\(\int (c+d x^2) (a+b x^8)^p \, dx\) [311]

Optimal result

Integrand size = 17, antiderivative size = 96 \[ \int \left (c+d x^2\right ) \left (a+b x^8\right )^p \, dx=c x \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},-p,\frac {9}{8},-\frac {b x^8}{a}\right )+\frac {1}{3} d x^3 \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{8},-p,\frac {11}{8},-\frac {b x^8}{a}\right ) \] Output:

c*x*(b*x^8+a)^p*hypergeom([1/8, -p],[9/8],-b*x^8/a)/((1+b*x^8/a)^p)+1/3*d* 
x^3*(b*x^8+a)^p*hypergeom([3/8, -p],[11/8],-b*x^8/a)/((1+b*x^8/a)^p)
 

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.78 \[ \int \left (c+d x^2\right ) \left (a+b x^8\right )^p \, dx=\frac {1}{3} x \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \left (3 c \operatorname {Hypergeometric2F1}\left (\frac {1}{8},-p,\frac {9}{8},-\frac {b x^8}{a}\right )+d x^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{8},-p,\frac {11}{8},-\frac {b x^8}{a}\right )\right ) \] Input:

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

Output:

(x*(a + b*x^8)^p*(3*c*Hypergeometric2F1[1/8, -p, 9/8, -((b*x^8)/a)] + d*x^ 
2*Hypergeometric2F1[3/8, -p, 11/8, -((b*x^8)/a)]))/(3*(1 + (b*x^8)/a)^p)
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, 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.

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

\(\Big \downarrow \) 2424

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

\(\Big \downarrow \) 2009

\(\displaystyle c x \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},-p,\frac {9}{8},-\frac {b x^8}{a}\right )+\frac {1}{3} d x^3 \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{8},-p,\frac {11}{8},-\frac {b x^8}{a}\right )\)

Input:

Int[(c + d*x^2)*(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^3*(a + b*x^8)^p*Hypergeometric2F1[3/8, -p, 11/8, -((b*x^8 
)/a)])/(3*(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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 130.95 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.78 \[ \int \left (c+d x^2\right ) \left (a+b x^8\right )^p \, dx=\frac {a^{p} c x \Gamma \left (\frac {1}{8}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{8}, - p \\ \frac {9}{8} \end {matrix}\middle | {\frac {b x^{8} e^{i \pi }}{a}} \right )}}{8 \Gamma \left (\frac {9}{8}\right )} + \frac {a^{p} d x^{3} \Gamma \left (\frac {3}{8}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{8}, - p \\ \frac {11}{8} \end {matrix}\middle | {\frac {b x^{8} e^{i \pi }}{a}} \right )}}{8 \Gamma \left (\frac {11}{8}\right )} \] Input:

integrate((d*x**2+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**3*gamma(3/8)*hyper((3/8, -p), (11/8,), b*x**8*exp_ 
polar(I*pi)/a)/(8*gamma(11/8))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \left (c+d x^2\right ) \left (a+b x^8\right )^p \, dx=\frac {8 \left (b \,x^{8}+a \right )^{p} c p x +3 \left (b \,x^{8}+a \right )^{p} c x +8 \left (b \,x^{8}+a \right )^{p} d p \,x^{3}+\left (b \,x^{8}+a \right )^{p} d \,x^{3}+4096 \left (\int \frac {\left (b \,x^{8}+a \right )^{p}}{64 b \,p^{2} x^{8}+32 b p \,x^{8}+3 b \,x^{8}+64 a \,p^{2}+32 a p +3 a}d x \right ) a c \,p^{4}+3584 \left (\int \frac {\left (b \,x^{8}+a \right )^{p}}{64 b \,p^{2} x^{8}+32 b p \,x^{8}+3 b \,x^{8}+64 a \,p^{2}+32 a p +3 a}d x \right ) a c \,p^{3}+960 \left (\int \frac {\left (b \,x^{8}+a \right )^{p}}{64 b \,p^{2} x^{8}+32 b p \,x^{8}+3 b \,x^{8}+64 a \,p^{2}+32 a p +3 a}d x \right ) a c \,p^{2}+72 \left (\int \frac {\left (b \,x^{8}+a \right )^{p}}{64 b \,p^{2} x^{8}+32 b p \,x^{8}+3 b \,x^{8}+64 a \,p^{2}+32 a p +3 a}d x \right ) a c p +4096 \left (\int \frac {\left (b \,x^{8}+a \right )^{p} x^{2}}{64 b \,p^{2} x^{8}+32 b p \,x^{8}+3 b \,x^{8}+64 a \,p^{2}+32 a p +3 a}d x \right ) a d \,p^{4}+2560 \left (\int \frac {\left (b \,x^{8}+a \right )^{p} x^{2}}{64 b \,p^{2} x^{8}+32 b p \,x^{8}+3 b \,x^{8}+64 a \,p^{2}+32 a p +3 a}d x \right ) a d \,p^{3}+448 \left (\int \frac {\left (b \,x^{8}+a \right )^{p} x^{2}}{64 b \,p^{2} x^{8}+32 b p \,x^{8}+3 b \,x^{8}+64 a \,p^{2}+32 a p +3 a}d x \right ) a d \,p^{2}+24 \left (\int \frac {\left (b \,x^{8}+a \right )^{p} x^{2}}{64 b \,p^{2} x^{8}+32 b p \,x^{8}+3 b \,x^{8}+64 a \,p^{2}+32 a p +3 a}d x \right ) a d p}{64 p^{2}+32 p +3} \] Input:

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

Output:

(8*(a + b*x**8)**p*c*p*x + 3*(a + b*x**8)**p*c*x + 8*(a + b*x**8)**p*d*p*x 
**3 + (a + b*x**8)**p*d*x**3 + 4096*int((a + b*x**8)**p/(64*a*p**2 + 32*a* 
p + 3*a + 64*b*p**2*x**8 + 32*b*p*x**8 + 3*b*x**8),x)*a*c*p**4 + 3584*int( 
(a + b*x**8)**p/(64*a*p**2 + 32*a*p + 3*a + 64*b*p**2*x**8 + 32*b*p*x**8 + 
 3*b*x**8),x)*a*c*p**3 + 960*int((a + b*x**8)**p/(64*a*p**2 + 32*a*p + 3*a 
 + 64*b*p**2*x**8 + 32*b*p*x**8 + 3*b*x**8),x)*a*c*p**2 + 72*int((a + b*x* 
*8)**p/(64*a*p**2 + 32*a*p + 3*a + 64*b*p**2*x**8 + 32*b*p*x**8 + 3*b*x**8 
),x)*a*c*p + 4096*int(((a + b*x**8)**p*x**2)/(64*a*p**2 + 32*a*p + 3*a + 6 
4*b*p**2*x**8 + 32*b*p*x**8 + 3*b*x**8),x)*a*d*p**4 + 2560*int(((a + b*x** 
8)**p*x**2)/(64*a*p**2 + 32*a*p + 3*a + 64*b*p**2*x**8 + 32*b*p*x**8 + 3*b 
*x**8),x)*a*d*p**3 + 448*int(((a + b*x**8)**p*x**2)/(64*a*p**2 + 32*a*p + 
3*a + 64*b*p**2*x**8 + 32*b*p*x**8 + 3*b*x**8),x)*a*d*p**2 + 24*int(((a + 
b*x**8)**p*x**2)/(64*a*p**2 + 32*a*p + 3*a + 64*b*p**2*x**8 + 32*b*p*x**8 
+ 3*b*x**8),x)*a*d*p)/(64*p**2 + 32*p + 3)