\(\int x^{-1-n (3+p)} (a+b x^n)^p (c+d x^n) \, dx\) [515]

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

Optimal result

Integrand size = 27, antiderivative size = 137 \[ \int x^{-1-n (3+p)} \left (a+b x^n\right )^p \left (c+d x^n\right ) \, dx=-\frac {b (2 b c-a d (3+p)) x^{-n (1+p)} \left (a+b x^n\right )^{1+p}}{a^3 n (1+p) (2+p) (3+p)}+\frac {(2 b c-a d (3+p)) x^{-n (2+p)} \left (a+b x^n\right )^{1+p}}{a^2 n (2+p) (3+p)}-\frac {c x^{-n (3+p)} \left (a+b x^n\right )^{1+p}}{a n (3+p)} \] Output:

-b*(2*b*c-a*d*(3+p))*(a+b*x^n)^(p+1)/a^3/n/(p+1)/(2+p)/(3+p)/(x^(n*(p+1))) 
+(2*b*c-a*d*(3+p))*(a+b*x^n)^(p+1)/a^2/n/(2+p)/(3+p)/(x^(n*(2+p)))-c*(a+b* 
x^n)^(p+1)/a/n/(3+p)/(x^(n*(3+p)))
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.19 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.77 \[ \int x^{-1-n (3+p)} \left (a+b x^n\right )^p \left (c+d x^n\right ) \, dx=-\frac {x^{-n (3+p)} \left (a+b x^n\right )^p \left (1+\frac {b x^n}{a}\right )^{-p} \left (c (2+p) \operatorname {Hypergeometric2F1}\left (-3-p,-p,-2-p,-\frac {b x^n}{a}\right )+d (3+p) x^n \operatorname {Hypergeometric2F1}\left (-2-p,-p,-1-p,-\frac {b x^n}{a}\right )\right )}{n (2+p) (3+p)} \] Input:

Integrate[x^(-1 - n*(3 + p))*(a + b*x^n)^p*(c + d*x^n),x]
 

Output:

-(((a + b*x^n)^p*(c*(2 + p)*Hypergeometric2F1[-3 - p, -p, -2 - p, -((b*x^n 
)/a)] + d*(3 + p)*x^n*Hypergeometric2F1[-2 - p, -p, -1 - p, -((b*x^n)/a)]) 
)/(n*(2 + p)*(3 + p)*x^(n*(3 + p))*(1 + (b*x^n)/a)^p))
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.20, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {959, 803, 803, 796}

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.

\(\displaystyle \int x^{-n (p+3)-1} \left (c+d x^n\right ) \left (a+b x^n\right )^p \, dx\)

\(\Big \downarrow \) 959

\(\displaystyle \frac {(2 b c-a d (p+3)) \int x^{-n (p+3)-1} \left (b x^n+a\right )^pdx}{2 b}-\frac {d x^{-n (p+3)} \left (a+b x^n\right )^{p+1}}{2 b n}\)

\(\Big \downarrow \) 803

\(\displaystyle \frac {(2 b c-a d (p+3)) \left (-\frac {2 b \int x^{-n (p+2)-1} \left (b x^n+a\right )^pdx}{a (p+3)}-\frac {x^{-n (p+3)} \left (a+b x^n\right )^{p+1}}{a n (p+3)}\right )}{2 b}-\frac {d x^{-n (p+3)} \left (a+b x^n\right )^{p+1}}{2 b n}\)

\(\Big \downarrow \) 803

\(\displaystyle \frac {(2 b c-a d (p+3)) \left (-\frac {2 b \left (-\frac {b \int x^{-n (p+1)-1} \left (b x^n+a\right )^pdx}{a (p+2)}-\frac {x^{-n (p+2)} \left (a+b x^n\right )^{p+1}}{a n (p+2)}\right )}{a (p+3)}-\frac {x^{-n (p+3)} \left (a+b x^n\right )^{p+1}}{a n (p+3)}\right )}{2 b}-\frac {d x^{-n (p+3)} \left (a+b x^n\right )^{p+1}}{2 b n}\)

\(\Big \downarrow \) 796

\(\displaystyle \frac {(2 b c-a d (p+3)) \left (-\frac {2 b \left (\frac {b x^{-n (p+1)} \left (a+b x^n\right )^{p+1}}{a^2 n (p+1) (p+2)}-\frac {x^{-n (p+2)} \left (a+b x^n\right )^{p+1}}{a n (p+2)}\right )}{a (p+3)}-\frac {x^{-n (p+3)} \left (a+b x^n\right )^{p+1}}{a n (p+3)}\right )}{2 b}-\frac {d x^{-n (p+3)} \left (a+b x^n\right )^{p+1}}{2 b n}\)

Input:

Int[x^(-1 - n*(3 + p))*(a + b*x^n)^p*(c + d*x^n),x]
 

Output:

-1/2*(d*(a + b*x^n)^(1 + p))/(b*n*x^(n*(3 + p))) + ((2*b*c - a*d*(3 + p))* 
(-((a + b*x^n)^(1 + p)/(a*n*(3 + p)*x^(n*(3 + p)))) - (2*b*((b*(a + b*x^n) 
^(1 + p))/(a^2*n*(1 + p)*(2 + p)*x^(n*(1 + p))) - (a + b*x^n)^(1 + p)/(a*n 
*(2 + p)*x^(n*(2 + p)))))/(a*(3 + p))))/(2*b)
 

Defintions of rubi rules used

rule 796
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* 
x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, 
 p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]
 

rule 803
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*(( 
a + b*x^n)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + n*(p + 1) + 1)/(a*(m + 1 
)))   Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x] && I 
LtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]
 

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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(487\) vs. \(2(140)=280\).

Time = 2.20 (sec) , antiderivative size = 488, normalized size of antiderivative = 3.56

method result size
parallelrisch \(\frac {x \,x^{3 n} x^{-n p -3 n -1} \left (a +b \,x^{n}\right )^{p} a \,b^{3} d p -x \,x^{2 n} x^{-n p -3 n -1} \left (a +b \,x^{n}\right )^{p} a^{2} b^{2} d \,p^{2}+3 x \,x^{3 n} x^{-n p -3 n -1} \left (a +b \,x^{n}\right )^{p} a \,b^{3} d -2 x \,x^{3 n} x^{-n p -3 n -1} \left (a +b \,x^{n}\right )^{p} b^{4} c -3 x \,x^{2 n} x^{-n p -3 n -1} \left (a +b \,x^{n}\right )^{p} a^{2} b^{2} d p +2 x \,x^{2 n} x^{-n p -3 n -1} \left (a +b \,x^{n}\right )^{p} a \,b^{3} c p -x \,x^{n} x^{-n p -3 n -1} \left (a +b \,x^{n}\right )^{p} a^{3} b d \,p^{2}-x \,x^{n} x^{-n p -3 n -1} \left (a +b \,x^{n}\right )^{p} a^{2} b^{2} c \,p^{2}-4 x \,x^{n} x^{-n p -3 n -1} \left (a +b \,x^{n}\right )^{p} a^{3} b d p -x \,x^{n} x^{-n p -3 n -1} \left (a +b \,x^{n}\right )^{p} a^{2} b^{2} c p -x \,x^{-n p -3 n -1} \left (a +b \,x^{n}\right )^{p} a^{3} b c \,p^{2}-3 x \,x^{n} x^{-n p -3 n -1} \left (a +b \,x^{n}\right )^{p} a^{3} b d -3 x \,x^{-n p -3 n -1} \left (a +b \,x^{n}\right )^{p} a^{3} b c p -2 x \,x^{-n p -3 n -1} \left (a +b \,x^{n}\right )^{p} a^{3} b c}{b \left (3+p \right ) \left (p^{2}+3 p +2\right ) a^{3} n}\) \(488\)

Input:

int(x^(-1-n*(3+p))*(a+b*x^n)^p*(c+d*x^n),x,method=_RETURNVERBOSE)
 

Output:

(x*(x^n)^3*x^(-n*p-3*n-1)*(a+b*x^n)^p*a*b^3*d*p-x*(x^n)^2*x^(-n*p-3*n-1)*( 
a+b*x^n)^p*a^2*b^2*d*p^2+3*x*(x^n)^3*x^(-n*p-3*n-1)*(a+b*x^n)^p*a*b^3*d-2* 
x*(x^n)^3*x^(-n*p-3*n-1)*(a+b*x^n)^p*b^4*c-3*x*(x^n)^2*x^(-n*p-3*n-1)*(a+b 
*x^n)^p*a^2*b^2*d*p+2*x*(x^n)^2*x^(-n*p-3*n-1)*(a+b*x^n)^p*a*b^3*c*p-x*x^n 
*x^(-n*p-3*n-1)*(a+b*x^n)^p*a^3*b*d*p^2-x*x^n*x^(-n*p-3*n-1)*(a+b*x^n)^p*a 
^2*b^2*c*p^2-4*x*x^n*x^(-n*p-3*n-1)*(a+b*x^n)^p*a^3*b*d*p-x*x^n*x^(-n*p-3* 
n-1)*(a+b*x^n)^p*a^2*b^2*c*p-x*x^(-n*p-3*n-1)*(a+b*x^n)^p*a^3*b*c*p^2-3*x* 
x^n*x^(-n*p-3*n-1)*(a+b*x^n)^p*a^3*b*d-3*x*x^(-n*p-3*n-1)*(a+b*x^n)^p*a^3* 
b*c*p-2*x*x^(-n*p-3*n-1)*(a+b*x^n)^p*a^3*b*c)/b/(3+p)/(p^2+3*p+2)/a^3/n
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.61 \[ \int x^{-1-n (3+p)} \left (a+b x^n\right )^p \left (c+d x^n\right ) \, dx=\frac {{\left ({\left (a b^{2} d p - 2 \, b^{3} c + 3 \, a b^{2} d\right )} x x^{-n p - 3 \, n - 1} x^{3 \, n} - {\left (a^{2} b d p^{2} - {\left (2 \, a b^{2} c - 3 \, a^{2} b d\right )} p\right )} x x^{-n p - 3 \, n - 1} x^{2 \, n} - {\left (3 \, a^{3} d + {\left (a^{2} b c + a^{3} d\right )} p^{2} + {\left (a^{2} b c + 4 \, a^{3} d\right )} p\right )} x x^{-n p - 3 \, n - 1} x^{n} - {\left (a^{3} c p^{2} + 3 \, a^{3} c p + 2 \, a^{3} c\right )} x x^{-n p - 3 \, n - 1}\right )} {\left (b x^{n} + a\right )}^{p}}{a^{3} n p^{3} + 6 \, a^{3} n p^{2} + 11 \, a^{3} n p + 6 \, a^{3} n} \] Input:

integrate(x^(-1-n*(3+p))*(a+b*x^n)^p*(c+d*x^n),x, algorithm="fricas")
 

Output:

((a*b^2*d*p - 2*b^3*c + 3*a*b^2*d)*x*x^(-n*p - 3*n - 1)*x^(3*n) - (a^2*b*d 
*p^2 - (2*a*b^2*c - 3*a^2*b*d)*p)*x*x^(-n*p - 3*n - 1)*x^(2*n) - (3*a^3*d 
+ (a^2*b*c + a^3*d)*p^2 + (a^2*b*c + 4*a^3*d)*p)*x*x^(-n*p - 3*n - 1)*x^n 
- (a^3*c*p^2 + 3*a^3*c*p + 2*a^3*c)*x*x^(-n*p - 3*n - 1))*(b*x^n + a)^p/(a 
^3*n*p^3 + 6*a^3*n*p^2 + 11*a^3*n*p + 6*a^3*n)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 648 vs. \(2 (110) = 220\).

Time = 95.26 (sec) , antiderivative size = 648, normalized size of antiderivative = 4.73 \[ \int x^{-1-n (3+p)} \left (a+b x^n\right )^p \left (c+d x^n\right ) \, dx=\frac {a^{2} a^{p} a^{- p - 3} b^{p + 3} c p^{2} \left (\frac {a x^{- n}}{b} + 1\right )^{p + 3} \Gamma \left (- p - 3\right )}{a^{2} n \Gamma \left (- p\right ) + 2 a b n x^{n} \Gamma \left (- p\right ) + b^{2} n x^{2 n} \Gamma \left (- p\right )} + \frac {3 a^{2} a^{p} a^{- p - 3} b^{p + 3} c p \left (\frac {a x^{- n}}{b} + 1\right )^{p + 3} \Gamma \left (- p - 3\right )}{a^{2} n \Gamma \left (- p\right ) + 2 a b n x^{n} \Gamma \left (- p\right ) + b^{2} n x^{2 n} \Gamma \left (- p\right )} + \frac {2 a^{2} a^{p} a^{- p - 3} b^{p + 3} c \left (\frac {a x^{- n}}{b} + 1\right )^{p + 3} \Gamma \left (- p - 3\right )}{a^{2} n \Gamma \left (- p\right ) + 2 a b n x^{n} \Gamma \left (- p\right ) + b^{2} n x^{2 n} \Gamma \left (- p\right )} - \frac {2 a a^{p} a^{- p - 3} b b^{p + 3} c p x^{n} \left (\frac {a x^{- n}}{b} + 1\right )^{p + 3} \Gamma \left (- p - 3\right )}{a^{2} n \Gamma \left (- p\right ) + 2 a b n x^{n} \Gamma \left (- p\right ) + b^{2} n x^{2 n} \Gamma \left (- p\right )} - \frac {2 a a^{p} a^{- p - 3} b b^{p + 3} c x^{n} \left (\frac {a x^{- n}}{b} + 1\right )^{p + 3} \Gamma \left (- p - 3\right )}{a^{2} n \Gamma \left (- p\right ) + 2 a b n x^{n} \Gamma \left (- p\right ) + b^{2} n x^{2 n} \Gamma \left (- p\right )} - \frac {a a^{p} a^{- p - 2} b^{p + 2} d p x^{- n} \left (\frac {a x^{- n}}{b} + 1\right )^{p + 1} \Gamma \left (- p - 2\right )}{b n \Gamma \left (- p\right )} - \frac {a a^{p} a^{- p - 2} b^{p + 2} d x^{- n} \left (\frac {a x^{- n}}{b} + 1\right )^{p + 1} \Gamma \left (- p - 2\right )}{b n \Gamma \left (- p\right )} + \frac {2 a^{p} a^{- p - 3} b^{2} b^{p + 3} c x^{2 n} \left (\frac {a x^{- n}}{b} + 1\right )^{p + 3} \Gamma \left (- p - 3\right )}{a^{2} n \Gamma \left (- p\right ) + 2 a b n x^{n} \Gamma \left (- p\right ) + b^{2} n x^{2 n} \Gamma \left (- p\right )} + \frac {a^{p} a^{- p - 2} b^{p + 2} d \left (\frac {a x^{- n}}{b} + 1\right )^{p + 1} \Gamma \left (- p - 2\right )}{n \Gamma \left (- p\right )} \] Input:

integrate(x**(-1-n*(3+p))*(a+b*x**n)**p*(c+d*x**n),x)
 

Output:

a**2*a**p*a**(-p - 3)*b**(p + 3)*c*p**2*(a/(b*x**n) + 1)**(p + 3)*gamma(-p 
 - 3)/(a**2*n*gamma(-p) + 2*a*b*n*x**n*gamma(-p) + b**2*n*x**(2*n)*gamma(- 
p)) + 3*a**2*a**p*a**(-p - 3)*b**(p + 3)*c*p*(a/(b*x**n) + 1)**(p + 3)*gam 
ma(-p - 3)/(a**2*n*gamma(-p) + 2*a*b*n*x**n*gamma(-p) + b**2*n*x**(2*n)*ga 
mma(-p)) + 2*a**2*a**p*a**(-p - 3)*b**(p + 3)*c*(a/(b*x**n) + 1)**(p + 3)* 
gamma(-p - 3)/(a**2*n*gamma(-p) + 2*a*b*n*x**n*gamma(-p) + b**2*n*x**(2*n) 
*gamma(-p)) - 2*a*a**p*a**(-p - 3)*b*b**(p + 3)*c*p*x**n*(a/(b*x**n) + 1)* 
*(p + 3)*gamma(-p - 3)/(a**2*n*gamma(-p) + 2*a*b*n*x**n*gamma(-p) + b**2*n 
*x**(2*n)*gamma(-p)) - 2*a*a**p*a**(-p - 3)*b*b**(p + 3)*c*x**n*(a/(b*x**n 
) + 1)**(p + 3)*gamma(-p - 3)/(a**2*n*gamma(-p) + 2*a*b*n*x**n*gamma(-p) + 
 b**2*n*x**(2*n)*gamma(-p)) - a*a**p*a**(-p - 2)*b**(p + 2)*d*p*(a/(b*x**n 
) + 1)**(p + 1)*gamma(-p - 2)/(b*n*x**n*gamma(-p)) - a*a**p*a**(-p - 2)*b* 
*(p + 2)*d*(a/(b*x**n) + 1)**(p + 1)*gamma(-p - 2)/(b*n*x**n*gamma(-p)) + 
2*a**p*a**(-p - 3)*b**2*b**(p + 3)*c*x**(2*n)*(a/(b*x**n) + 1)**(p + 3)*ga 
mma(-p - 3)/(a**2*n*gamma(-p) + 2*a*b*n*x**n*gamma(-p) + b**2*n*x**(2*n)*g 
amma(-p)) + a**p*a**(-p - 2)*b**(p + 2)*d*(a/(b*x**n) + 1)**(p + 1)*gamma( 
-p - 2)/(n*gamma(-p))
 

Maxima [F]

\[ \int x^{-1-n (3+p)} \left (a+b x^n\right )^p \left (c+d x^n\right ) \, dx=\int { {\left (d x^{n} + c\right )} {\left (b x^{n} + a\right )}^{p} x^{-n {\left (p + 3\right )} - 1} \,d x } \] Input:

integrate(x^(-1-n*(3+p))*(a+b*x^n)^p*(c+d*x^n),x, algorithm="maxima")
 

Output:

integrate((d*x^n + c)*(b*x^n + a)^p*x^(-n*(p + 3) - 1), x)
 

Giac [F(-2)]

Exception generated. \[ \int x^{-1-n (3+p)} \left (a+b x^n\right )^p \left (c+d x^n\right ) \, dx=\text {Exception raised: TypeError} \] Input:

integrate(x^(-1-n*(3+p))*(a+b*x^n)^p*(c+d*x^n),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{-2,[0,0,2,2,1,1,1,0,1]%%%}+%%%{-2,[0,0,2,2,1,1,0,0,1]%%%}+ 
%%%{1,[0,
 

Mupad [F(-1)]

Timed out. \[ \int x^{-1-n (3+p)} \left (a+b x^n\right )^p \left (c+d x^n\right ) \, dx=\int \frac {{\left (a+b\,x^n\right )}^p\,\left (c+d\,x^n\right )}{x^{n\,\left (p+3\right )+1}} \,d x \] Input:

int(((a + b*x^n)^p*(c + d*x^n))/x^(n*(p + 3) + 1),x)
 

Output:

int(((a + b*x^n)^p*(c + d*x^n))/x^(n*(p + 3) + 1), x)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.43 \[ \int x^{-1-n (3+p)} \left (a+b x^n\right )^p \left (c+d x^n\right ) \, dx=\frac {\left (x^{n} b +a \right )^{p} \left (x^{3 n} a \,b^{2} d p +3 x^{3 n} a \,b^{2} d -2 x^{3 n} b^{3} c -x^{2 n} a^{2} b d \,p^{2}-3 x^{2 n} a^{2} b d p +2 x^{2 n} a \,b^{2} c p -x^{n} a^{3} d \,p^{2}-4 x^{n} a^{3} d p -3 x^{n} a^{3} d -x^{n} a^{2} b c \,p^{2}-x^{n} a^{2} b c p -a^{3} c \,p^{2}-3 a^{3} c p -2 a^{3} c \right )}{x^{n p +3 n} a^{3} n \left (p^{3}+6 p^{2}+11 p +6\right )} \] Input:

int(x^(-1-n*(3+p))*(a+b*x^n)^p*(c+d*x^n),x)
 

Output:

((x**n*b + a)**p*(x**(3*n)*a*b**2*d*p + 3*x**(3*n)*a*b**2*d - 2*x**(3*n)*b 
**3*c - x**(2*n)*a**2*b*d*p**2 - 3*x**(2*n)*a**2*b*d*p + 2*x**(2*n)*a*b**2 
*c*p - x**n*a**3*d*p**2 - 4*x**n*a**3*d*p - 3*x**n*a**3*d - x**n*a**2*b*c* 
p**2 - x**n*a**2*b*c*p - a**3*c*p**2 - 3*a**3*c*p - 2*a**3*c))/(x**(n*p + 
3*n)*a**3*n*(p**3 + 6*p**2 + 11*p + 6))