\(\int \frac {(c+d x^4)^2 (a+b x^8)^p}{x^3} \, dx\) [16]

Optimal result

Integrand size = 22, antiderivative size = 155 \[ \int \frac {\left (c+d x^4\right )^2 \left (a+b x^8\right )^p}{x^3} \, dx=\frac {d^2 \left (a+b x^8\right )^{1+p}}{2 b (3+4 p) x^2}-\frac {\left (a d^2+b c^2 (3+4 p)\right ) \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},-p,\frac {3}{4},-\frac {b x^8}{a}\right )}{2 b (3+4 p) x^2}+c d x^2 \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^8}{a}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.69 \[ \int \frac {\left (c+d x^4\right )^2 \left (a+b x^8\right )^p}{x^3} \, dx=\frac {\left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \left (-3 c^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},-p,\frac {3}{4},-\frac {b x^8}{a}\right )+d x^4 \left (6 c \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^8}{a}\right )+d x^4 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b x^8}{a}\right )\right )\right )}{6 x^2} \] Input:

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

Output:

((a + b*x^8)^p*(-3*c^2*Hypergeometric2F1[-1/4, -p, 3/4, -((b*x^8)/a)] + d* 
x^4*(6*c*Hypergeometric2F1[1/4, -p, 5/4, -((b*x^8)/a)] + d*x^4*Hypergeomet 
ric2F1[3/4, -p, 7/4, -((b*x^8)/a)])))/(6*x^2*(1 + (b*x^8)/a)^p)
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.01, 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.

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

\(\Big \downarrow \) 1815

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

\(\Big \downarrow \) 1675

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(-((c^2*(a + b*x^8)^p*Hypergeometric2F1[-1/4, -p, 3/4, -((b*x^8)/a)])/(x^2 
*(1 + (b*x^8)/a)^p)) + (2*c*d*x^2*(a + b*x^8)^p*Hypergeometric2F1[1/4, -p, 
 5/4, -((b*x^8)/a)])/(1 + (b*x^8)/a)^p + (d^2*x^6*(a + b*x^8)^p*Hypergeome 
tric2F1[3/4, -p, 7/4, -((b*x^8)/a)])/(3*(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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(16*(a + b*x**8)**p*a*d**2*p**2 + 4*(a + b*x**8)**p*a*d**2*p + 16*(a + b*x 
**8)**p*b*c**2*p**2 + 16*(a + b*x**8)**p*b*c**2*p + 3*(a + b*x**8)**p*b*c* 
*2 + 32*(a + b*x**8)**p*b*c*d*p**2*x**4 + 16*(a + b*x**8)**p*b*c*d*p*x**4 
- 6*(a + b*x**8)**p*b*c*d*x**4 + 16*(a + b*x**8)**p*b*d**2*p**2*x**8 - (a 
+ b*x**8)**p*b*d**2*x**8 + 2048*int((a + b*x**8)**p/(64*a*p**3*x**3 + 48*a 
*p**2*x**3 - 4*a*p*x**3 - 3*a*x**3 + 64*b*p**3*x**11 + 48*b*p**2*x**11 - 4 
*b*p*x**11 - 3*b*x**11),x)*a**2*d**2*p**5*x**2 + 2048*int((a + b*x**8)**p/ 
(64*a*p**3*x**3 + 48*a*p**2*x**3 - 4*a*p*x**3 - 3*a*x**3 + 64*b*p**3*x**11 
 + 48*b*p**2*x**11 - 4*b*p*x**11 - 3*b*x**11),x)*a**2*d**2*p**4*x**2 + 256 
*int((a + b*x**8)**p/(64*a*p**3*x**3 + 48*a*p**2*x**3 - 4*a*p*x**3 - 3*a*x 
**3 + 64*b*p**3*x**11 + 48*b*p**2*x**11 - 4*b*p*x**11 - 3*b*x**11),x)*a**2 
*d**2*p**3*x**2 - 128*int((a + b*x**8)**p/(64*a*p**3*x**3 + 48*a*p**2*x**3 
 - 4*a*p*x**3 - 3*a*x**3 + 64*b*p**3*x**11 + 48*b*p**2*x**11 - 4*b*p*x**11 
 - 3*b*x**11),x)*a**2*d**2*p**2*x**2 - 24*int((a + b*x**8)**p/(64*a*p**3*x 
**3 + 48*a*p**2*x**3 - 4*a*p*x**3 - 3*a*x**3 + 64*b*p**3*x**11 + 48*b*p**2 
*x**11 - 4*b*p*x**11 - 3*b*x**11),x)*a**2*d**2*p*x**2 + 8192*int((a + b*x* 
*8)**p/(64*a*p**3*x**3 + 48*a*p**2*x**3 - 4*a*p*x**3 - 3*a*x**3 + 64*b*p** 
3*x**11 + 48*b*p**2*x**11 - 4*b*p*x**11 - 3*b*x**11),x)*a*b*c**2*p**6*x**2 
 + 14336*int((a + b*x**8)**p/(64*a*p**3*x**3 + 48*a*p**2*x**3 - 4*a*p*x**3 
 - 3*a*x**3 + 64*b*p**3*x**11 + 48*b*p**2*x**11 - 4*b*p*x**11 - 3*b*x**...