\(\int \frac {(c^2-d^2 x^2)^p (A+B x+C x^2+D x^3)}{(c+d x)^{3/2}} \, dx\) [245]

Optimal result

Integrand size = 39, antiderivative size = 342 \[ \int \frac {\left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{3/2}} \, dx=-\frac {\left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \left (c^2-d^2 x^2\right )^{1+p}}{c d^4 (1-2 p) (c+d x)^{3/2}}-\frac {2 (C d (5+4 p)-c D (9+8 p)) \left (c^2-d^2 x^2\right )^{1+p}}{d^4 (3+4 p) (5+4 p) \sqrt {c+d x}}-\frac {2 D \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{1+p}}{d^4 (5+4 p)}-\frac {2^{-\frac {3}{2}+p} \left (3 c^3 D (23+24 p)+3 B c d^2 \left (15+32 p+16 p^2\right )-c^2 C d \left (55+84 p+32 p^2\right )-A d^3 \left (15+92 p+144 p^2+64 p^3\right )\right ) \left (1+\frac {d x}{c}\right )^{-\frac {1}{2}-p} \left (c^2-d^2 x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-p,1+p,2+p,\frac {c-d x}{2 c}\right )}{c^2 d^4 (1-2 p) (1+p) (3+4 p) (5+4 p) \sqrt {c+d x}} \] Output:

-(A*d^3-B*c*d^2+C*c^2*d-D*c^3)*(-d^2*x^2+c^2)^(p+1)/c/d^4/(1-2*p)/(d*x+c)^ 
(3/2)-2*(C*d*(5+4*p)-c*D*(9+8*p))*(-d^2*x^2+c^2)^(p+1)/d^4/(3+4*p)/(5+4*p) 
/(d*x+c)^(1/2)-2*D*(d*x+c)^(1/2)*(-d^2*x^2+c^2)^(p+1)/d^4/(5+4*p)-2^(-3/2+ 
p)*(3*c^3*D*(23+24*p)+3*B*c*d^2*(16*p^2+32*p+15)-c^2*C*d*(32*p^2+84*p+55)- 
A*d^3*(64*p^3+144*p^2+92*p+15))*(1+d*x/c)^(-1/2-p)*(-d^2*x^2+c^2)^(p+1)*hy 
pergeom([p+1, 1/2-p],[2+p],1/2*(-d*x+c)/c)/c^2/d^4/(1-2*p)/(p+1)/(3+4*p)/( 
5+4*p)/(d*x+c)^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 3.63 (sec) , antiderivative size = 313, normalized size of antiderivative = 0.92 \[ \int \frac {\left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{3/2}} \, dx=\frac {\left (1+\frac {d x}{c}\right )^{\frac {1}{2}-p} \left (1-\frac {d^2 x^2}{c^2}\right )^{-p} \left (6 B d (1+p) x^2 (c-d x)^p (c+d x)^p \left (1+\frac {d x}{c}\right )^p \operatorname {AppellF1}\left (2,-p,\frac {3}{2}-p,3,\frac {d x}{c},-\frac {d x}{c}\right )+4 C d (1+p) x^3 (c-d x)^p (c+d x)^p \left (1+\frac {d x}{c}\right )^p \operatorname {AppellF1}\left (3,-p,\frac {3}{2}-p,4,\frac {d x}{c},-\frac {d x}{c}\right )+3 d D (1+p) x^4 (c-d x)^p (c+d x)^p \left (1+\frac {d x}{c}\right )^p \operatorname {AppellF1}\left (4,-p,\frac {3}{2}-p,5,\frac {d x}{c},-\frac {d x}{c}\right )-3\ 2^{\frac {1}{2}+p} A (c-d x) \left (c^2-d^2 x^2\right )^p \left (1-\frac {d^2 x^2}{c^2}\right )^p \operatorname {Hypergeometric2F1}\left (\frac {3}{2}-p,1+p,2+p,\frac {c-d x}{2 c}\right )\right )}{12 c d (1+p) \sqrt {c+d x}} \] Input:

Integrate[((c^2 - d^2*x^2)^p*(A + B*x + C*x^2 + D*x^3))/(c + d*x)^(3/2),x]
 

Output:

((1 + (d*x)/c)^(1/2 - p)*(6*B*d*(1 + p)*x^2*(c - d*x)^p*(c + d*x)^p*(1 + ( 
d*x)/c)^p*AppellF1[2, -p, 3/2 - p, 3, (d*x)/c, -((d*x)/c)] + 4*C*d*(1 + p) 
*x^3*(c - d*x)^p*(c + d*x)^p*(1 + (d*x)/c)^p*AppellF1[3, -p, 3/2 - p, 4, ( 
d*x)/c, -((d*x)/c)] + 3*d*D*(1 + p)*x^4*(c - d*x)^p*(c + d*x)^p*(1 + (d*x) 
/c)^p*AppellF1[4, -p, 3/2 - p, 5, (d*x)/c, -((d*x)/c)] - 3*2^(1/2 + p)*A*( 
c - d*x)*(c^2 - d^2*x^2)^p*(1 - (d^2*x^2)/c^2)^p*Hypergeometric2F1[3/2 - p 
, 1 + p, 2 + p, (c - d*x)/(2*c)]))/(12*c*d*(1 + p)*Sqrt[c + d*x]*(1 - (d^2 
*x^2)/c^2)^p)
 

Rubi [A] (verified)

Time = 1.50 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.205, Rules used = {2170, 27, 2170, 27, 671, 474, 473, 79}

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.

\(\displaystyle \int \frac {\left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{3/2}} \, dx\)

\(\Big \downarrow \) 2170

\(\displaystyle -\frac {2 \int -\frac {\left (c^2-d^2 x^2\right )^p \left ((C d (4 p+5)-c D (8 p+9)) x^2 d^4-\left (c^2 D (4 p+3)-B d^2 (4 p+5)\right ) x d^3+\left (D c^3+A d^3 (4 p+5)\right ) d^2\right )}{2 (c+d x)^{3/2}}dx}{d^5 (4 p+5)}-\frac {2 D \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{p+1}}{d^4 (4 p+5)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\left (c^2-d^2 x^2\right )^p \left ((C d (4 p+5)-c D (8 p+9)) x^2 d^4-\left (c^2 D (4 p+3)-B d^2 (4 p+5)\right ) x d^3+\left (D c^3+A d^3 (4 p+5)\right ) d^2\right )}{(c+d x)^{3/2}}dx}{d^5 (4 p+5)}-\frac {2 D \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{p+1}}{d^4 (4 p+5)}\)

\(\Big \downarrow \) 2170

\(\displaystyle \frac {-\frac {2 \int -\frac {d^6 \left (12 D (p+1) c^3-C d (4 p+5) c^2+A d^3 \left (16 p^2+32 p+15\right )-d \left (-D \left (16 p^2+44 p+27\right ) c^2+4 C d \left (4 p^2+9 p+5\right ) c-B d^2 \left (16 p^2+32 p+15\right )\right ) x\right ) \left (c^2-d^2 x^2\right )^p}{2 (c+d x)^{3/2}}dx}{d^4 (4 p+3)}-\frac {2 d \left (c^2-d^2 x^2\right )^{p+1} (C d (4 p+5)-c D (8 p+9))}{(4 p+3) \sqrt {c+d x}}}{d^5 (4 p+5)}-\frac {2 D \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{p+1}}{d^4 (4 p+5)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {d^2 \int \frac {\left (12 D (p+1) c^3-C d (4 p+5) c^2+A d^3 \left (16 p^2+32 p+15\right )-d \left (-D \left (16 p^2+44 p+27\right ) c^2+4 C d \left (4 p^2+9 p+5\right ) c-B d^2 \left (16 p^2+32 p+15\right )\right ) x\right ) \left (c^2-d^2 x^2\right )^p}{(c+d x)^{3/2}}dx}{4 p+3}-\frac {2 d \left (c^2-d^2 x^2\right )^{p+1} (C d (4 p+5)-c D (8 p+9))}{(4 p+3) \sqrt {c+d x}}}{d^5 (4 p+5)}-\frac {2 D \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{p+1}}{d^4 (4 p+5)}\)

\(\Big \downarrow \) 671

\(\displaystyle \frac {\frac {d^2 \left (\frac {\left (-A d^3 \left (64 p^3+144 p^2+92 p+15\right )+3 B c d^2 \left (16 p^2+32 p+15\right )+3 c^3 D (24 p+23)-c^2 C d \left (32 p^2+84 p+55\right )\right ) \int \frac {\left (c^2-d^2 x^2\right )^p}{\sqrt {c+d x}}dx}{2 c (1-2 p)}-\frac {\left (16 p^2+32 p+15\right ) \left (c^2-d^2 x^2\right )^{p+1} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (1-2 p) (c+d x)^{3/2}}\right )}{4 p+3}-\frac {2 d \left (c^2-d^2 x^2\right )^{p+1} (C d (4 p+5)-c D (8 p+9))}{(4 p+3) \sqrt {c+d x}}}{d^5 (4 p+5)}-\frac {2 D \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{p+1}}{d^4 (4 p+5)}\)

\(\Big \downarrow \) 474

\(\displaystyle \frac {\frac {d^2 \left (\frac {\sqrt {\frac {d x}{c}+1} \left (-A d^3 \left (64 p^3+144 p^2+92 p+15\right )+3 B c d^2 \left (16 p^2+32 p+15\right )+3 c^3 D (24 p+23)-c^2 C d \left (32 p^2+84 p+55\right )\right ) \int \frac {\left (c^2-d^2 x^2\right )^p}{\sqrt {\frac {d x}{c}+1}}dx}{2 c (1-2 p) \sqrt {c+d x}}-\frac {\left (16 p^2+32 p+15\right ) \left (c^2-d^2 x^2\right )^{p+1} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (1-2 p) (c+d x)^{3/2}}\right )}{4 p+3}-\frac {2 d \left (c^2-d^2 x^2\right )^{p+1} (C d (4 p+5)-c D (8 p+9))}{(4 p+3) \sqrt {c+d x}}}{d^5 (4 p+5)}-\frac {2 D \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{p+1}}{d^4 (4 p+5)}\)

\(\Big \downarrow \) 473

\(\displaystyle \frac {\frac {d^2 \left (\frac {\left (\frac {d x}{c}+1\right )^{-p-\frac {1}{2}} \left (c^2-c d x\right )^{-p-1} \left (c^2-d^2 x^2\right )^{p+1} \left (-A d^3 \left (64 p^3+144 p^2+92 p+15\right )+3 B c d^2 \left (16 p^2+32 p+15\right )+3 c^3 D (24 p+23)-c^2 C d \left (32 p^2+84 p+55\right )\right ) \int \left (\frac {d x}{c}+1\right )^{p-\frac {1}{2}} \left (c^2-c d x\right )^pdx}{2 c (1-2 p) \sqrt {c+d x}}-\frac {\left (16 p^2+32 p+15\right ) \left (c^2-d^2 x^2\right )^{p+1} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (1-2 p) (c+d x)^{3/2}}\right )}{4 p+3}-\frac {2 d \left (c^2-d^2 x^2\right )^{p+1} (C d (4 p+5)-c D (8 p+9))}{(4 p+3) \sqrt {c+d x}}}{d^5 (4 p+5)}-\frac {2 D \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{p+1}}{d^4 (4 p+5)}\)

\(\Big \downarrow \) 79

\(\displaystyle \frac {\frac {d^2 \left (-\frac {2^{p-\frac {3}{2}} \left (c^2-d^2 x^2\right )^{p+1} \left (\frac {d x}{c}+1\right )^{-p-\frac {1}{2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-p,p+1,p+2,\frac {c-d x}{2 c}\right ) \left (-A d^3 \left (64 p^3+144 p^2+92 p+15\right )+3 B c d^2 \left (16 p^2+32 p+15\right )+3 c^3 D (24 p+23)-c^2 C d \left (32 p^2+84 p+55\right )\right )}{c^2 d (1-2 p) (p+1) \sqrt {c+d x}}-\frac {\left (16 p^2+32 p+15\right ) \left (c^2-d^2 x^2\right )^{p+1} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (1-2 p) (c+d x)^{3/2}}\right )}{4 p+3}-\frac {2 d \left (c^2-d^2 x^2\right )^{p+1} (C d (4 p+5)-c D (8 p+9))}{(4 p+3) \sqrt {c+d x}}}{d^5 (4 p+5)}-\frac {2 D \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{p+1}}{d^4 (4 p+5)}\)

Input:

Int[((c^2 - d^2*x^2)^p*(A + B*x + C*x^2 + D*x^3))/(c + d*x)^(3/2),x]
 

Output:

(-2*D*Sqrt[c + d*x]*(c^2 - d^2*x^2)^(1 + p))/(d^4*(5 + 4*p)) + ((-2*d*(C*d 
*(5 + 4*p) - c*D*(9 + 8*p))*(c^2 - d^2*x^2)^(1 + p))/((3 + 4*p)*Sqrt[c + d 
*x]) + (d^2*(-(((c^2*C*d - B*c*d^2 + A*d^3 - c^3*D)*(15 + 32*p + 16*p^2)*( 
c^2 - d^2*x^2)^(1 + p))/(c*d*(1 - 2*p)*(c + d*x)^(3/2))) - (2^(-3/2 + p)*( 
3*c^3*D*(23 + 24*p) + 3*B*c*d^2*(15 + 32*p + 16*p^2) - c^2*C*d*(55 + 84*p 
+ 32*p^2) - A*d^3*(15 + 92*p + 144*p^2 + 64*p^3))*(1 + (d*x)/c)^(-1/2 - p) 
*(c^2 - d^2*x^2)^(1 + p)*Hypergeometric2F1[1/2 - p, 1 + p, 2 + p, (c - d*x 
)/(2*c)])/(c^2*d*(1 - 2*p)*(1 + p)*Sqrt[c + d*x])))/(3 + 4*p))/(d^5*(5 + 4 
*p))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 79

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( 
a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 
, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] 
&&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] 
 ||  !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
 

rule 473

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
c^(n - 1)*((a + b*x^2)^(p + 1)/((1 + d*(x/c))^(p + 1)*(a/c + (b*x)/d)^(p + 
1)))   Int[(1 + d*(x/c))^(n + p)*(a/c + (b/d)*x)^p, x], x] /; FreeQ[{a, b, 
c, d, n}, x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[n] || GtQ[c, 0]) &&  !Gt 
Q[a, 0] &&  !(IntegerQ[n] && (IntegerQ[3*p] || IntegerQ[4*p]))
 

rule 474

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

rule 671

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

rule 2170

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x) 
^(m + q - 1)*((a + b*x^2)^(p + 1)/(b*e^(q - 1)*(m + q + 2*p + 1))), x] + Si 
mp[1/(b*e^q*(m + q + 2*p + 1))   Int[(d + e*x)^m*(a + b*x^2)^p*ExpandToSum[ 
b*e^q*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*(d + e*x)^q - 2*e*f*(m + 
 p + q)*(d + e*x)^(q - 2)*(a*e - b*d*x), x], x], x] /; NeQ[m + q + 2*p + 1, 
 0]] /; FreeQ[{a, b, d, e, m, p}, x] && PolyQ[Pq, x] && EqQ[b*d^2 + a*e^2, 
0] &&  !IGtQ[m, 0]
 

Maple [F]

\[\int \frac {\left (-d^{2} x^{2}+c^{2}\right )^{p} \left (D x^{3}+C \,x^{2}+B x +A \right )}{\left (d x +c \right )^{\frac {3}{2}}}d x\]

Input:

int((-d^2*x^2+c^2)^p*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(3/2),x)
 

Output:

int((-d^2*x^2+c^2)^p*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(3/2),x)
 

Fricas [F]

\[ \int \frac {\left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{3/2}} \, dx=\int { \frac {{\left (D x^{3} + C x^{2} + B x + A\right )} {\left (-d^{2} x^{2} + c^{2}\right )}^{p}}{{\left (d x + c\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((-d^2*x^2+c^2)^p*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(3/2),x, algorithm= 
"fricas")
 

Output:

integral((D*x^3 + C*x^2 + B*x + A)*sqrt(d*x + c)*(-d^2*x^2 + c^2)^p/(d^2*x 
^2 + 2*c*d*x + c^2), x)
 

Sympy [F]

\[ \int \frac {\left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{3/2}} \, dx=\int \frac {\left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{p} \left (A + B x + C x^{2} + D x^{3}\right )}{\left (c + d x\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((-d**2*x**2+c**2)**p*(D*x**3+C*x**2+B*x+A)/(d*x+c)**(3/2),x)
 

Output:

Integral((-(-c + d*x)*(c + d*x))**p*(A + B*x + C*x**2 + D*x**3)/(c + d*x)* 
*(3/2), x)
 

Maxima [F]

\[ \int \frac {\left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{3/2}} \, dx=\int { \frac {{\left (D x^{3} + C x^{2} + B x + A\right )} {\left (-d^{2} x^{2} + c^{2}\right )}^{p}}{{\left (d x + c\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((-d^2*x^2+c^2)^p*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(3/2),x, algorithm= 
"maxima")
 

Output:

integrate((D*x^3 + C*x^2 + B*x + A)*(-d^2*x^2 + c^2)^p/(d*x + c)^(3/2), x)
 

Giac [F]

\[ \int \frac {\left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{3/2}} \, dx=\int { \frac {{\left (D x^{3} + C x^{2} + B x + A\right )} {\left (-d^{2} x^{2} + c^{2}\right )}^{p}}{{\left (d x + c\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((-d^2*x^2+c^2)^p*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(3/2),x, algorithm= 
"giac")
 

Output:

integrate((D*x^3 + C*x^2 + B*x + A)*(-d^2*x^2 + c^2)^p/(d*x + c)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{3/2}} \, dx=\int \frac {{\left (c^2-d^2\,x^2\right )}^p\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{{\left (c+d\,x\right )}^{3/2}} \,d x \] Input:

int(((c^2 - d^2*x^2)^p*(A + B*x + C*x^2 + x^3*D))/(c + d*x)^(3/2),x)
 

Output:

int(((c^2 - d^2*x^2)^p*(A + B*x + C*x^2 + x^3*D))/(c + d*x)^(3/2), x)
 

Reduce [F]

\[ \int \frac {\left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{3/2}} \, dx=\text {too large to display} \] Input:

int((-d^2*x^2+c^2)^p*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(3/2),x)
                                                                                    
                                                                                    
 

Output:

(2*( - 64*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*a*d**2*p**3 - 144*sqrt(c + d 
*x)*(c**2 - d**2*x**2)**p*a*d**2*p**2 - 92*sqrt(c + d*x)*(c**2 - d**2*x**2 
)**p*a*d**2*p - 15*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*a*d**2 + 32*sqrt(c 
+ d*x)*(c**2 - d**2*x**2)**p*b*c*d*p**2 + 64*sqrt(c + d*x)*(c**2 - d**2*x* 
*2)**p*b*c*d*p + 30*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*b*c*d + 16*sqrt(c 
+ d*x)*(c**2 - d**2*x**2)**p*b*d**2*p**2*x + 32*sqrt(c + d*x)*(c**2 - d**2 
*x**2)**p*b*d**2*p*x + 15*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*b*d**2*x - 3 
2*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*c**3*p**2 - 16*sqrt(c + d*x)*(c**2 - 
 d**2*x**2)**p*c**3*p + 8*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*c**3 - 16*sq 
rt(c + d*x)*(c**2 - d**2*x**2)**p*c**2*d*p**2*x - 8*sqrt(c + d*x)*(c**2 - 
d**2*x**2)**p*c**2*d*p*x + 4*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*c**2*d*x 
- 4*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*c*d**2*p*x**2 - sqrt(c + d*x)*(c** 
2 - d**2*x**2)**p*c*d**2*x**2 + 16*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*d** 
3*p**2*x**3 + 16*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*d**3*p*x**3 + 3*sqrt( 
c + d*x)*(c**2 - d**2*x**2)**p*d**3*x**3 - 8192*int((sqrt(c + d*x)*(c**2 - 
 d**2*x**2)**p*x)/(64*c**3*p**3 + 144*c**3*p**2 + 92*c**3*p + 15*c**3 + 64 
*c**2*d*p**3*x + 144*c**2*d*p**2*x + 92*c**2*d*p*x + 15*c**2*d*x - 64*c*d* 
*2*p**3*x**2 - 144*c*d**2*p**2*x**2 - 92*c*d**2*p*x**2 - 15*c*d**2*x**2 - 
64*d**3*p**3*x**3 - 144*d**3*p**2*x**3 - 92*d**3*p*x**3 - 15*d**3*x**3),x) 
*a*c*d**4*p**7 - 36864*int((sqrt(c + d*x)*(c**2 - d**2*x**2)**p*x)/(64*...