\(\int \frac {(a+b x^8)^p}{(c+d x^4)^3} \, dx\) [308]

Optimal result

Integrand size = 19, antiderivative size = 255 \[ \int \frac {\left (a+b x^8\right )^p}{\left (c+d x^4\right )^3} \, dx=\frac {x \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{8},-p,3,\frac {9}{8},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{c^3}-\frac {3 d x^5 \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{8},-p,3,\frac {13}{8},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{5 c^4}+\frac {d^2 x^9 \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {9}{8},-p,3,\frac {17}{8},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{3 c^5}-\frac {d^3 x^{13} \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {13}{8},-p,3,\frac {21}{8},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{13 c^6} \] Output:

x*(b*x^8+a)^p*AppellF1(1/8,3,-p,9/8,d^2*x^8/c^2,-b*x^8/a)/c^3/((1+b*x^8/a) 
^p)-3/5*d*x^5*(b*x^8+a)^p*AppellF1(5/8,3,-p,13/8,d^2*x^8/c^2,-b*x^8/a)/c^4 
/((1+b*x^8/a)^p)+1/3*d^2*x^9*(b*x^8+a)^p*AppellF1(9/8,3,-p,17/8,d^2*x^8/c^ 
2,-b*x^8/a)/c^5/((1+b*x^8/a)^p)-1/13*d^3*x^13*(b*x^8+a)^p*AppellF1(13/8,3, 
-p,21/8,d^2*x^8/c^2,-b*x^8/a)/c^6/((1+b*x^8/a)^p)
 

Mathematica [F]

\[ \int \frac {\left (a+b x^8\right )^p}{\left (c+d x^4\right )^3} \, dx=\int \frac {\left (a+b x^8\right )^p}{\left (c+d x^4\right )^3} \, dx \] Input:

Integrate[(a + b*x^8)^p/(c + d*x^4)^3,x]
 

Output:

Integrate[(a + b*x^8)^p/(c + d*x^4)^3, x]
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {1768, 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.

\(\displaystyle \int \frac {\left (a+b x^8\right )^p}{\left (c+d x^4\right )^3} \, dx\)

\(\Big \downarrow \) 1768

\(\displaystyle \int \left (\frac {3 c d^2 x^8 \left (a+b x^8\right )^p}{\left (c^2-d^2 x^8\right )^3}-\frac {3 c^2 d x^4 \left (a+b x^8\right )^p}{\left (c^2-d^2 x^8\right )^3}+\frac {d^3 x^{12} \left (a+b x^8\right )^p}{\left (d^2 x^8-c^2\right )^3}+\frac {c^3 \left (a+b x^8\right )^p}{\left (c^2-d^2 x^8\right )^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {d^3 x^{13} \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {13}{8},-p,3,\frac {21}{8},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{13 c^6}+\frac {d^2 x^9 \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {9}{8},-p,3,\frac {17}{8},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{3 c^5}-\frac {3 d x^5 \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{8},-p,3,\frac {13}{8},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{5 c^4}+\frac {x \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{8},-p,3,\frac {9}{8},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{c^3}\)

Input:

Int[(a + b*x^8)^p/(c + d*x^4)^3,x]
 

Output:

(x*(a + b*x^8)^p*AppellF1[1/8, -p, 3, 9/8, -((b*x^8)/a), (d^2*x^8)/c^2])/( 
c^3*(1 + (b*x^8)/a)^p) - (3*d*x^5*(a + b*x^8)^p*AppellF1[5/8, -p, 3, 13/8, 
 -((b*x^8)/a), (d^2*x^8)/c^2])/(5*c^4*(1 + (b*x^8)/a)^p) + (d^2*x^9*(a + b 
*x^8)^p*AppellF1[9/8, -p, 3, 17/8, -((b*x^8)/a), (d^2*x^8)/c^2])/(3*c^5*(1 
 + (b*x^8)/a)^p) - (d^3*x^13*(a + b*x^8)^p*AppellF1[13/8, -p, 3, 21/8, -(( 
b*x^8)/a), (d^2*x^8)/c^2])/(13*c^6*(1 + (b*x^8)/a)^p)
 

Defintions of rubi rules used

rule 1768

Int[((d_) + (e_.)*(x_)^(n_))^(q_)*((a_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] 
 :> Int[ExpandIntegrand[(a + c*x^(2*n))^p, (d/(d^2 - e^2*x^(2*n)) - e*(x^n/ 
(d^2 - e^2*x^(2*n))))^(-q), x], x] /; FreeQ[{a, c, d, e, n, p}, x] && EqQ[n 
2, 2*n] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[q, 0]
 

rule 2009

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

Maple [F]

\[\int \frac {\left (b \,x^{8}+a \right )^{p}}{\left (x^{4} d +c \right )^{3}}d x\]

Input:

int((b*x^8+a)^p/(d*x^4+c)^3,x)
 

Output:

int((b*x^8+a)^p/(d*x^4+c)^3,x)
 

Fricas [F]

\[ \int \frac {\left (a+b x^8\right )^p}{\left (c+d x^4\right )^3} \, dx=\int { \frac {{\left (b x^{8} + a\right )}^{p}}{{\left (d x^{4} + c\right )}^{3}} \,d x } \] Input:

integrate((b*x^8+a)^p/(d*x^4+c)^3,x, algorithm="fricas")
 

Output:

integral((b*x^8 + a)^p/(d^3*x^12 + 3*c*d^2*x^8 + 3*c^2*d*x^4 + c^3), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^8\right )^p}{\left (c+d x^4\right )^3} \, dx=\text {Timed out} \] Input:

integrate((b*x**8+a)**p/(d*x**4+c)**3,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (a+b x^8\right )^p}{\left (c+d x^4\right )^3} \, dx=\int { \frac {{\left (b x^{8} + a\right )}^{p}}{{\left (d x^{4} + c\right )}^{3}} \,d x } \] Input:

integrate((b*x^8+a)^p/(d*x^4+c)^3,x, algorithm="maxima")
 

Output:

integrate((b*x^8 + a)^p/(d*x^4 + c)^3, x)
 

Giac [F]

\[ \int \frac {\left (a+b x^8\right )^p}{\left (c+d x^4\right )^3} \, dx=\int { \frac {{\left (b x^{8} + a\right )}^{p}}{{\left (d x^{4} + c\right )}^{3}} \,d x } \] Input:

integrate((b*x^8+a)^p/(d*x^4+c)^3,x, algorithm="giac")
 

Output:

integrate((b*x^8 + a)^p/(d*x^4 + c)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^8\right )^p}{\left (c+d x^4\right )^3} \, dx=\int \frac {{\left (b\,x^8+a\right )}^p}{{\left (d\,x^4+c\right )}^3} \,d x \] Input:

int((a + b*x^8)^p/(c + d*x^4)^3,x)
 

Output:

int((a + b*x^8)^p/(c + d*x^4)^3, x)
 

Reduce [F]

\[ \int \frac {\left (a+b x^8\right )^p}{\left (c+d x^4\right )^3} \, dx=\int \frac {\left (b \,x^{8}+a \right )^{p}}{d^{3} x^{12}+3 c \,d^{2} x^{8}+3 c^{2} d \,x^{4}+c^{3}}d x \] Input:

int((b*x^8+a)^p/(d*x^4+c)^3,x)
 

Output:

int((a + b*x**8)**p/(c**3 + 3*c**2*d*x**4 + 3*c*d**2*x**8 + d**3*x**12),x)