\(\int \frac {(a+b x^8)^p}{x (c+d x^4)^2} \, dx\) [39]

Optimal result

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

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

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.01, 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.

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

\(\Big \downarrow \) 1803

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

\(\Big \downarrow \) 622

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

((d^2*(a + b*x^8)^(1 + p))/(2*(b*c^2 + a*d^2)*(c^2 - d^2*x^8)) - (2*d*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^3 
*(1 + (b*x^8)/a)^p) + (d^2*(a*d^2 + b*c^2*(1 - p))*(a + b*x^8)^(1 + p)*Hyp 
ergeometric2F1[1, 1 + p, 2 + p, (d^2*(a + b*x^8))/(b*c^2 + a*d^2)])/(2*c^2 
*(b*c^2 + a*d^2)^2*(1 + p)) - ((a + b*x^8)^(1 + p)*Hypergeometric2F1[1, 1 
+ p, 2 + p, 1 + (b*x^8)/a])/(2*a*c^2*(1 + p)) + (b*d^2*(a + b*x^8)^(1 + p) 
*Hypergeometric2F1[2, 1 + p, 2 + p, (d^2*(a + b*x^8))/(b*c^2 + a*d^2)])/(2 
*(b*c^2 + a*d^2)^2*(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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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