c} \\ & = -\frac {(2 (b c-a d)) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^3} \, dx}{c^2}+\frac {(a p) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^2} \, dx}{c}+\frac {(3 (b c-a d) p) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{c q+d x^2+c p x^3} \, dx}{c}+\frac {(2 d (b c-a d)) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x} \, dx}{c^3 q}-\frac {\left (2 d^2 (b c-a d)\right ) \int \frac {x \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{c q+d x^2+c p x^3} \, dx}{c^3 q}-\frac {(2 d (b c-a d) p) \int \frac {x^2 \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{c q+d x^2+c p x^3} \, dx}{c^2 q}-\frac {(2 a q) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5} \, dx}{c} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.16 (sec) , antiderivative size = 398, normalized size of antiderivative = 1.59 \[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5 \left (c q+d x^2+c p x^3\right )} \, dx=\frac {a c^2 q \sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}+2 b c^2 x^2 \sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}-2 a c d x^2 \sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}+a c^2 p x^3 \sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}-4 (-b c+a d) \sqrt {-d^2+2 c^2 p q} x^4 \arctan \left (\frac {\sqrt {-d^2+2 c^2 p q} x^2}{d x^2+c \left (q+p x^3+\sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}\right )}\right )+4 b c d x^4 \log (x)-4 a d^2 x^4 \log (x)+4 a c^2 p q x^4 \log (x)-2 b c d x^4 \log \left (q+p x^3+\sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}\right )+2 a d^2 x^4 \log \left (q+p x^3+\sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}\right )-2 a c^2 p q x^4 \log \left (q+p x^3+\sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}\right )}{2 c^3 x^4} \]

[In]

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]

[Out]

(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)

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.18

method result size
pseudoelliptic \(-\frac {2 \left (-\frac {c^{2} \left (\left (a p \,x^{3}+2 b \,x^{2}+a q \right ) c -2 a d \,x^{2}\right ) \sqrt {\frac {-2 c^{2} p q +d^{2}}{c^{2}}}\, \sqrt {\frac {p^{2} x^{6}-2 q \,x^{3} \left (-1+x \right ) p +q^{2}}{x^{2}}}}{4}+\left (\frac {\left (a \,c^{2} p q -a \,d^{2}+b c d \right ) c \ln \left (\frac {p \,x^{3}+\sqrt {\frac {p^{2} x^{6}-2 q \,x^{3} \left (-1+x \right ) p +q^{2}}{x^{2}}}\, x +q}{x^{2}}\right ) \sqrt {\frac {-2 c^{2} p q +d^{2}}{c^{2}}}}{2}+\left (c^{2} p q -\frac {d^{2}}{2}\right ) \left (a d -b c \right ) \left (\ln \left (\frac {-2 c p q \,x^{2}-d p \,x^{3}+\sqrt {\frac {-2 c^{2} p q +d^{2}}{c^{2}}}\, \sqrt {\frac {p^{2} x^{6}-2 q \,x^{3} \left (-1+x \right ) p +q^{2}}{x^{2}}}\, c x -d q}{c p \,x^{3}+d \,x^{2}+c q}\right )+\ln \left (2\right )\right )\right ) x^{3}\right )}{\sqrt {\frac {-2 c^{2} p q +d^{2}}{c^{2}}}\, c^{4} x^{3}}\) \(296\)

[In]

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)

[Out]

-2*(-1/4*c^2*((a*p*x^3+2*b*x^2+a*q)*c-2*a*d*x^2)*((-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)+(1/2*(a*c^2*p*q-a*d^2+b*c*d)*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)*((-2*
c^2*p*q+d^2)/c^2)^(1/2)+(c^2*p*q-1/2*d^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)))*x^3)/((-2*c^2*p*q+d^2)/c^2)^(1/2
)/c^4/x^3

Fricas [F(-1)]

Timed out. \[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5 \left (c q+d x^2+c p x^3\right )} \, dx=\text {Timed out} \]

[In]

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, a
lgorithm="fricas")

[Out]

Timed out

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5 \left (c q+d x^2+c p x^3\right )} \, dx=\text {Timed out} \]

[In]

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)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5 \left (c q+d x^2+c p x^3\right )} \, dx=\int { \frac {\sqrt {p^{2} x^{6} - 2 \, p q x^{4} + 2 \, p q x^{3} + q^{2}} {\left (a p x^{3} + b x^{2} + a q\right )} {\left (p x^{3} - 2 \, q\right )}}{{\left (c p x^{3} + d x^{2} + c q\right )} x^{5}} \,d x } \]

[In]

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, a
lgorithm="maxima")

[Out]

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)

Giac [F]

\[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5 \left (c q+d x^2+c p x^3\right )} \, dx=\int { \frac {\sqrt {p^{2} x^{6} - 2 \, p q x^{4} + 2 \, p q x^{3} + q^{2}} {\left (a p x^{3} + b x^{2} + a q\right )} {\left (p x^{3} - 2 \, q\right )}}{{\left (c p x^{3} + d x^{2} + c q\right )} x^{5}} \,d x } \]

[In]

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, a
lgorithm="giac")

[Out]

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)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (-2 q+p x^3\right ) \left (a q+b x^2+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5 \left (c q+d x^2+c p x^3\right )} \, dx=\int -\frac {\left (2\,q-p\,x^3\right )\,\left (a\,p\,x^3+b\,x^2+a\,q\right )\,\sqrt {p^2\,x^6-2\,p\,q\,x^4+2\,p\,q\,x^3+q^2}}{x^5\,\left (c\,p\,x^3+d\,x^2+c\,q\right )} \,d x \]

[In]

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)

[Out]

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)