\(\int \frac {(e x)^m (A+B x^3)}{\sqrt [3]{a+b x^3}} \, dx\) [392]

Optimal result

Integrand size = 24, antiderivative size = 115 \[ \int \frac {(e x)^m \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\frac {B (e x)^{1+m} \left (a+b x^3\right )^{2/3}}{b e (3+m)}+\frac {\left (\frac {A}{1+m}-\frac {a B}{b (3+m)}\right ) (e x)^{1+m} \sqrt [3]{1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1+m}{3},\frac {4+m}{3},-\frac {b x^3}{a}\right )}{e \sqrt [3]{a+b x^3}} \] Output:

B*(e*x)^(1+m)*(b*x^3+a)^(2/3)/b/e/(3+m)+(A/(1+m)-a*B/b/(3+m))*(e*x)^(1+m)* 
(1+b*x^3/a)^(1/3)*hypergeom([1/3, 1/3+1/3*m],[4/3+1/3*m],-b*x^3/a)/e/(b*x^ 
3+a)^(1/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.96 \[ \int \frac {(e x)^m \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\frac {x (e x)^m \sqrt [3]{1+\frac {b x^3}{a}} \left (A (4+m) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1+m}{3},\frac {4+m}{3},-\frac {b x^3}{a}\right )+B (1+m) x^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {4+m}{3},\frac {7+m}{3},-\frac {b x^3}{a}\right )\right )}{(1+m) (4+m) \sqrt [3]{a+b x^3}} \] Input:

Integrate[((e*x)^m*(A + B*x^3))/(a + b*x^3)^(1/3),x]
 

Output:

(x*(e*x)^m*(1 + (b*x^3)/a)^(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)*(a + b*x^3)^(1/3))
 

Rubi [A] (verified)

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.

\(\displaystyle \int \frac {\left (A+B x^3\right ) (e x)^m}{\sqrt [3]{a+b x^3}} \, dx\)

\(\Big \downarrow \) 959

\(\displaystyle \left (A-\frac {a B (m+1)}{b (m+3)}\right ) \int \frac {(e x)^m}{\sqrt [3]{b x^3+a}}dx+\frac {B \left (a+b x^3\right )^{2/3} (e x)^{m+1}}{b e (m+3)}\)

\(\Big \downarrow \) 889

\(\displaystyle \frac {\sqrt [3]{\frac {b x^3}{a}+1} \left (A-\frac {a B (m+1)}{b (m+3)}\right ) \int \frac {(e x)^m}{\sqrt [3]{\frac {b x^3}{a}+1}}dx}{\sqrt [3]{a+b x^3}}+\frac {B \left (a+b x^3\right )^{2/3} (e x)^{m+1}}{b e (m+3)}\)

\(\Big \downarrow \) 888

\(\displaystyle \frac {\sqrt [3]{\frac {b x^3}{a}+1} (e x)^{m+1} \left (A-\frac {a B (m+1)}{b (m+3)}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {m+1}{3},\frac {m+4}{3},-\frac {b x^3}{a}\right )}{e (m+1) \sqrt [3]{a+b x^3}}+\frac {B \left (a+b x^3\right )^{2/3} (e x)^{m+1}}{b e (m+3)}\)

Input:

Int[((e*x)^m*(A + B*x^3))/(a + b*x^3)^(1/3),x]
 

Output:

(B*(e*x)^(1 + m)*(a + b*x^3)^(2/3))/(b*e*(3 + m)) + ((A - (a*B*(1 + m))/(b 
*(3 + m)))*(e*x)^(1 + m)*(1 + (b*x^3)/a)^(1/3)*Hypergeometric2F1[1/3, (1 + 
 m)/3, (4 + m)/3, -((b*x^3)/a)])/(e*(1 + m)*(a + b*x^3)^(1/3))
 

Defintions of rubi rules used

rule 888

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])
 

rule 889

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])
 

rule 959

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]
 

Maple [F]

\[\int \frac {\left (e x \right )^{m} \left (B \,x^{3}+A \right )}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x\]

Input:

int((e*x)^m*(B*x^3+A)/(b*x^3+a)^(1/3),x)
 

Output:

int((e*x)^m*(B*x^3+A)/(b*x^3+a)^(1/3),x)
 

Fricas [F]

\[ \int \frac {(e x)^m \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} \left (e x\right )^{m}}{{\left (b x^{3} + a\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate((e*x)^m*(B*x^3+A)/(b*x^3+a)^(1/3),x, algorithm="fricas")
 

Output:

integral((B*x^3 + A)*(e*x)^m/(b*x^3 + a)^(1/3), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.92 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.02 \[ \int \frac {(e x)^m \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\frac {A e^{m} x^{m + 1} \Gamma \left (\frac {m}{3} + \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {m}{3} + \frac {1}{3} \\ \frac {m}{3} + \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {m}{3} + \frac {4}{3}\right )} + \frac {B e^{m} x^{m + 4} \Gamma \left (\frac {m}{3} + \frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {m}{3} + \frac {4}{3} \\ \frac {m}{3} + \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [3]{a} \Gamma \left (\frac {m}{3} + \frac {7}{3}\right )} \] Input:

integrate((e*x)**m*(B*x**3+A)/(b*x**3+a)**(1/3),x)
 

Output:

A*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*a**(1/3)*gamma(m/3 + 4/3)) + B*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*a**(1/3)*gamma(m/3 + 7/3))
                                                                                    
                                                                                    
 

Maxima [F]

\[ \int \frac {(e x)^m \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} \left (e x\right )^{m}}{{\left (b x^{3} + a\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate((e*x)^m*(B*x^3+A)/(b*x^3+a)^(1/3),x, algorithm="maxima")
 

Output:

integrate((B*x^3 + A)*(e*x)^m/(b*x^3 + a)^(1/3), x)
 

Giac [F]

\[ \int \frac {(e x)^m \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} \left (e x\right )^{m}}{{\left (b x^{3} + a\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate((e*x)^m*(B*x^3+A)/(b*x^3+a)^(1/3),x, algorithm="giac")
 

Output:

integrate((B*x^3 + A)*(e*x)^m/(b*x^3 + a)^(1/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^m \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=\int \frac {\left (B\,x^3+A\right )\,{\left (e\,x\right )}^m}{{\left (b\,x^3+a\right )}^{1/3}} \,d x \] 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)
 

Reduce [F]

\[ \int \frac {(e x)^m \left (A+B x^3\right )}{\sqrt [3]{a+b x^3}} \, dx=e^{m} \left (\left (\int \frac {x^{m}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x \right ) a +\left (\int \frac {x^{m} x^{3}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}}}d x \right ) b \right ) \] Input:

int((e*x)^m*(B*x^3+A)/(b*x^3+a)^(1/3),x)
 

Output:

e**m*(int(x**m/(a + b*x**3)**(1/3),x)*a + int((x**m*x**3)/(a + b*x**3)**(1 
/3),x)*b)