\(\int \frac {(a+b x^8)^p}{x^2 (c+d x^4)} \, dx\) [33]

Optimal result

Integrand size = 22, antiderivative size = 126 \[ \int \frac {\left (a+b x^8\right )^p}{x^2 \left (c+d x^4\right )} \, dx=-\frac {\left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {AppellF1}\left (-\frac {1}{8},-p,1,\frac {7}{8},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{c x}-\frac {d x^3 \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{8},-p,1,\frac {11}{8},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{3 c^2} \] Output:

-(b*x^8+a)^p*AppellF1(-1/8,1,-p,7/8,d^2*x^8/c^2,-b*x^8/a)/c/x/((1+b*x^8/a) 
^p)-1/3*d*x^3*(b*x^8+a)^p*AppellF1(3/8,1,-p,11/8,d^2*x^8/c^2,-b*x^8/a)/c^2 
/((1+b*x^8/a)^p)
 

Mathematica [F]

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

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

Output:

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

Rubi [F]

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.

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

\(\Big \downarrow \) 1888

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

Input:

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

Output:

$Aborted
 

Defintions of rubi rules used

rule 1888

Int[((f_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^ 
(n_))^(q_.), x_Symbol] :> Unintegrable[(f*x)^m*(d + e*x^n)^q*(a + c*x^(2*n) 
)^p, x] /; FreeQ[{a, c, d, e, f, m, n, p, q}, x] && EqQ[n2, 2*n]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((b*x^8 + a)^p/(d*x^6 + c*x^2), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+b x^8\right )^p}{x^2 \left (c+d x^4\right )} \, dx=\int \frac {\left (b \,x^{8}+a \right )^{p}}{d \,x^{6}+c \,x^{2}}d x \] Input:

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

Output:

int((a + b*x**8)**p/(c*x**2 + d*x**6),x)