\(\int \frac {x^3 (a+b x^8)^p}{(c+d x^4)^2} \, dx\) [37]

Optimal result

Integrand size = 22, antiderivative size = 198 \[ \int \frac {x^3 \left (a+b x^8\right )^p}{\left (c+d x^4\right )^2} \, dx=\frac {x^4 \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,2,\frac {3}{2},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{4 c^2}+\frac {d^2 x^{12} \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-p,2,\frac {5}{2},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{12 c^4}-\frac {b c d \left (a+b x^8\right )^{1+p} \operatorname {Hypergeometric2F1}\left (2,1+p,2+p,\frac {d^2 \left (a+b x^8\right )}{b c^2+a d^2}\right )}{4 \left (b c^2+a d^2\right )^2 (1+p)} \] Output:

1/4*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^2/((1+b* 
x^8/a)^p)+1/12*d^2*x^12*(b*x^8+a)^p*AppellF1(3/2,2,-p,5/2,d^2*x^8/c^2,-b*x 
^8/a)/c^4/((1+b*x^8/a)^p)-1/4*b*c*d*(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 [A] (verified)

Time = 0.70 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.80 \[ \int \frac {x^3 \left (a+b x^8\right )^p}{\left (c+d x^4\right )^2} \, dx=\frac {\left (\frac {d \left (-\sqrt {-\frac {a}{b}}+x^4\right )}{c+d x^4}\right )^{-p} \left (\frac {d \left (\sqrt {-\frac {a}{b}}+x^4\right )}{c+d x^4}\right )^{-p} \left (a+b x^8\right )^p \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,\frac {c-\sqrt {-\frac {a}{b}} d}{c+d x^4},\frac {c+\sqrt {-\frac {a}{b}} d}{c+d x^4}\right )}{4 d (-1+2 p) \left (c+d x^4\right )} \] Input:

Integrate[(x^3*(a + b*x^8)^p)/(c + d*x^4)^2,x]
 

Output:

((a + b*x^8)^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)])/(4*d*(-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))
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1799, 505, 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.

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

\(\Big \downarrow \) 1799

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

\(\Big \downarrow \) 505

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{4} \left (\frac {x^4 \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,2,\frac {3}{2},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{c^2}+\frac {d^2 x^{12} \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-p,2,\frac {5}{2},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{3 c^4}-\frac {b c d \left (a+b x^8\right )^{p+1} \operatorname {Hypergeometric2F1}\left (2,p+1,p+2,\frac {d^2 \left (b x^8+a\right )}{b c^2+a d^2}\right )}{(p+1) \left (a d^2+b c^2\right )^2}\right )\)

Input:

Int[(x^3*(a + b*x^8)^p)/(c + d*x^4)^2,x]
 

Output:

((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^2*(1 + (b*x^8)/a)^p) + (d^2*x^12*(a + b*x^8)^p*AppellF1[3/2, -p, 2, 5 
/2, -((b*x^8)/a), (d^2*x^8)/c^2])/(3*c^4*(1 + (b*x^8)/a)^p) - (b*c*d*(a + 
b*x^8)^(1 + p)*Hypergeometric2F1[2, 1 + p, 2 + p, (d^2*(a + b*x^8))/(b*c^2 
 + a*d^2)])/((b*c^2 + a*d^2)^2*(1 + p)))/4
 

Defintions of rubi rules used

rule 505

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[E 
xpandIntegrand[(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, p}, x] && ILtQ[n, -1] && PosQ[a/b]
 

rule 1799

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q 
_.), x_Symbol] :> Simp[1/n   Subst[Int[(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] && EqQ[Simplif 
y[m - n + 1], 0]
 

rule 2009

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

integrate(x^3*(b*x^8+a)^p/(d*x^4+c)^2,x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate(x^3*(b*x^8+a)^p/(d*x^4+c)^2,x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate(x^3*(b*x^8+a)^p/(d*x^4+c)^2,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

((a + b*x**8)**p*x**4 - 16*int(((a + b*x**8)**p*x**15)/(2*a*c**2*p + a*c** 
2 + 4*a*c*d*p*x**4 + 2*a*c*d*x**4 + 2*a*d**2*p*x**8 + a*d**2*x**8 + 2*b*c* 
*2*p*x**8 + b*c**2*x**8 + 4*b*c*d*p*x**12 + 2*b*c*d*x**12 + 2*b*d**2*p*x** 
16 + b*d**2*x**16),x)*b*c*d*p**2 - 8*int(((a + b*x**8)**p*x**15)/(2*a*c**2 
*p + a*c**2 + 4*a*c*d*p*x**4 + 2*a*c*d*x**4 + 2*a*d**2*p*x**8 + a*d**2*x** 
8 + 2*b*c**2*p*x**8 + b*c**2*x**8 + 4*b*c*d*p*x**12 + 2*b*c*d*x**12 + 2*b* 
d**2*p*x**16 + b*d**2*x**16),x)*b*c*d*p - 16*int(((a + b*x**8)**p*x**15)/( 
2*a*c**2*p + a*c**2 + 4*a*c*d*p*x**4 + 2*a*c*d*x**4 + 2*a*d**2*p*x**8 + a* 
d**2*x**8 + 2*b*c**2*p*x**8 + b*c**2*x**8 + 4*b*c*d*p*x**12 + 2*b*c*d*x**1 
2 + 2*b*d**2*p*x**16 + b*d**2*x**16),x)*b*d**2*p**2*x**4 - 8*int(((a + b*x 
**8)**p*x**15)/(2*a*c**2*p + a*c**2 + 4*a*c*d*p*x**4 + 2*a*c*d*x**4 + 2*a* 
d**2*p*x**8 + a*d**2*x**8 + 2*b*c**2*p*x**8 + b*c**2*x**8 + 4*b*c*d*p*x**1 
2 + 2*b*c*d*x**12 + 2*b*d**2*p*x**16 + b*d**2*x**16),x)*b*d**2*p*x**4 + 16 
*int(((a + b*x**8)**p*x**3)/(2*a*c**2*p + a*c**2 + 4*a*c*d*p*x**4 + 2*a*c* 
d*x**4 + 2*a*d**2*p*x**8 + a*d**2*x**8 + 2*b*c**2*p*x**8 + b*c**2*x**8 + 4 
*b*c*d*p*x**12 + 2*b*c*d*x**12 + 2*b*d**2*p*x**16 + b*d**2*x**16),x)*a*c** 
2*p**2 + 8*int(((a + b*x**8)**p*x**3)/(2*a*c**2*p + a*c**2 + 4*a*c*d*p*x** 
4 + 2*a*c*d*x**4 + 2*a*d**2*p*x**8 + a*d**2*x**8 + 2*b*c**2*p*x**8 + b*c** 
2*x**8 + 4*b*c*d*p*x**12 + 2*b*c*d*x**12 + 2*b*d**2*p*x**16 + b*d**2*x**16 
),x)*a*c**2*p + 16*int(((a + b*x**8)**p*x**3)/(2*a*c**2*p + a*c**2 + 4*...