\(\int (c+d x) (a+b x^8)^p \, dx\) [317]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [F]
Fricas [F]
Sympy [C] (verification not implemented)
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 15, antiderivative size = 96 \[ \int (c+d x) \left (a+b x^8\right )^p \, dx=c x \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},-p,\frac {9}{8},-\frac {b x^8}{a}\right )+\frac {1}{2} d x^2 \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^8}{a}\right ) \] Output:

c*x*(b*x^8+a)^p*hypergeom([1/8, -p],[9/8],-b*x^8/a)/((1+b*x^8/a)^p)+1/2*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 = 0.45 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.76 \[ \int (c+d x) \left (a+b x^8\right )^p \, dx=\frac {1}{2} x \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \left (2 c \operatorname {Hypergeometric2F1}\left (\frac {1}{8},-p,\frac {9}{8},-\frac {b x^8}{a}\right )+d x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^8}{a}\right )\right ) \] Input:

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

Output:

(x*(a + b*x^8)^p*(2*c*Hypergeometric2F1[1/8, -p, 9/8, -((b*x^8)/a)] + d*x* 
Hypergeometric2F1[1/4, -p, 5/4, -((b*x^8)/a)]))/(2*(1 + (b*x^8)/a)^p)
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2424, 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.

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

\(\Big \downarrow \) 2424

\(\displaystyle \int \left (c \left (a+b x^8\right )^p+d x \left (a+b x^8\right )^p\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle c x \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},-p,\frac {9}{8},-\frac {b x^8}{a}\right )+\frac {1}{2} d x^2 \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^8}{a}\right )\)

Input:

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

Output:

(c*x*(a + b*x^8)^p*Hypergeometric2F1[1/8, -p, 9/8, -((b*x^8)/a)])/(1 + (b* 
x^8)/a)^p + (d*x^2*(a + b*x^8)^p*Hypergeometric2F1[1/4, -p, 5/4, -((b*x^8) 
/a)])/(2*(1 + (b*x^8)/a)^p)
 

Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2424
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, 
 x], j, k}, Int[Sum[x^j*Sum[Coeff[Pq, x, j + k*(n/2)]*x^(k*(n/2)), {k, 0, 2 
*((q - j)/n) + 1}]*(a + b*x^n)^p, {j, 0, n/2 - 1}], x]] /; FreeQ[{a, b, p}, 
 x] && PolyQ[Pq, x] && IGtQ[n/2, 0] &&  !PolyQ[Pq, x^(n/2)]
 
Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 92.31 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.78 \[ \int (c+d x) \left (a+b x^8\right )^p \, dx=\frac {a^{p} c x \Gamma \left (\frac {1}{8}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{8}, - p \\ \frac {9}{8} \end {matrix}\middle | {\frac {b x^{8} e^{i \pi }}{a}} \right )}}{8 \Gamma \left (\frac {9}{8}\right )} + \frac {a^{p} d x^{2} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, - p \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{8} e^{i \pi }}{a}} \right )}}{8 \Gamma \left (\frac {5}{4}\right )} \] Input:

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

Output:

a**p*c*x*gamma(1/8)*hyper((1/8, -p), (9/8,), b*x**8*exp_polar(I*pi)/a)/(8* 
gamma(9/8)) + a**p*d*x**2*gamma(1/4)*hyper((1/4, -p), (5/4,), b*x**8*exp_p 
olar(I*pi)/a)/(8*gamma(5/4))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (c+d x) \left (a+b x^8\right )^p \, dx=\frac {8 \left (b \,x^{8}+a \right )^{p} c p x +2 \left (b \,x^{8}+a \right )^{p} c x +8 \left (b \,x^{8}+a \right )^{p} d p \,x^{2}+\left (b \,x^{8}+a \right )^{p} d \,x^{2}+2048 \left (\int \frac {\left (b \,x^{8}+a \right )^{p}}{32 b \,p^{2} x^{8}+12 b p \,x^{8}+b \,x^{8}+32 a \,p^{2}+12 a p +a}d x \right ) a c \,p^{4}+1280 \left (\int \frac {\left (b \,x^{8}+a \right )^{p}}{32 b \,p^{2} x^{8}+12 b p \,x^{8}+b \,x^{8}+32 a \,p^{2}+12 a p +a}d x \right ) a c \,p^{3}+256 \left (\int \frac {\left (b \,x^{8}+a \right )^{p}}{32 b \,p^{2} x^{8}+12 b p \,x^{8}+b \,x^{8}+32 a \,p^{2}+12 a p +a}d x \right ) a c \,p^{2}+16 \left (\int \frac {\left (b \,x^{8}+a \right )^{p}}{32 b \,p^{2} x^{8}+12 b p \,x^{8}+b \,x^{8}+32 a \,p^{2}+12 a p +a}d x \right ) a c p +2048 \left (\int \frac {\left (b \,x^{8}+a \right )^{p} x}{32 b \,p^{2} x^{8}+12 b p \,x^{8}+b \,x^{8}+32 a \,p^{2}+12 a p +a}d x \right ) a d \,p^{4}+1024 \left (\int \frac {\left (b \,x^{8}+a \right )^{p} x}{32 b \,p^{2} x^{8}+12 b p \,x^{8}+b \,x^{8}+32 a \,p^{2}+12 a p +a}d x \right ) a d \,p^{3}+160 \left (\int \frac {\left (b \,x^{8}+a \right )^{p} x}{32 b \,p^{2} x^{8}+12 b p \,x^{8}+b \,x^{8}+32 a \,p^{2}+12 a p +a}d x \right ) a d \,p^{2}+8 \left (\int \frac {\left (b \,x^{8}+a \right )^{p} x}{32 b \,p^{2} x^{8}+12 b p \,x^{8}+b \,x^{8}+32 a \,p^{2}+12 a p +a}d x \right ) a d p}{64 p^{2}+24 p +2} \] Input:

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

Output:

(8*(a + b*x**8)**p*c*p*x + 2*(a + b*x**8)**p*c*x + 8*(a + b*x**8)**p*d*p*x 
**2 + (a + b*x**8)**p*d*x**2 + 2048*int((a + b*x**8)**p/(32*a*p**2 + 12*a* 
p + a + 32*b*p**2*x**8 + 12*b*p*x**8 + b*x**8),x)*a*c*p**4 + 1280*int((a + 
 b*x**8)**p/(32*a*p**2 + 12*a*p + a + 32*b*p**2*x**8 + 12*b*p*x**8 + b*x** 
8),x)*a*c*p**3 + 256*int((a + b*x**8)**p/(32*a*p**2 + 12*a*p + a + 32*b*p* 
*2*x**8 + 12*b*p*x**8 + b*x**8),x)*a*c*p**2 + 16*int((a + b*x**8)**p/(32*a 
*p**2 + 12*a*p + a + 32*b*p**2*x**8 + 12*b*p*x**8 + b*x**8),x)*a*c*p + 204 
8*int(((a + b*x**8)**p*x)/(32*a*p**2 + 12*a*p + a + 32*b*p**2*x**8 + 12*b* 
p*x**8 + b*x**8),x)*a*d*p**4 + 1024*int(((a + b*x**8)**p*x)/(32*a*p**2 + 1 
2*a*p + a + 32*b*p**2*x**8 + 12*b*p*x**8 + b*x**8),x)*a*d*p**3 + 160*int(( 
(a + b*x**8)**p*x)/(32*a*p**2 + 12*a*p + a + 32*b*p**2*x**8 + 12*b*p*x**8 
+ b*x**8),x)*a*d*p**2 + 8*int(((a + b*x**8)**p*x)/(32*a*p**2 + 12*a*p + a 
+ 32*b*p**2*x**8 + 12*b*p*x**8 + b*x**8),x)*a*d*p)/(2*(32*p**2 + 12*p + 1) 
)