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

Optimal result

Integrand size = 20, antiderivative size = 93 \[ \int \frac {\left (c+d x^4\right ) \left (a+b x^8\right )^p}{x} \, dx=\frac {1}{4} d x^4 \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^8}{a}\right )-\frac {c \left (a+b x^8\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b x^8}{a}\right )}{8 a (1+p)} \] Output:

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

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.96 \[ \int \frac {\left (c+d x^4\right ) \left (a+b x^8\right )^p}{x} \, dx=\frac {1}{8} \left (a+b x^8\right )^p \left (2 d x^4 \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^8}{a}\right )-\frac {c \left (a+b x^8\right ) \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b x^8}{a}\right )}{a (1+p)}\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1803, 542, 238, 237, 243, 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.

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

\(\Big \downarrow \) 1803

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

\(\Big \downarrow \) 542

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

\(\Big \downarrow \) 238

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

\(\Big \downarrow \) 237

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 75

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

Input:

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

Output:

((d*x^4*(a + b*x^8)^p*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^8)/a)])/(1 + 
(b*x^8)/a)^p - (c*(a + b*x^8)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, 1 
 + (b*x^8)/a])/(2*a*(1 + p)))/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 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 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 542

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

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 70.63 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.73 \[ \int \frac {\left (c+d x^4\right ) \left (a+b x^8\right )^p}{x} \, dx=\frac {a^{p} d x^{4} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{8} e^{i \pi }}{a}} \right )}}{4} - \frac {b^{p} c x^{8 p} \Gamma \left (- p\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, - p \\ 1 - p \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{8}}} \right )}}{8 \Gamma \left (1 - p\right )} \] Input:

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

Output:

a**p*d*x**4*hyper((1/2, -p), (3/2,), b*x**8*exp_polar(I*pi)/a)/4 - b**p*c* 
x**(8*p)*gamma(-p)*hyper((-p, -p), (1 - p,), a*exp_polar(I*pi)/(b*x**8))/( 
8*gamma(1 - p))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (c+d x^4\right ) \left (a+b x^8\right )^p}{x} \, dx=\frac {2 \left (b \,x^{8}+a \right )^{p} c p +\left (b \,x^{8}+a \right )^{p} c +2 \left (b \,x^{8}+a \right )^{p} d p \,x^{4}+32 \left (\int \frac {\left (b \,x^{8}+a \right )^{p}}{2 b p \,x^{9}+b \,x^{9}+2 a p x +a x}d x \right ) a c \,p^{3}+32 \left (\int \frac {\left (b \,x^{8}+a \right )^{p}}{2 b p \,x^{9}+b \,x^{9}+2 a p x +a x}d x \right ) a c \,p^{2}+8 \left (\int \frac {\left (b \,x^{8}+a \right )^{p}}{2 b p \,x^{9}+b \,x^{9}+2 a p x +a x}d x \right ) a c p +32 \left (\int \frac {\left (b \,x^{8}+a \right )^{p} x^{3}}{2 b p \,x^{8}+b \,x^{8}+2 a p +a}d x \right ) a d \,p^{3}+16 \left (\int \frac {\left (b \,x^{8}+a \right )^{p} x^{3}}{2 b p \,x^{8}+b \,x^{8}+2 a p +a}d x \right ) a d \,p^{2}}{8 p \left (2 p +1\right )} \] Input:

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

Output:

(2*(a + b*x**8)**p*c*p + (a + b*x**8)**p*c + 2*(a + b*x**8)**p*d*p*x**4 + 
32*int((a + b*x**8)**p/(2*a*p*x + a*x + 2*b*p*x**9 + b*x**9),x)*a*c*p**3 + 
 32*int((a + b*x**8)**p/(2*a*p*x + a*x + 2*b*p*x**9 + b*x**9),x)*a*c*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*c*p + 3 
2*int(((a + b*x**8)**p*x**3)/(2*a*p + a + 2*b*p*x**8 + b*x**8),x)*a*d*p**3 
 + 16*int(((a + b*x**8)**p*x**3)/(2*a*p + a + 2*b*p*x**8 + b*x**8),x)*a*d* 
p**2)/(8*p*(2*p + 1))