\(\int \frac {x^7 (a+b x^8)^p}{(c+d x^4)^2} \, dx\) [35]

Optimal result

Integrand size = 22, antiderivative size = 287 \[ \int \frac {x^7 \left (a+b x^8\right )^p}{\left (c+d x^4\right )^2} \, dx=\frac {c \left (a+b x^8\right )^{1+p}}{4 \left (b c^2+a d^2\right ) \left (c+d x^4\right )}+\frac {\left (a d^2+b c^2 (1+2 p)\right ) x^4 \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{4 c d \left (b c^2+a d^2\right )}-\frac {b c (1+2 p) x^4 \left (a+b x^8\right )^p \left (1+\frac {b x^8}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^8}{a}\right )}{4 d \left (b c^2+a d^2\right )}-\frac {\left (a d^2+b c^2 (1+2 p)\right ) \left (a+b x^8\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {d^2 \left (a+b x^8\right )}{b c^2+a d^2}\right )}{8 \left (b c^2+a d^2\right )^2 (1+p)} \] Output:

1/4*c*(b*x^8+a)^(p+1)/(a*d^2+b*c^2)/(d*x^4+c)+1/4*(a*d^2+b*c^2*(1+2*p))*x^ 
4*(b*x^8+a)^p*AppellF1(1/2,1,-p,3/2,d^2*x^8/c^2,-b*x^8/a)/c/d/(a*d^2+b*c^2 
)/((1+b*x^8/a)^p)-1/4*b*c*(1+2*p)*x^4*(b*x^8+a)^p*hypergeom([1/2, -p],[3/2 
],-b*x^8/a)/d/(a*d^2+b*c^2)/((1+b*x^8/a)^p)-1/8*(a*d^2+b*c^2*(1+2*p))*(b*x 
^8+a)^(p+1)*hypergeom([1, p+1],[2+p],d^2*(b*x^8+a)/(a*d^2+b*c^2))/(a*d^2+b 
*c^2)^2/(p+1)
 

Mathematica [A] (warning: unable to verify)

Time = 0.79 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.85 \[ \int \frac {x^7 \left (a+b x^8\right )^p}{\left (c+d x^4\right )^2} \, dx=\frac {\left (\frac {d \left (-\sqrt {-\frac {a}{b}}+x^4\right )}{c+d x^4}\right )^{-p} \left (\frac {d \left (\sqrt {-\frac {a}{b}}+x^4\right )}{c+d x^4}\right )^{-p} \left (a+b x^8\right )^p \left (-2 c p \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,\frac {c-\sqrt {-\frac {a}{b}} d}{c+d x^4},\frac {c+\sqrt {-\frac {a}{b}} d}{c+d x^4}\right )+(-1+2 p) \left (c+d x^4\right ) \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {c-\sqrt {-\frac {a}{b}} d}{c+d x^4},\frac {c+\sqrt {-\frac {a}{b}} d}{c+d x^4}\right )\right )}{8 d^2 p (-1+2 p) \left (c+d x^4\right )} \] Input:

Integrate[(x^7*(a + b*x^8)^p)/(c + d*x^4)^2,x]
 

Output:

((a + b*x^8)^p*(-2*c*p*AppellF1[1 - 2*p, -p, -p, 2 - 2*p, (c - Sqrt[-(a/b) 
]*d)/(c + d*x^4), (c + Sqrt[-(a/b)]*d)/(c + d*x^4)] + (-1 + 2*p)*(c + d*x^ 
4)*AppellF1[-2*p, -p, -p, 1 - 2*p, (c - Sqrt[-(a/b)]*d)/(c + d*x^4), (c + 
Sqrt[-(a/b)]*d)/(c + d*x^4)]))/(8*d^2*p*(-1 + 2*p)*((d*(-Sqrt[-(a/b)] + x^ 
4))/(c + d*x^4))^p*((d*(Sqrt[-(a/b)] + x^4))/(c + d*x^4))^p*(c + d*x^4))
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 259, normalized size of antiderivative = 0.90, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {1803, 594, 25, 719, 238, 237, 504, 334, 333, 353, 78}

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.

\(\displaystyle \int \frac {x^7 \left (a+b x^8\right )^p}{\left (c+d x^4\right )^2} \, dx\)

\(\Big \downarrow \) 1803

\(\displaystyle \frac {1}{4} \int \frac {x^4 \left (b x^8+a\right )^p}{\left (d x^4+c\right )^2}dx^4\)

\(\Big \downarrow \) 594

\(\displaystyle \frac {1}{4} \left (\frac {c \left (a+b x^8\right )^{p+1}}{\left (c+d x^4\right ) \left (a d^2+b c^2\right )}-\frac {\int -\frac {\left (a d-b c (2 p+1) x^4\right ) \left (b x^8+a\right )^p}{d x^4+c}dx^4}{a d^2+b c^2}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{4} \left (\frac {\int \frac {\left (a d-b c (2 p+1) x^4\right ) \left (b x^8+a\right )^p}{d x^4+c}dx^4}{a d^2+b c^2}+\frac {c \left (a+b x^8\right )^{p+1}}{\left (c+d x^4\right ) \left (a d^2+b c^2\right )}\right )\)

\(\Big \downarrow \) 719

\(\displaystyle \frac {1}{4} \left (\frac {\frac {\left (a d^2+b c^2 (2 p+1)\right ) \int \frac {\left (b x^8+a\right )^p}{d x^4+c}dx^4}{d}-\frac {b c (2 p+1) \int \left (b x^8+a\right )^pdx^4}{d}}{a d^2+b c^2}+\frac {c \left (a+b x^8\right )^{p+1}}{\left (c+d x^4\right ) \left (a d^2+b c^2\right )}\right )\)

\(\Big \downarrow \) 238

\(\displaystyle \frac {1}{4} \left (\frac {\frac {\left (a d^2+b c^2 (2 p+1)\right ) \int \frac {\left (b x^8+a\right )^p}{d x^4+c}dx^4}{d}-\frac {b c (2 p+1) \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \int \left (\frac {b x^8}{a}+1\right )^pdx^4}{d}}{a d^2+b c^2}+\frac {c \left (a+b x^8\right )^{p+1}}{\left (c+d x^4\right ) \left (a d^2+b c^2\right )}\right )\)

\(\Big \downarrow \) 237

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

\(\Big \downarrow \) 504

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

\(\Big \downarrow \) 334

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

\(\Big \downarrow \) 333

\(\displaystyle \frac {1}{4} \left (\frac {\frac {\left (a d^2+b c^2 (2 p+1)\right ) \left (\frac {x^4 \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{c}-d \int \frac {x^4 \left (b x^8+a\right )^p}{c^2-d^2 x^8}dx^4\right )}{d}-\frac {b c (2 p+1) x^4 \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^8}{a}\right )}{d}}{a d^2+b c^2}+\frac {c \left (a+b x^8\right )^{p+1}}{\left (c+d x^4\right ) \left (a d^2+b c^2\right )}\right )\)

\(\Big \downarrow \) 353

\(\displaystyle \frac {1}{4} \left (\frac {\frac {\left (a d^2+b c^2 (2 p+1)\right ) \left (\frac {x^4 \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{c}-\frac {1}{2} d \int \frac {\left (b x^8+a\right )^p}{c^2-d^2 x^8}dx^8\right )}{d}-\frac {b c (2 p+1) x^4 \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^8}{a}\right )}{d}}{a d^2+b c^2}+\frac {c \left (a+b x^8\right )^{p+1}}{\left (c+d x^4\right ) \left (a d^2+b c^2\right )}\right )\)

\(\Big \downarrow \) 78

\(\displaystyle \frac {1}{4} \left (\frac {\frac {\left (a d^2+b c^2 (2 p+1)\right ) \left (\frac {x^4 \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,1,\frac {3}{2},-\frac {b x^8}{a},\frac {d^2 x^8}{c^2}\right )}{c}-\frac {d \left (a+b x^8\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {d^2 \left (b x^8+a\right )}{b c^2+a d^2}\right )}{2 (p+1) \left (a d^2+b c^2\right )}\right )}{d}-\frac {b c (2 p+1) x^4 \left (a+b x^8\right )^p \left (\frac {b x^8}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^8}{a}\right )}{d}}{a d^2+b c^2}+\frac {c \left (a+b x^8\right )^{p+1}}{\left (c+d x^4\right ) \left (a d^2+b c^2\right )}\right )\)

Input:

Int[(x^7*(a + b*x^8)^p)/(c + d*x^4)^2,x]
 

Output:

((c*(a + b*x^8)^(1 + p))/((b*c^2 + a*d^2)*(c + d*x^4)) + (-((b*c*(1 + 2*p) 
*x^4*(a + b*x^8)^p*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^8)/a)])/(d*(1 + 
(b*x^8)/a)^p)) + ((a*d^2 + b*c^2*(1 + 2*p))*((x^4*(a + b*x^8)^p*AppellF1[1 
/2, -p, 1, 3/2, -((b*x^8)/a), (d^2*x^8)/c^2])/(c*(1 + (b*x^8)/a)^p) - (d*( 
a + b*x^8)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, (d^2*(a + b*x^8))/(b 
*c^2 + a*d^2)])/(2*(b*c^2 + a*d^2)*(1 + p))))/d)/(b*c^2 + a*d^2))/4
 

Defintions of rubi rules used

rule 25

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

rule 78

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b 
*c - a*d)^n*((a + b*x)^(m + 1)/(b^(n + 1)*(m + 1)))*Hypergeometric2F1[-n, m 
 + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m}, x] 
 &&  !IntegerQ[m] && IntegerQ[n]
 

rule 237

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[- 
p, 1/2, 1/2 + 1, (-b)*(x^2/a)], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*p 
] && GtQ[a, 0]
 

rule 238

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^2) 
^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(1 + b*(x^2/a))^p, x], x] / 
; FreeQ[{a, b, p}, x] &&  !IntegerQ[2*p] &&  !GtQ[a, 0]
 

rule 333

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F 
reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 
0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 334

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

rule 353

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

rule 504

Int[((a_) + (b_.)*(x_)^2)^(p_)/((c_) + (d_.)*(x_)), x_Symbol] :> Simp[c   I 
nt[(a + b*x^2)^p/(c^2 - d^2*x^2), x], x] - Simp[d   Int[x*((a + b*x^2)^p/(c 
^2 - d^2*x^2)), x], x] /; FreeQ[{a, b, c, d, p}, x]
 

rule 594

Int[(x_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[(-c)*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/((n + 1)*(b*c^2 + a*d^2))) 
, x] + Simp[1/((n + 1)*(b*c^2 + a*d^2))   Int[(c + d*x)^(n + 1)*(a + b*x^2) 
^p*(a*d*(n + 1) + b*c*(n + 2*p + 3)*x), x], x] /; FreeQ[{a, b, c, d, p}, x] 
 && LtQ[n, -1] && NeQ[b*c^2 + a*d^2, 0]
 

rule 719

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[g/e   Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] + 
Simp[(e*f - d*g)/e   Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, 
d, e, f, g, m, p}, x] &&  !IGtQ[m, 0]
 

rule 1803

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q 
_.), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(d + e*x 
)^q*(a + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, c, d, e, m, n, p, q}, x] && 
 EqQ[n2, 2*n] && IntegerQ[Simplify[(m + 1)/n]]
 

Maple [F]

\[\int \frac {x^{7} \left (b \,x^{8}+a \right )^{p}}{\left (x^{4} d +c \right )^{2}}d x\]

Input:

int(x^7*(b*x^8+a)^p/(d*x^4+c)^2,x)
 

Output:

int(x^7*(b*x^8+a)^p/(d*x^4+c)^2,x)
 

Fricas [F]

\[ \int \frac {x^7 \left (a+b x^8\right )^p}{\left (c+d x^4\right )^2} \, dx=\int { \frac {{\left (b x^{8} + a\right )}^{p} x^{7}}{{\left (d x^{4} + c\right )}^{2}} \,d x } \] Input:

integrate(x^7*(b*x^8+a)^p/(d*x^4+c)^2,x, algorithm="fricas")
 

Output:

integral((b*x^8 + a)^p*x^7/(d^2*x^8 + 2*c*d*x^4 + c^2), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^7 \left (a+b x^8\right )^p}{\left (c+d x^4\right )^2} \, dx=\text {Timed out} \] Input:

integrate(x**7*(b*x**8+a)**p/(d*x**4+c)**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {x^7 \left (a+b x^8\right )^p}{\left (c+d x^4\right )^2} \, dx=\int { \frac {{\left (b x^{8} + a\right )}^{p} x^{7}}{{\left (d x^{4} + c\right )}^{2}} \,d x } \] Input:

integrate(x^7*(b*x^8+a)^p/(d*x^4+c)^2,x, algorithm="maxima")
 

Output:

integrate((b*x^8 + a)^p*x^7/(d*x^4 + c)^2, x)
 

Giac [F]

\[ \int \frac {x^7 \left (a+b x^8\right )^p}{\left (c+d x^4\right )^2} \, dx=\int { \frac {{\left (b x^{8} + a\right )}^{p} x^{7}}{{\left (d x^{4} + c\right )}^{2}} \,d x } \] Input:

integrate(x^7*(b*x^8+a)^p/(d*x^4+c)^2,x, algorithm="giac")
 

Output:

integrate((b*x^8 + a)^p*x^7/(d*x^4 + c)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^7 \left (a+b x^8\right )^p}{\left (c+d x^4\right )^2} \, dx=\int \frac {x^7\,{\left (b\,x^8+a\right )}^p}{{\left (d\,x^4+c\right )}^2} \,d x \] Input:

int((x^7*(a + b*x^8)^p)/(c + d*x^4)^2,x)
 

Output:

int((x^7*(a + b*x^8)^p)/(c + d*x^4)^2, x)
 

Reduce [F]

\[ \int \frac {x^7 \left (a+b x^8\right )^p}{\left (c+d x^4\right )^2} \, dx=\text {too large to display} \] Input:

int(x^7*(b*x^8+a)^p/(d*x^4+c)^2,x)
 

Output:

(2*(a + b*x**8)**p*a*c*p + (a + b*x**8)**p*a*c - 2*(a + b*x**8)**p*a*d*p*x 
**4 + (a + b*x**8)**p*a*d*x**4 + 32*int(((a + b*x**8)**p*x**15)/(2*a*c**2* 
p + a*c**2 + 4*a*c*d*p*x**4 + 2*a*c*d*x**4 + 2*a*d**2*p*x**8 + a*d**2*x**8 
 + 2*b*c**2*p*x**8 + b*c**2*x**8 + 4*b*c*d*p*x**12 + 2*b*c*d*x**12 + 2*b*d 
**2*p*x**16 + b*d**2*x**16),x)*a*b*c*d**2*p**3 - 8*int(((a + b*x**8)**p*x* 
*15)/(2*a*c**2*p + a*c**2 + 4*a*c*d*p*x**4 + 2*a*c*d*x**4 + 2*a*d**2*p*x** 
8 + a*d**2*x**8 + 2*b*c**2*p*x**8 + b*c**2*x**8 + 4*b*c*d*p*x**12 + 2*b*c* 
d*x**12 + 2*b*d**2*p*x**16 + b*d**2*x**16),x)*a*b*c*d**2*p + 32*int(((a + 
b*x**8)**p*x**15)/(2*a*c**2*p + a*c**2 + 4*a*c*d*p*x**4 + 2*a*c*d*x**4 + 2 
*a*d**2*p*x**8 + a*d**2*x**8 + 2*b*c**2*p*x**8 + b*c**2*x**8 + 4*b*c*d*p*x 
**12 + 2*b*c*d*x**12 + 2*b*d**2*p*x**16 + b*d**2*x**16),x)*a*b*d**3*p**3*x 
**4 - 8*int(((a + b*x**8)**p*x**15)/(2*a*c**2*p + a*c**2 + 4*a*c*d*p*x**4 
+ 2*a*c*d*x**4 + 2*a*d**2*p*x**8 + a*d**2*x**8 + 2*b*c**2*p*x**8 + b*c**2* 
x**8 + 4*b*c*d*p*x**12 + 2*b*c*d*x**12 + 2*b*d**2*p*x**16 + b*d**2*x**16), 
x)*a*b*d**3*p*x**4 + 32*int(((a + b*x**8)**p*x**15)/(2*a*c**2*p + a*c**2 + 
 4*a*c*d*p*x**4 + 2*a*c*d*x**4 + 2*a*d**2*p*x**8 + a*d**2*x**8 + 2*b*c**2* 
p*x**8 + b*c**2*x**8 + 4*b*c*d*p*x**12 + 2*b*c*d*x**12 + 2*b*d**2*p*x**16 
+ b*d**2*x**16),x)*b**2*c**3*p**3 + 32*int(((a + b*x**8)**p*x**15)/(2*a*c* 
*2*p + a*c**2 + 4*a*c*d*p*x**4 + 2*a*c*d*x**4 + 2*a*d**2*p*x**8 + a*d**2*x 
**8 + 2*b*c**2*p*x**8 + b*c**2*x**8 + 4*b*c*d*p*x**12 + 2*b*c*d*x**12 +...