\(\int (c+d x^2)^3 (a+b x^8)^p \, dx\) [309]

Optimal result

Integrand size = 19, antiderivative size = 203 \[ \int \left (c+d x^2\right )^3 \left (a+b x^8\right )^p \, dx=c^3 x \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},-p,\frac {9}{8},-\frac {b x^8}{a}\right )+c^2 d x^3 \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{8},-p,\frac {11}{8},-\frac {b x^8}{a}\right )+\frac {3}{5} c d^2 x^5 \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{8},-p,\frac {13}{8},-\frac {b x^8}{a}\right )+\frac {1}{7} d^3 x^7 \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {7}{8},-p,\frac {15}{8},-\frac {b x^8}{a}\right ) \] Output:

c^3*x*(b*x^8+a)^p*hypergeom([1/8, -p],[9/8],-b*x^8/a)/((1+b*x^8/a)^p)+c^2* 
d*x^3*(b*x^8+a)^p*hypergeom([3/8, -p],[11/8],-b*x^8/a)/((1+b*x^8/a)^p)+3/5 
*c*d^2*x^5*(b*x^8+a)^p*hypergeom([5/8, -p],[13/8],-b*x^8/a)/((1+b*x^8/a)^p 
)+1/7*d^3*x^7*(b*x^8+a)^p*hypergeom([7/8, -p],[15/8],-b*x^8/a)/((1+b*x^8/a 
)^p)
 

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.67 \[ \int \left (c+d x^2\right )^3 \left (a+b x^8\right )^p \, dx=\frac {1}{35} x \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \left (35 c^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{8},-p,\frac {9}{8},-\frac {b x^8}{a}\right )+d x^2 \left (35 c^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{8},-p,\frac {11}{8},-\frac {b x^8}{a}\right )+d x^2 \left (21 c \operatorname {Hypergeometric2F1}\left (\frac {5}{8},-p,\frac {13}{8},-\frac {b x^8}{a}\right )+5 d x^2 \operatorname {Hypergeometric2F1}\left (\frac {7}{8},-p,\frac {15}{8},-\frac {b x^8}{a}\right )\right )\right )\right ) \] Input:

Integrate[(c + d*x^2)^3*(a + b*x^8)^p,x]
 

Output:

(x*(a + b*x^8)^p*(35*c^3*Hypergeometric2F1[1/8, -p, 9/8, -((b*x^8)/a)] + d 
*x^2*(35*c^2*Hypergeometric2F1[3/8, -p, 11/8, -((b*x^8)/a)] + d*x^2*(21*c* 
Hypergeometric2F1[5/8, -p, 13/8, -((b*x^8)/a)] + 5*d*x^2*Hypergeometric2F1 
[7/8, -p, 15/8, -((b*x^8)/a)]))))/(35*(1 + (b*x^8)/a)^p)
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, 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.

\(\displaystyle \int \left (c+d x^2\right )^3 \left (a+b x^8\right )^p \, dx\)

\(\Big \downarrow \) 2424

\(\displaystyle \int \left (\left (c^3+3 c d^2 x^4\right ) \left (a+b x^8\right )^p+x^2 \left (3 c^2 d+d^3 x^4\right ) \left (a+b x^8\right )^p\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle c^3 x \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},-p,\frac {9}{8},-\frac {b x^8}{a}\right )+c^2 d x^3 \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{8},-p,\frac {11}{8},-\frac {b x^8}{a}\right )+\frac {3}{5} c d^2 x^5 \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{8},-p,\frac {13}{8},-\frac {b x^8}{a}\right )+\frac {1}{7} d^3 x^7 \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {7}{8},-p,\frac {15}{8},-\frac {b x^8}{a}\right )\)

Input:

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

Output:

(c^3*x*(a + b*x^8)^p*Hypergeometric2F1[1/8, -p, 9/8, -((b*x^8)/a)])/(1 + ( 
b*x^8)/a)^p + (c^2*d*x^3*(a + b*x^8)^p*Hypergeometric2F1[3/8, -p, 11/8, -( 
(b*x^8)/a)])/(1 + (b*x^8)/a)^p + (3*c*d^2*x^5*(a + b*x^8)^p*Hypergeometric 
2F1[5/8, -p, 13/8, -((b*x^8)/a)])/(5*(1 + (b*x^8)/a)^p) + (d^3*x^7*(a + b* 
x^8)^p*Hypergeometric2F1[7/8, -p, 15/8, -((b*x^8)/a)])/(7*(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]

\[\int \left (d \,x^{2}+c \right )^{3} \left (b \,x^{8}+a \right )^{p}d x\]

Input:

int((d*x^2+c)^3*(b*x^8+a)^p,x)
 

Output:

int((d*x^2+c)^3*(b*x^8+a)^p,x)
 

Fricas [F]

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

integrate((d*x^2+c)^3*(b*x^8+a)^p,x, algorithm="fricas")
 

Output:

integral((d^3*x^6 + 3*c*d^2*x^4 + 3*c^2*d*x^2 + c^3)*(b*x^8 + a)^p, x)
 

Sympy [F(-1)]

Timed out. \[ \int \left (c+d x^2\right )^3 \left (a+b x^8\right )^p \, dx=\text {Timed out} \] Input:

integrate((d*x**2+c)**3*(b*x**8+a)**p,x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((d*x^2+c)^3*(b*x^8+a)^p,x, algorithm="maxima")
 

Output:

integrate((d*x^2 + c)^3*(b*x^8 + a)^p, x)
 

Giac [F]

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

integrate((d*x^2+c)^3*(b*x^8+a)^p,x, algorithm="giac")
 

Output:

integrate((d*x^2 + c)^3*(b*x^8 + a)^p, x)
 

Mupad [F(-1)]

Timed out. \[ \int \left (c+d x^2\right )^3 \left (a+b x^8\right )^p \, dx=\int {\left (b\,x^8+a\right )}^p\,{\left (d\,x^2+c\right )}^3 \,d x \] Input:

int((a + b*x^8)^p*(c + d*x^2)^3,x)
 

Output:

int((a + b*x^8)^p*(c + d*x^2)^3, x)
 

Reduce [F]

\[ \int \left (c+d x^2\right )^3 \left (a+b x^8\right )^p \, dx=\text {too large to display} \] Input:

int((d*x^2+c)^3*(b*x^8+a)^p,x)
 

Output:

(512*(a + b*x**8)**p*c**3*p**3*x + 960*(a + b*x**8)**p*c**3*p**2*x + 568*( 
a + b*x**8)**p*c**3*p*x + 105*(a + b*x**8)**p*c**3*x + 1536*(a + b*x**8)** 
p*c**2*d*p**3*x**3 + 2496*(a + b*x**8)**p*c**2*d*p**2*x**3 + 1128*(a + b*x 
**8)**p*c**2*d*p*x**3 + 105*(a + b*x**8)**p*c**2*d*x**3 + 1536*(a + b*x**8 
)**p*c*d**2*p**3*x**5 + 2112*(a + b*x**8)**p*c*d**2*p**2*x**5 + 744*(a + b 
*x**8)**p*c*d**2*p*x**5 + 63*(a + b*x**8)**p*c*d**2*x**5 + 512*(a + b*x**8 
)**p*d**3*p**3*x**7 + 576*(a + b*x**8)**p*d**3*p**2*x**7 + 184*(a + b*x**8 
)**p*d**3*p*x**7 + 15*(a + b*x**8)**p*d**3*x**7 + 16777216*int((a + b*x**8 
)**p/(4096*a*p**4 + 8192*a*p**3 + 5504*a*p**2 + 1408*a*p + 105*a + 4096*b* 
p**4*x**8 + 8192*b*p**3*x**8 + 5504*b*p**2*x**8 + 1408*b*p*x**8 + 105*b*x* 
*8),x)*a*c**3*p**8 + 65011712*int((a + b*x**8)**p/(4096*a*p**4 + 8192*a*p* 
*3 + 5504*a*p**2 + 1408*a*p + 105*a + 4096*b*p**4*x**8 + 8192*b*p**3*x**8 
+ 5504*b*p**2*x**8 + 1408*b*p*x**8 + 105*b*x**8),x)*a*c**3*p**7 + 10407116 
8*int((a + b*x**8)**p/(4096*a*p**4 + 8192*a*p**3 + 5504*a*p**2 + 1408*a*p 
+ 105*a + 4096*b*p**4*x**8 + 8192*b*p**3*x**8 + 5504*b*p**2*x**8 + 1408*b* 
p*x**8 + 105*b*x**8),x)*a*c**3*p**6 + 88702976*int((a + b*x**8)**p/(4096*a 
*p**4 + 8192*a*p**3 + 5504*a*p**2 + 1408*a*p + 105*a + 4096*b*p**4*x**8 + 
8192*b*p**3*x**8 + 5504*b*p**2*x**8 + 1408*b*p*x**8 + 105*b*x**8),x)*a*c** 
3*p**5 + 43134976*int((a + b*x**8)**p/(4096*a*p**4 + 8192*a*p**3 + 5504*a* 
p**2 + 1408*a*p + 105*a + 4096*b*p**4*x**8 + 8192*b*p**3*x**8 + 5504*b*...