\(\int \frac {x^3 (a+b x^8)^p}{c+d x^4} \, dx\) [25]

Optimal result

Integrand size = 22, antiderivative size = 130 \[ \int \frac {x^3 \left (a+b x^8\right )^p}{c+d x^4} \, dx=\frac {x^4 \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{4 c}-\frac {d \left (a+b x^8\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {d^2 \left (a+b x^8\right )}{b c^2+a d^2}\right )}{8 \left (b c^2+a d^2\right ) (1+p)} \] Output:

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

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.10 \[ \int \frac {x^3 \left (a+b x^8\right )^p}{c+d x^4} \, dx=\frac {\left (\frac {d \left (-\sqrt {-\frac {a}{b}}+x^4\right )}{c+d x^4}\right )^{-p} \left (\frac {d \left (\sqrt {-\frac {a}{b}}+x^4\right )}{c+d x^4}\right )^{-p} \left (a+b x^8\right )^p \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {c-\sqrt {-\frac {a}{b}} d}{c+d x^4},\frac {c+\sqrt {-\frac {a}{b}} d}{c+d x^4}\right )}{8 d p} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1799, 504, 334, 333, 353, 78}

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.

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

\(\Big \downarrow \) 1799

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

\(\Big \downarrow \) 504

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

\(\Big \downarrow \) 334

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

\(\Big \downarrow \) 333

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

\(\Big \downarrow \) 353

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

\(\Big \downarrow \) 78

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 78

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b 
*c - a*d)^n*((a + b*x)^(m + 1)/(b^(n + 1)*(m + 1)))*Hypergeometric2F1[-n, m 
 + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m}, x] 
 &&  !IntegerQ[m] && IntegerQ[n]
 

rule 333

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F 
reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 
0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 334

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[ 
(1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, p, q}, x] && 
NeQ[b*c - a*d, 0] &&  !(IntegerQ[p] || GtQ[a, 0])
 

rule 353

Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] 
 :> Simp[1/2   Subst[Int[(a + b*x)^p*(c + d*x)^q, x], x, x^2], x] /; FreeQ[ 
{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0]
 

rule 504

Int[((a_) + (b_.)*(x_)^2)^(p_)/((c_) + (d_.)*(x_)), x_Symbol] :> Simp[c   I 
nt[(a + b*x^2)^p/(c^2 - d^2*x^2), x], x] - Simp[d   Int[x*((a + b*x^2)^p/(c 
^2 - d^2*x^2)), x], x] /; FreeQ[{a, b, c, d, p}, x]
 

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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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