\(\int \frac {x^{-1-n (1+p)} (a+b x^n)^p}{(c+d x^n)^2} \, dx\) [530]

Optimal result

Integrand size = 29, antiderivative size = 191 \[ \int \frac {x^{-1-n (1+p)} \left (a+b x^n\right )^p}{\left (c+d x^n\right )^2} \, dx=-\frac {d (b c-a d (2+p)) x^{-n p} \left (a+b x^n\right )^{1+p}}{a c^2 (b c-a d) n (1+p) \left (c+d x^n\right )}-\frac {x^{-n (1+p)} \left (a+b x^n\right )^{1+p}}{a c n (1+p) \left (c+d x^n\right )}+\frac {d (2 b c-a d (2+p)) x^{-n p} \left (a+b x^n\right )^p \operatorname {Hypergeometric2F1}\left (1,-p,1-p,\frac {(b c-a d) x^n}{c \left (a+b x^n\right )}\right )}{c^3 (b c-a d) n p} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 5 in optimal.

Time = 0.78 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.48 \[ \int \frac {x^{-1-n (1+p)} \left (a+b x^n\right )^p}{\left (c+d x^n\right )^2} \, dx=-\frac {x^{-n (1+p)} \left (a+b x^n\right )^{-1+p} \left (c \left (2-3 p+p^2\right ) \left (a+b x^n\right ) \left (c^2 (-1+p)^2+2 c d (-1+p)^2 x^n-d^2 (-1+3 p) x^{2 n}\right ) \Phi \left (\frac {(b c-a d) x^n}{c \left (a+b x^n\right )},1,1-p\right )-c \left (2-3 p+p^2\right ) \left (a+b x^n\right ) \left (c^2 p^2+2 c d \left (-1-p+p^2\right ) x^n-d^2 (2+3 p) x^{2 n}\right ) \Phi \left (\frac {(b c-a d) x^n}{c \left (a+b x^n\right )},1,-p\right )+(b c-a d) x^n \left (c+d x^n\right )^2 \, _3F_2\left (2,2,1-p;1,3-p;\frac {(b c-a d) x^n}{c \left (a+b x^n\right )}\right )\right )}{c^4 n (-2+p) (-1+p) (1+p) \left (c+d x^n\right )} \] Input:

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

Output:

-(((a + b*x^n)^(-1 + p)*(c*(2 - 3*p + p^2)*(a + b*x^n)*(c^2*(-1 + p)^2 + 2 
*c*d*(-1 + p)^2*x^n - d^2*(-1 + 3*p)*x^(2*n))*HurwitzLerchPhi[((b*c - a*d) 
*x^n)/(c*(a + b*x^n)), 1, 1 - p] - c*(2 - 3*p + p^2)*(a + b*x^n)*(c^2*p^2 
+ 2*c*d*(-1 - p + p^2)*x^n - d^2*(2 + 3*p)*x^(2*n))*HurwitzLerchPhi[((b*c 
- a*d)*x^n)/(c*(a + b*x^n)), 1, -p] + (b*c - a*d)*x^n*(c + d*x^n)^2*Hyperg 
eometricPFQ[{2, 2, 1 - p}, {1, 3 - p}, ((b*c - a*d)*x^n)/(c*(a + b*x^n))]) 
)/(c^4*n*(-2 + p)*(-1 + p)*(1 + p)*x^(n*(1 + p))*(c + d*x^n)))
 

Rubi [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 5 in optimal.

Time = 1.60 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.55, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {1013, 1012}

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.

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

\(\Big \downarrow \) 1013

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

\(\Big \downarrow \) 1012

\(\displaystyle -\frac {x^{-n (p+1)} \left (a+b x^n\right )^{p-1} \left (x^n (b c-a d) \left (c+d x^n\right )^2 \, _3F_2\left (2,2,1-p;1,3-p;\frac {(b c-a d) x^n}{c \left (b x^n+a\right )}\right )-c (1-p) (2-p) \left (a+b x^n\right ) \left (c^2 p^2-2 c d \left (-p^2+p+1\right ) x^n-d^2 (3 p+2) x^{2 n}\right ) \Phi \left (\frac {(b c-a d) x^n}{c \left (b x^n+a\right )},1,-p\right )+c (1-p) (2-p) \left (a+b x^n\right ) \left (c^2 (1-p)^2+2 c d (1-p)^2 x^n+d^2 (1-3 p) x^{2 n}\right ) \Phi \left (\frac {(b c-a d) x^n}{c \left (b x^n+a\right )},1,1-p\right )\right )}{c^5 n (1-p) (2-p) (p+1) \left (\frac {d x^n}{c}+1\right )}\)

Input:

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

Output:

-(((a + b*x^n)^(-1 + p)*(c*(1 - p)*(2 - p)*(a + b*x^n)*(c^2*(1 - p)^2 + 2* 
c*d*(1 - p)^2*x^n + d^2*(1 - 3*p)*x^(2*n))*HurwitzLerchPhi[((b*c - a*d)*x^ 
n)/(c*(a + b*x^n)), 1, 1 - p] - c*(1 - p)*(2 - p)*(a + b*x^n)*(c^2*p^2 - 2 
*c*d*(1 + p - p^2)*x^n - d^2*(2 + 3*p)*x^(2*n))*HurwitzLerchPhi[((b*c - a* 
d)*x^n)/(c*(a + b*x^n)), 1, -p] + (b*c - a*d)*x^n*(c + d*x^n)^2*Hypergeome 
tricPFQ[{2, 2, 1 - p}, {1, 3 - p}, ((b*c - a*d)*x^n)/(c*(a + b*x^n))]))/(c 
^5*n*(1 - p)*(2 - p)*(1 + p)*x^(n*(1 + p))*(1 + (d*x^n)/c)))
 

Defintions of rubi rules used

rule 1012

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m 
+ 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, 
 b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n 
 - 1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 1013

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^ 
n/a))^FracPart[p])   Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; 
 FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] & 
& NeQ[m, n - 1] &&  !(IntegerQ[p] || GtQ[a, 0])
 

Maple [F]

\[\int \frac {x^{-1-n \left (p +1\right )} \left (a +b \,x^{n}\right )^{p}}{\left (c +d \,x^{n}\right )^{2}}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int \frac {x^{-1-n (1+p)} \left (a+b x^n\right )^p}{\left (c+d x^n\right )^2} \, dx=\int { \frac {{\left (b x^{n} + a\right )}^{p} x^{-n {\left (p + 1\right )} - 1}}{{\left (d x^{n} + c\right )}^{2}} \,d x } \] Input:

integrate(x^(-1-n*(p+1))*(a+b*x^n)^p/(c+d*x^n)^2,x, algorithm="fricas")
 

Output:

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

Sympy [F(-2)]

Exception generated. \[ \int \frac {x^{-1-n (1+p)} \left (a+b x^n\right )^p}{\left (c+d x^n\right )^2} \, dx=\text {Exception raised: HeuristicGCDFailed} \] Input:

integrate(x**(-1-n*(p+1))*(a+b*x**n)**p/(c+d*x**n)**2,x)
 

Output:

Exception raised: HeuristicGCDFailed >> no luck
 

Maxima [F]

\[ \int \frac {x^{-1-n (1+p)} \left (a+b x^n\right )^p}{\left (c+d x^n\right )^2} \, dx=\int { \frac {{\left (b x^{n} + a\right )}^{p} x^{-n {\left (p + 1\right )} - 1}}{{\left (d x^{n} + c\right )}^{2}} \,d x } \] Input:

integrate(x^(-1-n*(p+1))*(a+b*x^n)^p/(c+d*x^n)^2,x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {x^{-1-n (1+p)} \left (a+b x^n\right )^p}{\left (c+d x^n\right )^2} \, dx=\int { \frac {{\left (b x^{n} + a\right )}^{p} x^{-n {\left (p + 1\right )} - 1}}{{\left (d x^{n} + c\right )}^{2}} \,d x } \] Input:

integrate(x^(-1-n*(p+1))*(a+b*x^n)^p/(c+d*x^n)^2,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{-1-n (1+p)} \left (a+b x^n\right )^p}{\left (c+d x^n\right )^2} \, dx=\int \frac {{\left (a+b\,x^n\right )}^p}{x^{n\,\left (p+1\right )+1}\,{\left (c+d\,x^n\right )}^2} \,d x \] Input:

int((a + b*x^n)^p/(x^(n*(p + 1) + 1)*(c + d*x^n)^2),x)
 

Output:

int((a + b*x^n)^p/(x^(n*(p + 1) + 1)*(c + d*x^n)^2), x)
 

Reduce [F]

\[ \int \frac {x^{-1-n (1+p)} \left (a+b x^n\right )^p}{\left (c+d x^n\right )^2} \, dx=\text {too large to display} \] Input:

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

Output:

( - 2*x**(2*n)*(x**n*b + a)**p*b**2*d + 2*x**n*(x**n*b + a)**p*a*b*d*p - 2 
*x**n*(x**n*b + a)**p*b**2*c - (x**n*b + a)**p*a**2*d*p**2 - (x**n*b + a)* 
*p*a**2*d*p + 2*(x**n*b + a)**p*a*b*c*p - x**(n*p + 2*n)*int((x**n*b + a)* 
*p/(x**(n*p + 3*n)*a*b*d**3*p*x + x**(n*p + 3*n)*a*b*d**3*x - 2*x**(n*p + 
3*n)*b**2*c*d**2*x + x**(n*p + 2*n)*a**2*d**3*p*x + x**(n*p + 2*n)*a**2*d* 
*3*x + 2*x**(n*p + 2*n)*a*b*c*d**2*p*x - 4*x**(n*p + 2*n)*b**2*c**2*d*x + 
2*x**(n*p + n)*a**2*c*d**2*p*x + 2*x**(n*p + n)*a**2*c*d**2*x + x**(n*p + 
n)*a*b*c**2*d*p*x - 3*x**(n*p + n)*a*b*c**2*d*x - 2*x**(n*p + n)*b**2*c**3 
*x + x**(n*p)*a**2*c**2*d*p*x + x**(n*p)*a**2*c**2*d*x - 2*x**(n*p)*a*b*c* 
*3*x),x)*a**4*d**4*n*p**4 - 4*x**(n*p + 2*n)*int((x**n*b + a)**p/(x**(n*p 
+ 3*n)*a*b*d**3*p*x + x**(n*p + 3*n)*a*b*d**3*x - 2*x**(n*p + 3*n)*b**2*c* 
d**2*x + x**(n*p + 2*n)*a**2*d**3*p*x + x**(n*p + 2*n)*a**2*d**3*x + 2*x** 
(n*p + 2*n)*a*b*c*d**2*p*x - 4*x**(n*p + 2*n)*b**2*c**2*d*x + 2*x**(n*p + 
n)*a**2*c*d**2*p*x + 2*x**(n*p + n)*a**2*c*d**2*x + x**(n*p + n)*a*b*c**2* 
d*p*x - 3*x**(n*p + n)*a*b*c**2*d*x - 2*x**(n*p + n)*b**2*c**3*x + x**(n*p 
)*a**2*c**2*d*p*x + x**(n*p)*a**2*c**2*d*x - 2*x**(n*p)*a*b*c**3*x),x)*a** 
4*d**4*n*p**3 - 5*x**(n*p + 2*n)*int((x**n*b + a)**p/(x**(n*p + 3*n)*a*b*d 
**3*p*x + x**(n*p + 3*n)*a*b*d**3*x - 2*x**(n*p + 3*n)*b**2*c*d**2*x + x** 
(n*p + 2*n)*a**2*d**3*p*x + x**(n*p + 2*n)*a**2*d**3*x + 2*x**(n*p + 2*n)* 
a*b*c*d**2*p*x - 4*x**(n*p + 2*n)*b**2*c**2*d*x + 2*x**(n*p + n)*a**2*c...