\(\int \frac {(a+b x^8)^p}{x^5 (c+d x^4)^2} \, dx\) [41]

Optimal result

Integrand size = 22, antiderivative size = 310 \[ \int \frac {\left (a+b x^8\right )^p}{x^5 \left (c+d x^4\right )^2} \, dx=-\frac {d^3 \left (a+b x^8\right )^{1+p}}{4 c \left (b c^2+a d^2\right ) \left (c^2-d^2 x^8\right )}-\frac {\left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {AppellF1}\left (-\frac {1}{2},-p,2,\frac {1}{2},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{4 c^2 x^4}+\frac {d^2 x^4 \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,2,\frac {3}{2},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{4 c^4}-\frac {d^3 \left (a d^2+b c^2 (1-p)\right ) \left (a+b x^8\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {d^2 \left (a+b x^8\right )}{b c^2+a d^2}\right )}{4 c^3 \left (b c^2+a d^2\right )^2 (1+p)}+\frac {d \left (a+b x^8\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b x^8}{a}\right )}{4 a c^3 (1+p)} \] Output:

-1/4*d^3*(b*x^8+a)^(p+1)/c/(a*d^2+b*c^2)/(-d^2*x^8+c^2)-1/4*(b*x^8+a)^p*Ap 
pellF1(-1/2,2,-p,1/2,d^2*x^8/c^2,-b*x^8/a)/c^2/x^4/((1+b*x^8/a)^p)+1/4*d^2 
*x^4*(b*x^8+a)^p*AppellF1(1/2,2,-p,3/2,d^2*x^8/c^2,-b*x^8/a)/c^4/((1+b*x^8 
/a)^p)-1/4*d^3*(a*d^2+b*c^2*(1-p))*(b*x^8+a)^(p+1)*hypergeom([1, p+1],[2+p 
],d^2*(b*x^8+a)/(a*d^2+b*c^2))/c^3/(a*d^2+b*c^2)^2/(p+1)+1/4*d*(b*x^8+a)^( 
p+1)*hypergeom([1, p+1],[2+p],1+b*x^8/a)/a/c^3/(p+1)
 

Mathematica [A] (warning: unable to verify)

Time = 2.20 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.10 \[ \int \frac {\left (a+b x^8\right )^p}{x^5 \left (c+d x^4\right )^2} \, dx=\frac {d \left (\frac {d \left (-\sqrt {-\frac {a}{b}}+x^4\right )}{c+d x^4}\right )^{-p} \left (\frac {d \left (\sqrt {-\frac {a}{b}}+x^4\right )}{c+d x^4}\right )^{-p} \left (a+b x^8\right )^p \left (c p \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,\frac {c-\sqrt {-\frac {a}{b}} d}{c+d x^4},\frac {c+\sqrt {-\frac {a}{b}} d}{c+d x^4}\right )+(-1+2 p) \left (c+d x^4\right ) \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {c-\sqrt {-\frac {a}{b}} d}{c+d x^4},\frac {c+\sqrt {-\frac {a}{b}} d}{c+d x^4}\right )\right )}{4 c^3 p (-1+2 p) \left (c+d x^4\right )}+\frac {\left (a+b x^8\right )^p \left (-\frac {c \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b x^8}{a}\right )}{x^4}-\frac {d \left (1+\frac {a}{b x^8}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-p,-p,1-p,-\frac {a}{b x^8}\right )}{p}\right )}{4 c^3} \] Input:

Integrate[(a + b*x^8)^p/(x^5*(c + d*x^4)^2),x]
 

Output:

(d*(a + b*x^8)^p*(c*p*AppellF1[1 - 2*p, -p, -p, 2 - 2*p, (c - Sqrt[-(a/b)] 
*d)/(c + d*x^4), (c + Sqrt[-(a/b)]*d)/(c + d*x^4)] + (-1 + 2*p)*(c + d*x^4 
)*AppellF1[-2*p, -p, -p, 1 - 2*p, (c - Sqrt[-(a/b)]*d)/(c + d*x^4), (c + S 
qrt[-(a/b)]*d)/(c + d*x^4)]))/(4*c^3*p*(-1 + 2*p)*((d*(-Sqrt[-(a/b)] + x^4 
))/(c + d*x^4))^p*((d*(Sqrt[-(a/b)] + x^4))/(c + d*x^4))^p*(c + d*x^4)) + 
((a + b*x^8)^p*(-((c*Hypergeometric2F1[-1/2, -p, 1/2, -((b*x^8)/a)])/(x^4* 
(1 + (b*x^8)/a)^p)) - (d*Hypergeometric2F1[-p, -p, 1 - p, -(a/(b*x^8))])/( 
p*(1 + a/(b*x^8))^p)))/(4*c^3)
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 302, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1803, 622, 2009}

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.

\(\displaystyle \int \frac {\left (a+b x^8\right )^p}{x^5 \left (c+d x^4\right )^2} \, dx\)

\(\Big \downarrow \) 1803

\(\displaystyle \frac {1}{4} \int \frac {\left (b x^8+a\right )^p}{x^8 \left (d x^4+c\right )^2}dx^4\)

\(\Big \downarrow \) 622

\(\displaystyle \frac {1}{4} \int \left (-\frac {2 c d \left (b x^8+a\right )^p}{x^4 \left (c^2-d^2 x^8\right )^2}+\frac {c^2 \left (b x^8+a\right )^p}{x^8 \left (c^2-d^2 x^8\right )^2}+\frac {d^2 \left (b x^8+a\right )^p}{\left (d^2 x^8-c^2\right )^2}\right )dx^4\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{4} \left (-\frac {\left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {AppellF1}\left (-\frac {1}{2},-p,2,\frac {1}{2},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{c^2 x^4}+\frac {d^2 x^4 \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,2,\frac {3}{2},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{c^4}+\frac {d \left (a+b x^8\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b x^8}{a}+1\right )}{a c^3 (p+1)}-\frac {d^3 \left (a+b x^8\right )^{p+1}}{c \left (c^2-d^2 x^8\right ) \left (a d^2+b c^2\right )}-\frac {d^3 \left (a+b x^8\right )^{p+1} \left (a d^2+b c^2 (1-p)\right ) \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {d^2 \left (b x^8+a\right )}{b c^2+a d^2}\right )}{c^3 (p+1) \left (a d^2+b c^2\right )^2}\right )\)

Input:

Int[(a + b*x^8)^p/(x^5*(c + d*x^4)^2),x]
 

Output:

(-((d^3*(a + b*x^8)^(1 + p))/(c*(b*c^2 + a*d^2)*(c^2 - d^2*x^8))) - ((a + 
b*x^8)^p*AppellF1[-1/2, -p, 2, 1/2, -((b*x^8)/a), (d^2*x^8)/c^2])/(c^2*x^4 
*(1 + (b*x^8)/a)^p) + (d^2*x^4*(a + b*x^8)^p*AppellF1[1/2, -p, 2, 3/2, -(( 
b*x^8)/a), (d^2*x^8)/c^2])/(c^4*(1 + (b*x^8)/a)^p) - (d^3*(a*d^2 + b*c^2*( 
1 - p))*(a + b*x^8)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, (d^2*(a + b 
*x^8))/(b*c^2 + a*d^2)])/(c^3*(b*c^2 + a*d^2)^2*(1 + p)) + (d*(a + b*x^8)^ 
(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, 1 + (b*x^8)/a])/(a*c^3*(1 + p)) 
)/4
 

Defintions of rubi rules used

rule 622

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> Int[ExpandIntegrand[x^m*(a + b*x^2)^p, (c/(c^2 - d^2*x^2) - d*(x/(c^2 
 - d^2*x^2)))^(-n), x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[n, -1]
 

rule 1803

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q 
_.), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(d + e*x 
)^q*(a + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, c, d, e, m, n, p, q}, x] && 
 EqQ[n2, 2*n] && IntegerQ[Simplify[(m + 1)/n]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [F]

\[\int \frac {\left (b \,x^{8}+a \right )^{p}}{x^{5} \left (x^{4} d +c \right )^{2}}d x\]

Input:

int((b*x^8+a)^p/x^5/(d*x^4+c)^2,x)
 

Output:

int((b*x^8+a)^p/x^5/(d*x^4+c)^2,x)
 

Fricas [F]

\[ \int \frac {\left (a+b x^8\right )^p}{x^5 \left (c+d x^4\right )^2} \, dx=\int { \frac {{\left (b x^{8} + a\right )}^{p}}{{\left (d x^{4} + c\right )}^{2} x^{5}} \,d x } \] Input:

integrate((b*x^8+a)^p/x^5/(d*x^4+c)^2,x, algorithm="fricas")
 

Output:

integral((b*x^8 + a)^p/(d^2*x^13 + 2*c*d*x^9 + c^2*x^5), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^8\right )^p}{x^5 \left (c+d x^4\right )^2} \, dx=\text {Timed out} \] Input:

integrate((b*x**8+a)**p/x**5/(d*x**4+c)**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (a+b x^8\right )^p}{x^5 \left (c+d x^4\right )^2} \, dx=\int { \frac {{\left (b x^{8} + a\right )}^{p}}{{\left (d x^{4} + c\right )}^{2} x^{5}} \,d x } \] Input:

integrate((b*x^8+a)^p/x^5/(d*x^4+c)^2,x, algorithm="maxima")
 

Output:

integrate((b*x^8 + a)^p/((d*x^4 + c)^2*x^5), x)
 

Giac [F]

\[ \int \frac {\left (a+b x^8\right )^p}{x^5 \left (c+d x^4\right )^2} \, dx=\int { \frac {{\left (b x^{8} + a\right )}^{p}}{{\left (d x^{4} + c\right )}^{2} x^{5}} \,d x } \] Input:

integrate((b*x^8+a)^p/x^5/(d*x^4+c)^2,x, algorithm="giac")
 

Output:

integrate((b*x^8 + a)^p/((d*x^4 + c)^2*x^5), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^8\right )^p}{x^5 \left (c+d x^4\right )^2} \, dx=\int \frac {{\left (b\,x^8+a\right )}^p}{x^5\,{\left (d\,x^4+c\right )}^2} \,d x \] Input:

int((a + b*x^8)^p/(x^5*(c + d*x^4)^2),x)
 

Output:

int((a + b*x^8)^p/(x^5*(c + d*x^4)^2), x)
 

Reduce [F]

\[ \int \frac {\left (a+b x^8\right )^p}{x^5 \left (c+d x^4\right )^2} \, dx=\int \frac {\left (b \,x^{8}+a \right )^{p}}{d^{2} x^{13}+2 c d \,x^{9}+c^{2} x^{5}}d x \] Input:

int((b*x^8+a)^p/x^5/(d*x^4+c)^2,x)
 

Output:

int((a + b*x**8)**p/(c**2*x**5 + 2*c*d*x**9 + d**2*x**13),x)