Integrand size = 24, antiderivative size = 132 \[ \int (e x)^m \left (a+b x^3\right )^{3/2} \left (A+B x^3\right ) \, dx=\frac {2 B (e x)^{1+m} \left (a+b x^3\right )^{5/2}}{b e (17+2 m)}-\frac {a (2 a B (1+m)-A b (17+2 m)) (e x)^{1+m} \sqrt {a+b x^3} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1+m}{3},\frac {4+m}{3},-\frac {b x^3}{a}\right )}{b e (1+m) (17+2 m) \sqrt {1+\frac {b x^3}{a}}} \]
2*B*(e*x)^(1+m)*(b*x^3+a)^(5/2)/b/e/(17+2*m)-a*(2*a*B*(1+m)-A*b*(17+2*m))* (e*x)^(1+m)*hypergeom([-3/2, 1/3+1/3*m],[4/3+1/3*m],-b*x^3/a)*(b*x^3+a)^(1 /2)/b/e/(1+m)/(17+2*m)/(1+b*x^3/a)^(1/2)
Time = 0.34 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.84 \[ \int (e x)^m \left (a+b x^3\right )^{3/2} \left (A+B x^3\right ) \, dx=\frac {a x (e x)^m \sqrt {a+b x^3} \left (A (4+m) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1+m}{3},\frac {4+m}{3},-\frac {b x^3}{a}\right )+B (1+m) x^3 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {4+m}{3},\frac {7+m}{3},-\frac {b x^3}{a}\right )\right )}{(1+m) (4+m) \sqrt {1+\frac {b x^3}{a}}} \]
(a*x*(e*x)^m*Sqrt[a + b*x^3]*(A*(4 + m)*Hypergeometric2F1[-3/2, (1 + m)/3, (4 + m)/3, -((b*x^3)/a)] + B*(1 + m)*x^3*Hypergeometric2F1[-3/2, (4 + m)/ 3, (7 + m)/3, -((b*x^3)/a)]))/((1 + m)*(4 + m)*Sqrt[1 + (b*x^3)/a])
Time = 0.27 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, 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.
\(\displaystyle \int \left (a+b x^3\right )^{3/2} \left (A+B x^3\right ) (e x)^m \, dx\) |
\(\Big \downarrow \) 959 |
\(\displaystyle \frac {2 B \left (a+b x^3\right )^{5/2} (e x)^{m+1}}{b e (2 m+17)}-\frac {(2 a B (m+1)-A b (2 m+17)) \int (e x)^m \left (b x^3+a\right )^{3/2}dx}{b (2 m+17)}\) |
\(\Big \downarrow \) 889 |
\(\displaystyle \frac {2 B \left (a+b x^3\right )^{5/2} (e x)^{m+1}}{b e (2 m+17)}-\frac {a \sqrt {a+b x^3} (2 a B (m+1)-A b (2 m+17)) \int (e x)^m \left (\frac {b x^3}{a}+1\right )^{3/2}dx}{b (2 m+17) \sqrt {\frac {b x^3}{a}+1}}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {2 B \left (a+b x^3\right )^{5/2} (e x)^{m+1}}{b e (2 m+17)}-\frac {a \sqrt {a+b x^3} (e x)^{m+1} (2 a B (m+1)-A b (2 m+17)) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {m+1}{3},\frac {m+4}{3},-\frac {b x^3}{a}\right )}{b e (m+1) (2 m+17) \sqrt {\frac {b x^3}{a}+1}}\) |
(2*B*(e*x)^(1 + m)*(a + b*x^3)^(5/2))/(b*e*(17 + 2*m)) - (a*(2*a*B*(1 + m) - A*b*(17 + 2*m))*(e*x)^(1 + m)*Sqrt[a + b*x^3]*Hypergeometric2F1[-3/2, ( 1 + m)/3, (4 + m)/3, -((b*x^3)/a)])/(b*e*(1 + m)*(17 + 2*m)*Sqrt[1 + (b*x^ 3)/a])
3.6.1.3.1 Defintions of rubi rules used
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]
\[\int \left (e x \right )^{m} \left (b \,x^{3}+a \right )^{\frac {3}{2}} \left (x^{3} B +A \right )d x\]
\[ \int (e x)^m \left (a+b x^3\right )^{3/2} \left (A+B x^3\right ) \, dx=\int { {\left (B x^{3} + A\right )} {\left (b x^{3} + a\right )}^{\frac {3}{2}} \left (e x\right )^{m} \,d x } \]
Result contains complex when optimal does not.
Time = 6.89 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.86 \[ \int (e x)^m \left (a+b x^3\right )^{3/2} \left (A+B x^3\right ) \, dx=\frac {A a^{\frac {3}{2}} e^{m} x^{m + 1} \Gamma \left (\frac {m}{3} + \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{3} + \frac {1}{3} \\ \frac {m}{3} + \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {m}{3} + \frac {4}{3}\right )} + \frac {A \sqrt {a} b e^{m} x^{m + 4} \Gamma \left (\frac {m}{3} + \frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{3} + \frac {4}{3} \\ \frac {m}{3} + \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {m}{3} + \frac {7}{3}\right )} + \frac {B a^{\frac {3}{2}} e^{m} x^{m + 4} \Gamma \left (\frac {m}{3} + \frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{3} + \frac {4}{3} \\ \frac {m}{3} + \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {m}{3} + \frac {7}{3}\right )} + \frac {B \sqrt {a} b e^{m} x^{m + 7} \Gamma \left (\frac {m}{3} + \frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {m}{3} + \frac {7}{3} \\ \frac {m}{3} + \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \Gamma \left (\frac {m}{3} + \frac {10}{3}\right )} \]
A*a**(3/2)*e**m*x**(m + 1)*gamma(m/3 + 1/3)*hyper((-1/2, m/3 + 1/3), (m/3 + 4/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(m/3 + 4/3)) + A*sqrt(a)*b*e**m *x**(m + 4)*gamma(m/3 + 4/3)*hyper((-1/2, m/3 + 4/3), (m/3 + 7/3,), b*x**3 *exp_polar(I*pi)/a)/(3*gamma(m/3 + 7/3)) + B*a**(3/2)*e**m*x**(m + 4)*gamm a(m/3 + 4/3)*hyper((-1/2, m/3 + 4/3), (m/3 + 7/3,), b*x**3*exp_polar(I*pi) /a)/(3*gamma(m/3 + 7/3)) + B*sqrt(a)*b*e**m*x**(m + 7)*gamma(m/3 + 7/3)*hy per((-1/2, m/3 + 7/3), (m/3 + 10/3,), b*x**3*exp_polar(I*pi)/a)/(3*gamma(m /3 + 10/3))
\[ \int (e x)^m \left (a+b x^3\right )^{3/2} \left (A+B x^3\right ) \, dx=\int { {\left (B x^{3} + A\right )} {\left (b x^{3} + a\right )}^{\frac {3}{2}} \left (e x\right )^{m} \,d x } \]
\[ \int (e x)^m \left (a+b x^3\right )^{3/2} \left (A+B x^3\right ) \, dx=\int { {\left (B x^{3} + A\right )} {\left (b x^{3} + a\right )}^{\frac {3}{2}} \left (e x\right )^{m} \,d x } \]
Timed out. \[ \int (e x)^m \left (a+b x^3\right )^{3/2} \left (A+B x^3\right ) \, dx=\int \left (B\,x^3+A\right )\,{\left (e\,x\right )}^m\,{\left (b\,x^3+a\right )}^{3/2} \,d x \]