Integrand size = 61, antiderivative size = 159 \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x \left (b x^4+a \left (q+p x^3\right )^2\right )} \, dx=\frac {\sqrt {b+2 a p q} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {b+2 a p q} x^4}{a q^2+2 a p q x^3+b x^4+a p^2 x^6+\left (a q+a p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}\right )}{a \sqrt {b}}-\frac {2 \log (x)}{a}+\frac {\log \left (q+p x^3+\sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}\right )}{a} \]
(2*a*p*q+b)^(1/2)*arctanh(b^(1/2)*(2*a*p*q+b)^(1/2)*x^4/(a*q^2+2*a*p*q*x^3 +b*x^4+a*p^2*x^6+(a*p*x^3+a*q)*(p^2*x^6-2*p*q*x^4+2*p*q*x^3+q^2)^(1/2)))/a /b^(1/2)-2*ln(x)/a+ln(q+p*x^3+(p^2*x^6-2*p*q*x^4+2*p*q*x^3+q^2)^(1/2))/a
Time = 0.69 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.81 \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x \left (b x^4+a \left (q+p x^3\right )^2\right )} \, dx=\frac {\frac {\sqrt {b+2 a p q} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {b+2 a p q} x^4}{b x^4+a \left (q+p x^3\right ) \left (q+p x^3+\sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}\right )}\right )}{\sqrt {b}}-2 \log (x)+\log \left (q+p x^3+\sqrt {q^2-2 p q (-1+x) x^3+p^2 x^6}\right )}{a} \]
Integrate[((-2*q + p*x^3)*Sqrt[q^2 + 2*p*q*x^3 - 2*p*q*x^4 + p^2*x^6])/(x* (b*x^4 + a*(q + p*x^3)^2)),x]
((Sqrt[b + 2*a*p*q]*ArcTanh[(Sqrt[b]*Sqrt[b + 2*a*p*q]*x^4)/(b*x^4 + a*(q + p*x^3)*(q + p*x^3 + Sqrt[q^2 - 2*p*q*(-1 + x)*x^3 + p^2*x^6]))])/Sqrt[b] - 2*Log[x] + Log[q + p*x^3 + Sqrt[q^2 - 2*p*q*(-1 + x)*x^3 + p^2*x^6]])/a
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.
\(\displaystyle \int \frac {\left (p x^3-2 q\right ) \sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{x \left (a \left (p x^3+q\right )^2+b x^4\right )} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x^2 \sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2} \left (2 a p^2 x^3+5 a p q+2 b x\right )}{a q \left (a p^2 x^6+2 a p q x^3+a q^2+b x^4\right )}-\frac {2 \sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{a q x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 b \int \frac {x^3 \sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{a p^2 x^6+b x^4+2 a p q x^3+a q^2}dx}{a q}+\frac {2 p^2 \int \frac {x^5 \sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{a p^2 x^6+b x^4+2 a p q x^3+a q^2}dx}{q}+5 p \int \frac {x^2 \sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{a p^2 x^6+b x^4+2 a p q x^3+a q^2}dx-\frac {2 \int \frac {\sqrt {p^2 x^6-2 p q x^4+2 p q x^3+q^2}}{x}dx}{a q}\) |
Int[((-2*q + p*x^3)*Sqrt[q^2 + 2*p*q*x^3 - 2*p*q*x^4 + p^2*x^6])/(x*(b*x^4 + a*(q + p*x^3)^2)),x]
3.22.68.3.1 Defintions of rubi rules used
Time = 0.31 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.85
method | result | size |
pseudoelliptic | \(\frac {\sqrt {-a b}\, \ln \left (\frac {p \,x^{3}+\sqrt {\frac {p^{2} x^{6}-2 p q \,x^{3} \left (-1+x \right )+q^{2}}{x^{2}}}\, x +q}{x^{2}}\right ) \sqrt {\frac {-2 a p q -b}{a}}+\left (\ln \left (\frac {\sqrt {\frac {p^{2} x^{6}-2 p q \,x^{3} \left (-1+x \right )+q^{2}}{x^{2}}}\, \sqrt {\frac {-2 a p q -b}{a}}\, a x +\left (p \,x^{3}+q \right ) \sqrt {-a b}-2 a p q \,x^{2}}{\left (p \,x^{3}+q \right ) a -\sqrt {-a b}\, x^{2}}\right )-\ln \left (\frac {-2 a p q \,x^{2}-\sqrt {-a b}\, p \,x^{3}+\sqrt {\frac {p^{2} x^{6}-2 p q \,x^{3} \left (-1+x \right )+q^{2}}{x^{2}}}\, \sqrt {\frac {-2 a p q -b}{a}}\, a x -\sqrt {-a b}\, q}{\left (p \,x^{3}+q \right ) a +\sqrt {-a b}\, x^{2}}\right )\right ) \left (a p q +\frac {b}{2}\right )}{\sqrt {\frac {-2 a p q -b}{a}}\, \sqrt {-a b}\, a}\) | \(294\) |
int((p*x^3-2*q)*(p^2*x^6-2*p*q*x^4+2*p*q*x^3+q^2)^(1/2)/x/(b*x^4+a*(p*x^3+ q)^2),x,method=_RETURNVERBOSE)
1/((-2*a*p*q-b)/a)^(1/2)/(-a*b)^(1/2)*((-a*b)^(1/2)*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)*((-2*a*p*q-b)/a)^(1/2)+(ln((((p^2 *x^6-2*p*q*x^3*(-1+x)+q^2)/x^2)^(1/2)*((-2*a*p*q-b)/a)^(1/2)*a*x+(p*x^3+q) *(-a*b)^(1/2)-2*a*p*q*x^2)/((p*x^3+q)*a-(-a*b)^(1/2)*x^2))-ln((-2*a*p*q*x^ 2-(-a*b)^(1/2)*p*x^3+((p^2*x^6-2*p*q*x^3*(-1+x)+q^2)/x^2)^(1/2)*((-2*a*p*q -b)/a)^(1/2)*a*x-(-a*b)^(1/2)*q)/((p*x^3+q)*a+(-a*b)^(1/2)*x^2)))*(a*p*q+1 /2*b))/a
Timed out. \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x \left (b x^4+a \left (q+p x^3\right )^2\right )} \, dx=\text {Timed out} \]
integrate((p*x^3-2*q)*(p^2*x^6-2*p*q*x^4+2*p*q*x^3+q^2)^(1/2)/x/(b*x^4+a*( p*x^3+q)^2),x, algorithm="fricas")
\[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x \left (b x^4+a \left (q+p x^3\right )^2\right )} \, dx=\int \frac {\left (p x^{3} - 2 q\right ) \sqrt {p^{2} x^{6} - 2 p q x^{4} + 2 p q x^{3} + q^{2}}}{x \left (a p^{2} x^{6} + 2 a p q x^{3} + a q^{2} + b x^{4}\right )}\, dx \]
integrate((p*x**3-2*q)*(p**2*x**6-2*p*q*x**4+2*p*q*x**3+q**2)**(1/2)/x/(b* x**4+a*(p*x**3+q)**2),x)
Integral((p*x**3 - 2*q)*sqrt(p**2*x**6 - 2*p*q*x**4 + 2*p*q*x**3 + q**2)/( x*(a*p**2*x**6 + 2*a*p*q*x**3 + a*q**2 + b*x**4)), x)
\[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x \left (b x^4+a \left (q+p x^3\right )^2\right )} \, dx=\int { \frac {\sqrt {p^{2} x^{6} - 2 \, p q x^{4} + 2 \, p q x^{3} + q^{2}} {\left (p x^{3} - 2 \, q\right )}}{{\left (b x^{4} + {\left (p x^{3} + q\right )}^{2} a\right )} x} \,d x } \]
integrate((p*x^3-2*q)*(p^2*x^6-2*p*q*x^4+2*p*q*x^3+q^2)^(1/2)/x/(b*x^4+a*( p*x^3+q)^2),x, algorithm="maxima")
integrate(sqrt(p^2*x^6 - 2*p*q*x^4 + 2*p*q*x^3 + q^2)*(p*x^3 - 2*q)/((b*x^ 4 + (p*x^3 + q)^2*a)*x), x)
\[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x \left (b x^4+a \left (q+p x^3\right )^2\right )} \, dx=\int { \frac {\sqrt {p^{2} x^{6} - 2 \, p q x^{4} + 2 \, p q x^{3} + q^{2}} {\left (p x^{3} - 2 \, q\right )}}{{\left (b x^{4} + {\left (p x^{3} + q\right )}^{2} a\right )} x} \,d x } \]
integrate((p*x^3-2*q)*(p^2*x^6-2*p*q*x^4+2*p*q*x^3+q^2)^(1/2)/x/(b*x^4+a*( p*x^3+q)^2),x, algorithm="giac")
integrate(sqrt(p^2*x^6 - 2*p*q*x^4 + 2*p*q*x^3 + q^2)*(p*x^3 - 2*q)/((b*x^ 4 + (p*x^3 + q)^2*a)*x), x)
Timed out. \[ \int \frac {\left (-2 q+p x^3\right ) \sqrt {q^2+2 p q x^3-2 p q x^4+p^2 x^6}}{x \left (b x^4+a \left (q+p x^3\right )^2\right )} \, dx=\int -\frac {\left (2\,q-p\,x^3\right )\,\sqrt {p^2\,x^6-2\,p\,q\,x^4+2\,p\,q\,x^3+q^2}}{x\,\left (a\,{\left (p\,x^3+q\right )}^2+b\,x^4\right )} \,d x \]
int(-((2*q - p*x^3)*(p^2*x^6 + q^2 + 2*p*q*x^3 - 2*p*q*x^4)^(1/2))/(x*(a*( q + p*x^3)^2 + b*x^4)),x)