\(\int (e x)^{3/2} (a+b x^n)^{3/2} (A+B x^n) \, dx\) [377]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [F]
Fricas [F(-2)]
Sympy [F(-1)]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 26, antiderivative size = 123 \[ \int (e x)^{3/2} \left (a+b x^n\right )^{3/2} \left (A+B x^n\right ) \, dx=\frac {2 B (e x)^{5/2} \left (a+b x^n\right )^{5/2}}{5 b e (1+n)}-\frac {2 a (a B-A b (1+n)) (e x)^{5/2} \sqrt {a+b x^n} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {5}{2 n},1+\frac {5}{2 n},-\frac {b x^n}{a}\right )}{5 b e (1+n) \sqrt {1+\frac {b x^n}{a}}} \] Output:

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

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.97 \[ \int (e x)^{3/2} \left (a+b x^n\right )^{3/2} \left (A+B x^n\right ) \, dx=\frac {2 a x (e x)^{3/2} \sqrt {a+b x^n} \left (A (5+2 n) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {5}{2 n},1+\frac {5}{2 n},-\frac {b x^n}{a}\right )+5 B x^n \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {\frac {5}{2}+n}{n},2+\frac {5}{2 n},-\frac {b x^n}{a}\right )\right )}{5 (5+2 n) \sqrt {1+\frac {b x^n}{a}}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.95, 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.

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

\(\Big \downarrow \) 959

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

\(\Big \downarrow \) 889

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

\(\Big \downarrow \) 888

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F(-2)]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

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

Output:

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

Reduce [F]

\[ \int (e x)^{3/2} \left (a+b x^n\right )^{3/2} \left (A+B x^n\right ) \, dx=\frac {\sqrt {e}\, e \left (6 x^{2 n +\frac {1}{2}} \sqrt {x^{n} b +a}\, b^{2} n^{2} x^{2}+40 x^{2 n +\frac {1}{2}} \sqrt {x^{n} b +a}\, b^{2} n \,x^{2}+50 x^{2 n +\frac {1}{2}} \sqrt {x^{n} b +a}\, b^{2} x^{2}+22 x^{n +\frac {1}{2}} \sqrt {x^{n} b +a}\, a b \,n^{2} x^{2}+130 x^{n +\frac {1}{2}} \sqrt {x^{n} b +a}\, a b n \,x^{2}+100 x^{n +\frac {1}{2}} \sqrt {x^{n} b +a}\, a b \,x^{2}+46 \sqrt {x}\, \sqrt {x^{n} b +a}\, a^{2} n^{2} x^{2}+90 \sqrt {x}\, \sqrt {x^{n} b +a}\, a^{2} n \,x^{2}+50 \sqrt {x}\, \sqrt {x^{n} b +a}\, a^{2} x^{2}+45 \left (\int \frac {\sqrt {x}\, \sqrt {x^{n} b +a}\, x}{3 x^{n} b \,n^{3}+23 x^{n} b \,n^{2}+45 x^{n} b n +25 x^{n} b +3 a \,n^{3}+23 a \,n^{2}+45 a n +25 a}d x \right ) a^{3} n^{6}+345 \left (\int \frac {\sqrt {x}\, \sqrt {x^{n} b +a}\, x}{3 x^{n} b \,n^{3}+23 x^{n} b \,n^{2}+45 x^{n} b n +25 x^{n} b +3 a \,n^{3}+23 a \,n^{2}+45 a n +25 a}d x \right ) a^{3} n^{5}+675 \left (\int \frac {\sqrt {x}\, \sqrt {x^{n} b +a}\, x}{3 x^{n} b \,n^{3}+23 x^{n} b \,n^{2}+45 x^{n} b n +25 x^{n} b +3 a \,n^{3}+23 a \,n^{2}+45 a n +25 a}d x \right ) a^{3} n^{4}+375 \left (\int \frac {\sqrt {x}\, \sqrt {x^{n} b +a}\, x}{3 x^{n} b \,n^{3}+23 x^{n} b \,n^{2}+45 x^{n} b n +25 x^{n} b +3 a \,n^{3}+23 a \,n^{2}+45 a n +25 a}d x \right ) a^{3} n^{3}\right )}{15 n^{3}+115 n^{2}+225 n +125} \] Input:

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

Output:

(sqrt(e)*e*(6*x**((4*n + 1)/2)*sqrt(x**n*b + a)*b**2*n**2*x**2 + 40*x**((4 
*n + 1)/2)*sqrt(x**n*b + a)*b**2*n*x**2 + 50*x**((4*n + 1)/2)*sqrt(x**n*b 
+ a)*b**2*x**2 + 22*x**((2*n + 1)/2)*sqrt(x**n*b + a)*a*b*n**2*x**2 + 130* 
x**((2*n + 1)/2)*sqrt(x**n*b + a)*a*b*n*x**2 + 100*x**((2*n + 1)/2)*sqrt(x 
**n*b + a)*a*b*x**2 + 46*sqrt(x)*sqrt(x**n*b + a)*a**2*n**2*x**2 + 90*sqrt 
(x)*sqrt(x**n*b + a)*a**2*n*x**2 + 50*sqrt(x)*sqrt(x**n*b + a)*a**2*x**2 + 
 45*int((sqrt(x)*sqrt(x**n*b + a)*x)/(3*x**n*b*n**3 + 23*x**n*b*n**2 + 45* 
x**n*b*n + 25*x**n*b + 3*a*n**3 + 23*a*n**2 + 45*a*n + 25*a),x)*a**3*n**6 
+ 345*int((sqrt(x)*sqrt(x**n*b + a)*x)/(3*x**n*b*n**3 + 23*x**n*b*n**2 + 4 
5*x**n*b*n + 25*x**n*b + 3*a*n**3 + 23*a*n**2 + 45*a*n + 25*a),x)*a**3*n** 
5 + 675*int((sqrt(x)*sqrt(x**n*b + a)*x)/(3*x**n*b*n**3 + 23*x**n*b*n**2 + 
 45*x**n*b*n + 25*x**n*b + 3*a*n**3 + 23*a*n**2 + 45*a*n + 25*a),x)*a**3*n 
**4 + 375*int((sqrt(x)*sqrt(x**n*b + a)*x)/(3*x**n*b*n**3 + 23*x**n*b*n**2 
 + 45*x**n*b*n + 25*x**n*b + 3*a*n**3 + 23*a*n**2 + 45*a*n + 25*a),x)*a**3 
*n**3))/(5*(3*n**3 + 23*n**2 + 45*n + 25))