\(\int \frac {(c+d x)^{9/2} (A+B x+C x^2+D x^3)}{(c^2-d^2 x^2)^{5/2}} \, dx\) [227]

Optimal result

Integrand size = 41, antiderivative size = 336 \[ \int \frac {(c+d x)^{9/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {\left (c^2 C d+B c d^2+A d^3+c^3 D\right ) (c+d x)^{9/2}}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\left (5 c^2 C d+3 B c d^2+A d^3+7 c^3 D\right ) (c+d x)^{7/2}}{2 c^2 d^4 \sqrt {c^2-d^2 x^2}}-\frac {16 \left (203 c^2 C d+105 B c d^2+35 A d^3+321 c^3 D\right ) \sqrt {c^2-d^2 x^2}}{105 d^4 \sqrt {c+d x}}-\frac {4 \left (203 c^2 C d+105 B c d^2+35 A d^3+321 c^3 D\right ) \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{105 c d^4}-\frac {\left (203 c^2 C d+105 B c d^2+35 A d^3+321 c^3 D\right ) (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}{70 c^2 d^4}-\frac {2 D (c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}{7 d^4} \] Output:

1/3*(A*d^3+B*c*d^2+C*c^2*d+D*c^3)*(d*x+c)^(9/2)/c/d^4/(-d^2*x^2+c^2)^(3/2) 
-1/2*(A*d^3+3*B*c*d^2+5*C*c^2*d+7*D*c^3)*(d*x+c)^(7/2)/c^2/d^4/(-d^2*x^2+c 
^2)^(1/2)-16/105*(35*A*d^3+105*B*c*d^2+203*C*c^2*d+321*D*c^3)*(-d^2*x^2+c^ 
2)^(1/2)/d^4/(d*x+c)^(1/2)-4/105*(35*A*d^3+105*B*c*d^2+203*C*c^2*d+321*D*c 
^3)*(d*x+c)^(1/2)*(-d^2*x^2+c^2)^(1/2)/c/d^4-1/70*(35*A*d^3+105*B*c*d^2+20 
3*C*c^2*d+321*D*c^3)*(d*x+c)^(3/2)*(-d^2*x^2+c^2)^(1/2)/c^2/d^4-2/7*D*(d*x 
+c)^(5/2)*(-d^2*x^2+c^2)^(1/2)/d^4
 

Mathematica [A] (verified)

Time = 2.90 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.50 \[ \int \frac {(c+d x)^{9/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {2 \sqrt {c+d x} \left (3632 c^5 D+8 c^4 d (287 C-681 D x)+2 c^3 d^2 (595 B+3 x (-574 C+227 D x))+6 c d^4 x (-105 A+x (70 B+3 x (7 C+4 D x)))+d^5 x^2 (105 A+x (35 B+3 x (7 C+5 D x)))+c^2 d^3 (385 A+x (-1785 B+x (861 C+227 D x)))\right )}{105 d^4 (-c+d x) \sqrt {c^2-d^2 x^2}} \] Input:

Integrate[((c + d*x)^(9/2)*(A + B*x + C*x^2 + D*x^3))/(c^2 - d^2*x^2)^(5/2 
),x]
 

Output:

(2*Sqrt[c + d*x]*(3632*c^5*D + 8*c^4*d*(287*C - 681*D*x) + 2*c^3*d^2*(595* 
B + 3*x*(-574*C + 227*D*x)) + 6*c*d^4*x*(-105*A + x*(70*B + 3*x*(7*C + 4*D 
*x))) + d^5*x^2*(105*A + x*(35*B + 3*x*(7*C + 5*D*x))) + c^2*d^3*(385*A + 
x*(-1785*B + x*(861*C + 227*D*x)))))/(105*d^4*(-c + d*x)*Sqrt[c^2 - d^2*x^ 
2])
 

Rubi [A] (verified)

Time = 1.28 (sec) , antiderivative size = 307, normalized size of antiderivative = 0.91, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {2166, 27, 2166, 27, 672, 459, 459, 458}

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.

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

\(\Big \downarrow \) 2166

\(\displaystyle \frac {(c+d x)^{9/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\int \frac {3 (c+d x)^{7/2} \left (\frac {2 c D x^2}{d}+\frac {2 c (C d+c D) x}{d^2}+\frac {3 D c^3+3 C d c^2+3 B d^2 c+A d^3}{d^3}\right )}{2 \left (c^2-d^2 x^2\right )^{3/2}}dx}{3 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(c+d x)^{9/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\int \frac {(c+d x)^{7/2} \left (\frac {2 c D x^2}{d}+\frac {2 c (C d+c D) x}{d^2}+A+\frac {3 c \left (D c^2+C d c+B d^2\right )}{d^3}\right )}{\left (c^2-d^2 x^2\right )^{3/2}}dx}{2 c}\)

\(\Big \downarrow \) 2166

\(\displaystyle \frac {(c+d x)^{9/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {(c+d x)^{7/2} \left (A d^3+3 B c d^2+7 c^3 D+5 c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {\int \frac {(c+d x)^{5/2} \left (43 D c^3+29 C d c^2+4 d D x c^2+15 B d^2 c+5 A d^3\right )}{2 d^3 \sqrt {c^2-d^2 x^2}}dx}{c}}{2 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(c+d x)^{9/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {(c+d x)^{7/2} \left (A d^3+3 B c d^2+7 c^3 D+5 c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {\int \frac {(c+d x)^{5/2} \left (43 D c^3+29 C d c^2+4 d D x c^2+15 B d^2 c+5 A d^3\right )}{\sqrt {c^2-d^2 x^2}}dx}{2 c d^3}}{2 c}\)

\(\Big \downarrow \) 672

\(\displaystyle \frac {(c+d x)^{9/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {(c+d x)^{7/2} \left (A d^3+3 B c d^2+7 c^3 D+5 c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {\frac {1}{7} \left (35 A d^3+105 B c d^2+321 c^3 D+203 c^2 C d\right ) \int \frac {(c+d x)^{5/2}}{\sqrt {c^2-d^2 x^2}}dx-\frac {8 c^2 D (c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}{7 d}}{2 c d^3}}{2 c}\)

\(\Big \downarrow \) 459

\(\displaystyle \frac {(c+d x)^{9/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {(c+d x)^{7/2} \left (A d^3+3 B c d^2+7 c^3 D+5 c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {\frac {1}{7} \left (35 A d^3+105 B c d^2+321 c^3 D+203 c^2 C d\right ) \left (\frac {8}{5} c \int \frac {(c+d x)^{3/2}}{\sqrt {c^2-d^2 x^2}}dx-\frac {2 (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}{5 d}\right )-\frac {8 c^2 D (c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}{7 d}}{2 c d^3}}{2 c}\)

\(\Big \downarrow \) 459

\(\displaystyle \frac {(c+d x)^{9/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {(c+d x)^{7/2} \left (A d^3+3 B c d^2+7 c^3 D+5 c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {\frac {1}{7} \left (35 A d^3+105 B c d^2+321 c^3 D+203 c^2 C d\right ) \left (\frac {8}{5} c \left (\frac {4}{3} c \int \frac {\sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}dx-\frac {2 \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{3 d}\right )-\frac {2 (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}{5 d}\right )-\frac {8 c^2 D (c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}{7 d}}{2 c d^3}}{2 c}\)

\(\Big \downarrow \) 458

\(\displaystyle \frac {(c+d x)^{9/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {(c+d x)^{7/2} \left (A d^3+3 B c d^2+7 c^3 D+5 c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {\frac {1}{7} \left (\frac {8}{5} c \left (-\frac {8 c \sqrt {c^2-d^2 x^2}}{3 d \sqrt {c+d x}}-\frac {2 \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{3 d}\right )-\frac {2 (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}{5 d}\right ) \left (35 A d^3+105 B c d^2+321 c^3 D+203 c^2 C d\right )-\frac {8 c^2 D (c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}{7 d}}{2 c d^3}}{2 c}\)

Input:

Int[((c + d*x)^(9/2)*(A + B*x + C*x^2 + D*x^3))/(c^2 - d^2*x^2)^(5/2),x]
 

Output:

((c^2*C*d + B*c*d^2 + A*d^3 + c^3*D)*(c + d*x)^(9/2))/(3*c*d^4*(c^2 - d^2* 
x^2)^(3/2)) - (((5*c^2*C*d + 3*B*c*d^2 + A*d^3 + 7*c^3*D)*(c + d*x)^(7/2)) 
/(c*d^4*Sqrt[c^2 - d^2*x^2]) - ((-8*c^2*D*(c + d*x)^(5/2)*Sqrt[c^2 - d^2*x 
^2])/(7*d) + ((203*c^2*C*d + 105*B*c*d^2 + 35*A*d^3 + 321*c^3*D)*((-2*(c + 
 d*x)^(3/2)*Sqrt[c^2 - d^2*x^2])/(5*d) + (8*c*((-8*c*Sqrt[c^2 - d^2*x^2])/ 
(3*d*Sqrt[c + d*x]) - (2*Sqrt[c + d*x]*Sqrt[c^2 - d^2*x^2])/(3*d)))/5))/7) 
/(2*c*d^3))/(2*c)
 

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 458

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

rule 459

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

rule 672

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_ 
), x_Symbol] :> Simp[g*(d + e*x)^m*((a + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), 
 x] + Simp[(m*(d*g + e*f) + 2*e*f*(p + 1))/(e*(m + 2*p + 2))   Int[(d + e*x 
)^m*(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] && NeQ[m + 2*p + 2, 0]
 

rule 2166

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{Qx = PolynomialQuotient[Pq, a*e + b*d*x, x], R = PolynomialRemainde 
r[Pq, a*e + b*d*x, x]}, Simp[(-d)*R*(d + e*x)^m*((a + b*x^2)^(p + 1)/(2*a*e 
*(p + 1))), x] + Simp[d/(2*a*(p + 1))   Int[(d + e*x)^(m - 1)*(a + b*x^2)^( 
p + 1)*ExpandToSum[2*a*e*(p + 1)*Qx + R*(m + 2*p + 2), x], x], x]] /; FreeQ 
[{a, b, d, e}, x] && PolyQ[Pq, x] && EqQ[b*d^2 + a*e^2, 0] && ILtQ[p + 1/2, 
 0] && GtQ[m, 0]
 

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.60

method	result	size
gosper	\(-\frac {2 \left (-d x +c \right ) \left (15 D d^{5} x^{5}+21 C \,d^{5} x^{4}+72 D c \,d^{4} x^{4}+35 B \,d^{5} x^{3}+126 C c \,d^{4} x^{3}+227 D c^{2} d^{3} x^{3}+105 A \,d^{5} x^{2}+420 B c \,d^{4} x^{2}+861 C \,c^{2} d^{3} x^{2}+1362 D c^{3} d^{2} x^{2}-630 A c \,d^{4} x -1785 B \,c^{2} d^{3} x -3444 C \,c^{3} d^{2} x -5448 D c^{4} d x +385 A \,c^{2} d^{3}+1190 B \,c^{3} d^{2}+2296 C \,c^{4} d +3632 D c^{5}\right ) \left (d x +c \right )^{\frac {5}{2}}}{105 d^{4} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}\)	\(203\)
orering	\(-\frac {2 \left (-d x +c \right ) \left (15 D d^{5} x^{5}+21 C \,d^{5} x^{4}+72 D c \,d^{4} x^{4}+35 B \,d^{5} x^{3}+126 C c \,d^{4} x^{3}+227 D c^{2} d^{3} x^{3}+105 A \,d^{5} x^{2}+420 B c \,d^{4} x^{2}+861 C \,c^{2} d^{3} x^{2}+1362 D c^{3} d^{2} x^{2}-630 A c \,d^{4} x -1785 B \,c^{2} d^{3} x -3444 C \,c^{3} d^{2} x -5448 D c^{4} d x +385 A \,c^{2} d^{3}+1190 B \,c^{3} d^{2}+2296 C \,c^{4} d +3632 D c^{5}\right ) \left (d x +c \right )^{\frac {5}{2}}}{105 d^{4} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}\)	\(203\)
default	\(-\frac {2 \sqrt {-d^{2} x^{2}+c^{2}}\, \left (15 D d^{5} x^{5}+21 C \,d^{5} x^{4}+72 D c \,d^{4} x^{4}+35 B \,d^{5} x^{3}+126 C c \,d^{4} x^{3}+227 D c^{2} d^{3} x^{3}+105 A \,d^{5} x^{2}+420 B c \,d^{4} x^{2}+861 C \,c^{2} d^{3} x^{2}+1362 D c^{3} d^{2} x^{2}-630 A c \,d^{4} x -1785 B \,c^{2} d^{3} x -3444 C \,c^{3} d^{2} x -5448 D c^{4} d x +385 A \,c^{2} d^{3}+1190 B \,c^{3} d^{2}+2296 C \,c^{4} d +3632 D c^{5}\right )}{105 \sqrt {d x +c}\, \left (-d x +c \right )^{2} d^{4}}\)	\(205\)

Input:

int((d*x+c)^(9/2)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(5/2),x,method=_RETUR 
NVERBOSE)
 

Output:

-2/105*(-d*x+c)*(15*D*d^5*x^5+21*C*d^5*x^4+72*D*c*d^4*x^4+35*B*d^5*x^3+126 
*C*c*d^4*x^3+227*D*c^2*d^3*x^3+105*A*d^5*x^2+420*B*c*d^4*x^2+861*C*c^2*d^3 
*x^2+1362*D*c^3*d^2*x^2-630*A*c*d^4*x-1785*B*c^2*d^3*x-3444*C*c^3*d^2*x-54 
48*D*c^4*d*x+385*A*c^2*d^3+1190*B*c^3*d^2+2296*C*c^4*d+3632*D*c^5)*(d*x+c) 
^(5/2)/d^4/(-d^2*x^2+c^2)^(5/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.65 \[ \int \frac {(c+d x)^{9/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (15 \, D d^{5} x^{5} + 3632 \, D c^{5} + 2296 \, C c^{4} d + 1190 \, B c^{3} d^{2} + 385 \, A c^{2} d^{3} + 3 \, {\left (24 \, D c d^{4} + 7 \, C d^{5}\right )} x^{4} + {\left (227 \, D c^{2} d^{3} + 126 \, C c d^{4} + 35 \, B d^{5}\right )} x^{3} + 3 \, {\left (454 \, D c^{3} d^{2} + 287 \, C c^{2} d^{3} + 140 \, B c d^{4} + 35 \, A d^{5}\right )} x^{2} - 3 \, {\left (1816 \, D c^{4} d + 1148 \, C c^{3} d^{2} + 595 \, B c^{2} d^{3} + 210 \, A c d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{105 \, {\left (d^{7} x^{3} - c d^{6} x^{2} - c^{2} d^{5} x + c^{3} d^{4}\right )}} \] Input:

integrate((d*x+c)^(9/2)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(5/2),x, algori 
thm="fricas")
 

Output:

-2/105*(15*D*d^5*x^5 + 3632*D*c^5 + 2296*C*c^4*d + 1190*B*c^3*d^2 + 385*A* 
c^2*d^3 + 3*(24*D*c*d^4 + 7*C*d^5)*x^4 + (227*D*c^2*d^3 + 126*C*c*d^4 + 35 
*B*d^5)*x^3 + 3*(454*D*c^3*d^2 + 287*C*c^2*d^3 + 140*B*c*d^4 + 35*A*d^5)*x 
^2 - 3*(1816*D*c^4*d + 1148*C*c^3*d^2 + 595*B*c^2*d^3 + 210*A*c*d^4)*x)*sq 
rt(-d^2*x^2 + c^2)*sqrt(d*x + c)/(d^7*x^3 - c*d^6*x^2 - c^2*d^5*x + c^3*d^ 
4)
                                                                                    
                                                                                    
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(c+d x)^{9/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((d*x+c)**(9/2)*(D*x**3+C*x**2+B*x+A)/(-d**2*x**2+c**2)**(5/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 240, normalized size of antiderivative = 0.71 \[ \int \frac {(c+d x)^{9/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {2 \, {\left (3 \, d^{2} x^{2} - 18 \, c d x + 11 \, c^{2}\right )} A}{3 \, {\left (d^{2} x - c d\right )} \sqrt {-d x + c}} + \frac {2 \, {\left (d^{3} x^{3} + 12 \, c d^{2} x^{2} - 51 \, c^{2} d x + 34 \, c^{3}\right )} B}{3 \, {\left (d^{3} x - c d^{2}\right )} \sqrt {-d x + c}} + \frac {2 \, {\left (3 \, d^{4} x^{4} + 18 \, c d^{3} x^{3} + 123 \, c^{2} d^{2} x^{2} - 492 \, c^{3} d x + 328 \, c^{4}\right )} C}{15 \, {\left (d^{4} x - c d^{3}\right )} \sqrt {-d x + c}} + \frac {2 \, {\left (15 \, d^{5} x^{5} + 72 \, c d^{4} x^{4} + 227 \, c^{2} d^{3} x^{3} + 1362 \, c^{3} d^{2} x^{2} - 5448 \, c^{4} d x + 3632 \, c^{5}\right )} D}{105 \, {\left (d^{5} x - c d^{4}\right )} \sqrt {-d x + c}} \] Input:

integrate((d*x+c)^(9/2)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(5/2),x, algori 
thm="maxima")
 

Output:

2/3*(3*d^2*x^2 - 18*c*d*x + 11*c^2)*A/((d^2*x - c*d)*sqrt(-d*x + c)) + 2/3 
*(d^3*x^3 + 12*c*d^2*x^2 - 51*c^2*d*x + 34*c^3)*B/((d^3*x - c*d^2)*sqrt(-d 
*x + c)) + 2/15*(3*d^4*x^4 + 18*c*d^3*x^3 + 123*c^2*d^2*x^2 - 492*c^3*d*x 
+ 328*c^4)*C/((d^4*x - c*d^3)*sqrt(-d*x + c)) + 2/105*(15*d^5*x^5 + 72*c*d 
^4*x^4 + 227*c^2*d^3*x^3 + 1362*c^3*d^2*x^2 - 5448*c^4*d*x + 3632*c^5)*D/( 
(d^5*x - c*d^4)*sqrt(-d*x + c))
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 297, normalized size of antiderivative = 0.88 \[ \int \frac {(c+d x)^{9/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (\frac {140 \, {\left (12 \, {\left (d x - c\right )} D c^{4} + D c^{5} + 9 \, {\left (d x - c\right )} C c^{3} d + C c^{4} d + 6 \, {\left (d x - c\right )} B c^{2} d^{2} + B c^{3} d^{2} + 3 \, {\left (d x - c\right )} A c d^{3} + A c^{2} d^{3}\right )}}{{\left (d x - c\right )} \sqrt {-d x + c} d^{3}} + \frac {15 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} D d^{18} + 147 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} D c d^{18} - 665 \, {\left (-d x + c\right )}^{\frac {3}{2}} D c^{2} d^{18} + 2625 \, \sqrt {-d x + c} D c^{3} d^{18} + 21 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} C d^{19} - 210 \, {\left (-d x + c\right )}^{\frac {3}{2}} C c d^{19} + 1365 \, \sqrt {-d x + c} C c^{2} d^{19} - 35 \, {\left (-d x + c\right )}^{\frac {3}{2}} B d^{20} + 525 \, \sqrt {-d x + c} B c d^{20} + 105 \, \sqrt {-d x + c} A d^{21}}{d^{21}}\right )}}{105 \, d} \] Input:

integrate((d*x+c)^(9/2)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(5/2),x, algori 
thm="giac")
 

Output:

-2/105*(140*(12*(d*x - c)*D*c^4 + D*c^5 + 9*(d*x - c)*C*c^3*d + C*c^4*d + 
6*(d*x - c)*B*c^2*d^2 + B*c^3*d^2 + 3*(d*x - c)*A*c*d^3 + A*c^2*d^3)/((d*x 
 - c)*sqrt(-d*x + c)*d^3) + (15*(d*x - c)^3*sqrt(-d*x + c)*D*d^18 + 147*(d 
*x - c)^2*sqrt(-d*x + c)*D*c*d^18 - 665*(-d*x + c)^(3/2)*D*c^2*d^18 + 2625 
*sqrt(-d*x + c)*D*c^3*d^18 + 21*(d*x - c)^2*sqrt(-d*x + c)*C*d^19 - 210*(- 
d*x + c)^(3/2)*C*c*d^19 + 1365*sqrt(-d*x + c)*C*c^2*d^19 - 35*(-d*x + c)^( 
3/2)*B*d^20 + 525*sqrt(-d*x + c)*B*c*d^20 + 105*sqrt(-d*x + c)*A*d^21)/d^2 
1)/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^{9/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\int \frac {{\left (c+d\,x\right )}^{9/2}\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{{\left (c^2-d^2\,x^2\right )}^{5/2}} \,d x \] Input:

int(((c + d*x)^(9/2)*(A + B*x + C*x^2 + x^3*D))/(c^2 - d^2*x^2)^(5/2),x)
 

Output:

int(((c + d*x)^(9/2)*(A + B*x + C*x^2 + x^3*D))/(c^2 - d^2*x^2)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.40 \[ \int \frac {(c+d x)^{9/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {-\frac {2}{7} d^{5} x^{5}-\frac {62}{35} c \,d^{4} x^{4}-\frac {2}{3} b \,d^{4} x^{3}-\frac {706}{105} c^{2} d^{3} x^{3}-2 a \,d^{4} x^{2}-8 b c \,d^{3} x^{2}-\frac {1482}{35} c^{3} d^{2} x^{2}+12 a c \,d^{3} x +34 b \,c^{2} d^{2} x +\frac {5928}{35} c^{4} d x -\frac {22}{3} a \,c^{2} d^{2}-\frac {68}{3} b \,c^{3} d -\frac {3952}{35} c^{5}}{\sqrt {-d x +c}\, d^{3} \left (-d x +c \right )} \] Input:

int((d*x+c)^(9/2)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(5/2),x)
 

Output:

(2*( - 385*a*c**2*d**2 + 630*a*c*d**3*x - 105*a*d**4*x**2 - 1190*b*c**3*d 
+ 1785*b*c**2*d**2*x - 420*b*c*d**3*x**2 - 35*b*d**4*x**3 - 5928*c**5 + 88 
92*c**4*d*x - 2223*c**3*d**2*x**2 - 353*c**2*d**3*x**3 - 93*c*d**4*x**4 - 
15*d**5*x**5))/(105*sqrt(c - d*x)*d**3*(c - d*x))