Integrand size = 22, antiderivative size = 86 \[ \int \frac {x^3 \left (A+B x^3\right )}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {B x^4 \sqrt [3]{a+b x^3}}{5 b}+\frac {(5 A b-4 a B) x^4 \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {4}{3},\frac {7}{3},-\frac {b x^3}{a}\right )}{20 b \left (a+b x^3\right )^{2/3}} \] Output:
1/5*B*x^4*(b*x^3+a)^(1/3)/b+1/20*(5*A*b-4*B*a)*x^4*(1+b*x^3/a)^(2/3)*hyper geom([2/3, 4/3],[7/3],-b*x^3/a)/b/(b*x^3+a)^(2/3)
Time = 10.05 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.93 \[ \int \frac {x^3 \left (A+B x^3\right )}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {\left (1+\frac {b x^3}{a}\right )^{2/3} \left (7 A x^4 \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {4}{3},\frac {7}{3},-\frac {b x^3}{a}\right )+4 B x^7 \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {7}{3},\frac {10}{3},-\frac {b x^3}{a}\right )\right )}{28 \left (a+b x^3\right )^{2/3}} \] Input:
Integrate[(x^3*(A + B*x^3))/(a + b*x^3)^(2/3),x]
Output:
((1 + (b*x^3)/a)^(2/3)*(7*A*x^4*Hypergeometric2F1[2/3, 4/3, 7/3, -((b*x^3) /a)] + 4*B*x^7*Hypergeometric2F1[2/3, 7/3, 10/3, -((b*x^3)/a)]))/(28*(a + b*x^3)^(2/3))
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.
\(\displaystyle \int \frac {x^3 \left (A+B x^3\right )}{\left (a+b x^3\right )^{2/3}} \, dx\) |
\(\Big \downarrow \) 959 |
\(\displaystyle \frac {(5 A b-4 a B) \int \frac {x^3}{\left (b x^3+a\right )^{2/3}}dx}{5 b}+\frac {B x^4 \sqrt [3]{a+b x^3}}{5 b}\) |
\(\Big \downarrow \) 889 |
\(\displaystyle \frac {\left (\frac {b x^3}{a}+1\right )^{2/3} (5 A b-4 a B) \int \frac {x^3}{\left (\frac {b x^3}{a}+1\right )^{2/3}}dx}{5 b \left (a+b x^3\right )^{2/3}}+\frac {B x^4 \sqrt [3]{a+b x^3}}{5 b}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {x^4 \left (\frac {b x^3}{a}+1\right )^{2/3} (5 A b-4 a B) \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {4}{3},\frac {7}{3},-\frac {b x^3}{a}\right )}{20 b \left (a+b x^3\right )^{2/3}}+\frac {B x^4 \sqrt [3]{a+b x^3}}{5 b}\) |
Input:
Int[(x^3*(A + B*x^3))/(a + b*x^3)^(2/3),x]
Output:
(B*x^4*(a + b*x^3)^(1/3))/(5*b) + ((5*A*b - 4*a*B)*x^4*(1 + (b*x^3)/a)^(2/ 3)*Hypergeometric2F1[2/3, 4/3, 7/3, -((b*x^3)/a)])/(20*b*(a + b*x^3)^(2/3) )
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]
\[\int \frac {x^{3} \left (B \,x^{3}+A \right )}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x\]
Input:
int(x^3*(B*x^3+A)/(b*x^3+a)^(2/3),x)
Output:
int(x^3*(B*x^3+A)/(b*x^3+a)^(2/3),x)
\[ \int \frac {x^3 \left (A+B x^3\right )}{\left (a+b x^3\right )^{2/3}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} x^{3}}{{\left (b x^{3} + a\right )}^{\frac {2}{3}}} \,d x } \] Input:
integrate(x^3*(B*x^3+A)/(b*x^3+a)^(2/3),x, algorithm="fricas")
Output:
integral((B*x^6 + A*x^3)/(b*x^3 + a)^(2/3), x)
Result contains complex when optimal does not.
Time = 1.68 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.93 \[ \int \frac {x^3 \left (A+B x^3\right )}{\left (a+b x^3\right )^{2/3}} \, dx=\frac {A x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {7}{3}\right )} + \frac {B x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}} \Gamma \left (\frac {10}{3}\right )} \] Input:
integrate(x**3*(B*x**3+A)/(b*x**3+a)**(2/3),x)
Output:
A*x**4*gamma(4/3)*hyper((2/3, 4/3), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*a **(2/3)*gamma(7/3)) + B*x**7*gamma(7/3)*hyper((2/3, 7/3), (10/3,), b*x**3* exp_polar(I*pi)/a)/(3*a**(2/3)*gamma(10/3))
\[ \int \frac {x^3 \left (A+B x^3\right )}{\left (a+b x^3\right )^{2/3}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} x^{3}}{{\left (b x^{3} + a\right )}^{\frac {2}{3}}} \,d x } \] Input:
integrate(x^3*(B*x^3+A)/(b*x^3+a)^(2/3),x, algorithm="maxima")
Output:
integrate((B*x^3 + A)*x^3/(b*x^3 + a)^(2/3), x)
\[ \int \frac {x^3 \left (A+B x^3\right )}{\left (a+b x^3\right )^{2/3}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} x^{3}}{{\left (b x^{3} + a\right )}^{\frac {2}{3}}} \,d x } \] Input:
integrate(x^3*(B*x^3+A)/(b*x^3+a)^(2/3),x, algorithm="giac")
Output:
integrate((B*x^3 + A)*x^3/(b*x^3 + a)^(2/3), x)
Timed out. \[ \int \frac {x^3 \left (A+B x^3\right )}{\left (a+b x^3\right )^{2/3}} \, dx=\int \frac {x^3\,\left (B\,x^3+A\right )}{{\left (b\,x^3+a\right )}^{2/3}} \,d x \] Input:
int((x^3*(A + B*x^3))/(a + b*x^3)^(2/3),x)
Output:
int((x^3*(A + B*x^3))/(a + b*x^3)^(2/3), x)
\[ \int \frac {x^3 \left (A+B x^3\right )}{\left (a+b x^3\right )^{2/3}} \, dx=\left (\int \frac {x^{6}}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) b +\left (\int \frac {x^{3}}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) a \] Input:
int(x^3*(B*x^3+A)/(b*x^3+a)^(2/3),x)
Output:
int(x**6/(a + b*x**3)**(2/3),x)*b + int(x**3/(a + b*x**3)**(2/3),x)*a