\(\int \frac {(-2 q+p x^3) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} (b x^6+a (q+p x^3)^3)}{x^{11}} \, dx\) [2393]

Optimal result

Integrand size = 59, antiderivative size = 192 \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (b x^6+a \left (q+p x^3\right )^3\right )}{x^{11}} \, dx=\frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (6 a q^4+24 a p q^3 x^3-4 a p q^3 x^4+15 b q x^6+36 a p^2 q^2 x^6-8 a p^2 q^2 x^7-16 a p^2 q^2 x^8+15 b p x^9+24 a p^3 q x^9-4 a p^3 q x^{10}+6 a p^4 x^{12}\right )}{30 x^{10}}+2 b p q \log (x)-b p q \log \left (q+p x^3+\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}\right ) \]

[Out]

1/30*(p^2*x^6-2*p*q*x^4+2*p*q*x^3+q^2)^(1/2)*(6*a*p^4*x^12-4*a*p^3*q*x^10+24*a*p^3*q*x^9-16*a*p^2*q^2*x^8-8*a*
p^2*q^2*x^7+36*a*p^2*q^2*x^6+15*b*p*x^9-4*a*p*q^3*x^4+24*a*p*q^3*x^3+15*b*q*x^6+6*a*q^4)/x^10+2*b*p*q*ln(x)-b*
p*q*ln(q+p*x^3+(p^2*x^6-2*p*q*x^4+2*p*q*x^3+q^2)^(1/2))

Rubi [F]

\[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (b x^6+a \left (q+p x^3\right )^3\right )}{x^{11}} \, dx=\int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (b x^6+a \left (q+p x^3\right )^3\right )}{x^{11}} \, dx \]

[In]

Int[((-2*q + p*x^3)*Sqrt[q^2 + 2*p*q*x^3 - 2*p*q*x^4 + p^2*x^6]*(b*x^6 + a*(q + p*x^3)^3))/x^11,x]

[Out]

-2*a*q^4*Defer[Int][Sqrt[q^2 + 2*p*q*x^3 - 2*p*q*x^4 + p^2*x^6]/x^11, x] - 5*a*p*q^3*Defer[Int][Sqrt[q^2 + 2*p
*q*x^3 - 2*p*q*x^4 + p^2*x^6]/x^8, x] - q*(2*b + 3*a*p^2*q)*Defer[Int][Sqrt[q^2 + 2*p*q*x^3 - 2*p*q*x^4 + p^2*
x^6]/x^5, x] + p*(b + a*p^2*q)*Defer[Int][Sqrt[q^2 + 2*p*q*x^3 - 2*p*q*x^4 + p^2*x^6]/x^2, x] + a*p^4*Defer[In
t][x*Sqrt[q^2 + 2*p*q*x^3 - 2*p*q*x^4 + p^2*x^6], x]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {2 a q^4 \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^{11}}-\frac {5 a p q^3 \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^8}-\frac {q \left (2 b+3 a p^2 q\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5}+\frac {p \left (b+a p^2 q\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^2}+a p^4 x \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}\right ) \, dx \\ & = \left (a p^4\right ) \int x \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \, dx-\left (5 a p q^3\right ) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^8} \, dx-\left (2 a q^4\right ) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^{11}} \, dx+\left (p \left (b+a p^2 q\right )\right ) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^2} \, dx-\left (q \left (2 b+3 a p^2 q\right )\right ) \int \frac {\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x^5} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.80 \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (b x^6+a \left (q+p x^3\right )^3\right )}{x^{11}} \, dx=\frac {\sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6} \left (15 b x^6 \left (q+p x^3\right )+2 a \left (3 q^4-2 p q^3 (-6+x) x^3-2 p^3 q (-6+x) x^9+3 p^4 x^{12}-2 p^2 q^2 x^6 \left (-9+2 x+4 x^2\right )\right )\right )}{30 x^{10}}+2 b p q \log (x)-b p q \log \left (q+p x^3+\sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}\right ) \]

[In]

Integrate[((-2*q + p*x^3)*Sqrt[q^2 + 2*p*q*x^3 - 2*p*q*x^4 + p^2*x^6]*(b*x^6 + a*(q + p*x^3)^3))/x^11,x]

[Out]

(Sqrt[q^2 - 2*p*q*(-1 + x)*x^3 + p^2*x^6]*(15*b*x^6*(q + p*x^3) + 2*a*(3*q^4 - 2*p*q^3*(-6 + x)*x^3 - 2*p^3*q*
(-6 + x)*x^9 + 3*p^4*x^12 - 2*p^2*q^2*x^6*(-9 + 2*x + 4*x^2))))/(30*x^10) + 2*b*p*q*Log[x] - b*p*q*Log[q + p*x
^3 + Sqrt[q^2 - 2*p*q*(-1 + x)*x^3 + p^2*x^6]]

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.98

method	result	size
pseudoelliptic	\(\frac {\left (6 a \,p^{4} x^{12}-4 a \,p^{3} q \,x^{10}+\left (24 q a \,p^{3}+15 b p \right ) x^{9}-16 a \,p^{2} q^{2} x^{8}-8 a \,p^{2} q^{2} x^{7}+\left (36 q^{2} a \,p^{2}+15 q b \right ) x^{6}-4 a p \,q^{3} x^{4}+24 a p \,q^{3} x^{3}+6 a \,q^{4}\right ) \sqrt {\frac {p^{2} x^{6}-2 p q \,x^{3} \left (-1+x \right )+q^{2}}{x^{2}}}-30 b p q \ln \left (\frac {p \,x^{3}+\sqrt {\frac {p^{2} x^{6}-2 p q \,x^{3} \left (-1+x \right )+q^{2}}{x^{2}}}\, x +q}{x^{2}}\right ) x^{9}}{30 x^{9}}\)	\(188\)

[In]

int((p*x^3-2*q)*(p^2*x^6-2*p*q*x^4+2*p*q*x^3+q^2)^(1/2)*(b*x^6+a*(p*x^3+q)^3)/x^11,x,method=_RETURNVERBOSE)

[Out]

1/30*((6*a*p^4*x^12-4*a*p^3*q*x^10+(24*a*p^3*q+15*b*p)*x^9-16*a*p^2*q^2*x^8-8*a*p^2*q^2*x^7+(36*a*p^2*q^2+15*b
*q)*x^6-4*a*p*q^3*x^4+24*a*p*q^3*x^3+6*a*q^4)*((p^2*x^6-2*p*q*x^3*(-1+x)+q^2)/x^2)^(1/2)-30*b*p*q*ln((p*x^3+((
p^2*x^6-2*p*q*x^3*(-1+x)+q^2)/x^2)^(1/2)*x+q)/x^2)*x^9)/x^9

Fricas [F(-1)]

Timed out. \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (b x^6+a \left (q+p x^3\right )^3\right )}{x^{11}} \, dx=\text {Timed out} \]

[In]

integrate((p*x^3-2*q)*(p^2*x^6-2*p*q*x^4+2*p*q*x^3+q^2)^(1/2)*(b*x^6+a*(p*x^3+q)^3)/x^11,x, algorithm="fricas"
)

[Out]

Timed out

Sympy [F]

\[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (b x^6+a \left (q+p x^3\right )^3\right )}{x^{11}} \, dx=\int \frac {\left (p x^{3} - 2 q\right ) \sqrt {p^{2} x^{6} - 2 p q x^{4} + 2 p q x^{3} + q^{2}} \left (a p^{3} x^{9} + 3 a p^{2} q x^{6} + 3 a p q^{2} x^{3} + a q^{3} + b x^{6}\right )}{x^{11}}\, dx \]

[In]

integrate((p*x**3-2*q)*(p**2*x**6-2*p*q*x**4+2*p*q*x**3+q**2)**(1/2)*(b*x**6+a*(p*x**3+q)**3)/x**11,x)

[Out]

Integral((p*x**3 - 2*q)*sqrt(p**2*x**6 - 2*p*q*x**4 + 2*p*q*x**3 + q**2)*(a*p**3*x**9 + 3*a*p**2*q*x**6 + 3*a*
p*q**2*x**3 + a*q**3 + b*x**6)/x**11, x)

Maxima [F]

\[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (b x^6+a \left (q+p x^3\right )^3\right )}{x^{11}} \, dx=\int { \frac {\sqrt {p^{2} x^{6} - 2 \, p q x^{4} + 2 \, p q x^{3} + q^{2}} {\left (b x^{6} + {\left (p x^{3} + q\right )}^{3} a\right )} {\left (p x^{3} - 2 \, q\right )}}{x^{11}} \,d x } \]

[In]

integrate((p*x^3-2*q)*(p^2*x^6-2*p*q*x^4+2*p*q*x^3+q^2)^(1/2)*(b*x^6+a*(p*x^3+q)^3)/x^11,x, algorithm="maxima"
)

[Out]

integrate(sqrt(p^2*x^6 - 2*p*q*x^4 + 2*p*q*x^3 + q^2)*(b*x^6 + (p*x^3 + q)^3*a)*(p*x^3 - 2*q)/x^11, x)

Giac [F]

\[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (b x^6+a \left (q+p x^3\right )^3\right )}{x^{11}} \, dx=\int { \frac {\sqrt {p^{2} x^{6} - 2 \, p q x^{4} + 2 \, p q x^{3} + q^{2}} {\left (b x^{6} + {\left (p x^{3} + q\right )}^{3} a\right )} {\left (p x^{3} - 2 \, q\right )}}{x^{11}} \,d x } \]

[In]

integrate((p*x^3-2*q)*(p^2*x^6-2*p*q*x^4+2*p*q*x^3+q^2)^(1/2)*(b*x^6+a*(p*x^3+q)^3)/x^11,x, algorithm="giac")

[Out]

integrate(sqrt(p^2*x^6 - 2*p*q*x^4 + 2*p*q*x^3 + q^2)*(b*x^6 + (p*x^3 + q)^3*a)*(p*x^3 - 2*q)/x^11, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6} \left (b x^6+a \left (q+p x^3\right )^3\right )}{x^{11}} \, dx=\int -\frac {\left (a\,{\left (p\,x^3+q\right )}^3+b\,x^6\right )\,\left (2\,q-p\,x^3\right )\,\sqrt {p^2\,x^6-2\,p\,q\,x^4+2\,p\,q\,x^3+q^2}}{x^{11}} \,d x \]

[In]

int(-((a*(q + p*x^3)^3 + b*x^6)*(2*q - p*x^3)*(p^2*x^6 + q^2 + 2*p*q*x^3 - 2*p*q*x^4)^(1/2))/x^11,x)

[Out]

int(-((a*(q + p*x^3)^3 + b*x^6)*(2*q - p*x^3)*(p^2*x^6 + q^2 + 2*p*q*x^3 - 2*p*q*x^4)^(1/2))/x^11, x)