\(\int \frac {(-q+2 p x^3) (a q+b x+a p x^3) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^4} \, dx\) [1875]

Optimal result

Integrand size = 56, antiderivative size = 129 \[ \int \frac {\left (-q+2 p x^3\right ) \left (a q+b x+a p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^4} \, dx=\frac {\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6} \left (2 a q^2+3 b q x-4 a p q x^2+4 a p q x^3+3 b p x^4+2 a p^2 x^6\right )}{6 x^3}+b p q \log (x)-b p q \log \left (q+p x^3+\sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}\right ) \]

[Out]

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))

Rubi [F]

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

[In]

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^4,x]

[Out]

2*b*p*Defer[Int][Sqrt[q^2 - 2*p*q*x^2 + 2*p*q*x^3 + p^2*x^6], x] - a*q^2*Defer[Int][Sqrt[q^2 - 2*p*q*x^2 + 2*p
*q*x^3 + p^2*x^6]/x^4, x] - b*q*Defer[Int][Sqrt[q^2 - 2*p*q*x^2 + 2*p*q*x^3 + p^2*x^6]/x^3, x] + a*p*q*Defer[I
nt][Sqrt[q^2 - 2*p*q*x^2 + 2*p*q*x^3 + p^2*x^6]/x, x] + 2*a*p^2*Defer[Int][x^2*Sqrt[q^2 - 2*p*q*x^2 + 2*p*q*x^
3 + p^2*x^6], x]

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

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.88 \[ \int \frac {\left (-q+2 p x^3\right ) \left (a q+b x+a p x^3\right ) \sqrt {q^2-2 p q x^2+2 p q x^3+p^2 x^6}}{x^4} \, dx=\frac {\sqrt {q^2+2 p q (-1+x) x^2+p^2 x^6} \left (3 b x \left (q+p x^3\right )+2 a \left (q^2+2 p q (-1+x) x^2+p^2 x^6\right )\right )}{6 x^3}+b p q \log (x)-b p q \log \left (q+p x^3+\sqrt {q^2+2 p q (-1+x) x^2+p^2 x^6}\right ) \]

[In]

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]

[Out]

(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]]

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.91

method	result	size
pseudoelliptic	\(\frac {\sqrt {p^{2} x^{6}+2 p q \,x^{2} \left (-1+x \right )+q^{2}}\, \left (2 a \,p^{2} x^{6}+4 a p q \,x^{3}+3 b p \,x^{4}-4 a p q \,x^{2}+2 a \,q^{2}+3 b q x \right )-6 b p q \ln \left (\frac {q +p \,x^{3}+\sqrt {p^{2} x^{6}+2 p q \,x^{2} \left (-1+x \right )+q^{2}}}{x}\right ) x^{3}}{6 x^{3}}\)	\(118\)

[In]

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)

[Out]

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

Fricas [F(-1)]

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

[In]

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")

[Out]

Timed out

Sympy [F]

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

[In]

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)

[Out]

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)

Maxima [F]

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

[In]

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")

[Out]

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)

Giac [F]

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

[In]

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")

[Out]

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)

Mupad [F(-1)]

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

[In]

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)

[Out]

-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)