Integrand size = 39, antiderivative size = 365 \[ \int \frac {\left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{7/2}} \, dx=-\frac {\left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \left (c^2-d^2 x^2\right )^{1+p}}{c d^4 (5-2 p) (c+d x)^{7/2}}-\frac {\left (7 B c d^2+c^3 D (27-8 p)+A d^3 (3-4 p)-c^2 C d (17-4 p)\right ) \left (c^2-d^2 x^2\right )^{1+p}}{2 c^2 d^4 (3-2 p) (5-2 p) (c+d x)^{5/2}}-\frac {2 D \left (c^2-d^2 x^2\right )^{1+p}}{d^4 (1+4 p) (c+d x)^{3/2}}+\frac {2^{-\frac {7}{2}+p} \left (21 c^3 D (13+8 p)-7 B c d^2 \left (1-16 p^2\right )-c^2 C d \left (43+180 p+32 p^2\right )-A d^3 \left (3-4 p-48 p^2+64 p^3\right )\right ) \left (1+\frac {d x}{c}\right )^{-\frac {1}{2}-p} \left (c^2-d^2 x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2}-p,1+p,2+p,\frac {c-d x}{2 c}\right )}{c^4 d^4 (3-2 p) (5-2 p) (1+p) (1+4 p) \sqrt {c+d x}} \] 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/(5-2*p)/(d*x+c)^ (7/2)-1/2*(7*B*c*d^2+c^3*D*(27-8*p)+A*d^3*(3-4*p)-c^2*C*d*(17-4*p))*(-d^2* x^2+c^2)^(p+1)/c^2/d^4/(3-2*p)/(5-2*p)/(d*x+c)^(5/2)-2*D*(-d^2*x^2+c^2)^(p +1)/d^4/(1+4*p)/(d*x+c)^(3/2)+2^(-7/2+p)*(21*c^3*D*(13+8*p)-7*B*c*d^2*(-16 *p^2+1)-c^2*C*d*(32*p^2+180*p+43)-A*d^3*(64*p^3-48*p^2-4*p+3))*(1+d*x/c)^( -1/2-p)*(-d^2*x^2+c^2)^(p+1)*hypergeom([p+1, 3/2-p],[2+p],1/2*(-d*x+c)/c)/ c^4/d^4/(3-2*p)/(5-2*p)/(p+1)/(1+4*p)/(d*x+c)^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 4.72 (sec) , antiderivative size = 313, normalized size of antiderivative = 0.86 \[ \int \frac {\left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{7/2}} \, dx=\frac {\left (1+\frac {d x}{c}\right )^{\frac {1}{2}-p} \left (1-\frac {d^2 x^2}{c^2}\right )^{-p} \left (24 B d (1+p) x^2 (c-d x)^p (c+d x)^p \left (1+\frac {d x}{c}\right )^p \operatorname {AppellF1}\left (2,-p,\frac {7}{2}-p,3,\frac {d x}{c},-\frac {d x}{c}\right )+16 C d (1+p) x^3 (c-d x)^p (c+d x)^p \left (1+\frac {d x}{c}\right )^p \operatorname {AppellF1}\left (3,-p,\frac {7}{2}-p,4,\frac {d x}{c},-\frac {d x}{c}\right )+12 d D (1+p) x^4 (c-d x)^p (c+d x)^p \left (1+\frac {d x}{c}\right )^p \operatorname {AppellF1}\left (4,-p,\frac {7}{2}-p,5,\frac {d x}{c},-\frac {d x}{c}\right )-3\ 2^{\frac {1}{2}+p} A (c-d x) \left (c^2-d^2 x^2\right )^p \left (1-\frac {d^2 x^2}{c^2}\right )^p \operatorname {Hypergeometric2F1}\left (\frac {7}{2}-p,1+p,2+p,\frac {c-d x}{2 c}\right )\right )}{48 c^3 d (1+p) \sqrt {c+d x}} \] Input:
Integrate[((c^2 - d^2*x^2)^p*(A + B*x + C*x^2 + D*x^3))/(c + d*x)^(7/2),x]
Output:
((1 + (d*x)/c)^(1/2 - p)*(24*B*d*(1 + p)*x^2*(c - d*x)^p*(c + d*x)^p*(1 + (d*x)/c)^p*AppellF1[2, -p, 7/2 - p, 3, (d*x)/c, -((d*x)/c)] + 16*C*d*(1 + p)*x^3*(c - d*x)^p*(c + d*x)^p*(1 + (d*x)/c)^p*AppellF1[3, -p, 7/2 - p, 4, (d*x)/c, -((d*x)/c)] + 12*d*D*(1 + p)*x^4*(c - d*x)^p*(c + d*x)^p*(1 + (d *x)/c)^p*AppellF1[4, -p, 7/2 - 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[7/2 - p, 1 + p, 2 + p, (c - d*x)/(2*c)]))/(48*c^3*d*(1 + p)*Sqrt[c + d*x]*(1 - (d^2*x^2)/c^2)^p)
Time = 1.52 (sec) , antiderivative size = 347, normalized size of antiderivative = 0.95, 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.
\(\displaystyle \int \frac {\left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{7/2}} \, dx\) |
\(\Big \downarrow \) 2170 |
\(\displaystyle -\frac {2 \int \frac {\left (c^2-d^2 x^2\right )^p \left ((c D (8 p+5)-C (4 p d+d)) x^2 d^4-\left (B d^2 (4 p+1)-c^2 D (4 p+7)\right ) x d^3+\left (3 c^3 D-A d^3 (4 p+1)\right ) d^2\right )}{2 (c+d x)^{7/2}}dx}{d^5 (4 p+1)}-\frac {2 D \left (c^2-d^2 x^2\right )^{p+1}}{d^4 (4 p+1) (c+d x)^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {\left (c^2-d^2 x^2\right )^p \left ((c D (8 p+5)-C (4 p d+d)) x^2 d^4-\left (B d^2 (4 p+1)-c^2 D (4 p+7)\right ) x d^3+\left (3 c^3 D-A d^3 (4 p+1)\right ) d^2\right )}{(c+d x)^{7/2}}dx}{d^5 (4 p+1)}-\frac {2 D \left (c^2-d^2 x^2\right )^{p+1}}{d^4 (4 p+1) (c+d x)^{3/2}}\) |
\(\Big \downarrow \) 2170 |
\(\displaystyle -\frac {\frac {2 \int \frac {d^6 \left (28 D (p+1) c^3-5 C d (4 p+1) c^2-A d^3 \left (1-16 p^2\right )-d \left (-D \left (16 p^2+28 p+27\right ) c^2+4 C d \left (4 p^2+5 p+1\right ) c+B d^2 \left (1-16 p^2\right )\right ) x\right ) \left (c^2-d^2 x^2\right )^p}{2 (c+d x)^{7/2}}dx}{d^4 (1-4 p)}+\frac {2 d \left (c^2-d^2 x^2\right )^{p+1} (c D (8 p+5)-C (4 d p+d))}{(1-4 p) (c+d x)^{5/2}}}{d^5 (4 p+1)}-\frac {2 D \left (c^2-d^2 x^2\right )^{p+1}}{d^4 (4 p+1) (c+d x)^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\frac {d^2 \int \frac {\left (28 D (p+1) c^3-5 C d (4 p+1) c^2-A d^3 \left (1-16 p^2\right )-d \left (-D \left (16 p^2+28 p+27\right ) c^2+4 C d \left (4 p^2+5 p+1\right ) c+B d^2 \left (1-16 p^2\right )\right ) x\right ) \left (c^2-d^2 x^2\right )^p}{(c+d x)^{7/2}}dx}{1-4 p}+\frac {2 d \left (c^2-d^2 x^2\right )^{p+1} (c D (8 p+5)-C (4 d p+d))}{(1-4 p) (c+d x)^{5/2}}}{d^5 (4 p+1)}-\frac {2 D \left (c^2-d^2 x^2\right )^{p+1}}{d^4 (4 p+1) (c+d x)^{3/2}}\) |
\(\Big \downarrow \) 671 |
\(\displaystyle -\frac {\frac {d^2 \left (\frac {\left (-A d^3 \left (64 p^3-48 p^2-4 p+3\right )-7 B c d^2 \left (1-16 p^2\right )+21 c^3 D (8 p+13)-c^2 C d \left (32 p^2+180 p+43\right )\right ) \int \frac {\left (c^2-d^2 x^2\right )^p}{(c+d x)^{5/2}}dx}{2 c (5-2 p)}+\frac {\left (1-16 p^2\right ) \left (c^2-d^2 x^2\right )^{p+1} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (5-2 p) (c+d x)^{7/2}}\right )}{1-4 p}+\frac {2 d \left (c^2-d^2 x^2\right )^{p+1} (c D (8 p+5)-C (4 d p+d))}{(1-4 p) (c+d x)^{5/2}}}{d^5 (4 p+1)}-\frac {2 D \left (c^2-d^2 x^2\right )^{p+1}}{d^4 (4 p+1) (c+d x)^{3/2}}\) |
\(\Big \downarrow \) 474 |
\(\displaystyle -\frac {\frac {d^2 \left (\frac {\sqrt {\frac {d x}{c}+1} \left (-A d^3 \left (64 p^3-48 p^2-4 p+3\right )-7 B c d^2 \left (1-16 p^2\right )+21 c^3 D (8 p+13)-c^2 C d \left (32 p^2+180 p+43\right )\right ) \int \frac {\left (c^2-d^2 x^2\right )^p}{\left (\frac {d x}{c}+1\right )^{5/2}}dx}{2 c^3 (5-2 p) \sqrt {c+d x}}+\frac {\left (1-16 p^2\right ) \left (c^2-d^2 x^2\right )^{p+1} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (5-2 p) (c+d x)^{7/2}}\right )}{1-4 p}+\frac {2 d \left (c^2-d^2 x^2\right )^{p+1} (c D (8 p+5)-C (4 d p+d))}{(1-4 p) (c+d x)^{5/2}}}{d^5 (4 p+1)}-\frac {2 D \left (c^2-d^2 x^2\right )^{p+1}}{d^4 (4 p+1) (c+d x)^{3/2}}\) |
\(\Big \downarrow \) 473 |
\(\displaystyle -\frac {\frac {d^2 \left (\frac {\left (\frac {d x}{c}+1\right )^{-p-\frac {1}{2}} \left (c^2-d^2 x^2\right )^{p+1} \left (c^2-c d x\right )^{-p-1} \left (-A d^3 \left (64 p^3-48 p^2-4 p+3\right )-7 B c d^2 \left (1-16 p^2\right )+21 c^3 D (8 p+13)-c^2 C d \left (32 p^2+180 p+43\right )\right ) \int \left (\frac {d x}{c}+1\right )^{p-\frac {5}{2}} \left (c^2-c d x\right )^pdx}{2 c^3 (5-2 p) \sqrt {c+d x}}+\frac {\left (1-16 p^2\right ) \left (c^2-d^2 x^2\right )^{p+1} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (5-2 p) (c+d x)^{7/2}}\right )}{1-4 p}+\frac {2 d \left (c^2-d^2 x^2\right )^{p+1} (c D (8 p+5)-C (4 d p+d))}{(1-4 p) (c+d x)^{5/2}}}{d^5 (4 p+1)}-\frac {2 D \left (c^2-d^2 x^2\right )^{p+1}}{d^4 (4 p+1) (c+d x)^{3/2}}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle -\frac {\frac {d^2 \left (\frac {\left (1-16 p^2\right ) \left (c^2-d^2 x^2\right )^{p+1} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (5-2 p) (c+d x)^{7/2}}-\frac {2^{p-\frac {7}{2}} \left (\frac {d x}{c}+1\right )^{-p-\frac {1}{2}} \left (c^2-d^2 x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (\frac {5}{2}-p,p+1,p+2,\frac {c-d x}{2 c}\right ) \left (-A d^3 \left (64 p^3-48 p^2-4 p+3\right )-7 B c d^2 \left (1-16 p^2\right )+21 c^3 D (8 p+13)-c^2 C d \left (32 p^2+180 p+43\right )\right )}{c^4 d (5-2 p) (p+1) \sqrt {c+d x}}\right )}{1-4 p}+\frac {2 d \left (c^2-d^2 x^2\right )^{p+1} (c D (8 p+5)-C (4 d p+d))}{(1-4 p) (c+d x)^{5/2}}}{d^5 (4 p+1)}-\frac {2 D \left (c^2-d^2 x^2\right )^{p+1}}{d^4 (4 p+1) (c+d x)^{3/2}}\) |
Input:
Int[((c^2 - d^2*x^2)^p*(A + B*x + C*x^2 + D*x^3))/(c + d*x)^(7/2),x]
Output:
(-2*D*(c^2 - d^2*x^2)^(1 + p))/(d^4*(1 + 4*p)*(c + d*x)^(3/2)) - ((2*d*(c* D*(5 + 8*p) - C*(d + 4*d*p))*(c^2 - d^2*x^2)^(1 + p))/((1 - 4*p)*(c + d*x) ^(5/2)) + (d^2*(((c^2*C*d - B*c*d^2 + A*d^3 - c^3*D)*(1 - 16*p^2)*(c^2 - d ^2*x^2)^(1 + p))/(c*d*(5 - 2*p)*(c + d*x)^(7/2)) - (2^(-7/2 + p)*(21*c^3*D *(13 + 8*p) - 7*B*c*d^2*(1 - 16*p^2) - c^2*C*d*(43 + 180*p + 32*p^2) - A*d ^3*(3 - 4*p - 48*p^2 + 64*p^3))*(1 + (d*x)/c)^(-1/2 - p)*(c^2 - d^2*x^2)^( 1 + p)*Hypergeometric2F1[5/2 - p, 1 + p, 2 + p, (c - d*x)/(2*c)])/(c^4*d*( 5 - 2*p)*(1 + p)*Sqrt[c + d*x])))/(1 - 4*p))/(d^5*(1 + 4*p))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]))
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]))
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])
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]
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]
\[\int \frac {\left (-d^{2} x^{2}+c^{2}\right )^{p} \left (D x^{3}+C \,x^{2}+B x +A \right )}{\left (d x +c \right )^{\frac {7}{2}}}d x\]
Input:
int((-d^2*x^2+c^2)^p*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(7/2),x)
Output:
int((-d^2*x^2+c^2)^p*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(7/2),x)
\[ \int \frac {\left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{7/2}} \, dx=\int { \frac {{\left (D x^{3} + C x^{2} + B x + A\right )} {\left (-d^{2} x^{2} + c^{2}\right )}^{p}}{{\left (d x + c\right )}^{\frac {7}{2}}} \,d x } \] Input:
integrate((-d^2*x^2+c^2)^p*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(7/2),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^4*x ^4 + 4*c*d^3*x^3 + 6*c^2*d^2*x^2 + 4*c^3*d*x + c^4), x)
\[ \int \frac {\left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{7/2}} \, dx=\int \frac {\left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{p} \left (A + B x + C x^{2} + D x^{3}\right )}{\left (c + d x\right )^{\frac {7}{2}}}\, dx \] Input:
integrate((-d**2*x**2+c**2)**p*(D*x**3+C*x**2+B*x+A)/(d*x+c)**(7/2),x)
Output:
Integral((-(-c + d*x)*(c + d*x))**p*(A + B*x + C*x**2 + D*x**3)/(c + d*x)* *(7/2), x)
\[ \int \frac {\left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{7/2}} \, dx=\int { \frac {{\left (D x^{3} + C x^{2} + B x + A\right )} {\left (-d^{2} x^{2} + c^{2}\right )}^{p}}{{\left (d x + c\right )}^{\frac {7}{2}}} \,d x } \] Input:
integrate((-d^2*x^2+c^2)^p*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(7/2),x, algorithm= "maxima")
Output:
integrate((D*x^3 + C*x^2 + B*x + A)*(-d^2*x^2 + c^2)^p/(d*x + c)^(7/2), x)
\[ \int \frac {\left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{7/2}} \, dx=\int { \frac {{\left (D x^{3} + C x^{2} + B x + A\right )} {\left (-d^{2} x^{2} + c^{2}\right )}^{p}}{{\left (d x + c\right )}^{\frac {7}{2}}} \,d x } \] Input:
integrate((-d^2*x^2+c^2)^p*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(7/2),x, algorithm= "giac")
Output:
integrate((D*x^3 + C*x^2 + B*x + A)*(-d^2*x^2 + c^2)^p/(d*x + c)^(7/2), x)
Timed out. \[ \int \frac {\left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{7/2}} \, dx=\int \frac {{\left (c^2-d^2\,x^2\right )}^p\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{{\left (c+d\,x\right )}^{7/2}} \,d x \] Input:
int(((c^2 - d^2*x^2)^p*(A + B*x + C*x^2 + x^3*D))/(c + d*x)^(7/2),x)
Output:
int(((c^2 - d^2*x^2)^p*(A + B*x + C*x^2 + x^3*D))/(c + d*x)^(7/2), x)
\[ \int \frac {\left (c^2-d^2 x^2\right )^p \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{7/2}} \, dx=\text {too large to display} \] Input:
int((-d^2*x^2+c^2)^p*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(7/2),x)
Output:
(2*( - 64*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*a*d**2*p**3 + 48*sqrt(c + d* x)*(c**2 - d**2*x**2)**p*a*d**2*p**2 + 4*sqrt(c + d*x)*(c**2 - d**2*x**2)* *p*a*d**2*p - 3*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 - 2*sqrt(c + d*x)*(c**2 - d**2*x**2)* *p*b*c*d + 80*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*b*d**2*p**2*x - 5*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*b*d**2*x - 32*sqrt(c + d*x)*(c**2 - d**2*x** 2)**p*c**3*p**2 + 48*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*c**3*p + 40*sqrt( c + d*x)*(c**2 - d**2*x**2)**p*c**3 - 80*sqrt(c + d*x)*(c**2 - d**2*x**2)* *p*c**2*d*p**2*x + 120*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*c**2*d*p*x + 10 0*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*c**2*d*x - 100*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*c*d**2*p*x**2 + 75*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*c*d* *2*x**2 + 80*sqrt(c + d*x)*(c**2 - d**2*x**2)**p*d**3*p**2*x**3 - 80*sqrt( c + d*x)*(c**2 - d**2*x**2)**p*d**3*p*x**3 + 15*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**5*p**3 - 48*c**5*p**2 - 4*c**5*p + 3*c**5 + 192*c**4*d*p**3*x - 144*c* *4*d*p**2*x - 12*c**4*d*p*x + 9*c**4*d*x + 128*c**3*d**2*p**3*x**2 - 96*c* *3*d**2*p**2*x**2 - 8*c**3*d**2*p*x**2 + 6*c**3*d**2*x**2 - 128*c**2*d**3* p**3*x**3 + 96*c**2*d**3*p**2*x**3 + 8*c**2*d**3*p*x**3 - 6*c**2*d**3*x**3 - 192*c*d**4*p**3*x**4 + 144*c*d**4*p**2*x**4 + 12*c*d**4*p*x**4 - 9*c*d* *4*x**4 - 64*d**5*p**3*x**5 + 48*d**5*p**2*x**5 + 4*d**5*p*x**5 - 3*d**...