Integrand size = 22, antiderivative size = 86 \[ \int x^3 \sqrt [3]{a+b x^3} \left (A+B x^3\right ) \, dx=\frac {B x^4 \left (a+b x^3\right )^{4/3}}{8 b}+\frac {(2 A b-a B) x^4 \sqrt [3]{a+b x^3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {4}{3},\frac {7}{3},-\frac {b x^3}{a}\right )}{8 b \sqrt [3]{1+\frac {b x^3}{a}}} \] Output:
1/8*B*x^4*(b*x^3+a)^(4/3)/b+1/8*(2*A*b-B*a)*x^4*(b*x^3+a)^(1/3)*hypergeom( [-1/3, 4/3],[7/3],-b*x^3/a)/b/(1+b*x^3/a)^(1/3)
Time = 6.76 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.93 \[ \int x^3 \sqrt [3]{a+b x^3} \left (A+B x^3\right ) \, dx=\frac {\sqrt [3]{a+b x^3} \left (7 A x^4 \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {4}{3},\frac {7}{3},-\frac {b x^3}{a}\right )+4 B x^7 \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {7}{3},\frac {10}{3},-\frac {b x^3}{a}\right )\right )}{28 \sqrt [3]{1+\frac {b x^3}{a}}} \] Input:
Integrate[x^3*(a + b*x^3)^(1/3)*(A + B*x^3),x]
Output:
((a + b*x^3)^(1/3)*(7*A*x^4*Hypergeometric2F1[-1/3, 4/3, 7/3, -((b*x^3)/a) ] + 4*B*x^7*Hypergeometric2F1[-1/3, 7/3, 10/3, -((b*x^3)/a)]))/(28*(1 + (b *x^3)/a)^(1/3))
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.
\(\displaystyle \int x^3 \sqrt [3]{a+b x^3} \left (A+B x^3\right ) \, dx\) |
\(\Big \downarrow \) 959 |
\(\displaystyle \frac {(2 A b-a B) \int x^3 \sqrt [3]{b x^3+a}dx}{2 b}+\frac {B x^4 \left (a+b x^3\right )^{4/3}}{8 b}\) |
\(\Big \downarrow \) 889 |
\(\displaystyle \frac {\sqrt [3]{a+b x^3} (2 A b-a B) \int x^3 \sqrt [3]{\frac {b x^3}{a}+1}dx}{2 b \sqrt [3]{\frac {b x^3}{a}+1}}+\frac {B x^4 \left (a+b x^3\right )^{4/3}}{8 b}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {x^4 \sqrt [3]{a+b x^3} (2 A b-a B) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {4}{3},\frac {7}{3},-\frac {b x^3}{a}\right )}{8 b \sqrt [3]{\frac {b x^3}{a}+1}}+\frac {B x^4 \left (a+b x^3\right )^{4/3}}{8 b}\) |
Input:
Int[x^3*(a + b*x^3)^(1/3)*(A + B*x^3),x]
Output:
(B*x^4*(a + b*x^3)^(4/3))/(8*b) + ((2*A*b - a*B)*x^4*(a + b*x^3)^(1/3)*Hyp ergeometric2F1[-1/3, 4/3, 7/3, -((b*x^3)/a)])/(8*b*(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[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]
\[\int x^{3} \left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (B \,x^{3}+A \right )d x\]
Input:
int(x^3*(b*x^3+a)^(1/3)*(B*x^3+A),x)
Output:
int(x^3*(b*x^3+a)^(1/3)*(B*x^3+A),x)
\[ \int x^3 \sqrt [3]{a+b x^3} \left (A+B x^3\right ) \, dx=\int { {\left (B x^{3} + A\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}} x^{3} \,d x } \] Input:
integrate(x^3*(b*x^3+a)^(1/3)*(B*x^3+A),x, algorithm="fricas")
Output:
integral((B*x^6 + A*x^3)*(b*x^3 + a)^(1/3), x)
Result contains complex when optimal does not.
Time = 1.19 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.97 \[ \int x^3 \sqrt [3]{a+b x^3} \left (A+B x^3\right ) \, dx=\frac {A \sqrt [3]{a} x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} + \frac {B \sqrt [3]{a} x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {10}{3}\right )} \] Input:
integrate(x**3*(b*x**3+a)**(1/3)*(B*x**3+A),x)
Output:
A*a**(1/3)*x**4*gamma(4/3)*hyper((-1/3, 4/3), (7/3,), b*x**3*exp_polar(I*p i)/a)/(3*gamma(7/3)) + B*a**(1/3)*x**7*gamma(7/3)*hyper((-1/3, 7/3), (10/3 ,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(10/3))
\[ \int x^3 \sqrt [3]{a+b x^3} \left (A+B x^3\right ) \, dx=\int { {\left (B x^{3} + A\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}} x^{3} \,d x } \] Input:
integrate(x^3*(b*x^3+a)^(1/3)*(B*x^3+A),x, algorithm="maxima")
Output:
integrate((B*x^3 + A)*(b*x^3 + a)^(1/3)*x^3, x)
\[ \int x^3 \sqrt [3]{a+b x^3} \left (A+B x^3\right ) \, dx=\int { {\left (B x^{3} + A\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}} x^{3} \,d x } \] Input:
integrate(x^3*(b*x^3+a)^(1/3)*(B*x^3+A),x, algorithm="giac")
Output:
integrate((B*x^3 + A)*(b*x^3 + a)^(1/3)*x^3, x)
Timed out. \[ \int x^3 \sqrt [3]{a+b x^3} \left (A+B x^3\right ) \, dx=\int x^3\,\left (B\,x^3+A\right )\,{\left (b\,x^3+a\right )}^{1/3} \,d x \] Input:
int(x^3*(A + B*x^3)*(a + b*x^3)^(1/3),x)
Output:
int(x^3*(A + B*x^3)*(a + b*x^3)^(1/3), x)
\[ \int x^3 \sqrt [3]{a+b x^3} \left (A+B x^3\right ) \, dx=\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2} x +9 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a b \,x^{4}+5 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{2} x^{7}-2 \left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}}}d x \right ) a^{3}}{40 b} \] Input:
int(x^3*(b*x^3+a)^(1/3)*(B*x^3+A),x)
Output:
(2*(a + b*x**3)**(1/3)*a**2*x + 9*(a + b*x**3)**(1/3)*a*b*x**4 + 5*(a + b* x**3)**(1/3)*b**2*x**7 - 2*int((a + b*x**3)**(1/3)/(a + b*x**3),x)*a**3)/( 40*b)