\(\int \frac {(e x)^m (A+B x^3)}{(a+b x^3)^{3/2}} \, dx\) [387]

Optimal result

Integrand size = 24, antiderivative size = 132 \[ \int \frac {(e x)^m \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=-\frac {2 B (e x)^{1+m}}{b e (1-2 m) \sqrt {a+b x^3}}+\frac {(2 a B (1+m)+A (b-2 b m)) (e x)^{1+m} \sqrt {1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1+m}{3},\frac {4+m}{3},-\frac {b x^3}{a}\right )}{a b e (1-2 m) (1+m) \sqrt {a+b x^3}} \] Output:

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

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.86 \[ \int \frac {(e x)^m \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\frac {x (e x)^m \sqrt {1+\frac {b x^3}{a}} \left (A (4+m) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1+m}{3},\frac {4+m}{3},-\frac {b x^3}{a}\right )+B (1+m) x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {4+m}{3},\frac {7+m}{3},-\frac {b x^3}{a}\right )\right )}{a (1+m) (4+m) \sqrt {a+b x^3}} \] Input:

Integrate[((e*x)^m*(A + B*x^3))/(a + b*x^3)^(3/2),x]
 

Output:

(x*(e*x)^m*Sqrt[1 + (b*x^3)/a]*(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)]))/(a*(1 + m)*(4 + m)*Sqrt[a + b*x^3])
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.01, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {957, 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.

\(\displaystyle \int \frac {\left (A+B x^3\right ) (e x)^m}{\left (a+b x^3\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 957

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

\(\Big \downarrow \) 889

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

\(\Big \downarrow \) 888

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

Input:

Int[((e*x)^m*(A + B*x^3))/(a + b*x^3)^(3/2),x]
 

Output:

(2*(A*b - a*B)*(e*x)^(1 + m))/(3*a*b*e*Sqrt[a + b*x^3]) + ((2*a*B*(1 + m) 
+ A*(b - 2*b*m))*(e*x)^(1 + m)*Sqrt[1 + (b*x^3)/a]*Hypergeometric2F1[1/2, 
(1 + m)/3, (4 + m)/3, -((b*x^3)/a)])/(3*a*b*e*(1 + m)*Sqrt[a + b*x^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 957

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_)), x_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a 
*b*e*n*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b*n* 
(p + 1))   Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, 
 m, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && N 
eQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] && LeQ[-1 
, m, (-n)*(p + 1)]))
 

Maple [F]

\[\int \frac {\left (e x \right )^{m} \left (B \,x^{3}+A \right )}{\left (b \,x^{3}+a \right )^{\frac {3}{2}}}d x\]

Input:

int((e*x)^m*(B*x^3+A)/(b*x^3+a)^(3/2),x)
 

Output:

int((e*x)^m*(B*x^3+A)/(b*x^3+a)^(3/2),x)
 

Fricas [F]

\[ \int \frac {(e x)^m \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} \left (e x\right )^{m}}{{\left (b x^{3} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((e*x)^m*(B*x^3+A)/(b*x^3+a)^(3/2),x, algorithm="fricas")
 

Output:

integral((B*x^3 + A)*sqrt(b*x^3 + a)*(e*x)^m/(b^2*x^6 + 2*a*b*x^3 + a^2), 
x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 25.78 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.89 \[ \int \frac {(e x)^m \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\frac {A e^{m} x^{m + 1} \Gamma \left (\frac {m}{3} + \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {m}{3} + \frac {1}{3} \\ \frac {m}{3} + \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {3}{2}} \Gamma \left (\frac {m}{3} + \frac {4}{3}\right )} + \frac {B e^{m} x^{m + 4} \Gamma \left (\frac {m}{3} + \frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {m}{3} + \frac {4}{3} \\ \frac {m}{3} + \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {3}{2}} \Gamma \left (\frac {m}{3} + \frac {7}{3}\right )} \] Input:

integrate((e*x)**m*(B*x**3+A)/(b*x**3+a)**(3/2),x)
 

Output:

A*e**m*x**(m + 1)*gamma(m/3 + 1/3)*hyper((3/2, m/3 + 1/3), (m/3 + 4/3,), b 
*x**3*exp_polar(I*pi)/a)/(3*a**(3/2)*gamma(m/3 + 4/3)) + B*e**m*x**(m + 4) 
*gamma(m/3 + 4/3)*hyper((3/2, m/3 + 4/3), (m/3 + 7/3,), b*x**3*exp_polar(I 
*pi)/a)/(3*a**(3/2)*gamma(m/3 + 7/3))
                                                                                    
                                                                                    
 

Maxima [F]

\[ \int \frac {(e x)^m \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} \left (e x\right )^{m}}{{\left (b x^{3} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((e*x)^m*(B*x^3+A)/(b*x^3+a)^(3/2),x, algorithm="maxima")
 

Output:

integrate((B*x^3 + A)*(e*x)^m/(b*x^3 + a)^(3/2), x)
 

Giac [F]

\[ \int \frac {(e x)^m \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} \left (e x\right )^{m}}{{\left (b x^{3} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((e*x)^m*(B*x^3+A)/(b*x^3+a)^(3/2),x, algorithm="giac")
 

Output:

integrate((B*x^3 + A)*(e*x)^m/(b*x^3 + a)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^m \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\int \frac {\left (B\,x^3+A\right )\,{\left (e\,x\right )}^m}{{\left (b\,x^3+a\right )}^{3/2}} \,d x \] Input:

int(((A + B*x^3)*(e*x)^m)/(a + b*x^3)^(3/2),x)
 

Output:

int(((A + B*x^3)*(e*x)^m)/(a + b*x^3)^(3/2), x)
 

Reduce [F]

\[ \int \frac {(e x)^m \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=e^{m} \left (\int \frac {x^{m} \sqrt {b \,x^{3}+a}}{b \,x^{3}+a}d x \right ) \] Input:

int((e*x)^m*(B*x^3+A)/(b*x^3+a)^(3/2),x)
 

Output:

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