\(\int \frac {(c+d x^4)^2 (a+b x^8)^p}{x} \, dx\) [15]

Optimal result

Integrand size = 22, antiderivative size = 122 \[ \int \frac {\left (c+d x^4\right )^2 \left (a+b x^8\right )^p}{x} \, dx=\frac {d^2 \left (a+b x^8\right )^{1+p}}{8 b (1+p)}+\frac {1}{2} c d x^4 \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^8}{a}\right )-\frac {c^2 \left (a+b x^8\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b x^8}{a}\right )}{8 a (1+p)} \] Output:

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

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.84 \[ \int \frac {\left (c+d x^4\right )^2 \left (a+b x^8\right )^p}{x} \, dx=\frac {1}{8} \left (a+b x^8\right )^p \left (4 c d x^4 \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^8}{a}\right )+\frac {\left (a+b x^8\right ) \left (a d^2-b c^2 \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b x^8}{a}\right )\right )}{a b (1+p)}\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 124, 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, 543, 27, 238, 237, 354, 90, 75}

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.

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

\(\Big \downarrow \) 1803

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

\(\Big \downarrow \) 543

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 238

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

\(\Big \downarrow \) 237

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

\(\Big \downarrow \) 354

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 75

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 0]
 

rule 237

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[- 
p, 1/2, 1/2 + 1, (-b)*(x^2/a)], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*p 
] && GtQ[a, 0]
 

rule 238

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2) 
^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(1 + b*(x^2/a))^p, x], x] / 
; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*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 543

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> Module[{k}, Int[x^m*Sum[Binomial[n, 2*k]*c^(n - 2*k)*d^(2*k)*x^(2*k), 
 {k, 0, n/2}]*(a + b*x^2)^p, x] + Int[x^(m + 1)*Sum[Binomial[n, 2*k + 1]*c^ 
(n - 2*k - 1)*d^(2*k + 1)*x^(2*k), {k, 0, (n - 1)/2}]*(a + b*x^2)^p, x]] /; 
 FreeQ[{a, b, c, d, p}, x] && IGtQ[n, 1] && IntegerQ[m] &&  !IntegerQ[2*p] 
&&  !(EqQ[m, 1] && EqQ[b*c^2 + a*d^2, 0])
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [A] (verification not implemented)

Time = 85.05 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.90 \[ \int \frac {\left (c+d x^4\right )^2 \left (a+b x^8\right )^p}{x} \, dx=\frac {a^{p} c d x^{4} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{8} e^{i \pi }}{a}} \right )}}{2} - \frac {b^{p} c^{2} x^{8 p} \Gamma \left (- p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, - p \\ 1 - p \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{8}}} \right )}}{8 \Gamma \left (1 - p\right )} + d^{2} \left (\begin {cases} \frac {a^{p} x^{8}}{8} & \text {for}\: b = 0 \\\frac {\begin {cases} \frac {\left (a + b x^{8}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (a + b x^{8} \right )} & \text {otherwise} \end {cases}}{8 b} & \text {otherwise} \end {cases}\right ) \] Input:

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

Output:

a**p*c*d*x**4*hyper((1/2, -p), (3/2,), b*x**8*exp_polar(I*pi)/a)/2 - b**p* 
c**2*x**(8*p)*gamma(-p)*hyper((-p, -p), (1 - p,), a*exp_polar(I*pi)/(b*x** 
8))/(8*gamma(1 - p)) + d**2*Piecewise((a**p*x**8/8, Eq(b, 0)), (Piecewise( 
((a + b*x**8)**(p + 1)/(p + 1), Ne(p, -1)), (log(a + b*x**8), True))/(8*b) 
, True))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (c+d x^4\right )^2 \left (a+b x^8\right )^p}{x} \, dx=\frac {2 \left (b \,x^{8}+a \right )^{p} a \,d^{2} p^{2}+\left (b \,x^{8}+a \right )^{p} a \,d^{2} p +2 \left (b \,x^{8}+a \right )^{p} b \,c^{2} p^{2}+3 \left (b \,x^{8}+a \right )^{p} b \,c^{2} p +\left (b \,x^{8}+a \right )^{p} b \,c^{2}+4 \left (b \,x^{8}+a \right )^{p} b c d \,p^{2} x^{4}+4 \left (b \,x^{8}+a \right )^{p} b c d p \,x^{4}+2 \left (b \,x^{8}+a \right )^{p} b \,d^{2} p^{2} x^{8}+\left (b \,x^{8}+a \right )^{p} b \,d^{2} p \,x^{8}+32 \left (\int \frac {\left (b \,x^{8}+a \right )^{p}}{2 b p \,x^{9}+b \,x^{9}+2 a p x +a x}d x \right ) a b \,c^{2} p^{4}+64 \left (\int \frac {\left (b \,x^{8}+a \right )^{p}}{2 b p \,x^{9}+b \,x^{9}+2 a p x +a x}d x \right ) a b \,c^{2} p^{3}+40 \left (\int \frac {\left (b \,x^{8}+a \right )^{p}}{2 b p \,x^{9}+b \,x^{9}+2 a p x +a x}d x \right ) a b \,c^{2} p^{2}+8 \left (\int \frac {\left (b \,x^{8}+a \right )^{p}}{2 b p \,x^{9}+b \,x^{9}+2 a p x +a x}d x \right ) a b \,c^{2} p +64 \left (\int \frac {\left (b \,x^{8}+a \right )^{p} x^{3}}{2 b p \,x^{8}+b \,x^{8}+2 a p +a}d x \right ) a b c d \,p^{4}+96 \left (\int \frac {\left (b \,x^{8}+a \right )^{p} x^{3}}{2 b p \,x^{8}+b \,x^{8}+2 a p +a}d x \right ) a b c d \,p^{3}+32 \left (\int \frac {\left (b \,x^{8}+a \right )^{p} x^{3}}{2 b p \,x^{8}+b \,x^{8}+2 a p +a}d x \right ) a b c d \,p^{2}}{8 b p \left (2 p^{2}+3 p +1\right )} \] Input:

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

Output:

(2*(a + b*x**8)**p*a*d**2*p**2 + (a + b*x**8)**p*a*d**2*p + 2*(a + b*x**8) 
**p*b*c**2*p**2 + 3*(a + b*x**8)**p*b*c**2*p + (a + b*x**8)**p*b*c**2 + 4* 
(a + b*x**8)**p*b*c*d*p**2*x**4 + 4*(a + b*x**8)**p*b*c*d*p*x**4 + 2*(a + 
b*x**8)**p*b*d**2*p**2*x**8 + (a + b*x**8)**p*b*d**2*p*x**8 + 32*int((a + 
b*x**8)**p/(2*a*p*x + a*x + 2*b*p*x**9 + b*x**9),x)*a*b*c**2*p**4 + 64*int 
((a + b*x**8)**p/(2*a*p*x + a*x + 2*b*p*x**9 + b*x**9),x)*a*b*c**2*p**3 + 
40*int((a + b*x**8)**p/(2*a*p*x + a*x + 2*b*p*x**9 + b*x**9),x)*a*b*c**2*p 
**2 + 8*int((a + b*x**8)**p/(2*a*p*x + a*x + 2*b*p*x**9 + b*x**9),x)*a*b*c 
**2*p + 64*int(((a + b*x**8)**p*x**3)/(2*a*p + a + 2*b*p*x**8 + b*x**8),x) 
*a*b*c*d*p**4 + 96*int(((a + b*x**8)**p*x**3)/(2*a*p + a + 2*b*p*x**8 + b* 
x**8),x)*a*b*c*d*p**3 + 32*int(((a + b*x**8)**p*x**3)/(2*a*p + a + 2*b*p*x 
**8 + b*x**8),x)*a*b*c*d*p**2)/(8*b*p*(2*p**2 + 3*p + 1))