\(\int (c+d x)^3 (a+b x^8)^p \, dx\) [315]

Optimal result

Integrand size = 17, antiderivative size = 203 \[ \int (c+d x)^3 \left (a+b x^8\right )^p \, dx=c^3 x \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},-p,\frac {9}{8},-\frac {b x^8}{a}\right )+\frac {3}{2} c^2 d x^2 \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^8}{a}\right )+c d^2 x^3 \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{8},-p,\frac {11}{8},-\frac {b x^8}{a}\right )+\frac {1}{4} d^3 x^4 \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^8}{a}\right ) \] 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)+3/2* 
c^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)+ 
c*d^2*x^3*(b*x^8+a)^p*hypergeom([3/8, -p],[11/8],-b*x^8/a)/((1+b*x^8/a)^p) 
+1/4*d^3*x^4*(b*x^8+a)^p*hypergeom([1/2, -p],[3/2],-b*x^8/a)/((1+b*x^8/a)^ 
p)
 

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.64 \[ \int (c+d x)^3 \left (a+b x^8\right )^p \, dx=\frac {1}{4} x \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \left (4 c^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{8},-p,\frac {9}{8},-\frac {b x^8}{a}\right )+d x \left (6 c^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^8}{a}\right )+d x \left (4 c \operatorname {Hypergeometric2F1}\left (\frac {3}{8},-p,\frac {11}{8},-\frac {b x^8}{a}\right )+d x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^8}{a}\right )\right )\right )\right ) \] Input:

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

Output:

(x*(a + b*x^8)^p*(4*c^3*Hypergeometric2F1[1/8, -p, 9/8, -((b*x^8)/a)] + d* 
x*(6*c^2*Hypergeometric2F1[1/4, -p, 5/4, -((b*x^8)/a)] + d*x*(4*c*Hypergeo 
metric2F1[3/8, -p, 11/8, -((b*x^8)/a)] + d*x*Hypergeometric2F1[1/2, -p, 3/ 
2, -((b*x^8)/a)]))))/(4*(1 + (b*x^8)/a)^p)
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, 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.

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

\(\Big \downarrow \) 2424

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

\(\Big \downarrow \) 2009

\(\displaystyle c^3 x \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},-p,\frac {9}{8},-\frac {b x^8}{a}\right )+\frac {3}{2} c^2 d x^2 \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^8}{a}\right )+c d^2 x^3 \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{8},-p,\frac {11}{8},-\frac {b x^8}{a}\right )+\frac {1}{4} d^3 x^4 \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^8}{a}\right )\)

Input:

Int[(c + d*x)^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 + (3*c^2*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) + (c*d^2*x^3*(a + b*x^8)^p*Hypergeomet 
ric2F1[3/8, -p, 11/8, -((b*x^8)/a)])/(1 + (b*x^8)/a)^p + (d^3*x^4*(a + b*x 
^8)^p*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^8)/a)])/(4*(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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(256*(a + b*x**8)**p*c**3*p**3*x + 288*(a + b*x**8)**p*c**3*p**2*x + 104*( 
a + b*x**8)**p*c**3*p*x + 12*(a + b*x**8)**p*c**3*x + 768*(a + b*x**8)**p* 
c**2*d*p**3*x**2 + 768*(a + b*x**8)**p*c**2*d*p**2*x**2 + 228*(a + b*x**8) 
**p*c**2*d*p*x**2 + 18*(a + b*x**8)**p*c**2*d*x**2 + 768*(a + b*x**8)**p*c 
*d**2*p**3*x**3 + 672*(a + b*x**8)**p*c*d**2*p**2*x**3 + 168*(a + b*x**8)* 
*p*c*d**2*p*x**3 + 12*(a + b*x**8)**p*c*d**2*x**3 + 256*(a + b*x**8)**p*d* 
*3*p**3*x**4 + 192*(a + b*x**8)**p*d**3*p**2*x**4 + 44*(a + b*x**8)**p*d** 
3*p*x**4 + 3*(a + b*x**8)**p*d**3*x**4 + 1048576*int((a + b*x**8)**p/(512* 
a*p**4 + 640*a*p**3 + 280*a*p**2 + 50*a*p + 3*a + 512*b*p**4*x**8 + 640*b* 
p**3*x**8 + 280*b*p**2*x**8 + 50*b*p*x**8 + 3*b*x**8),x)*a*c**3*p**8 + 249 
0368*int((a + b*x**8)**p/(512*a*p**4 + 640*a*p**3 + 280*a*p**2 + 50*a*p + 
3*a + 512*b*p**4*x**8 + 640*b*p**3*x**8 + 280*b*p**2*x**8 + 50*b*p*x**8 + 
3*b*x**8),x)*a*c**3*p**7 + 2473984*int((a + b*x**8)**p/(512*a*p**4 + 640*a 
*p**3 + 280*a*p**2 + 50*a*p + 3*a + 512*b*p**4*x**8 + 640*b*p**3*x**8 + 28 
0*b*p**2*x**8 + 50*b*p*x**8 + 3*b*x**8),x)*a*c**3*p**6 + 1329152*int((a + 
b*x**8)**p/(512*a*p**4 + 640*a*p**3 + 280*a*p**2 + 50*a*p + 3*a + 512*b*p* 
*4*x**8 + 640*b*p**3*x**8 + 280*b*p**2*x**8 + 50*b*p*x**8 + 3*b*x**8),x)*a 
*c**3*p**5 + 415744*int((a + b*x**8)**p/(512*a*p**4 + 640*a*p**3 + 280*a*p 
**2 + 50*a*p + 3*a + 512*b*p**4*x**8 + 640*b*p**3*x**8 + 280*b*p**2*x**8 + 
 50*b*p*x**8 + 3*b*x**8),x)*a*c**3*p**4 + 75392*int((a + b*x**8)**p/(51...