Integrand size = 22, antiderivative size = 85 \[ \int \frac {\sqrt [3]{a+b x^3} \left (A+B x^3\right )}{x^9} \, dx=-\frac {A \left (a+b x^3\right )^{4/3}}{8 a x^8}+\frac {(A b-2 a B) \sqrt [3]{a+b x^3} \operatorname {Hypergeometric2F1}\left (-\frac {5}{3},-\frac {1}{3},-\frac {2}{3},-\frac {b x^3}{a}\right )}{10 a x^5 \sqrt [3]{1+\frac {b x^3}{a}}} \] Output:
-1/8*A*(b*x^3+a)^(4/3)/a/x^8+1/10*(A*b-2*B*a)*(b*x^3+a)^(1/3)*hypergeom([- 5/3, -1/3],[-2/3],-b*x^3/a)/a/x^5/(1+b*x^3/a)^(1/3)
Time = 10.10 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.92 \[ \int \frac {\sqrt [3]{a+b x^3} \left (A+B x^3\right )}{x^9} \, dx=\frac {\sqrt [3]{a+b x^3} \left (-5 A \left (a+b x^3\right )+\frac {4 (A b-2 a B) x^3 \operatorname {Hypergeometric2F1}\left (-\frac {5}{3},-\frac {1}{3},-\frac {2}{3},-\frac {b x^3}{a}\right )}{\sqrt [3]{1+\frac {b x^3}{a}}}\right )}{40 a x^8} \] Input:
Integrate[((a + b*x^3)^(1/3)*(A + B*x^3))/x^9,x]
Output:
((a + b*x^3)^(1/3)*(-5*A*(a + b*x^3) + (4*(A*b - 2*a*B)*x^3*Hypergeometric 2F1[-5/3, -1/3, -2/3, -((b*x^3)/a)])/(1 + (b*x^3)/a)^(1/3)))/(40*a*x^8)
Time = 0.36 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {955, 889, 888}
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.
\(\displaystyle \int \frac {\sqrt [3]{a+b x^3} \left (A+B x^3\right )}{x^9} \, dx\) |
\(\Big \downarrow \) 955 |
\(\displaystyle -\frac {(A b-2 a B) \int \frac {\sqrt [3]{b x^3+a}}{x^6}dx}{2 a}-\frac {A \left (a+b x^3\right )^{4/3}}{8 a x^8}\) |
\(\Big \downarrow \) 889 |
\(\displaystyle -\frac {\sqrt [3]{a+b x^3} (A b-2 a B) \int \frac {\sqrt [3]{\frac {b x^3}{a}+1}}{x^6}dx}{2 a \sqrt [3]{\frac {b x^3}{a}+1}}-\frac {A \left (a+b x^3\right )^{4/3}}{8 a x^8}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {\sqrt [3]{a+b x^3} (A b-2 a B) \operatorname {Hypergeometric2F1}\left (-\frac {5}{3},-\frac {1}{3},-\frac {2}{3},-\frac {b x^3}{a}\right )}{10 a x^5 \sqrt [3]{\frac {b x^3}{a}+1}}-\frac {A \left (a+b x^3\right )^{4/3}}{8 a x^8}\) |
Input:
Int[((a + b*x^3)^(1/3)*(A + B*x^3))/x^9,x]
Output:
-1/8*(A*(a + b*x^3)^(4/3))/(a*x^8) + ((A*b - 2*a*B)*(a + b*x^3)^(1/3)*Hype rgeometric2F1[-5/3, -1/3, -2/3, -((b*x^3)/a)])/(10*a*x^5*(1 + (b*x^3)/a)^( 1/3))
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p *((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 , (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0] && (ILt Q[p, 0] || GtQ[a, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^I ntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]) Int[(c*x) ^m*(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0 ] && !(ILtQ[p, 0] || GtQ[a, 0])
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[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)) Int[(e *x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b* c - a*d, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) && !ILtQ[p, -1]
\[\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (B \,x^{3}+A \right )}{x^{9}}d x\]
Input:
int((b*x^3+a)^(1/3)*(B*x^3+A)/x^9,x)
Output:
int((b*x^3+a)^(1/3)*(B*x^3+A)/x^9,x)
\[ \int \frac {\sqrt [3]{a+b x^3} \left (A+B x^3\right )}{x^9} \, dx=\int { \frac {{\left (B x^{3} + A\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x^{9}} \,d x } \] Input:
integrate((b*x^3+a)^(1/3)*(B*x^3+A)/x^9,x, algorithm="fricas")
Output:
integral((B*x^3 + A)*(b*x^3 + a)^(1/3)/x^9, x)
Result contains complex when optimal does not.
Time = 1.59 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.11 \[ \int \frac {\sqrt [3]{a+b x^3} \left (A+B x^3\right )}{x^9} \, dx=\frac {A \sqrt [3]{a} \Gamma \left (- \frac {8}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {8}{3}, - \frac {1}{3} \\ - \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 x^{8} \Gamma \left (- \frac {5}{3}\right )} + \frac {B \sqrt [3]{b} \Gamma \left (- \frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{3}}} \right )}}{3 x^{4} \Gamma \left (- \frac {1}{3}\right )} \] Input:
integrate((b*x**3+a)**(1/3)*(B*x**3+A)/x**9,x)
Output:
A*a**(1/3)*gamma(-8/3)*hyper((-8/3, -1/3), (-5/3,), b*x**3*exp_polar(I*pi) /a)/(3*x**8*gamma(-5/3)) + B*b**(1/3)*gamma(-4/3)*hyper((-1/3, 4/3), (7/3, ), a*exp_polar(I*pi)/(b*x**3))/(3*x**4*gamma(-1/3))
\[ \int \frac {\sqrt [3]{a+b x^3} \left (A+B x^3\right )}{x^9} \, dx=\int { \frac {{\left (B x^{3} + A\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x^{9}} \,d x } \] Input:
integrate((b*x^3+a)^(1/3)*(B*x^3+A)/x^9,x, algorithm="maxima")
Output:
integrate((B*x^3 + A)*(b*x^3 + a)^(1/3)/x^9, x)
\[ \int \frac {\sqrt [3]{a+b x^3} \left (A+B x^3\right )}{x^9} \, dx=\int { \frac {{\left (B x^{3} + A\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x^{9}} \,d x } \] Input:
integrate((b*x^3+a)^(1/3)*(B*x^3+A)/x^9,x, algorithm="giac")
Output:
integrate((B*x^3 + A)*(b*x^3 + a)^(1/3)/x^9, x)
Timed out. \[ \int \frac {\sqrt [3]{a+b x^3} \left (A+B x^3\right )}{x^9} \, dx=\int \frac {\left (B\,x^3+A\right )\,{\left (b\,x^3+a\right )}^{1/3}}{x^9} \,d x \] Input:
int(((A + B*x^3)*(a + b*x^3)^(1/3))/x^9,x)
Output:
int(((A + B*x^3)*(a + b*x^3)^(1/3))/x^9, x)
\[ \int \frac {\sqrt [3]{a+b x^3} \left (A+B x^3\right )}{x^9} \, dx=\frac {-3 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a -7 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b \,x^{3}+4 \left (\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{b \,x^{12}+a \,x^{9}}d x \right ) a^{2} x^{8}}{28 x^{8}} \] Input:
int((b*x^3+a)^(1/3)*(B*x^3+A)/x^9,x)
Output:
( - 3*(a + b*x**3)**(1/3)*a - 7*(a + b*x**3)**(1/3)*b*x**3 + 4*int((a + b* x**3)**(1/3)/(a*x**9 + b*x**12),x)*a**2*x**8)/(28*x**8)