\(\int \frac {(c+d x^4)^2 (a+b x^8)^p}{x^4} \, dx\) [22]

Optimal result

Integrand size = 22, antiderivative size = 152 \[ \int \frac {\left (c+d x^4\right )^2 \left (a+b x^8\right )^p}{x^4} \, dx=\frac {d^2 \left (a+b x^8\right )^{1+p}}{b (5+8 p) x^3}-\frac {\left (3 a d^2+b c^2 (5+8 p)\right ) \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {3}{8},-p,\frac {5}{8},-\frac {b x^8}{a}\right )}{3 b (5+8 p) x^3}+2 c d x \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},-p,\frac {9}{8},-\frac {b x^8}{a}\right ) \] Output:

d^2*(b*x^8+a)^(p+1)/b/(5+8*p)/x^3-1/3*(3*a*d^2+b*c^2*(5+8*p))*(b*x^8+a)^p* 
hypergeom([-3/8, -p],[5/8],-b*x^8/a)/b/(5+8*p)/x^3/((1+b*x^8/a)^p)+2*c*d*x 
*(b*x^8+a)^p*hypergeom([1/8, -p],[9/8],-b*x^8/a)/((1+b*x^8/a)^p)
 

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.71 \[ \int \frac {\left (c+d x^4\right )^2 \left (a+b x^8\right )^p}{x^4} \, dx=\frac {\left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \left (-5 c^2 \operatorname {Hypergeometric2F1}\left (-\frac {3}{8},-p,\frac {5}{8},-\frac {b x^8}{a}\right )+3 d x^4 \left (10 c \operatorname {Hypergeometric2F1}\left (\frac {1}{8},-p,\frac {9}{8},-\frac {b x^8}{a}\right )+d x^4 \operatorname {Hypergeometric2F1}\left (\frac {5}{8},-p,\frac {13}{8},-\frac {b x^8}{a}\right )\right )\right )}{15 x^3} \] Input:

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

Output:

((a + b*x^8)^p*(-5*c^2*Hypergeometric2F1[-3/8, -p, 5/8, -((b*x^8)/a)] + 3* 
d*x^4*(10*c*Hypergeometric2F1[1/8, -p, 9/8, -((b*x^8)/a)] + d*x^4*Hypergeo 
metric2F1[5/8, -p, 13/8, -((b*x^8)/a)])))/(15*x^3*(1 + (b*x^8)/a)^p)
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1865, 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.

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

\(\Big \downarrow \) 1865

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {c^2 \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {3}{8},-p,\frac {5}{8},-\frac {b x^8}{a}\right )}{3 x^3}+2 c d x \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},-p,\frac {9}{8},-\frac {b x^8}{a}\right )+\frac {1}{5} d^2 x^5 \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{8},-p,\frac {13}{8},-\frac {b x^8}{a}\right )\)

Input:

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

Output:

-1/3*(c^2*(a + b*x^8)^p*Hypergeometric2F1[-3/8, -p, 5/8, -((b*x^8)/a)])/(x 
^3*(1 + (b*x^8)/a)^p) + (2*c*d*x*(a + b*x^8)^p*Hypergeometric2F1[1/8, -p, 
9/8, -((b*x^8)/a)])/(1 + (b*x^8)/a)^p + (d^2*x^5*(a + b*x^8)^p*Hypergeomet 
ric2F1[5/8, -p, 13/8, -((b*x^8)/a)])/(5*(1 + (b*x^8)/a)^p)
 

Defintions of rubi rules used

rule 1865

Int[((f_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^ 
(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^(2*n))^p, (f*x)^m*(d 
 + e*x^n)^q, x], x] /; FreeQ[{a, c, d, e, f, m, q}, x] && EqQ[n2, 2*n] && I 
GtQ[n, 0] && IGtQ[q, 0]
 

rule 2009

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(64*(a + b*x**8)**p*a*d**2*p**2 + 8*(a + b*x**8)**p*a*d**2*p + 64*(a + b*x 
**8)**p*b*c**2*p**2 + 48*(a + b*x**8)**p*b*c**2*p + 5*(a + b*x**8)**p*b*c* 
*2 + 128*(a + b*x**8)**p*b*c*d*p**2*x**4 + 32*(a + b*x**8)**p*b*c*d*p*x**4 
 - 30*(a + b*x**8)**p*b*c*d*x**4 + 64*(a + b*x**8)**p*b*d**2*p**2*x**8 - 1 
6*(a + b*x**8)**p*b*d**2*p*x**8 - 3*(a + b*x**8)**p*b*d**2*x**8 + 98304*in 
t((a + b*x**8)**p/(512*a*p**3*x**4 + 192*a*p**2*x**4 - 104*a*p*x**4 - 15*a 
*x**4 + 512*b*p**3*x**12 + 192*b*p**2*x**12 - 104*b*p*x**12 - 15*b*x**12), 
x)*a**2*d**2*p**5*x**3 + 49152*int((a + b*x**8)**p/(512*a*p**3*x**4 + 192* 
a*p**2*x**4 - 104*a*p*x**4 - 15*a*x**4 + 512*b*p**3*x**12 + 192*b*p**2*x** 
12 - 104*b*p*x**12 - 15*b*x**12),x)*a**2*d**2*p**4*x**3 - 15360*int((a + b 
*x**8)**p/(512*a*p**3*x**4 + 192*a*p**2*x**4 - 104*a*p*x**4 - 15*a*x**4 + 
512*b*p**3*x**12 + 192*b*p**2*x**12 - 104*b*p*x**12 - 15*b*x**12),x)*a**2* 
d**2*p**3*x**3 - 5376*int((a + b*x**8)**p/(512*a*p**3*x**4 + 192*a*p**2*x* 
*4 - 104*a*p*x**4 - 15*a*x**4 + 512*b*p**3*x**12 + 192*b*p**2*x**12 - 104* 
b*p*x**12 - 15*b*x**12),x)*a**2*d**2*p**2*x**3 - 360*int((a + b*x**8)**p/( 
512*a*p**3*x**4 + 192*a*p**2*x**4 - 104*a*p*x**4 - 15*a*x**4 + 512*b*p**3* 
x**12 + 192*b*p**2*x**12 - 104*b*p*x**12 - 15*b*x**12),x)*a**2*d**2*p*x**3 
 + 262144*int((a + b*x**8)**p/(512*a*p**3*x**4 + 192*a*p**2*x**4 - 104*a*p 
*x**4 - 15*a*x**4 + 512*b*p**3*x**12 + 192*b*p**2*x**12 - 104*b*p*x**12 - 
15*b*x**12),x)*a*b*c**2*p**6*x**3 + 294912*int((a + b*x**8)**p/(512*a*p...