Integrand size = 22, antiderivative size = 86 \[ \int \frac {x^4 \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\frac {B x^5 \left (a+b x^3\right )^{2/3}}{7 b}+\frac {(7 A b-5 a B) x^5 \sqrt [3]{1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {5}{3},\frac {8}{3},-\frac {b x^3}{a}\right )}{35 b \sqrt [3]{a+b x^3}} \] Output:
1/7*B*x^5*(b*x^3+a)^(2/3)/b+1/35*(7*A*b-5*B*a)*x^5*(1+b*x^3/a)^(1/3)*hyper geom([1/3, 5/3],[8/3],-b*x^3/a)/b/(b*x^3+a)^(1/3)
Time = 10.05 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.93 \[ \int \frac {x^4 \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\frac {\sqrt [3]{1+\frac {b x^3}{a}} \left (8 A x^5 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {5}{3},\frac {8}{3},-\frac {b x^3}{a}\right )+5 B x^8 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {8}{3},\frac {11}{3},-\frac {b x^3}{a}\right )\right )}{40 \sqrt [3]{a+b x^3}} \] Input:
Integrate[(x^4*(A + B*x^3))/(a + b*x^3)^(1/3),x]
Output:
((1 + (b*x^3)/a)^(1/3)*(8*A*x^5*Hypergeometric2F1[1/3, 5/3, 8/3, -((b*x^3) /a)] + 5*B*x^8*Hypergeometric2F1[1/3, 8/3, 11/3, -((b*x^3)/a)]))/(40*(a + b*x^3)^(1/3))
Time = 0.35 (sec) , antiderivative size = 86, 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 = {959, 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.
\(\displaystyle \int \frac {x^4 \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx\) |
\(\Big \downarrow \) 959 |
\(\displaystyle \frac {(7 A b-5 a B) \int \frac {x^4}{\sqrt [3]{b x^3+a}}dx}{7 b}+\frac {B x^5 \left (a+b x^3\right )^{2/3}}{7 b}\) |
\(\Big \downarrow \) 889 |
\(\displaystyle \frac {\sqrt [3]{\frac {b x^3}{a}+1} (7 A b-5 a B) \int \frac {x^4}{\sqrt [3]{\frac {b x^3}{a}+1}}dx}{7 b \sqrt [3]{a+b x^3}}+\frac {B x^5 \left (a+b x^3\right )^{2/3}}{7 b}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {x^5 \sqrt [3]{\frac {b x^3}{a}+1} (7 A b-5 a B) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {5}{3},\frac {8}{3},-\frac {b x^3}{a}\right )}{35 b \sqrt [3]{a+b x^3}}+\frac {B x^5 \left (a+b x^3\right )^{2/3}}{7 b}\) |
Input:
Int[(x^4*(A + B*x^3))/(a + b*x^3)^(1/3),x]
Output:
(B*x^5*(a + b*x^3)^(2/3))/(7*b) + ((7*A*b - 5*a*B)*x^5*(1 + (b*x^3)/a)^(1/ 3)*Hypergeometric2F1[1/3, 5/3, 8/3, -((b*x^3)/a)])/(35*b*(a + b*x^3)^(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[d*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p + 1) + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m + n*(p + 1) + 1)) Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[m + n*(p + 1) + 1, 0]
\[\int \frac {x^{4} \left (B \,x^{3}+A \right )}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x\]
Input:
int(x^4*(B*x^3+A)/(b*x^3+a)^(1/3),x)
Output:
int(x^4*(B*x^3+A)/(b*x^3+a)^(1/3),x)
\[ \int \frac {x^4 \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} x^{4}}{{\left (b x^{3} + a\right )}^{\frac {1}{3}}} \,d x } \] Input:
integrate(x^4*(B*x^3+A)/(b*x^3+a)^(1/3),x, algorithm="fricas")
Output:
integral((B*x^7 + A*x^4)/(b*x^3 + a)^(1/3), x)
Result contains complex when optimal does not.
Time = 1.54 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.93 \[ \int \frac {x^4 \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\frac {A x^{5} \Gamma \left (\frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {5}{3} \\ \frac {8}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {8}{3}\right )} + \frac {B x^{8} \Gamma \left (\frac {8}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {8}{3} \\ \frac {11}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {11}{3}\right )} \] Input:
integrate(x**4*(B*x**3+A)/(b*x**3+a)**(1/3),x)
Output:
A*x**5*gamma(5/3)*hyper((1/3, 5/3), (8/3,), b*x**3*exp_polar(I*pi)/a)/(3*a **(1/3)*gamma(8/3)) + B*x**8*gamma(8/3)*hyper((1/3, 8/3), (11/3,), b*x**3* exp_polar(I*pi)/a)/(3*a**(1/3)*gamma(11/3))
\[ \int \frac {x^4 \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} x^{4}}{{\left (b x^{3} + a\right )}^{\frac {1}{3}}} \,d x } \] Input:
integrate(x^4*(B*x^3+A)/(b*x^3+a)^(1/3),x, algorithm="maxima")
Output:
integrate((B*x^3 + A)*x^4/(b*x^3 + a)^(1/3), x)
\[ \int \frac {x^4 \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} x^{4}}{{\left (b x^{3} + a\right )}^{\frac {1}{3}}} \,d x } \] Input:
integrate(x^4*(B*x^3+A)/(b*x^3+a)^(1/3),x, algorithm="giac")
Output:
integrate((B*x^3 + A)*x^4/(b*x^3 + a)^(1/3), x)
Timed out. \[ \int \frac {x^4 \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\int \frac {x^4\,\left (B\,x^3+A\right )}{{\left (b\,x^3+a\right )}^{1/3}} \,d x \] Input:
int((x^4*(A + B*x^3))/(a + b*x^3)^(1/3),x)
Output:
int((x^4*(A + B*x^3))/(a + b*x^3)^(1/3), x)
\[ \int \frac {x^4 \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\left (\int \frac {x^{7}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x \right ) b +\left (\int \frac {x^{4}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x \right ) a \] Input:
int(x^4*(B*x^3+A)/(b*x^3+a)^(1/3),x)
Output:
int(x**7/(a + b*x**3)**(1/3),x)*b + int(x**4/(a + b*x**3)**(1/3),x)*a