Integrand size = 24, antiderivative size = 115 \[ \int (e x)^m \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx=\frac {B (e x)^{1+m} \left (a+b x^3\right )^{5/3}}{b e (6+m)}+\frac {\left (\frac {A}{1+m}-\frac {a B}{b (6+m)}\right ) (e x)^{1+m} \left (a+b x^3\right )^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1+m}{3},\frac {4+m}{3},-\frac {b x^3}{a}\right )}{e \left (1+\frac {b x^3}{a}\right )^{2/3}} \] Output:
B*(e*x)^(1+m)*(b*x^3+a)^(5/3)/b/e/(6+m)+(A/(1+m)-a*B/b/(6+m))*(e*x)^(1+m)* (b*x^3+a)^(2/3)*hypergeom([-2/3, 1/3+1/3*m],[4/3+1/3*m],-b*x^3/a)/e/(1+b*x ^3/a)^(2/3)
Time = 0.32 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.96 \[ \int (e x)^m \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx=\frac {x (e x)^m \left (a+b x^3\right )^{2/3} \left (A (4+m) \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {1+m}{3},\frac {4+m}{3},-\frac {b x^3}{a}\right )+B (1+m) x^3 \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {4+m}{3},\frac {7+m}{3},-\frac {b x^3}{a}\right )\right )}{(1+m) (4+m) \left (1+\frac {b x^3}{a}\right )^{2/3}} \] Input:
Integrate[(e*x)^m*(a + b*x^3)^(2/3)*(A + B*x^3),x]
Output:
(x*(e*x)^m*(a + b*x^3)^(2/3)*(A*(4 + m)*Hypergeometric2F1[-2/3, (1 + m)/3, (4 + m)/3, -((b*x^3)/a)] + B*(1 + m)*x^3*Hypergeometric2F1[-2/3, (4 + m)/ 3, (7 + m)/3, -((b*x^3)/a)]))/((1 + m)*(4 + m)*(1 + (b*x^3)/a)^(2/3))
Time = 0.43 (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.
\(\displaystyle \int \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) (e x)^m \, dx\) |
\(\Big \downarrow \) 959 |
\(\displaystyle \left (A-\frac {a B (m+1)}{b (m+6)}\right ) \int (e x)^m \left (b x^3+a\right )^{2/3}dx+\frac {B \left (a+b x^3\right )^{5/3} (e x)^{m+1}}{b e (m+6)}\) |
\(\Big \downarrow \) 889 |
\(\displaystyle \frac {\left (a+b x^3\right )^{2/3} \left (A-\frac {a B (m+1)}{b (m+6)}\right ) \int (e x)^m \left (\frac {b x^3}{a}+1\right )^{2/3}dx}{\left (\frac {b x^3}{a}+1\right )^{2/3}}+\frac {B \left (a+b x^3\right )^{5/3} (e x)^{m+1}}{b e (m+6)}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {\left (a+b x^3\right )^{2/3} (e x)^{m+1} \left (A-\frac {a B (m+1)}{b (m+6)}\right ) \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {m+1}{3},\frac {m+4}{3},-\frac {b x^3}{a}\right )}{e (m+1) \left (\frac {b x^3}{a}+1\right )^{2/3}}+\frac {B \left (a+b x^3\right )^{5/3} (e x)^{m+1}}{b e (m+6)}\) |
Input:
Int[(e*x)^m*(a + b*x^3)^(2/3)*(A + B*x^3),x]
Output:
(B*(e*x)^(1 + m)*(a + b*x^3)^(5/3))/(b*e*(6 + m)) + ((A - (a*B*(1 + m))/(b *(6 + m)))*(e*x)^(1 + m)*(a + b*x^3)^(2/3)*Hypergeometric2F1[-2/3, (1 + m) /3, (4 + m)/3, -((b*x^3)/a)])/(e*(1 + m)*(1 + (b*x^3)/a)^(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]
\[\int \left (e x \right )^{m} \left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (B \,x^{3}+A \right )d x\]
Input:
int((e*x)^m*(b*x^3+a)^(2/3)*(B*x^3+A),x)
Output:
int((e*x)^m*(b*x^3+a)^(2/3)*(B*x^3+A),x)
\[ \int (e x)^m \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx=\int { {\left (B x^{3} + A\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}} \left (e x\right )^{m} \,d x } \] Input:
integrate((e*x)^m*(b*x^3+a)^(2/3)*(B*x^3+A),x, algorithm="fricas")
Output:
integral((B*x^3 + A)*(b*x^3 + a)^(2/3)*(e*x)^m, x)
Result contains complex when optimal does not.
Time = 5.01 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.05 \[ \int (e x)^m \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx=\frac {A a^{\frac {2}{3}} e^{m} x^{m + 1} \Gamma \left (\frac {m}{3} + \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {m}{3} + \frac {1}{3} \\ \frac {m}{3} + \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {m}{3} + \frac {4}{3}\right )} + \frac {B a^{\frac {2}{3}} e^{m} x^{m + 4} \Gamma \left (\frac {m}{3} + \frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, \frac {m}{3} + \frac {4}{3} \\ \frac {m}{3} + \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {m}{3} + \frac {7}{3}\right )} \] Input:
integrate((e*x)**m*(b*x**3+a)**(2/3)*(B*x**3+A),x)
Output:
A*a**(2/3)*e**m*x**(m + 1)*gamma(m/3 + 1/3)*hyper((-2/3, m/3 + 1/3), (m/3 + 4/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(m/3 + 4/3)) + B*a**(2/3)*e**m* x**(m + 4)*gamma(m/3 + 4/3)*hyper((-2/3, m/3 + 4/3), (m/3 + 7/3,), b*x**3* exp_polar(I*pi)/a)/(3*gamma(m/3 + 7/3))
\[ \int (e x)^m \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx=\int { {\left (B x^{3} + A\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}} \left (e x\right )^{m} \,d x } \] Input:
integrate((e*x)^m*(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)*(e*x)^m, x)
\[ \int (e x)^m \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx=\int { {\left (B x^{3} + A\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}} \left (e x\right )^{m} \,d x } \] Input:
integrate((e*x)^m*(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)*(e*x)^m, x)
Timed out. \[ \int (e x)^m \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx=\int \left (B\,x^3+A\right )\,{\left (e\,x\right )}^m\,{\left (b\,x^3+a\right )}^{2/3} \,d x \] Input:
int((A + B*x^3)*(e*x)^m*(a + b*x^3)^(2/3),x)
Output:
int((A + B*x^3)*(e*x)^m*(a + b*x^3)^(2/3), x)
\[ \int (e x)^m \left (a+b x^3\right )^{2/3} \left (A+B x^3\right ) \, dx=\frac {e^{m} \left (x^{m} \left (b \,x^{3}+a \right )^{\frac {2}{3}} a m x +8 x^{m} \left (b \,x^{3}+a \right )^{\frac {2}{3}} a x +x^{m} \left (b \,x^{3}+a \right )^{\frac {2}{3}} b m \,x^{4}+3 x^{m} \left (b \,x^{3}+a \right )^{\frac {2}{3}} b \,x^{4}+10 \left (\int \frac {x^{m} \left (b \,x^{3}+a \right )^{\frac {2}{3}}}{b \,m^{2} x^{3}+9 b m \,x^{3}+18 b \,x^{3}+a \,m^{2}+9 a m +18 a}d x \right ) a^{2} m^{2}+90 \left (\int \frac {x^{m} \left (b \,x^{3}+a \right )^{\frac {2}{3}}}{b \,m^{2} x^{3}+9 b m \,x^{3}+18 b \,x^{3}+a \,m^{2}+9 a m +18 a}d x \right ) a^{2} m +180 \left (\int \frac {x^{m} \left (b \,x^{3}+a \right )^{\frac {2}{3}}}{b \,m^{2} x^{3}+9 b m \,x^{3}+18 b \,x^{3}+a \,m^{2}+9 a m +18 a}d x \right ) a^{2}\right )}{m^{2}+9 m +18} \] Input:
int((e*x)^m*(b*x^3+a)^(2/3)*(B*x^3+A),x)
Output:
(e**m*(x**m*(a + b*x**3)**(2/3)*a*m*x + 8*x**m*(a + b*x**3)**(2/3)*a*x + x **m*(a + b*x**3)**(2/3)*b*m*x**4 + 3*x**m*(a + b*x**3)**(2/3)*b*x**4 + 10* int((x**m*(a + b*x**3)**(2/3))/(a*m**2 + 9*a*m + 18*a + b*m**2*x**3 + 9*b* m*x**3 + 18*b*x**3),x)*a**2*m**2 + 90*int((x**m*(a + b*x**3)**(2/3))/(a*m* *2 + 9*a*m + 18*a + b*m**2*x**3 + 9*b*m*x**3 + 18*b*x**3),x)*a**2*m + 180* int((x**m*(a + b*x**3)**(2/3))/(a*m**2 + 9*a*m + 18*a + b*m**2*x**3 + 9*b* m*x**3 + 18*b*x**3),x)*a**2))/(m**2 + 9*m + 18)