\(\int \frac {(a+b x^8)^p}{x (c+d x^4)} \, dx\) [27]

Optimal result

Integrand size = 22, antiderivative size = 180 \[ \int \frac {\left (a+b x^8\right )^p}{x \left (c+d x^4\right )} \, dx=-\frac {d x^4 \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{4 c^2}+\frac {d^2 \left (a+b x^8\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {d^2 \left (a+b x^8\right )}{b c^2+a d^2}\right )}{8 c \left (b c^2+a d^2\right ) (1+p)}-\frac {\left (a+b x^8\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b x^8}{a}\right )}{8 a c (1+p)} \] Output:

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

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {1803, 621, 334, 333, 354, 97, 75, 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.

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

\(\Big \downarrow \) 1803

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

\(\Big \downarrow \) 621

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

\(\Big \downarrow \) 334

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

\(\Big \downarrow \) 333

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 97

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

\(\Big \downarrow \) 75

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

\(\Big \downarrow \) 78

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

Input:

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

Output:

(-((d*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^2*(1 + (b*x^8)/a)^p)) + (c*((d^2*(a + b*x^8)^(1 + p)*Hypergeometr 
ic2F1[1, 1 + p, 2 + p, (d^2*(a + b*x^8))/(b*c^2 + a*d^2)])/(c^2*(b*c^2 + a 
*d^2)*(1 + p)) - ((a + b*x^8)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, 1 
 + (b*x^8)/a])/(a*c^2*(1 + p))))/2)/4
 

Defintions of rubi rules used

rule 75

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x 
)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))*Hypergeometric2F1[-m, n + 1, n + 2, 1 + 
 d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (IntegerQ[m] 
 || GtQ[-d/(b*c), 0])
 

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 97

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), 
x_] :> Simp[b/(b*c - a*d)   Int[(e + f*x)^p/(a + b*x), x], x] - Simp[d/(b*c 
 - a*d)   Int[(e + f*x)^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, p}, 
 x] &&  !IntegerQ[p]
 

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 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(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] && IntegerQ 
[(m - 1)/2]
 

rule 621

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

rule 1803

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((b*x^8 + a)^p/(d*x^5 + c*x), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+b x^8\right )^p}{x \left (c+d x^4\right )} \, dx=\int \frac {\left (b \,x^{8}+a \right )^{p}}{d \,x^{5}+c x}d x \] Input:

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

Output:

int((a + b*x**8)**p/(c*x + d*x**5),x)