\(\int (a+b x^8)^p (c+d x^8)^3 \, dx\) [295]

Optimal result

Integrand size = 19, antiderivative size = 287 \[ \int \left (a+b x^8\right )^p \left (c+d x^8\right )^3 \, dx=\frac {3 d \left (51 a^2 d^2-9 a b c d (25+8 p)+b^2 c^2 \left (425+336 p+64 p^2\right )\right ) x \left (a+b x^8\right )^{1+p}}{b^3 (9+8 p) (17+8 p) (25+8 p)}-\frac {d^2 (17 a d-3 b c (25+8 p)) x^9 \left (a+b x^8\right )^{1+p}}{b^2 (17+8 p) (25+8 p)}+\frac {d^3 x^{17} \left (a+b x^8\right )^{1+p}}{b (25+8 p)}+\frac {\left (b^3 c^3 (9+8 p) \left (425+336 p+64 p^2\right )-3 a d \left (51 a^2 d^2-9 a b c d (25+8 p)+b^2 c^2 \left (425+336 p+64 p^2\right )\right )\right ) x \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},-p,\frac {9}{8},-\frac {b x^8}{a}\right )}{b^3 (9+8 p) (17+8 p) (25+8 p)} \] Output:

3*d*(51*a^2*d^2-9*a*b*c*d*(25+8*p)+b^2*c^2*(64*p^2+336*p+425))*x*(b*x^8+a) 
^(p+1)/b^3/(9+8*p)/(17+8*p)/(25+8*p)-d^2*(17*a*d-3*b*c*(25+8*p))*x^9*(b*x^ 
8+a)^(p+1)/b^2/(17+8*p)/(25+8*p)+d^3*x^17*(b*x^8+a)^(p+1)/b/(25+8*p)+(b^3* 
c^3*(9+8*p)*(64*p^2+336*p+425)-3*a*d*(51*a^2*d^2-9*a*b*c*d*(25+8*p)+b^2*c^ 
2*(64*p^2+336*p+425)))*x*(b*x^8+a)^p*hypergeom([1/8, -p],[9/8],-b*x^8/a)/b 
^3/(9+8*p)/(17+8*p)/(25+8*p)/((1+b*x^8/a)^p)
 

Mathematica [A] (verified)

Time = 8.30 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.48 \[ \int \left (a+b x^8\right )^p \left (c+d x^8\right )^3 \, dx=\frac {\left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \left (1275 c^3 x \operatorname {Hypergeometric2F1}\left (\frac {1}{8},-p,\frac {9}{8},-\frac {b x^8}{a}\right )+d x^9 \left (425 c^2 \operatorname {Hypergeometric2F1}\left (\frac {9}{8},-p,\frac {17}{8},-\frac {b x^8}{a}\right )+3 d x^8 \left (75 c \operatorname {Hypergeometric2F1}\left (\frac {17}{8},-p,\frac {25}{8},-\frac {b x^8}{a}\right )+17 d x^8 \operatorname {Hypergeometric2F1}\left (\frac {25}{8},-p,\frac {33}{8},-\frac {b x^8}{a}\right )\right )\right )\right )}{1275} \] Input:

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

Output:

((a + b*x^8)^p*(1275*c^3*x*Hypergeometric2F1[1/8, -p, 9/8, -((b*x^8)/a)] + 
 d*x^9*(425*c^2*Hypergeometric2F1[9/8, -p, 17/8, -((b*x^8)/a)] + 3*d*x^8*( 
75*c*Hypergeometric2F1[17/8, -p, 25/8, -((b*x^8)/a)] + 17*d*x^8*Hypergeome 
tric2F1[25/8, -p, 33/8, -((b*x^8)/a)]))))/(1275*(1 + (b*x^8)/a)^p)
 

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {933, 25, 1025, 25, 913, 779, 778}

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.

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

\(\Big \downarrow \) 933

\(\displaystyle \frac {\int -\left (b x^8+a\right )^p \left (d x^8+c\right ) \left (d (17 a d-b c (8 p+41)) x^8+c (a d-b c (8 p+25))\right )dx}{b (8 p+25)}+\frac {d x \left (c+d x^8\right )^2 \left (a+b x^8\right )^{p+1}}{b (8 p+25)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {d x \left (c+d x^8\right )^2 \left (a+b x^8\right )^{p+1}}{b (8 p+25)}-\frac {\int \left (b x^8+a\right )^p \left (d x^8+c\right ) \left (d (17 a d-b c (8 p+41)) x^8+c (a d-b c (8 p+25))\right )dx}{b (8 p+25)}\)

\(\Big \downarrow \) 1025

\(\displaystyle \frac {d x \left (c+d x^8\right )^2 \left (a+b x^8\right )^{p+1}}{b (8 p+25)}-\frac {\frac {\int -\left (b x^8+a\right )^p \left (d \left (b^2 \left (64 p^2+400 p+753\right ) c^2-2 a b d (40 p+261) c+153 a^2 d^2\right ) x^8+c \left (b^2 \left (64 p^2+336 p+425\right ) c^2-2 a b d (8 p+29) c+17 a^2 d^2\right )\right )dx}{b (8 p+17)}+\frac {d x \left (c+d x^8\right ) \left (a+b x^8\right )^{p+1} (17 a d-b c (8 p+41))}{b (8 p+17)}}{b (8 p+25)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {d x \left (c+d x^8\right )^2 \left (a+b x^8\right )^{p+1}}{b (8 p+25)}-\frac {\frac {d x \left (c+d x^8\right ) \left (a+b x^8\right )^{p+1} (17 a d-b c (8 p+41))}{b (8 p+17)}-\frac {\int \left (b x^8+a\right )^p \left (d \left (b^2 \left (64 p^2+400 p+753\right ) c^2-2 a b d (40 p+261) c+153 a^2 d^2\right ) x^8+c \left (b^2 \left (64 p^2+336 p+425\right ) c^2-2 a b d (8 p+29) c+17 a^2 d^2\right )\right )dx}{b (8 p+17)}}{b (8 p+25)}\)

\(\Big \downarrow \) 913

\(\displaystyle \frac {d x \left (c+d x^8\right )^2 \left (a+b x^8\right )^{p+1}}{b (8 p+25)}-\frac {\frac {d x \left (c+d x^8\right ) \left (a+b x^8\right )^{p+1} (17 a d-b c (8 p+41))}{b (8 p+17)}-\frac {\frac {d x \left (a+b x^8\right )^{p+1} \left (153 a^2 d^2-2 a b c d (40 p+261)+b^2 c^2 \left (64 p^2+400 p+753\right )\right )}{b (8 p+9)}-\frac {\left (153 a^3 d^3-27 a^2 b c d^2 (8 p+25)+3 a b^2 c^2 d \left (64 p^2+336 p+425\right )-b^3 c^3 \left (512 p^3+3264 p^2+6424 p+3825\right )\right ) \int \left (b x^8+a\right )^pdx}{b (8 p+9)}}{b (8 p+17)}}{b (8 p+25)}\)

\(\Big \downarrow \) 779

\(\displaystyle \frac {d x \left (c+d x^8\right )^2 \left (a+b x^8\right )^{p+1}}{b (8 p+25)}-\frac {\frac {d x \left (c+d x^8\right ) \left (a+b x^8\right )^{p+1} (17 a d-b c (8 p+41))}{b (8 p+17)}-\frac {\frac {d x \left (a+b x^8\right )^{p+1} \left (153 a^2 d^2-2 a b c d (40 p+261)+b^2 c^2 \left (64 p^2+400 p+753\right )\right )}{b (8 p+9)}-\frac {\left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \left (153 a^3 d^3-27 a^2 b c d^2 (8 p+25)+3 a b^2 c^2 d \left (64 p^2+336 p+425\right )-b^3 c^3 \left (512 p^3+3264 p^2+6424 p+3825\right )\right ) \int \left (\frac {b x^8}{a}+1\right )^pdx}{b (8 p+9)}}{b (8 p+17)}}{b (8 p+25)}\)

\(\Big \downarrow \) 778

\(\displaystyle \frac {d x \left (c+d x^8\right )^2 \left (a+b x^8\right )^{p+1}}{b (8 p+25)}-\frac {\frac {d x \left (c+d x^8\right ) \left (a+b x^8\right )^{p+1} (17 a d-b c (8 p+41))}{b (8 p+17)}-\frac {\frac {d x \left (a+b x^8\right )^{p+1} \left (153 a^2 d^2-2 a b c d (40 p+261)+b^2 c^2 \left (64 p^2+400 p+753\right )\right )}{b (8 p+9)}-\frac {x \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \left (153 a^3 d^3-27 a^2 b c d^2 (8 p+25)+3 a b^2 c^2 d \left (64 p^2+336 p+425\right )-b^3 c^3 \left (512 p^3+3264 p^2+6424 p+3825\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{8},-p,\frac {9}{8},-\frac {b x^8}{a}\right )}{b (8 p+9)}}{b (8 p+17)}}{b (8 p+25)}\)

Input:

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

Output:

(d*x*(a + b*x^8)^(1 + p)*(c + d*x^8)^2)/(b*(25 + 8*p)) - ((d*(17*a*d - b*c 
*(41 + 8*p))*x*(a + b*x^8)^(1 + p)*(c + d*x^8))/(b*(17 + 8*p)) - ((d*(153* 
a^2*d^2 - 2*a*b*c*d*(261 + 40*p) + b^2*c^2*(753 + 400*p + 64*p^2))*x*(a + 
b*x^8)^(1 + p))/(b*(9 + 8*p)) - ((153*a^3*d^3 - 27*a^2*b*c*d^2*(25 + 8*p) 
+ 3*a*b^2*c^2*d*(425 + 336*p + 64*p^2) - b^3*c^3*(3825 + 6424*p + 3264*p^2 
 + 512*p^3))*x*(a + b*x^8)^p*Hypergeometric2F1[1/8, -p, 9/8, -((b*x^8)/a)] 
)/(b*(9 + 8*p)*(1 + (b*x^8)/a)^p))/(b*(17 + 8*p)))/(b*(25 + 8*p))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 778

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F 
1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p 
, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p] || 
GtQ[a, 0])
 

rule 779

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x 
^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p])   Int[(1 + b*(x^n/a))^p, x], x 
] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Si 
mplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])
 

rule 913

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Si 
mp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(p + 1) + 1))), x] - Simp[(a*d - b*c*(n*( 
p + 1) + 1))/(b*(n*(p + 1) + 1))   Int[(a + b*x^n)^p, x], x] /; FreeQ[{a, b 
, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]
 

rule 933

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[d*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q - 1)/(b*(n*(p + q) + 1))), 
x] + Simp[1/(b*(n*(p + q) + 1))   Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)*Sim 
p[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q 
- 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d 
, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntBinomialQ[ 
a, b, c, d, n, p, q, x]
 

rule 1025

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + ( 
f_.)*(x_)^(n_)), x_Symbol] :> Simp[f*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/( 
b*(n*(p + q + 1) + 1))), x] + Simp[1/(b*(n*(p + q + 1) + 1))   Int[(a + b*x 
^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f + b*e*n*(p + q + 1)) + (d*(b*e 
- a*f) + f*n*q*(b*c - a*d) + b*d*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[ 
{a, b, c, d, e, f, n, p}, x] && GtQ[q, 0] && NeQ[n*(p + q + 1) + 1, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((d^3*x^24 + 3*c*d^2*x^16 + 3*c^2*d*x^8 + c^3)*(b*x^8 + a)^p, x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(1224*(a + b*x**8)**p*a**3*d**3*p*x - 1728*(a + b*x**8)**p*a**2*b*c*d**2*p 
**2*x - 5400*(a + b*x**8)**p*a**2*b*c*d**2*p*x - 1088*(a + b*x**8)**p*a**2 
*b*d**3*p**2*x**9 - 136*(a + b*x**8)**p*a**2*b*d**3*p*x**9 + 1536*(a + b*x 
**8)**p*a*b**2*c**2*d*p**3*x + 8064*(a + b*x**8)**p*a*b**2*c**2*d*p**2*x + 
 10200*(a + b*x**8)**p*a*b**2*c**2*d*p*x + 1536*(a + b*x**8)**p*a*b**2*c*d 
**2*p**3*x**9 + 4992*(a + b*x**8)**p*a*b**2*c*d**2*p**2*x**9 + 600*(a + b* 
x**8)**p*a*b**2*c*d**2*p*x**9 + 512*(a + b*x**8)**p*a*b**2*d**3*p**3*x**17 
 + 640*(a + b*x**8)**p*a*b**2*d**3*p**2*x**17 + 72*(a + b*x**8)**p*a*b**2* 
d**3*p*x**17 + 512*(a + b*x**8)**p*b**3*c**3*p**3*x + 3264*(a + b*x**8)**p 
*b**3*c**3*p**2*x + 6424*(a + b*x**8)**p*b**3*c**3*p*x + 3825*(a + b*x**8) 
**p*b**3*c**3*x + 1536*(a + b*x**8)**p*b**3*c**2*d*p**3*x**9 + 8256*(a + b 
*x**8)**p*b**3*c**2*d*p**2*x**9 + 11208*(a + b*x**8)**p*b**3*c**2*d*p*x**9 
 + 1275*(a + b*x**8)**p*b**3*c**2*d*x**9 + 1536*(a + b*x**8)**p*b**3*c*d** 
2*p**3*x**17 + 6720*(a + b*x**8)**p*b**3*c*d**2*p**2*x**17 + 6216*(a + b*x 
**8)**p*b**3*c*d**2*p*x**17 + 675*(a + b*x**8)**p*b**3*c*d**2*x**17 + 512* 
(a + b*x**8)**p*b**3*d**3*p**3*x**25 + 1728*(a + b*x**8)**p*b**3*d**3*p**2 
*x**25 + 1432*(a + b*x**8)**p*b**3*d**3*p*x**25 + 153*(a + b*x**8)**p*b**3 
*d**3*x**25 - 5013504*int((a + b*x**8)**p/(4096*a*p**4 + 26624*a*p**3 + 54 
656*a*p**2 + 37024*a*p + 3825*a + 4096*b*p**4*x**8 + 26624*b*p**3*x**8 + 5 
4656*b*p**2*x**8 + 37024*b*p*x**8 + 3825*b*x**8),x)*a**4*d**3*p**5 - 32...