\(\int x^4 (a+b x^3)^{2/3} (A+B x^3) \, dx\) [320]

Optimal result

Integrand size = 22, antiderivative size = 86 \[ \int x^4 \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx=\frac {B x^5 \left (a+b x^3\right )^{5/3}}{10 b}+\frac {(2 A b-a B) x^5 \left (a+b x^3\right )^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {5}{3},\frac {8}{3},-\frac {b x^3}{a}\right )}{10 b \left (1+\frac {b x^3}{a}\right )^{2/3}} \] Output:

1/10*B*x^5*(b*x^3+a)^(5/3)/b+1/10*(2*A*b-B*a)*x^5*(b*x^3+a)^(2/3)*hypergeo 
m([-2/3, 5/3],[8/3],-b*x^3/a)/b/(1+b*x^3/a)^(2/3)
 

Mathematica [A] (verified)

Time = 7.96 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.93 \[ \int x^4 \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx=\frac {\left (a+b x^3\right )^{2/3} \left (8 A x^5 \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {5}{3},\frac {8}{3},-\frac {b x^3}{a}\right )+5 B x^8 \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {8}{3},\frac {11}{3},-\frac {b x^3}{a}\right )\right )}{40 \left (1+\frac {b x^3}{a}\right )^{2/3}} \] Input:

Integrate[x^4*(a + b*x^3)^(2/3)*(A + B*x^3),x]
 

Output:

((a + b*x^3)^(2/3)*(8*A*x^5*Hypergeometric2F1[-2/3, 5/3, 8/3, -((b*x^3)/a) 
] + 5*B*x^8*Hypergeometric2F1[-2/3, 8/3, 11/3, -((b*x^3)/a)]))/(40*(1 + (b 
*x^3)/a)^(2/3))
 

Rubi [A] (verified)

Time = 0.36 (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.

\(\displaystyle \int x^4 \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx\)

\(\Big \downarrow \) 959

\(\displaystyle \frac {(2 A b-a B) \int x^4 \left (b x^3+a\right )^{2/3}dx}{2 b}+\frac {B x^5 \left (a+b x^3\right )^{5/3}}{10 b}\)

\(\Big \downarrow \) 889

\(\displaystyle \frac {\left (a+b x^3\right )^{2/3} (2 A b-a B) \int x^4 \left (\frac {b x^3}{a}+1\right )^{2/3}dx}{2 b \left (\frac {b x^3}{a}+1\right )^{2/3}}+\frac {B x^5 \left (a+b x^3\right )^{5/3}}{10 b}\)

\(\Big \downarrow \) 888

\(\displaystyle \frac {x^5 \left (a+b x^3\right )^{2/3} (2 A b-a B) \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {5}{3},\frac {8}{3},-\frac {b x^3}{a}\right )}{10 b \left (\frac {b x^3}{a}+1\right )^{2/3}}+\frac {B x^5 \left (a+b x^3\right )^{5/3}}{10 b}\)

Input:

Int[x^4*(a + b*x^3)^(2/3)*(A + B*x^3),x]
 

Output:

(B*x^5*(a + b*x^3)^(5/3))/(10*b) + ((2*A*b - a*B)*x^5*(a + b*x^3)^(2/3)*Hy 
pergeometric2F1[-2/3, 5/3, 8/3, -((b*x^3)/a)])/(10*b*(1 + (b*x^3)/a)^(2/3) 
)
 

Defintions of rubi rules used

rule 888

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])
 

rule 889

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])
 

rule 959

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]
 

Maple [F]

\[\int x^{4} \left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (B \,x^{3}+A \right )d x\]

Input:

int(x^4*(b*x^3+a)^(2/3)*(B*x^3+A),x)
 

Output:

int(x^4*(b*x^3+a)^(2/3)*(B*x^3+A),x)
 

Fricas [F]

\[ \int x^4 \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx=\int { {\left (B x^{3} + A\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}} x^{4} \,d x } \] Input:

integrate(x^4*(b*x^3+a)^(2/3)*(B*x^3+A),x, algorithm="fricas")
 

Output:

integral((B*x^7 + A*x^4)*(b*x^3 + a)^(2/3), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.37 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.97 \[ \int x^4 \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx=\frac {A a^{\frac {2}{3}} x^{5} \Gamma \left (\frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {5}{3} \\ \frac {8}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {8}{3}\right )} + \frac {B a^{\frac {2}{3}} x^{8} \Gamma \left (\frac {8}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {8}{3} \\ \frac {11}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {11}{3}\right )} \] Input:

integrate(x**4*(b*x**3+a)**(2/3)*(B*x**3+A),x)
 

Output:

A*a**(2/3)*x**5*gamma(5/3)*hyper((-2/3, 5/3), (8/3,), b*x**3*exp_polar(I*p 
i)/a)/(3*gamma(8/3)) + B*a**(2/3)*x**8*gamma(8/3)*hyper((-2/3, 8/3), (11/3 
,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(11/3))
                                                                                    
                                                                                    
 

Maxima [F]

\[ \int x^4 \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx=\int { {\left (B x^{3} + A\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}} x^{4} \,d x } \] Input:

integrate(x^4*(b*x^3+a)^(2/3)*(B*x^3+A),x, algorithm="maxima")
 

Output:

integrate((B*x^3 + A)*(b*x^3 + a)^(2/3)*x^4, x)
 

Giac [F]

\[ \int x^4 \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx=\int { {\left (B x^{3} + A\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}} x^{4} \,d x } \] Input:

integrate(x^4*(b*x^3+a)^(2/3)*(B*x^3+A),x, algorithm="giac")
 

Output:

integrate((B*x^3 + A)*(b*x^3 + a)^(2/3)*x^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int x^4 \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx=\int x^4\,\left (B\,x^3+A\right )\,{\left (b\,x^3+a\right )}^{2/3} \,d x \] Input:

int(x^4*(A + B*x^3)*(a + b*x^3)^(2/3),x)
 

Output:

int(x^4*(A + B*x^3)*(a + b*x^3)^(2/3), x)
 

Reduce [F]

\[ \int x^4 \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx=\frac {5 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a^{2} x^{2}+24 \left (b \,x^{3}+a \right )^{\frac {2}{3}} a b \,x^{5}+14 \left (b \,x^{3}+a \right )^{\frac {2}{3}} b^{2} x^{8}-10 \left (\int \frac {x}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x \right ) a^{3}}{140 b} \] Input:

int(x^4*(b*x^3+a)^(2/3)*(B*x^3+A),x)
 

Output:

(5*(a + b*x**3)**(2/3)*a**2*x**2 + 24*(a + b*x**3)**(2/3)*a*b*x**5 + 14*(a 
 + b*x**3)**(2/3)*b**2*x**8 - 10*int(((a + b*x**3)**(2/3)*x)/(a + b*x**3), 
x)*a**3)/(140*b)