\(\int \frac {(a+b x^n)^{3/2} (A+B x^n)}{(e x)^{3/2}} \, dx\) [380]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [F]
Fricas [F(-2)]
Sympy [C] (verification not implemented)
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 26, antiderivative size = 116 \[ \int \frac {\left (a+b x^n\right )^{3/2} \left (A+B x^n\right )}{(e x)^{3/2}} \, dx=-\frac {2 B \left (a+b x^n\right )^{5/2}}{b e (1-5 n) \sqrt {e x}}-\frac {2 a \left (A-\frac {a B}{b-5 b n}\right ) \sqrt {a+b x^n} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {1}{2 n},1-\frac {1}{2 n},-\frac {b x^n}{a}\right )}{e \sqrt {e x} \sqrt {1+\frac {b x^n}{a}}} \] Output:

-2*B*(a+b*x^n)^(5/2)/b/e/(1-5*n)/(e*x)^(1/2)-2*a*(A-a*B/(-5*b*n+b))*(a+b*x 
^n)^(1/2)*hypergeom([-3/2, -1/2/n],[1-1/2/n],-b*x^n/a)/e/(e*x)^(1/2)/(1+b* 
x^n/a)^(1/2)
 

Mathematica [A] (verified)

Time = 0.94 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^n\right )^{3/2} \left (A+B x^n\right )}{(e x)^{3/2}} \, dx=\frac {2 a x \sqrt {a+b x^n} \left (B x^n \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1-\frac {1}{2 n},2-\frac {1}{2 n},-\frac {b x^n}{a}\right )+A (1-2 n) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {1}{2 n},1-\frac {1}{2 n},-\frac {b x^n}{a}\right )\right )}{(-1+2 n) (e x)^{3/2} \sqrt {1+\frac {b x^n}{a}}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, 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.

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

\(\Big \downarrow \) 959

\(\displaystyle \left (A-\frac {a B}{b-5 b n}\right ) \int \frac {\left (b x^n+a\right )^{3/2}}{(e x)^{3/2}}dx-\frac {2 B \left (a+b x^n\right )^{5/2}}{b e (1-5 n) \sqrt {e x}}\)

\(\Big \downarrow \) 889

\(\displaystyle \frac {a \sqrt {a+b x^n} \left (A-\frac {a B}{b-5 b n}\right ) \int \frac {\left (\frac {b x^n}{a}+1\right )^{3/2}}{(e x)^{3/2}}dx}{\sqrt {\frac {b x^n}{a}+1}}-\frac {2 B \left (a+b x^n\right )^{5/2}}{b e (1-5 n) \sqrt {e x}}\)

\(\Big \downarrow \) 888

\(\displaystyle -\frac {2 a \sqrt {a+b x^n} \left (A-\frac {a B}{b-5 b n}\right ) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {1}{2 n},1-\frac {1}{2 n},-\frac {b x^n}{a}\right )}{e \sqrt {e x} \sqrt {\frac {b x^n}{a}+1}}-\frac {2 B \left (a+b x^n\right )^{5/2}}{b e (1-5 n) \sqrt {e x}}\)

Input:

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

Output:

(-2*B*(a + b*x^n)^(5/2))/(b*e*(1 - 5*n)*Sqrt[e*x]) - (2*a*(A - (a*B)/(b - 
5*b*n))*Sqrt[a + b*x^n]*Hypergeometric2F1[-3/2, -1/2*1/n, 1 - 1/(2*n), -(( 
b*x^n)/a)])/(e*Sqrt[e*x]*Sqrt[1 + (b*x^n)/a])
 

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]

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

Input:

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

Output:

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

Fricas [F(-2)]

Exception generated. \[ \int \frac {\left (a+b x^n\right )^{3/2} \left (A+B x^n\right )}{(e x)^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (has polynomial part)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 16.64 (sec) , antiderivative size = 316, normalized size of antiderivative = 2.72 \[ \int \frac {\left (a+b x^n\right )^{3/2} \left (A+B x^n\right )}{(e x)^{3/2}} \, dx=\frac {A a a^{- \frac {1}{2 n}} a^{\frac {1}{2} + \frac {1}{2 n}} \Gamma \left (- \frac {1}{2 n}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{2 n} \\ 1 - \frac {1}{2 n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{e^{\frac {3}{2}} n \sqrt {x} \Gamma \left (1 - \frac {1}{2 n}\right )} + \frac {A a^{- \frac {1}{2} + \frac {1}{2 n}} a^{1 - \frac {1}{2 n}} b x^{n - \frac {1}{2}} \Gamma \left (1 - \frac {1}{2 n}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, 1 - \frac {1}{2 n} \\ 2 - \frac {1}{2 n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{e^{\frac {3}{2}} n \Gamma \left (2 - \frac {1}{2 n}\right )} + \frac {B a a^{- \frac {1}{2} + \frac {1}{2 n}} a^{1 - \frac {1}{2 n}} x^{n - \frac {1}{2}} \Gamma \left (1 - \frac {1}{2 n}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, 1 - \frac {1}{2 n} \\ 2 - \frac {1}{2 n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{e^{\frac {3}{2}} n \Gamma \left (2 - \frac {1}{2 n}\right )} + \frac {B a^{- \frac {3}{2} + \frac {1}{2 n}} a^{2 - \frac {1}{2 n}} b x^{2 n - \frac {1}{2}} \Gamma \left (2 - \frac {1}{2 n}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, 2 - \frac {1}{2 n} \\ 3 - \frac {1}{2 n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{e^{\frac {3}{2}} n \Gamma \left (3 - \frac {1}{2 n}\right )} \] Input:

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

Output:

A*a*a**(1/2 + 1/(2*n))*gamma(-1/(2*n))*hyper((-1/2, -1/(2*n)), (1 - 1/(2*n 
),), b*x**n*exp_polar(I*pi)/a)/(a**(1/(2*n))*e**(3/2)*n*sqrt(x)*gamma(1 - 
1/(2*n))) + A*a**(-1/2 + 1/(2*n))*a**(1 - 1/(2*n))*b*x**(n - 1/2)*gamma(1 
- 1/(2*n))*hyper((-1/2, 1 - 1/(2*n)), (2 - 1/(2*n),), b*x**n*exp_polar(I*p 
i)/a)/(e**(3/2)*n*gamma(2 - 1/(2*n))) + B*a*a**(-1/2 + 1/(2*n))*a**(1 - 1/ 
(2*n))*x**(n - 1/2)*gamma(1 - 1/(2*n))*hyper((-1/2, 1 - 1/(2*n)), (2 - 1/( 
2*n),), b*x**n*exp_polar(I*pi)/a)/(e**(3/2)*n*gamma(2 - 1/(2*n))) + B*a**( 
-3/2 + 1/(2*n))*a**(2 - 1/(2*n))*b*x**(2*n - 1/2)*gamma(2 - 1/(2*n))*hyper 
((-1/2, 2 - 1/(2*n)), (3 - 1/(2*n),), b*x**n*exp_polar(I*pi)/a)/(e**(3/2)* 
n*gamma(3 - 1/(2*n)))
                                                                                    
                                                                                    
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+b x^n\right )^{3/2} \left (A+B x^n\right )}{(e x)^{3/2}} \, dx=\frac {\sqrt {e}\, \left (6 x^{2 n} \sqrt {x^{n} b +a}\, b^{2} n^{2}-8 x^{2 n} \sqrt {x^{n} b +a}\, b^{2} n +2 x^{2 n} \sqrt {x^{n} b +a}\, b^{2}+22 x^{n} \sqrt {x^{n} b +a}\, a b \,n^{2}-26 x^{n} \sqrt {x^{n} b +a}\, a b n +4 x^{n} \sqrt {x^{n} b +a}\, a b +46 \sqrt {x^{n} b +a}\, a^{2} n^{2}-18 \sqrt {x^{n} b +a}\, a^{2} n +2 \sqrt {x^{n} b +a}\, a^{2}+225 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {x^{n} b +a}}{15 x^{n} b \,n^{3} x^{2}-23 x^{n} b \,n^{2} x^{2}+9 x^{n} b n \,x^{2}-x^{n} b \,x^{2}+15 a \,n^{3} x^{2}-23 a \,n^{2} x^{2}+9 a n \,x^{2}-a \,x^{2}}d x \right ) a^{3} n^{6}-345 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {x^{n} b +a}}{15 x^{n} b \,n^{3} x^{2}-23 x^{n} b \,n^{2} x^{2}+9 x^{n} b n \,x^{2}-x^{n} b \,x^{2}+15 a \,n^{3} x^{2}-23 a \,n^{2} x^{2}+9 a n \,x^{2}-a \,x^{2}}d x \right ) a^{3} n^{5}+135 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {x^{n} b +a}}{15 x^{n} b \,n^{3} x^{2}-23 x^{n} b \,n^{2} x^{2}+9 x^{n} b n \,x^{2}-x^{n} b \,x^{2}+15 a \,n^{3} x^{2}-23 a \,n^{2} x^{2}+9 a n \,x^{2}-a \,x^{2}}d x \right ) a^{3} n^{4}-15 \sqrt {x}\, \left (\int \frac {\sqrt {x}\, \sqrt {x^{n} b +a}}{15 x^{n} b \,n^{3} x^{2}-23 x^{n} b \,n^{2} x^{2}+9 x^{n} b n \,x^{2}-x^{n} b \,x^{2}+15 a \,n^{3} x^{2}-23 a \,n^{2} x^{2}+9 a n \,x^{2}-a \,x^{2}}d x \right ) a^{3} n^{3}\right )}{\sqrt {x}\, e^{2} \left (15 n^{3}-23 n^{2}+9 n -1\right )} \] Input:

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

Output:

(sqrt(e)*(6*x**(2*n)*sqrt(x**n*b + a)*b**2*n**2 - 8*x**(2*n)*sqrt(x**n*b + 
 a)*b**2*n + 2*x**(2*n)*sqrt(x**n*b + a)*b**2 + 22*x**n*sqrt(x**n*b + a)*a 
*b*n**2 - 26*x**n*sqrt(x**n*b + a)*a*b*n + 4*x**n*sqrt(x**n*b + a)*a*b + 4 
6*sqrt(x**n*b + a)*a**2*n**2 - 18*sqrt(x**n*b + a)*a**2*n + 2*sqrt(x**n*b 
+ a)*a**2 + 225*sqrt(x)*int((sqrt(x)*sqrt(x**n*b + a))/(15*x**n*b*n**3*x** 
2 - 23*x**n*b*n**2*x**2 + 9*x**n*b*n*x**2 - x**n*b*x**2 + 15*a*n**3*x**2 - 
 23*a*n**2*x**2 + 9*a*n*x**2 - a*x**2),x)*a**3*n**6 - 345*sqrt(x)*int((sqr 
t(x)*sqrt(x**n*b + a))/(15*x**n*b*n**3*x**2 - 23*x**n*b*n**2*x**2 + 9*x**n 
*b*n*x**2 - x**n*b*x**2 + 15*a*n**3*x**2 - 23*a*n**2*x**2 + 9*a*n*x**2 - a 
*x**2),x)*a**3*n**5 + 135*sqrt(x)*int((sqrt(x)*sqrt(x**n*b + a))/(15*x**n* 
b*n**3*x**2 - 23*x**n*b*n**2*x**2 + 9*x**n*b*n*x**2 - x**n*b*x**2 + 15*a*n 
**3*x**2 - 23*a*n**2*x**2 + 9*a*n*x**2 - a*x**2),x)*a**3*n**4 - 15*sqrt(x) 
*int((sqrt(x)*sqrt(x**n*b + a))/(15*x**n*b*n**3*x**2 - 23*x**n*b*n**2*x**2 
 + 9*x**n*b*n*x**2 - x**n*b*x**2 + 15*a*n**3*x**2 - 23*a*n**2*x**2 + 9*a*n 
*x**2 - a*x**2),x)*a**3*n**3))/(sqrt(x)*e**2*(15*n**3 - 23*n**2 + 9*n - 1) 
)