Integrand size = 24, antiderivative size = 115 \[ \int (e x)^m \sqrt [3]{a+b x^3} \left (A+B x^3\right ) \, dx=\frac {B (e x)^{1+m} \left (a+b x^3\right )^{4/3}}{b e (5+m)}+\frac {\left (\frac {A}{1+m}-\frac {a B}{b (5+m)}\right ) (e x)^{1+m} \sqrt [3]{a+b x^3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1+m}{3},\frac {4+m}{3},-\frac {b x^3}{a}\right )}{e \sqrt [3]{1+\frac {b x^3}{a}}} \] Output:
B*(e*x)^(1+m)*(b*x^3+a)^(4/3)/b/e/(5+m)+(A/(1+m)-a*B/b/(5+m))*(e*x)^(1+m)* (b*x^3+a)^(1/3)*hypergeom([-1/3, 1/3+1/3*m],[4/3+1/3*m],-b*x^3/a)/e/(1+b*x ^3/a)^(1/3)
Time = 0.28 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.96 \[ \int (e x)^m \sqrt [3]{a+b x^3} \left (A+B x^3\right ) \, dx=\frac {x (e x)^m \sqrt [3]{a+b x^3} \left (A (4+m) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1+m}{3},\frac {4+m}{3},-\frac {b x^3}{a}\right )+B (1+m) x^3 \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {4+m}{3},\frac {7+m}{3},-\frac {b x^3}{a}\right )\right )}{(1+m) (4+m) \sqrt [3]{1+\frac {b x^3}{a}}} \] Input:
Integrate[(e*x)^m*(a + b*x^3)^(1/3)*(A + B*x^3),x]
Output:
(x*(e*x)^m*(a + b*x^3)^(1/3)*(A*(4 + m)*Hypergeometric2F1[-1/3, (1 + m)/3, (4 + m)/3, -((b*x^3)/a)] + B*(1 + m)*x^3*Hypergeometric2F1[-1/3, (4 + m)/ 3, (7 + m)/3, -((b*x^3)/a)]))/((1 + m)*(4 + m)*(1 + (b*x^3)/a)^(1/3))
Time = 0.42 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.02, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, 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.
| \(\displaystyle \int \sqrt [3]{a+b x^3} \left (A+B x^3\right ) (e x)^m \, dx\) |
| \(\Big \downarrow \) 959 |
| \(\displaystyle \left (A-\frac {a B (m+1)}{b (m+5)}\right ) \int (e x)^m \sqrt [3]{b x^3+a}dx+\frac {B \left (a+b x^3\right )^{4/3} (e x)^{m+1}}{b e (m+5)}\) |
| \(\Big \downarrow \) 889 |
| \(\displaystyle \frac {\sqrt [3]{a+b x^3} \left (A-\frac {a B (m+1)}{b (m+5)}\right ) \int (e x)^m \sqrt [3]{\frac {b x^3}{a}+1}dx}{\sqrt [3]{\frac {b x^3}{a}+1}}+\frac {B \left (a+b x^3\right )^{4/3} (e x)^{m+1}}{b e (m+5)}\) |
| \(\Big \downarrow \) 888 |
| \(\displaystyle \frac {\sqrt [3]{a+b x^3} (e x)^{m+1} \left (A-\frac {a B (m+1)}{b (m+5)}\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {m+1}{3},\frac {m+4}{3},-\frac {b x^3}{a}\right )}{e (m+1) \sqrt [3]{\frac {b x^3}{a}+1}}+\frac {B \left (a+b x^3\right )^{4/3} (e x)^{m+1}}{b e (m+5)}\) |
Input:
Int[(e*x)^m*(a + b*x^3)^(1/3)*(A + B*x^3),x]
Output:
(B*(e*x)^(1 + m)*(a + b*x^3)^(4/3))/(b*e*(5 + m)) + ((A - (a*B*(1 + m))/(b *(5 + m)))*(e*x)^(1 + m)*(a + b*x^3)^(1/3)*Hypergeometric2F1[-1/3, (1 + m) /3, (4 + m)/3, -((b*x^3)/a)])/(e*(1 + m)*(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]
\[\int \left (e x \right )^{m} \left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (B \,x^{3}+A \right )d x\]
Input:
int((e*x)^m*(b*x^3+a)^(1/3)*(B*x^3+A),x)
Output:
int((e*x)^m*(b*x^3+a)^(1/3)*(B*x^3+A),x)
\[ \int (e x)^m \sqrt [3]{a+b x^3} \left (A+B x^3\right ) \, dx=\int { {\left (B x^{3} + A\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (e x\right )^{m} \,d x } \] Input:
integrate((e*x)^m*(b*x^3+a)^(1/3)*(B*x^3+A),x, algorithm="fricas")
Output:
integral((B*x^3 + A)*(b*x^3 + a)^(1/3)*(e*x)^m, x)
Result contains complex when optimal does not.
Time = 3.23 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.05 \[ \int (e x)^m \sqrt [3]{a+b x^3} \left (A+B x^3\right ) \, dx=\frac {A \sqrt [3]{a} e^{m} x^{m + 1} \Gamma \left (\frac {m}{3} + \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {m}{3} + \frac {1}{3} \\ \frac {m}{3} + \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {m}{3} + \frac {4}{3}\right )} + \frac {B \sqrt [3]{a} e^{m} x^{m + 4} \Gamma \left (\frac {m}{3} + \frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {m}{3} + \frac {4}{3} \\ \frac {m}{3} + \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {m}{3} + \frac {7}{3}\right )} \] Input:
integrate((e*x)**m*(b*x**3+a)**(1/3)*(B*x**3+A),x)
Output:
A*a**(1/3)*e**m*x**(m + 1)*gamma(m/3 + 1/3)*hyper((-1/3, m/3 + 1/3), (m/3 + 4/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(m/3 + 4/3)) + B*a**(1/3)*e**m* x**(m + 4)*gamma(m/3 + 4/3)*hyper((-1/3, m/3 + 4/3), (m/3 + 7/3,), b*x**3* exp_polar(I*pi)/a)/(3*gamma(m/3 + 7/3))
\[ \int (e x)^m \sqrt [3]{a+b x^3} \left (A+B x^3\right ) \, dx=\int { {\left (B x^{3} + A\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (e x\right )^{m} \,d x } \] Input:
integrate((e*x)^m*(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)*(e*x)^m, x)
\[ \int (e x)^m \sqrt [3]{a+b x^3} \left (A+B x^3\right ) \, dx=\int { {\left (B x^{3} + A\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (e x\right )^{m} \,d x } \] Input:
integrate((e*x)^m*(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)*(e*x)^m, x)
Timed out. \[ \int (e x)^m \sqrt [3]{a+b x^3} \left (A+B x^3\right ) \, dx=\int \left (B\,x^3+A\right )\,{\left (e\,x\right )}^m\,{\left (b\,x^3+a\right )}^{1/3} \,d x \] Input:
int((A + B*x^3)*(e*x)^m*(a + b*x^3)^(1/3),x)
Output:
int((A + B*x^3)*(e*x)^m*(a + b*x^3)^(1/3), x)
\[ \int (e x)^m \sqrt [3]{a+b x^3} \left (A+B x^3\right ) \, dx=\frac {e^{m} \left (x^{m} \left (b \,x^{3}+a \right )^{\frac {1}{3}} a m x +6 x^{m} \left (b \,x^{3}+a \right )^{\frac {1}{3}} a x +x^{m} \left (b \,x^{3}+a \right )^{\frac {1}{3}} b m \,x^{4}+2 x^{m} \left (b \,x^{3}+a \right )^{\frac {1}{3}} b \,x^{4}+4 \left (\int \frac {x^{m} \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{b \,m^{2} x^{3}+7 b m \,x^{3}+10 b \,x^{3}+a \,m^{2}+7 a m +10 a}d x \right ) a^{2} m^{2}+28 \left (\int \frac {x^{m} \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{b \,m^{2} x^{3}+7 b m \,x^{3}+10 b \,x^{3}+a \,m^{2}+7 a m +10 a}d x \right ) a^{2} m +40 \left (\int \frac {x^{m} \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{b \,m^{2} x^{3}+7 b m \,x^{3}+10 b \,x^{3}+a \,m^{2}+7 a m +10 a}d x \right ) a^{2}\right )}{m^{2}+7 m +10} \] Input:
int((e*x)^m*(b*x^3+a)^(1/3)*(B*x^3+A),x)
Output:
(e**m*(x**m*(a + b*x**3)**(1/3)*a*m*x + 6*x**m*(a + b*x**3)**(1/3)*a*x + x **m*(a + b*x**3)**(1/3)*b*m*x**4 + 2*x**m*(a + b*x**3)**(1/3)*b*x**4 + 4*i nt((x**m*(a + b*x**3)**(1/3))/(a*m**2 + 7*a*m + 10*a + b*m**2*x**3 + 7*b*m *x**3 + 10*b*x**3),x)*a**2*m**2 + 28*int((x**m*(a + b*x**3)**(1/3))/(a*m** 2 + 7*a*m + 10*a + b*m**2*x**3 + 7*b*m*x**3 + 10*b*x**3),x)*a**2*m + 40*in t((x**m*(a + b*x**3)**(1/3))/(a*m**2 + 7*a*m + 10*a + b*m**2*x**3 + 7*b*m* x**3 + 10*b*x**3),x)*a**2))/(m**2 + 7*m + 10)