Integrand size = 22, antiderivative size = 190 \[ \int \frac {\left (a+b x^8\right )^p}{x^4 \left (c+d x^4\right )^2} \, dx=-\frac {\left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {AppellF1}\left (-\frac {3}{8},-p,2,\frac {5}{8},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{3 c^2 x^3}-\frac {2 d x \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{8},-p,2,\frac {9}{8},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{c^3}+\frac {d^2 x^5 \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{8},-p,2,\frac {13}{8},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{5 c^4} \] Output:
-1/3*(b*x^8+a)^p*AppellF1(-3/8,2,-p,5/8,d^2*x^8/c^2,-b*x^8/a)/c^2/x^3/((1+ b*x^8/a)^p)-2*d*x*(b*x^8+a)^p*AppellF1(1/8,2,-p,9/8,d^2*x^8/c^2,-b*x^8/a)/ c^3/((1+b*x^8/a)^p)+1/5*d^2*x^5*(b*x^8+a)^p*AppellF1(5/8,2,-p,13/8,d^2*x^8 /c^2,-b*x^8/a)/c^4/((1+b*x^8/a)^p)
\[ \int \frac {\left (a+b x^8\right )^p}{x^4 \left (c+d x^4\right )^2} \, dx=\int \frac {\left (a+b x^8\right )^p}{x^4 \left (c+d x^4\right )^2} \, dx \] Input:
Integrate[(a + b*x^8)^p/(x^4*(c + d*x^4)^2),x]
Output:
Integrate[(a + b*x^8)^p/(x^4*(c + d*x^4)^2), x]
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.
\(\displaystyle \int \frac {\left (a+b x^8\right )^p}{x^4 \left (c+d x^4\right )^2} \, dx\) |
\(\Big \downarrow \) 1888 |
\(\displaystyle \int \frac {\left (a+b x^8\right )^p}{x^4 \left (c+d x^4\right )^2}dx\) |
Input:
Int[(a + b*x^8)^p/(x^4*(c + d*x^4)^2),x]
Output:
$Aborted
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]
\[\int \frac {\left (b \,x^{8}+a \right )^{p}}{x^{4} \left (x^{4} d +c \right )^{2}}d x\]
Input:
int((b*x^8+a)^p/x^4/(d*x^4+c)^2,x)
Output:
int((b*x^8+a)^p/x^4/(d*x^4+c)^2,x)
\[ \int \frac {\left (a+b x^8\right )^p}{x^4 \left (c+d x^4\right )^2} \, dx=\int { \frac {{\left (b x^{8} + a\right )}^{p}}{{\left (d x^{4} + c\right )}^{2} x^{4}} \,d x } \] Input:
integrate((b*x^8+a)^p/x^4/(d*x^4+c)^2,x, algorithm="fricas")
Output:
integral((b*x^8 + a)^p/(d^2*x^12 + 2*c*d*x^8 + c^2*x^4), x)
Timed out. \[ \int \frac {\left (a+b x^8\right )^p}{x^4 \left (c+d x^4\right )^2} \, dx=\text {Timed out} \] Input:
integrate((b*x**8+a)**p/x**4/(d*x**4+c)**2,x)
Output:
Timed out
\[ \int \frac {\left (a+b x^8\right )^p}{x^4 \left (c+d x^4\right )^2} \, dx=\int { \frac {{\left (b x^{8} + a\right )}^{p}}{{\left (d x^{4} + c\right )}^{2} x^{4}} \,d x } \] Input:
integrate((b*x^8+a)^p/x^4/(d*x^4+c)^2,x, algorithm="maxima")
Output:
integrate((b*x^8 + a)^p/((d*x^4 + c)^2*x^4), x)
\[ \int \frac {\left (a+b x^8\right )^p}{x^4 \left (c+d x^4\right )^2} \, dx=\int { \frac {{\left (b x^{8} + a\right )}^{p}}{{\left (d x^{4} + c\right )}^{2} x^{4}} \,d x } \] Input:
integrate((b*x^8+a)^p/x^4/(d*x^4+c)^2,x, algorithm="giac")
Output:
integrate((b*x^8 + a)^p/((d*x^4 + c)^2*x^4), x)
Timed out. \[ \int \frac {\left (a+b x^8\right )^p}{x^4 \left (c+d x^4\right )^2} \, dx=\int \frac {{\left (b\,x^8+a\right )}^p}{x^4\,{\left (d\,x^4+c\right )}^2} \,d x \] Input:
int((a + b*x^8)^p/(x^4*(c + d*x^4)^2),x)
Output:
int((a + b*x^8)^p/(x^4*(c + d*x^4)^2), x)
\[ \int \frac {\left (a+b x^8\right )^p}{x^4 \left (c+d x^4\right )^2} \, dx=\int \frac {\left (b \,x^{8}+a \right )^{p}}{d^{2} x^{12}+2 c d \,x^{8}+c^{2} x^{4}}d x \] Input:
int((b*x^8+a)^p/x^4/(d*x^4+c)^2,x)
Output:
int((a + b*x**8)**p/(c**2*x**4 + 2*c*d*x**8 + d**2*x**12),x)