\(\int \frac {(c+d x)^{7/2} (A+B x+C x^2+D x^3)}{(c^2-d^2 x^2)^{3/2}} \, dx\) [220]

Optimal result

Integrand size = 41, antiderivative size = 311 \[ \int \frac {(c+d x)^{7/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {\left (c^2 C d+B c d^2+A d^3+c^3 D\right ) (c+d x)^{7/2}}{c d^4 \sqrt {c^2-d^2 x^2}}+\frac {32 c \left (177 c^2 C d+147 B c d^2+105 A d^3+203 c^3 D\right ) \sqrt {c^2-d^2 x^2}}{315 d^4 \sqrt {c+d x}}+\frac {8 \left (177 c^2 C d+147 B c d^2+105 A d^3+203 c^3 D\right ) \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{315 d^4}+\frac {\left (177 c^2 C d+147 B c d^2+105 A d^3+203 c^3 D\right ) (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}{105 c d^4}+\frac {2 (9 C d+7 c D) (c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}{63 d^4}+\frac {2 D (c+d x)^{7/2} \sqrt {c^2-d^2 x^2}}{9 d^4} \] Output:

(A*d^3+B*c*d^2+C*c^2*d+D*c^3)*(d*x+c)^(7/2)/c/d^4/(-d^2*x^2+c^2)^(1/2)+32/ 
315*c*(105*A*d^3+147*B*c*d^2+177*C*c^2*d+203*D*c^3)*(-d^2*x^2+c^2)^(1/2)/d 
^4/(d*x+c)^(1/2)+8/315*(105*A*d^3+147*B*c*d^2+177*C*c^2*d+203*D*c^3)*(d*x+ 
c)^(1/2)*(-d^2*x^2+c^2)^(1/2)/d^4+1/105*(105*A*d^3+147*B*c*d^2+177*C*c^2*d 
+203*D*c^3)*(d*x+c)^(3/2)*(-d^2*x^2+c^2)^(1/2)/c/d^4+2/63*(9*C*d+7*D*c)*(d 
*x+c)^(5/2)*(-d^2*x^2+c^2)^(1/2)/d^4+2/9*D*(d*x+c)^(7/2)*(-d^2*x^2+c^2)^(1 
/2)/d^4
 

Mathematica [A] (verified)

Time = 2.83 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.53 \[ \int \frac {(c+d x)^{7/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {2 \sqrt {c^2-d^2 x^2} \left (-4592 c^5 D-8 c^4 d (501 C-287 D x)+2 c^3 d^2 (-1659 B+x (1002 C+287 D x))+d^5 x^2 (105 A+x (63 B+5 x (9 C+7 D x)))+2 c d^4 x (525 A+x (168 B+x (99 C+70 D x)))+c^2 d^3 (-2415 A+x (1659 B+x (501 C+287 D x)))\right )}{315 d^4 (-c+d x) \sqrt {c+d x}} \] Input:

Integrate[((c + d*x)^(7/2)*(A + B*x + C*x^2 + D*x^3))/(c^2 - d^2*x^2)^(3/2 
),x]
 

Output:

(2*Sqrt[c^2 - d^2*x^2]*(-4592*c^5*D - 8*c^4*d*(501*C - 287*D*x) + 2*c^3*d^ 
2*(-1659*B + x*(1002*C + 287*D*x)) + d^5*x^2*(105*A + x*(63*B + 5*x*(9*C + 
 7*D*x))) + 2*c*d^4*x*(525*A + x*(168*B + x*(99*C + 70*D*x))) + c^2*d^3*(- 
2415*A + x*(1659*B + x*(501*C + 287*D*x)))))/(315*d^4*(-c + d*x)*Sqrt[c + 
d*x])
 

Rubi [A] (verified)

Time = 1.24 (sec) , antiderivative size = 283, 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, 2170, 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.

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

\(\Big \downarrow \) 2166

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2170

\(\displaystyle \frac {(c+d x)^{7/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {-\frac {2 \int -\frac {d (c+d x)^{5/2} \left (77 D c^3+63 C d c^2+63 B d^2 c+2 d (9 C d+7 c D) x c+45 A d^3\right )}{2 \sqrt {c^2-d^2 x^2}}dx}{9 d^4}-\frac {4 c D \sqrt {c^2-d^2 x^2} (c+d x)^{7/2}}{9 d^4}}{2 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(c+d x)^{7/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {\frac {\int \frac {(c+d x)^{5/2} \left (77 D c^3+63 C d c^2+63 B d^2 c+2 d (9 C d+7 c D) x c+45 A d^3\right )}{\sqrt {c^2-d^2 x^2}}dx}{9 d^3}-\frac {4 c D (c+d x)^{7/2} \sqrt {c^2-d^2 x^2}}{9 d^4}}{2 c}\)

\(\Big \downarrow \) 672

\(\displaystyle \frac {(c+d x)^{7/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {\frac {\frac {3}{7} \left (105 A d^3+147 B c d^2+203 c^3 D+177 c^2 C d\right ) \int \frac {(c+d x)^{5/2}}{\sqrt {c^2-d^2 x^2}}dx-\frac {4 c (c+d x)^{5/2} \sqrt {c^2-d^2 x^2} (7 c D+9 C d)}{7 d}}{9 d^3}-\frac {4 c D (c+d x)^{7/2} \sqrt {c^2-d^2 x^2}}{9 d^4}}{2 c}\)

\(\Big \downarrow \) 459

\(\displaystyle \frac {(c+d x)^{7/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {\frac {\frac {3}{7} \left (105 A d^3+147 B c d^2+203 c^3 D+177 c^2 C d\right ) \left (\frac {8}{5} c \int \frac {(c+d x)^{3/2}}{\sqrt {c^2-d^2 x^2}}dx-\frac {2 (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}{5 d}\right )-\frac {4 c (c+d x)^{5/2} \sqrt {c^2-d^2 x^2} (7 c D+9 C d)}{7 d}}{9 d^3}-\frac {4 c D (c+d x)^{7/2} \sqrt {c^2-d^2 x^2}}{9 d^4}}{2 c}\)

\(\Big \downarrow \) 459

\(\displaystyle \frac {(c+d x)^{7/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {\frac {\frac {3}{7} \left (105 A d^3+147 B c d^2+203 c^3 D+177 c^2 C d\right ) \left (\frac {8}{5} c \left (\frac {4}{3} c \int \frac {\sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}dx-\frac {2 \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{3 d}\right )-\frac {2 (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}{5 d}\right )-\frac {4 c (c+d x)^{5/2} \sqrt {c^2-d^2 x^2} (7 c D+9 C d)}{7 d}}{9 d^3}-\frac {4 c D (c+d x)^{7/2} \sqrt {c^2-d^2 x^2}}{9 d^4}}{2 c}\)

\(\Big \downarrow \) 458

\(\displaystyle \frac {(c+d x)^{7/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {\frac {\frac {3}{7} \left (\frac {8}{5} c \left (-\frac {8 c \sqrt {c^2-d^2 x^2}}{3 d \sqrt {c+d x}}-\frac {2 \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{3 d}\right )-\frac {2 (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}{5 d}\right ) \left (105 A d^3+147 B c d^2+203 c^3 D+177 c^2 C d\right )-\frac {4 c (c+d x)^{5/2} \sqrt {c^2-d^2 x^2} (7 c D+9 C d)}{7 d}}{9 d^3}-\frac {4 c D (c+d x)^{7/2} \sqrt {c^2-d^2 x^2}}{9 d^4}}{2 c}\)

Input:

Int[((c + d*x)^(7/2)*(A + B*x + C*x^2 + D*x^3))/(c^2 - d^2*x^2)^(3/2),x]
 

Output:

((c^2*C*d + B*c*d^2 + A*d^3 + c^3*D)*(c + d*x)^(7/2))/(c*d^4*Sqrt[c^2 - d^ 
2*x^2]) - ((-4*c*D*(c + d*x)^(7/2)*Sqrt[c^2 - d^2*x^2])/(9*d^4) + ((-4*c*( 
9*C*d + 7*c*D)*(c + d*x)^(5/2)*Sqrt[c^2 - d^2*x^2])/(7*d) + (3*(177*c^2*C* 
d + 147*B*c*d^2 + 105*A*d^3 + 203*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)/(9*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]
 

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 [A] (verified)

Time = 0.38 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.65

method	result	size
gosper	\(\frac {2 \left (-d x +c \right ) \left (-35 D d^{5} x^{5}-45 C \,d^{5} x^{4}-140 D c \,d^{4} x^{4}-63 B \,d^{5} x^{3}-198 C c \,d^{4} x^{3}-287 D c^{2} d^{3} x^{3}-105 A \,d^{5} x^{2}-336 B c \,d^{4} x^{2}-501 C \,c^{2} d^{3} x^{2}-574 D c^{3} d^{2} x^{2}-1050 A c \,d^{4} x -1659 B \,c^{2} d^{3} x -2004 C \,c^{3} d^{2} x -2296 D c^{4} d x +2415 A \,c^{2} d^{3}+3318 B \,c^{3} d^{2}+4008 C \,c^{4} d +4592 D c^{5}\right ) \left (d x +c \right )^{\frac {3}{2}}}{315 d^{4} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}\)	\(203\)
orering	\(\frac {2 \left (-d x +c \right ) \left (-35 D d^{5} x^{5}-45 C \,d^{5} x^{4}-140 D c \,d^{4} x^{4}-63 B \,d^{5} x^{3}-198 C c \,d^{4} x^{3}-287 D c^{2} d^{3} x^{3}-105 A \,d^{5} x^{2}-336 B c \,d^{4} x^{2}-501 C \,c^{2} d^{3} x^{2}-574 D c^{3} d^{2} x^{2}-1050 A c \,d^{4} x -1659 B \,c^{2} d^{3} x -2004 C \,c^{3} d^{2} x -2296 D c^{4} d x +2415 A \,c^{2} d^{3}+3318 B \,c^{3} d^{2}+4008 C \,c^{4} d +4592 D c^{5}\right ) \left (d x +c \right )^{\frac {3}{2}}}{315 d^{4} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}\)	\(203\)
default	\(\frac {2 \sqrt {-d^{2} x^{2}+c^{2}}\, \left (-35 D d^{5} x^{5}-45 C \,d^{5} x^{4}-140 D c \,d^{4} x^{4}-63 B \,d^{5} x^{3}-198 C c \,d^{4} x^{3}-287 D c^{2} d^{3} x^{3}-105 A \,d^{5} x^{2}-336 B c \,d^{4} x^{2}-501 C \,c^{2} d^{3} x^{2}-574 D c^{3} d^{2} x^{2}-1050 A c \,d^{4} x -1659 B \,c^{2} d^{3} x -2004 C \,c^{3} d^{2} x -2296 D c^{4} d x +2415 A \,c^{2} d^{3}+3318 B \,c^{3} d^{2}+4008 C \,c^{4} d +4592 D c^{5}\right )}{315 \sqrt {d x +c}\, \left (-d x +c \right ) d^{4}}\)	\(205\)

Input:

int((d*x+c)^(7/2)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(3/2),x,method=_RETUR 
NVERBOSE)
 

Output:

2/315*(-d*x+c)*(-35*D*d^5*x^5-45*C*d^5*x^4-140*D*c*d^4*x^4-63*B*d^5*x^3-19 
8*C*c*d^4*x^3-287*D*c^2*d^3*x^3-105*A*d^5*x^2-336*B*c*d^4*x^2-501*C*c^2*d^ 
3*x^2-574*D*c^3*d^2*x^2-1050*A*c*d^4*x-1659*B*c^2*d^3*x-2004*C*c^3*d^2*x-2 
296*D*c^4*d*x+2415*A*c^2*d^3+3318*B*c^3*d^2+4008*C*c^4*d+4592*D*c^5)*(d*x+ 
c)^(3/2)/d^4/(-d^2*x^2+c^2)^(3/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.64 \[ \int \frac {(c+d x)^{7/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {2 \, {\left (35 \, D d^{5} x^{5} - 4592 \, D c^{5} - 4008 \, C c^{4} d - 3318 \, B c^{3} d^{2} - 2415 \, A c^{2} d^{3} + 5 \, {\left (28 \, D c d^{4} + 9 \, C d^{5}\right )} x^{4} + {\left (287 \, D c^{2} d^{3} + 198 \, C c d^{4} + 63 \, B d^{5}\right )} x^{3} + {\left (574 \, D c^{3} d^{2} + 501 \, C c^{2} d^{3} + 336 \, B c d^{4} + 105 \, A d^{5}\right )} x^{2} + {\left (2296 \, D c^{4} d + 2004 \, C c^{3} d^{2} + 1659 \, B c^{2} d^{3} + 1050 \, A c d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{315 \, {\left (d^{6} x^{2} - c^{2} d^{4}\right )}} \] Input:

integrate((d*x+c)^(7/2)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(3/2),x, algori 
thm="fricas")
 

Output:

2/315*(35*D*d^5*x^5 - 4592*D*c^5 - 4008*C*c^4*d - 3318*B*c^3*d^2 - 2415*A* 
c^2*d^3 + 5*(28*D*c*d^4 + 9*C*d^5)*x^4 + (287*D*c^2*d^3 + 198*C*c*d^4 + 63 
*B*d^5)*x^3 + (574*D*c^3*d^2 + 501*C*c^2*d^3 + 336*B*c*d^4 + 105*A*d^5)*x^ 
2 + (2296*D*c^4*d + 2004*C*c^3*d^2 + 1659*B*c^2*d^3 + 1050*A*c*d^4)*x)*sqr 
t(-d^2*x^2 + c^2)*sqrt(d*x + c)/(d^6*x^2 - c^2*d^4)
 

Sympy [F]

\[ \int \frac {(c+d x)^{7/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\int \frac {\left (c + d x\right )^{\frac {7}{2}} \left (A + B x + C x^{2} + D x^{3}\right )}{\left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((d*x+c)**(7/2)*(D*x**3+C*x**2+B*x+A)/(-d**2*x**2+c**2)**(3/2),x)
 

Output:

Integral((c + d*x)**(7/2)*(A + B*x + C*x**2 + D*x**3)/(-(-c + d*x)*(c + d* 
x))**(3/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.64 \[ \int \frac {(c+d x)^{7/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=-\frac {2 \, {\left (d^{2} x^{2} + 10 \, c d x - 23 \, c^{2}\right )} A}{3 \, \sqrt {-d x + c} d} - \frac {2 \, {\left (3 \, d^{3} x^{3} + 16 \, c d^{2} x^{2} + 79 \, c^{2} d x - 158 \, c^{3}\right )} B}{15 \, \sqrt {-d x + c} d^{2}} - \frac {2 \, {\left (15 \, d^{4} x^{4} + 66 \, c d^{3} x^{3} + 167 \, c^{2} d^{2} x^{2} + 668 \, c^{3} d x - 1336 \, c^{4}\right )} C}{105 \, \sqrt {-d x + c} d^{3}} - \frac {2 \, {\left (5 \, d^{5} x^{5} + 20 \, c d^{4} x^{4} + 41 \, c^{2} d^{3} x^{3} + 82 \, c^{3} d^{2} x^{2} + 328 \, c^{4} d x - 656 \, c^{5}\right )} D}{45 \, \sqrt {-d x + c} d^{4}} \] Input:

integrate((d*x+c)^(7/2)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(3/2),x, algori 
thm="maxima")
 

Output:

-2/3*(d^2*x^2 + 10*c*d*x - 23*c^2)*A/(sqrt(-d*x + c)*d) - 2/15*(3*d^3*x^3 
+ 16*c*d^2*x^2 + 79*c^2*d*x - 158*c^3)*B/(sqrt(-d*x + c)*d^2) - 2/105*(15* 
d^4*x^4 + 66*c*d^3*x^3 + 167*c^2*d^2*x^2 + 668*c^3*d*x - 1336*c^4)*C/(sqrt 
(-d*x + c)*d^3) - 2/45*(5*d^5*x^5 + 20*c*d^4*x^4 + 41*c^2*d^3*x^3 + 82*c^3 
*d^2*x^2 + 328*c^4*d*x - 656*c^5)*D/(sqrt(-d*x + c)*d^4)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 324, normalized size of antiderivative = 1.04 \[ \int \frac {(c+d x)^{7/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {2 \, {\left (\frac {1260 \, {\left (D c^{5} + C c^{4} d + B c^{3} d^{2} + A c^{2} d^{3}\right )}}{\sqrt {-d x + c} d^{3}} + \frac {35 \, {\left (d x - c\right )}^{4} \sqrt {-d x + c} D d^{24} + 315 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} D c d^{24} + 1197 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} D c^{2} d^{24} - 2625 \, {\left (-d x + c\right )}^{\frac {3}{2}} D c^{3} d^{24} + 5040 \, \sqrt {-d x + c} D c^{4} d^{24} + 45 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} C d^{25} + 378 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} C c d^{25} - 1365 \, {\left (-d x + c\right )}^{\frac {3}{2}} C c^{2} d^{25} + 3780 \, \sqrt {-d x + c} C c^{3} d^{25} + 63 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} B d^{26} - 525 \, {\left (-d x + c\right )}^{\frac {3}{2}} B c d^{26} + 2520 \, \sqrt {-d x + c} B c^{2} d^{26} - 105 \, {\left (-d x + c\right )}^{\frac {3}{2}} A d^{27} + 1260 \, \sqrt {-d x + c} A c d^{27}}{d^{27}}\right )}}{315 \, d} \] Input:

integrate((d*x+c)^(7/2)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(3/2),x, algori 
thm="giac")
 

Output:

2/315*(1260*(D*c^5 + C*c^4*d + B*c^3*d^2 + A*c^2*d^3)/(sqrt(-d*x + c)*d^3) 
 + (35*(d*x - c)^4*sqrt(-d*x + c)*D*d^24 + 315*(d*x - c)^3*sqrt(-d*x + c)* 
D*c*d^24 + 1197*(d*x - c)^2*sqrt(-d*x + c)*D*c^2*d^24 - 2625*(-d*x + c)^(3 
/2)*D*c^3*d^24 + 5040*sqrt(-d*x + c)*D*c^4*d^24 + 45*(d*x - c)^3*sqrt(-d*x 
 + c)*C*d^25 + 378*(d*x - c)^2*sqrt(-d*x + c)*C*c*d^25 - 1365*(-d*x + c)^( 
3/2)*C*c^2*d^25 + 3780*sqrt(-d*x + c)*C*c^3*d^25 + 63*(d*x - c)^2*sqrt(-d* 
x + c)*B*d^26 - 525*(-d*x + c)^(3/2)*B*c*d^26 + 2520*sqrt(-d*x + c)*B*c^2* 
d^26 - 105*(-d*x + c)^(3/2)*A*d^27 + 1260*sqrt(-d*x + c)*A*c*d^27)/d^27)/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^{7/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\int \frac {{\left (c+d\,x\right )}^{7/2}\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{{\left (c^2-d^2\,x^2\right )}^{3/2}} \,d x \] Input:

int(((c + d*x)^(7/2)*(A + B*x + C*x^2 + x^3*D))/(c^2 - d^2*x^2)^(3/2),x)
 

Output:

int(((c + d*x)^(7/2)*(A + B*x + C*x^2 + x^3*D))/(c^2 - d^2*x^2)^(3/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.41 \[ \int \frac {(c+d x)^{7/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {-\frac {2}{9} d^{5} x^{5}-\frac {74}{63} c \,d^{4} x^{4}-\frac {2}{5} b \,d^{4} x^{3}-\frac {194}{63} c^{2} d^{3} x^{3}-\frac {2}{3} a \,d^{4} x^{2}-\frac {32}{15} b c \,d^{3} x^{2}-\frac {430}{63} c^{3} d^{2} x^{2}-\frac {20}{3} a c \,d^{3} x -\frac {158}{15} b \,c^{2} d^{2} x -\frac {1720}{63} c^{4} d x +\frac {46}{3} a \,c^{2} d^{2}+\frac {316}{15} b \,c^{3} d +\frac {3440}{63} c^{5}}{\sqrt {-d x +c}\, d^{3}} \] Input:

int((d*x+c)^(7/2)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(3/2),x)
 

Output:

(2*(2415*a*c**2*d**2 - 1050*a*c*d**3*x - 105*a*d**4*x**2 + 3318*b*c**3*d - 
 1659*b*c**2*d**2*x - 336*b*c*d**3*x**2 - 63*b*d**4*x**3 + 8600*c**5 - 430 
0*c**4*d*x - 1075*c**3*d**2*x**2 - 485*c**2*d**3*x**3 - 185*c*d**4*x**4 - 
35*d**5*x**5))/(315*sqrt(c - d*x)*d**3)