\(\int \frac {x^{-1-n (3+p)} (a+b x^n)^p}{c+d x^n} \, dx\) [517]

Optimal result

Integrand size = 29, antiderivative size = 302 \[ \int \frac {x^{-1-n (3+p)} \left (a+b x^n\right )^p}{c+d x^n} \, dx=-\frac {d^2 x^{-n (1+p)} \left (a+b x^n\right )^{1+p}}{a c^3 n (1+p)}-\frac {b d x^{-n (1+p)} \left (a+b x^n\right )^{1+p}}{a^2 c^2 n (1+p) (2+p)}-\frac {2 b^2 x^{-n (1+p)} \left (a+b x^n\right )^{1+p}}{a^3 c n (1+p) (2+p) (3+p)}+\frac {d x^{-n (2+p)} \left (a+b x^n\right )^{1+p}}{a c^2 n (2+p)}+\frac {2 b x^{-n (2+p)} \left (a+b x^n\right )^{1+p}}{a^2 c n (2+p) (3+p)}-\frac {x^{-n (3+p)} \left (a+b x^n\right )^{1+p}}{a c n (3+p)}+\frac {d^3 x^{-n p} \left (a+b x^n\right )^p \operatorname {Hypergeometric2F1}\left (1,-p,1-p,\frac {(b c-a d) x^n}{c \left (a+b x^n\right )}\right )}{c^4 n p} \] Output:

-d^2*(a+b*x^n)^(p+1)/a/c^3/n/(p+1)/(x^(n*(p+1)))-b*d*(a+b*x^n)^(p+1)/a^2/c 
^2/n/(p+1)/(2+p)/(x^(n*(p+1)))-2*b^2*(a+b*x^n)^(p+1)/a^3/c/n/(p+1)/(2+p)/( 
3+p)/(x^(n*(p+1)))+d*(a+b*x^n)^(p+1)/a/c^2/n/(2+p)/(x^(n*(2+p)))+2*b*(a+b* 
x^n)^(p+1)/a^2/c/n/(2+p)/(3+p)/(x^(n*(2+p)))-(a+b*x^n)^(p+1)/a/c/n/(3+p)/( 
x^(n*(3+p)))+d^3*(a+b*x^n)^p*hypergeom([1, -p],[1-p],(-a*d+b*c)*x^n/c/(a+b 
*x^n))/c^4/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 = 4.37 (sec) , antiderivative size = 343, normalized size of antiderivative = 1.14 \[ \int \frac {x^{-1-n (3+p)} \left (a+b x^n\right )^p}{c+d x^n} \, dx=\frac {x^{-n (3+p)} \left (a+b x^n\right )^{-1+p} \left (c (-1+p) \left (a+b x^n\right ) \left (c^3 p \left (2+3 p+p^2\right )-3 c^2 d p (1+p) x^n+6 c d^2 p x^{2 n}-6 d^3 x^{3 n}\right ) \Phi \left (\frac {(b c-a d) x^n}{c \left (a+b x^n\right )},1,-p\right )+(b c-a d) x^n \left (c+d x^n\right ) \left (\left (c^2 \left (2+6 p+3 p^2\right )-c d (5+9 p) x^n+11 d^2 x^{2 n}\right ) \operatorname {Hypergeometric2F1}\left (2,1-p,2-p,\frac {(b c-a d) x^n}{c \left (a+b x^n\right )}\right )-3 \left (c+c p-2 d x^n\right ) \left (c+d x^n\right ) \, _3F_2\left (2,2,1-p;1,2-p;\frac {(b c-a d) x^n}{c \left (a+b x^n\right )}\right )+\left (c+d x^n\right )^2 \, _4F_3\left (2,2,2,1-p;1,1,2-p;\frac {(b c-a d) x^n}{c \left (a+b x^n\right )}\right )\right )\right )}{c^5 n (-1+p) (1+p) (2+p) (3+p)} \] Input:

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

Output:

((a + b*x^n)^(-1 + p)*(c*(-1 + p)*(a + b*x^n)*(c^3*p*(2 + 3*p + p^2) - 3*c 
^2*d*p*(1 + p)*x^n + 6*c*d^2*p*x^(2*n) - 6*d^3*x^(3*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)*((c^ 
2*(2 + 6*p + 3*p^2) - c*d*(5 + 9*p)*x^n + 11*d^2*x^(2*n))*Hypergeometric2F 
1[2, 1 - p, 2 - p, ((b*c - a*d)*x^n)/(c*(a + b*x^n))] - 3*(c + c*p - 2*d*x 
^n)*(c + d*x^n)*HypergeometricPFQ[{2, 2, 1 - p}, {1, 2 - p}, ((b*c - a*d)* 
x^n)/(c*(a + b*x^n))] + (c + d*x^n)^2*HypergeometricPFQ[{2, 2, 2, 1 - p}, 
{1, 1, 2 - p}, ((b*c - a*d)*x^n)/(c*(a + b*x^n))])))/(c^5*n*(-1 + p)*(1 + 
p)*(2 + p)*(3 + p)*x^(n*(3 + p)))
 

Rubi [C] (warning: unable to verify)

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

Time = 7.06 (sec) , antiderivative size = 347, normalized size of antiderivative = 1.15, 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.

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

\(\Big \downarrow \) 1013

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

\(\Big \downarrow \) 1012

\(\displaystyle \frac {x^{-n (p+3)} \left (a+b x^n\right )^{p-1} \left (c (1-p) \left (a+b x^n\right ) \left (c^3 p (p+1) (p+2)-3 c^2 d p (p+1) x^n+6 c d^2 p x^{2 n}-6 d^3 x^{3 n}\right ) \Phi \left (\frac {(b c-a d) x^n}{c \left (b x^n+a\right )},1,-p\right )-x^n (b c-a d) \left (c+d x^n\right ) \left (\left (c+d x^n\right )^2 \, _4F_3\left (2,2,2,1-p;1,1,2-p;\frac {(b c-a d) x^n}{c \left (b x^n+a\right )}\right )-3 \left (c+d x^n\right ) \left (c (p+1)-2 d x^n\right ) \, _3F_2\left (2,2,1-p;1,2-p;\frac {(b c-a d) x^n}{c \left (b x^n+a\right )}\right )+\left (c^2 \left (3 p^2+6 p+2\right )-c d (9 p+5) x^n+11 d^2 x^{2 n}\right ) \operatorname {Hypergeometric2F1}\left (2,1-p,2-p,\frac {(b c-a d) x^n}{c \left (b x^n+a\right )}\right )\right )\right )}{c^5 n (1-p) (p+1) (p+2) (p+3)}\)

Input:

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

Output:

((a + b*x^n)^(-1 + p)*(c*(1 - p)*(a + b*x^n)*(c^3*p*(1 + p)*(2 + p) - 3*c^ 
2*d*p*(1 + p)*x^n + 6*c*d^2*p*x^(2*n) - 6*d^3*x^(3*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)*((c^2 
*(2 + 6*p + 3*p^2) - c*d*(5 + 9*p)*x^n + 11*d^2*x^(2*n))*Hypergeometric2F1 
[2, 1 - p, 2 - p, ((b*c - a*d)*x^n)/(c*(a + b*x^n))] - 3*(c*(1 + p) - 2*d* 
x^n)*(c + d*x^n)*HypergeometricPFQ[{2, 2, 1 - p}, {1, 2 - p}, ((b*c - a*d) 
*x^n)/(c*(a + b*x^n))] + (c + d*x^n)^2*HypergeometricPFQ[{2, 2, 2, 1 - p}, 
 {1, 1, 2 - p}, ((b*c - a*d)*x^n)/(c*(a + b*x^n))])))/(c^5*n*(1 - p)*(1 + 
p)*(2 + p)*(3 + p)*x^(n*(3 + p)))
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-2)]

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

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

Output:

Exception raised: HeuristicGCDFailed >> no luck
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(x**(3*n)*(x**n*b + a)**p*a**2*b*d**2*p**2 + 5*x**(3*n)*(x**n*b + a)**p*a* 
*2*b*d**2*p + 6*x**(3*n)*(x**n*b + a)**p*a**2*b*d**2 - x**(3*n)*(x**n*b + 
a)**p*a*b**2*c*d*p**2 - 3*x**(3*n)*(x**n*b + a)**p*a*b**2*c*d*p - 2*x**(3* 
n)*(x**n*b + a)**p*b**3*c**2*p - x**(2*n)*(x**n*b + a)**p*a**3*d**2*p**3 - 
 5*x**(2*n)*(x**n*b + a)**p*a**3*d**2*p**2 - 6*x**(2*n)*(x**n*b + a)**p*a* 
*3*d**2*p + x**(2*n)*(x**n*b + a)**p*a**2*b*c*d*p**3 + 3*x**(2*n)*(x**n*b 
+ a)**p*a**2*b*c*d*p**2 + 2*x**(2*n)*(x**n*b + a)**p*a*b**2*c**2*p**2 + x* 
*n*(x**n*b + a)**p*a**3*c*d*p**3 + 4*x**n*(x**n*b + a)**p*a**3*c*d*p**2 + 
3*x**n*(x**n*b + a)**p*a**3*c*d*p - x**n*(x**n*b + a)**p*a**2*b*c**2*p**3 
- x**n*(x**n*b + a)**p*a**2*b*c**2*p**2 - (x**n*b + a)**p*a**3*c**2*p**3 - 
 3*(x**n*b + a)**p*a**3*c**2*p**2 - 2*(x**n*b + a)**p*a**3*c**2*p - x**(n* 
p + 3*n)*int((x**n*b + a)**p/(x**(n*p + 2*n)*b*d*x + x**(n*p + n)*a*d*x + 
x**(n*p + n)*b*c*x + x**(n*p)*a*c*x),x)*a**4*d**3*n*p**4 - 6*x**(n*p + 3*n 
)*int((x**n*b + a)**p/(x**(n*p + 2*n)*b*d*x + x**(n*p + n)*a*d*x + x**(n*p 
 + n)*b*c*x + x**(n*p)*a*c*x),x)*a**4*d**3*n*p**3 - 11*x**(n*p + 3*n)*int( 
(x**n*b + a)**p/(x**(n*p + 2*n)*b*d*x + x**(n*p + n)*a*d*x + x**(n*p + n)* 
b*c*x + x**(n*p)*a*c*x),x)*a**4*d**3*n*p**2 - 6*x**(n*p + 3*n)*int((x**n*b 
 + a)**p/(x**(n*p + 2*n)*b*d*x + x**(n*p + n)*a*d*x + x**(n*p + n)*b*c*x + 
 x**(n*p)*a*c*x),x)*a**4*d**3*n*p + x**(n*p + 3*n)*int((x**n*b + a)**p/(x* 
*(n*p + 2*n)*b*d*x + x**(n*p + n)*a*d*x + x**(n*p + n)*b*c*x + x**(n*p)...