\(\int \frac {(a+b x^8)^p}{x^3 (c+d x^4)} \, dx\) [28]

Optimal result

Integrand size = 22, antiderivative size = 128 \[ \int \frac {\left (a+b x^8\right )^p}{x^3 \left (c+d x^4\right )} \, dx=-\frac {\left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {AppellF1}\left (-\frac {1}{4},-p,1,\frac {3}{4},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{2 c x^2}-\frac {d x^2 \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{4},-p,1,\frac {5}{4},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{2 c^2} \] Output:

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

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.43, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1815, 1675, 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.

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

\(\Big \downarrow \) 1815

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

\(\Big \downarrow \) 1675

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (-\frac {d x^2 \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{4},-p,1,\frac {5}{4},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{c^2}+\frac {d^2 x^6 \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{4},-p,1,\frac {7}{4},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{3 c^3}-\frac {\left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},-p,\frac {3}{4},-\frac {b x^8}{a}\right )}{c x^2}\right )\)

Input:

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

Output:

(-((d*x^2*(a + b*x^8)^p*AppellF1[1/4, -p, 1, 5/4, -((b*x^8)/a), (d^2*x^8)/ 
c^2])/(c^2*(1 + (b*x^8)/a)^p)) + (d^2*x^6*(a + b*x^8)^p*AppellF1[3/4, -p, 
1, 7/4, -((b*x^8)/a), (d^2*x^8)/c^2])/(3*c^3*(1 + (b*x^8)/a)^p) - ((a + b* 
x^8)^p*Hypergeometric2F1[-1/4, -p, 3/4, -((b*x^8)/a)])/(c*x^2*(1 + (b*x^8) 
/a)^p))/2
 

Defintions of rubi rules used

rule 1675

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p 
_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + c*x^4)^p, 
x], x] /; FreeQ[{a, c, d, e, f, m, p, q}, x] && (IGtQ[p, 0] || IGtQ[q, 0] | 
| IntegersQ[m, q])
 

rule 1815

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_)*((d_) + (e_.)*(x_)^(n_))^(q_ 
.), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/ 
k - 1)*(d + e*x^(n/k))^q*(a + c*x^(2*(n/k)))^p, x], x, x^k], x] /; k != 1] 
/; FreeQ[{a, c, d, e, p, q}, x] && EqQ[n2, 2*n] && IGtQ[n, 0] && IntegerQ[m 
]
 

rule 2009

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((b*x^8 + a)^p/(d*x^7 + c*x^3), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+b x^8\right )^p}{x^3 \left (c+d x^4\right )} \, dx=\int \frac {\left (b \,x^{8}+a \right )^{p}}{d \,x^{7}+c \,x^{3}}d x \] Input:

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

Output:

int((a + b*x**8)**p/(c*x**3 + d*x**7),x)