Integrand size = 22, antiderivative size = 113 \[ \int \frac {\left (a+b x^3\right )^{2/3} \left (A+B x^3\right )}{x^6} \, dx=-\frac {B \left (a+b x^3\right )^{2/3}}{2 x^2}-\frac {A \left (a+b x^3\right )^{5/3}}{5 a x^5}+\frac {b^{2/3} B \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} b^{2/3} B \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right ) \] Output:
-1/2*B*(b*x^3+a)^(2/3)/x^2-1/5*A*(b*x^3+a)^(5/3)/a/x^5+1/3*b^(2/3)*B*arcta n(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))*3^(1/2)-1/2*b^(2/3)*B*ln(-b ^(1/3)*x+(b*x^3+a)^(1/3))
Time = 0.62 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.50 \[ \int \frac {\left (a+b x^3\right )^{2/3} \left (A+B x^3\right )}{x^6} \, dx=\frac {\left (a+b x^3\right )^{2/3} \left (-2 a A-2 A b x^3-5 a B x^3\right )}{10 a x^5}+\frac {b^{2/3} B \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )}{\sqrt {3}}-\frac {1}{3} b^{2/3} B \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )+\frac {1}{6} b^{2/3} B \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right ) \] Input:
Integrate[((a + b*x^3)^(2/3)*(A + B*x^3))/x^6,x]
Output:
((a + b*x^3)^(2/3)*(-2*a*A - 2*A*b*x^3 - 5*a*B*x^3))/(10*a*x^5) + (b^(2/3) *B*ArcTan[(Sqrt[3]*b^(1/3)*x)/(b^(1/3)*x + 2*(a + b*x^3)^(1/3))])/Sqrt[3] - (b^(2/3)*B*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])/3 + (b^(2/3)*B*Log[b^( 2/3)*x^2 + b^(1/3)*x*(a + b*x^3)^(1/3) + (a + b*x^3)^(2/3)])/6
Time = 0.37 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {953, 809, 769}
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.
\(\displaystyle \int \frac {\left (a+b x^3\right )^{2/3} \left (A+B x^3\right )}{x^6} \, dx\) |
\(\Big \downarrow \) 953 |
\(\displaystyle B \int \frac {\left (b x^3+a\right )^{2/3}}{x^3}dx-\frac {A \left (a+b x^3\right )^{5/3}}{5 a x^5}\) |
\(\Big \downarrow \) 809 |
\(\displaystyle B \left (b \int \frac {1}{\sqrt [3]{b x^3+a}}dx-\frac {\left (a+b x^3\right )^{2/3}}{2 x^2}\right )-\frac {A \left (a+b x^3\right )^{5/3}}{5 a x^5}\) |
\(\Big \downarrow \) 769 |
\(\displaystyle B \left (b \left (\frac {\arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{2 \sqrt [3]{b}}\right )-\frac {\left (a+b x^3\right )^{2/3}}{2 x^2}\right )-\frac {A \left (a+b x^3\right )^{5/3}}{5 a x^5}\) |
Input:
Int[((a + b*x^3)^(2/3)*(A + B*x^3))/x^6,x]
Output:
-1/5*(A*(a + b*x^3)^(5/3))/(a*x^5) + B*(-1/2*(a + b*x^3)^(2/3)/x^2 + b*(Ar cTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*b^(1/3)) - Lo g[-(b^(1/3)*x) + (a + b*x^3)^(1/3)]/(2*b^(1/3))))
Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + 2*Rt[b, 3]* (x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^ 3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1))), x] - Simp[b*n*(p/(c^n*(m + 1))) I nt[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && IGtQ [n, 0] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + n*p + n + 1)/n, 0] && IntB inomialQ[a, b, c, n, m, p, x]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Simp[d/e^n Int[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 1, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && G tQ[m + n, -1]))
Time = 1.23 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.25
method | result | size |
pseudoelliptic | \(\frac {5 a \left (-2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right )+\ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )-2 \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )\right ) B \,x^{5} b^{\frac {2}{3}}-6 \left (\left (A b +\frac {5 B a}{2}\right ) x^{3}+A a \right ) \left (b \,x^{3}+a \right )^{\frac {2}{3}}}{30 a \,x^{5}}\) | \(141\) |
Input:
int((b*x^3+a)^(2/3)*(B*x^3+A)/x^6,x,method=_RETURNVERBOSE)
Output:
1/30*(5*a*(-2*3^(1/2)*arctan(1/3*3^(1/2)*(b^(1/3)*x+2*(b*x^3+a)^(1/3))/b^( 1/3)/x)+ln((b^(2/3)*x^2+b^(1/3)*(b*x^3+a)^(1/3)*x+(b*x^3+a)^(2/3))/x^2)-2* ln((-b^(1/3)*x+(b*x^3+a)^(1/3))/x))*B*x^5*b^(2/3)-6*((A*b+5/2*B*a)*x^3+A*a )*(b*x^3+a)^(2/3))/a/x^5
Timed out. \[ \int \frac {\left (a+b x^3\right )^{2/3} \left (A+B x^3\right )}{x^6} \, dx=\text {Timed out} \] Input:
integrate((b*x^3+a)^(2/3)*(B*x^3+A)/x^6,x, algorithm="fricas")
Output:
Timed out
Result contains complex when optimal does not.
Time = 1.69 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a+b x^3\right )^{2/3} \left (A+B x^3\right )}{x^6} \, dx=\frac {A b^{\frac {2}{3}} \left (\frac {a}{b x^{3}} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {5}{3}\right )}{3 x^{3} \Gamma \left (- \frac {2}{3}\right )} + \frac {A b^{\frac {5}{3}} \left (\frac {a}{b x^{3}} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {5}{3}\right )}{3 a \Gamma \left (- \frac {2}{3}\right )} + \frac {B a^{\frac {2}{3}} \Gamma \left (- \frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, - \frac {2}{3} \\ \frac {1}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 x^{2} \Gamma \left (\frac {1}{3}\right )} \] Input:
integrate((b*x**3+a)**(2/3)*(B*x**3+A)/x**6,x)
Output:
A*b**(2/3)*(a/(b*x**3) + 1)**(2/3)*gamma(-5/3)/(3*x**3*gamma(-2/3)) + A*b* *(5/3)*(a/(b*x**3) + 1)**(2/3)*gamma(-5/3)/(3*a*gamma(-2/3)) + B*a**(2/3)* gamma(-2/3)*hyper((-2/3, -2/3), (1/3,), b*x**3*exp_polar(I*pi)/a)/(3*x**2* gamma(1/3))
Time = 0.11 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.20 \[ \int \frac {\left (a+b x^3\right )^{2/3} \left (A+B x^3\right )}{x^6} \, dx=-\frac {1}{6} \, {\left (2 \, \sqrt {3} b^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right ) - b^{\frac {2}{3}} \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) + 2 \, b^{\frac {2}{3}} \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) + \frac {3 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )} B - \frac {{\left (b x^{3} + a\right )}^{\frac {5}{3}} A}{5 \, a x^{5}} \] Input:
integrate((b*x^3+a)^(2/3)*(B*x^3+A)/x^6,x, algorithm="maxima")
Output:
-1/6*(2*sqrt(3)*b^(2/3)*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/ x)/b^(1/3)) - b^(2/3)*log(b^(2/3) + (b*x^3 + a)^(1/3)*b^(1/3)/x + (b*x^3 + a)^(2/3)/x^2) + 2*b^(2/3)*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x) + 3*(b*x^3 + a)^(2/3)/x^2)*B - 1/5*(b*x^3 + a)^(5/3)*A/(a*x^5)
\[ \int \frac {\left (a+b x^3\right )^{2/3} \left (A+B x^3\right )}{x^6} \, dx=\int { \frac {{\left (B x^{3} + A\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{6}} \,d x } \] Input:
integrate((b*x^3+a)^(2/3)*(B*x^3+A)/x^6,x, algorithm="giac")
Output:
integrate((B*x^3 + A)*(b*x^3 + a)^(2/3)/x^6, x)
Timed out. \[ \int \frac {\left (a+b x^3\right )^{2/3} \left (A+B x^3\right )}{x^6} \, dx=\int \frac {\left (B\,x^3+A\right )\,{\left (b\,x^3+a\right )}^{2/3}}{x^6} \,d x \] Input:
int(((A + B*x^3)*(a + b*x^3)^(2/3))/x^6,x)
Output:
int(((A + B*x^3)*(a + b*x^3)^(2/3))/x^6, x)
\[ \int \frac {\left (a+b x^3\right )^{2/3} \left (A+B x^3\right )}{x^6} \, dx=\frac {-\left (b \,x^{3}+a \right )^{\frac {2}{3}} a -\left (b \,x^{3}+a \right )^{\frac {2}{3}} b \,x^{3}+5 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{3}}d x \right ) b \,x^{5}}{5 x^{5}} \] Input:
int((b*x^3+a)^(2/3)*(B*x^3+A)/x^6,x)
Output:
( - (a + b*x**3)**(2/3)*a - (a + b*x**3)**(2/3)*b*x**3 + 5*int((a + b*x**3 )**(2/3)/x**3,x)*b*x**5)/(5*x**5)