\(\int (c+d x)^{3/2} \sqrt {c^2-d^2 x^2} (A+B x+C x^2+D x^3) \, dx\) [194]

Optimal result

Integrand size = 41, antiderivative size = 306 \[ \int (c+d x)^{3/2} \sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=-\frac {64 c^2 \left (117 c^2 C d+143 B c d^2+429 A d^3+71 c^3 D\right ) \left (c^2-d^2 x^2\right )^{3/2}}{45045 d^4 (c+d x)^{3/2}}-\frac {16 c \left (117 c^2 C d+143 B c d^2+429 A d^3+71 c^3 D\right ) \left (c^2-d^2 x^2\right )^{3/2}}{15015 d^4 \sqrt {c+d x}}-\frac {2 \left (117 c^2 C d+143 B c d^2+429 A d^3+71 c^3 D\right ) \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}{3003 d^4}+\frac {2 \left (78 c C d-143 B d^2-125 c^2 D\right ) (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}}{1287 d^4}-\frac {2 (13 C d-19 c D) (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^{3/2}}{143 d^4}-\frac {2 D (c+d x)^{7/2} \left (c^2-d^2 x^2\right )^{3/2}}{13 d^4} \] Output:

-64/45045*c^2*(429*A*d^3+143*B*c*d^2+117*C*c^2*d+71*D*c^3)*(-d^2*x^2+c^2)^ 
(3/2)/d^4/(d*x+c)^(3/2)-16/15015*c*(429*A*d^3+143*B*c*d^2+117*C*c^2*d+71*D 
*c^3)*(-d^2*x^2+c^2)^(3/2)/d^4/(d*x+c)^(1/2)-2/3003*(429*A*d^3+143*B*c*d^2 
+117*C*c^2*d+71*D*c^3)*(d*x+c)^(1/2)*(-d^2*x^2+c^2)^(3/2)/d^4+2/1287*(-143 
*B*d^2+78*C*c*d-125*D*c^2)*(d*x+c)^(3/2)*(-d^2*x^2+c^2)^(3/2)/d^4-2/143*(1 
3*C*d-19*D*c)*(d*x+c)^(5/2)*(-d^2*x^2+c^2)^(3/2)/d^4-2/13*D*(d*x+c)^(7/2)* 
(-d^2*x^2+c^2)^(3/2)/d^4
 

Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.55 \[ \int (c+d x)^{3/2} \sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=-\frac {2 (c-d x) \sqrt {c^2-d^2 x^2} \left (6896 c^5 D+24 c^4 d (403 C+431 D x)+2 c^3 d^2 \left (7579 B+7254 C x+6465 D x^2\right )+5 d^5 x^2 (1287 A+7 x (143 B+9 x (13 C+11 D x)))+6 c d^4 x (3861 A+5 x (572 B+7 x (65 C+54 D x)))+c^2 d^3 (30459 A+x (22737 B+5 x (3627 C+3017 D x)))\right )}{45045 d^4 \sqrt {c+d x}} \] Input:

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

Output:

(-2*(c - d*x)*Sqrt[c^2 - d^2*x^2]*(6896*c^5*D + 24*c^4*d*(403*C + 431*D*x) 
 + 2*c^3*d^2*(7579*B + 7254*C*x + 6465*D*x^2) + 5*d^5*x^2*(1287*A + 7*x*(1 
43*B + 9*x*(13*C + 11*D*x))) + 6*c*d^4*x*(3861*A + 5*x*(572*B + 7*x*(65*C 
+ 54*D*x))) + c^2*d^3*(30459*A + x*(22737*B + 5*x*(3627*C + 3017*D*x)))))/ 
(45045*d^4*Sqrt[c + d*x])
 

Rubi [A] (verified)

Time = 1.12 (sec) , antiderivative size = 274, normalized size of antiderivative = 0.90, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {2170, 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.

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

\(\Big \downarrow \) 2170

\(\displaystyle -\frac {2 \int -\frac {1}{2} (c+d x)^{3/2} \sqrt {c^2-d^2 x^2} \left ((13 C d-19 c D) x^2 d^4+\left (D c^2+13 B d^2\right ) x d^3+\left (7 D c^3+13 A d^3\right ) d^2\right )dx}{13 d^5}-\frac {2 D \left (c^2-d^2 x^2\right )^{3/2} (c+d x)^{7/2}}{13 d^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int (c+d x)^{3/2} \sqrt {c^2-d^2 x^2} \left ((13 C d-19 c D) x^2 d^4+\left (D c^2+13 B d^2\right ) x d^3+\left (7 D c^3+13 A d^3\right ) d^2\right )dx}{13 d^5}-\frac {2 D (c+d x)^{7/2} \left (c^2-d^2 x^2\right )^{3/2}}{13 d^4}\)

\(\Big \downarrow \) 2170

\(\displaystyle \frac {-\frac {2 \int -\frac {1}{2} d^6 (c+d x)^{3/2} \left (-18 D c^3+65 C d c^2+143 A d^3-d \left (-125 D c^2+78 C d c-143 B d^2\right ) x\right ) \sqrt {c^2-d^2 x^2}dx}{11 d^4}-\frac {2}{11} d \left (c^2-d^2 x^2\right )^{3/2} (c+d x)^{5/2} (13 C d-19 c D)}{13 d^5}-\frac {2 D (c+d x)^{7/2} \left (c^2-d^2 x^2\right )^{3/2}}{13 d^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{11} d^2 \int (c+d x)^{3/2} \left (-18 D c^3+65 C d c^2+143 A d^3-d \left (-125 D c^2+78 C d c-143 B d^2\right ) x\right ) \sqrt {c^2-d^2 x^2}dx-\frac {2}{11} d (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^{3/2} (13 C d-19 c D)}{13 d^5}-\frac {2 D (c+d x)^{7/2} \left (c^2-d^2 x^2\right )^{3/2}}{13 d^4}\)

\(\Big \downarrow \) 672

\(\displaystyle \frac {\frac {1}{11} d^2 \left (\frac {1}{3} \left (429 A d^3+143 B c d^2+71 c^3 D+117 c^2 C d\right ) \int (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}dx+\frac {2 (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2} \left (-143 B d^2-125 c^2 D+78 c C d\right )}{9 d}\right )-\frac {2}{11} d (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^{3/2} (13 C d-19 c D)}{13 d^5}-\frac {2 D (c+d x)^{7/2} \left (c^2-d^2 x^2\right )^{3/2}}{13 d^4}\)

\(\Big \downarrow \) 459

\(\displaystyle \frac {\frac {1}{11} d^2 \left (\frac {1}{3} \left (429 A d^3+143 B c d^2+71 c^3 D+117 c^2 C d\right ) \left (\frac {8}{7} c \int \sqrt {c+d x} \sqrt {c^2-d^2 x^2}dx-\frac {2 \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}{7 d}\right )+\frac {2 (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2} \left (-143 B d^2-125 c^2 D+78 c C d\right )}{9 d}\right )-\frac {2}{11} d (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^{3/2} (13 C d-19 c D)}{13 d^5}-\frac {2 D (c+d x)^{7/2} \left (c^2-d^2 x^2\right )^{3/2}}{13 d^4}\)

\(\Big \downarrow \) 459

\(\displaystyle \frac {\frac {1}{11} d^2 \left (\frac {1}{3} \left (429 A d^3+143 B c d^2+71 c^3 D+117 c^2 C d\right ) \left (\frac {8}{7} c \left (\frac {4}{5} c \int \frac {\sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}dx-\frac {2 \left (c^2-d^2 x^2\right )^{3/2}}{5 d \sqrt {c+d x}}\right )-\frac {2 \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}{7 d}\right )+\frac {2 (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2} \left (-143 B d^2-125 c^2 D+78 c C d\right )}{9 d}\right )-\frac {2}{11} d (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^{3/2} (13 C d-19 c D)}{13 d^5}-\frac {2 D (c+d x)^{7/2} \left (c^2-d^2 x^2\right )^{3/2}}{13 d^4}\)

\(\Big \downarrow \) 458

\(\displaystyle \frac {\frac {1}{11} d^2 \left (\frac {1}{3} \left (\frac {8}{7} c \left (-\frac {2 \left (c^2-d^2 x^2\right )^{3/2}}{5 d \sqrt {c+d x}}-\frac {8 c \left (c^2-d^2 x^2\right )^{3/2}}{15 d (c+d x)^{3/2}}\right )-\frac {2 \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}{7 d}\right ) \left (429 A d^3+143 B c d^2+71 c^3 D+117 c^2 C d\right )+\frac {2 (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2} \left (-143 B d^2-125 c^2 D+78 c C d\right )}{9 d}\right )-\frac {2}{11} d (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^{3/2} (13 C d-19 c D)}{13 d^5}-\frac {2 D (c+d x)^{7/2} \left (c^2-d^2 x^2\right )^{3/2}}{13 d^4}\)

Input:

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

Output:

(-2*D*(c + d*x)^(7/2)*(c^2 - d^2*x^2)^(3/2))/(13*d^4) + ((-2*d*(13*C*d - 1 
9*c*D)*(c + d*x)^(5/2)*(c^2 - d^2*x^2)^(3/2))/11 + (d^2*((2*(78*c*C*d - 14 
3*B*d^2 - 125*c^2*D)*(c + d*x)^(3/2)*(c^2 - d^2*x^2)^(3/2))/(9*d) + ((117* 
c^2*C*d + 143*B*c*d^2 + 429*A*d^3 + 71*c^3*D)*((-2*Sqrt[c + d*x]*(c^2 - d^ 
2*x^2)^(3/2))/(7*d) + (8*c*((-8*c*(c^2 - d^2*x^2)^(3/2))/(15*d*(c + d*x)^( 
3/2)) - (2*(c^2 - d^2*x^2)^(3/2))/(5*d*Sqrt[c + d*x])))/7))/3))/11)/(13*d^ 
5)
 

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 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 = 0.36 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.66

method	result	size
gosper	\(-\frac {2 \left (-d x +c \right ) \left (3465 D d^{5} x^{5}+4095 C \,d^{5} x^{4}+11340 D c \,d^{4} x^{4}+5005 B \,d^{5} x^{3}+13650 C c \,d^{4} x^{3}+15085 D c^{2} d^{3} x^{3}+6435 A \,d^{5} x^{2}+17160 B c \,d^{4} x^{2}+18135 C \,c^{2} d^{3} x^{2}+12930 D c^{3} d^{2} x^{2}+23166 A c \,d^{4} x +22737 B \,c^{2} d^{3} x +14508 C \,c^{3} d^{2} x +10344 D c^{4} d x +30459 A \,c^{2} d^{3}+15158 B \,c^{3} d^{2}+9672 C \,c^{4} d +6896 D c^{5}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{45045 d^{4} \sqrt {d x +c}}\)	\(203\)
default	\(-\frac {2 \left (-d x +c \right ) \left (3465 D d^{5} x^{5}+4095 C \,d^{5} x^{4}+11340 D c \,d^{4} x^{4}+5005 B \,d^{5} x^{3}+13650 C c \,d^{4} x^{3}+15085 D c^{2} d^{3} x^{3}+6435 A \,d^{5} x^{2}+17160 B c \,d^{4} x^{2}+18135 C \,c^{2} d^{3} x^{2}+12930 D c^{3} d^{2} x^{2}+23166 A c \,d^{4} x +22737 B \,c^{2} d^{3} x +14508 C \,c^{3} d^{2} x +10344 D c^{4} d x +30459 A \,c^{2} d^{3}+15158 B \,c^{3} d^{2}+9672 C \,c^{4} d +6896 D c^{5}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{45045 d^{4} \sqrt {d x +c}}\)	\(203\)
orering	\(-\frac {2 \left (-d x +c \right ) \left (3465 D d^{5} x^{5}+4095 C \,d^{5} x^{4}+11340 D c \,d^{4} x^{4}+5005 B \,d^{5} x^{3}+13650 C c \,d^{4} x^{3}+15085 D c^{2} d^{3} x^{3}+6435 A \,d^{5} x^{2}+17160 B c \,d^{4} x^{2}+18135 C \,c^{2} d^{3} x^{2}+12930 D c^{3} d^{2} x^{2}+23166 A c \,d^{4} x +22737 B \,c^{2} d^{3} x +14508 C \,c^{3} d^{2} x +10344 D c^{4} d x +30459 A \,c^{2} d^{3}+15158 B \,c^{3} d^{2}+9672 C \,c^{4} d +6896 D c^{5}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{45045 d^{4} \sqrt {d x +c}}\)	\(203\)

Input:

int((d*x+c)^(3/2)*(-d^2*x^2+c^2)^(1/2)*(D*x^3+C*x^2+B*x+A),x,method=_RETUR 
NVERBOSE)
 

Output:

-2/45045*(-d*x+c)*(3465*D*d^5*x^5+4095*C*d^5*x^4+11340*D*c*d^4*x^4+5005*B* 
d^5*x^3+13650*C*c*d^4*x^3+15085*D*c^2*d^3*x^3+6435*A*d^5*x^2+17160*B*c*d^4 
*x^2+18135*C*c^2*d^3*x^2+12930*D*c^3*d^2*x^2+23166*A*c*d^4*x+22737*B*c^2*d 
^3*x+14508*C*c^3*d^2*x+10344*D*c^4*d*x+30459*A*c^2*d^3+15158*B*c^3*d^2+967 
2*C*c^4*d+6896*D*c^5)*(-d^2*x^2+c^2)^(1/2)/d^4/(d*x+c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.78 \[ \int (c+d x)^{3/2} \sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {2 \, {\left (3465 \, D d^{6} x^{6} - 6896 \, D c^{6} - 9672 \, C c^{5} d - 15158 \, B c^{4} d^{2} - 30459 \, A c^{3} d^{3} + 315 \, {\left (25 \, D c d^{5} + 13 \, C d^{6}\right )} x^{5} + 35 \, {\left (107 \, D c^{2} d^{4} + 273 \, C c d^{5} + 143 \, B d^{6}\right )} x^{4} - 5 \, {\left (431 \, D c^{3} d^{3} - 897 \, C c^{2} d^{4} - 2431 \, B c d^{5} - 1287 \, A d^{6}\right )} x^{3} - 3 \, {\left (862 \, D c^{4} d^{2} + 1209 \, C c^{3} d^{3} - 1859 \, B c^{2} d^{4} - 5577 \, A c d^{5}\right )} x^{2} - {\left (3448 \, D c^{5} d + 4836 \, C c^{4} d^{2} + 7579 \, B c^{3} d^{3} - 7293 \, A c^{2} d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{45045 \, {\left (d^{5} x + c d^{4}\right )}} \] Input:

integrate((d*x+c)^(3/2)*(-d^2*x^2+c^2)^(1/2)*(D*x^3+C*x^2+B*x+A),x, algori 
thm="fricas")
 

Output:

2/45045*(3465*D*d^6*x^6 - 6896*D*c^6 - 9672*C*c^5*d - 15158*B*c^4*d^2 - 30 
459*A*c^3*d^3 + 315*(25*D*c*d^5 + 13*C*d^6)*x^5 + 35*(107*D*c^2*d^4 + 273* 
C*c*d^5 + 143*B*d^6)*x^4 - 5*(431*D*c^3*d^3 - 897*C*c^2*d^4 - 2431*B*c*d^5 
 - 1287*A*d^6)*x^3 - 3*(862*D*c^4*d^2 + 1209*C*c^3*d^3 - 1859*B*c^2*d^4 - 
5577*A*c*d^5)*x^2 - (3448*D*c^5*d + 4836*C*c^4*d^2 + 7579*B*c^3*d^3 - 7293 
*A*c^2*d^4)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)/(d^5*x + c*d^4)
                                                                                    
                                                                                    
 

Sympy [F]

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

integrate((d*x+c)**(3/2)*(-d**2*x**2+c**2)**(1/2)*(D*x**3+C*x**2+B*x+A),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 301, normalized size of antiderivative = 0.98 \[ \int (c+d x)^{3/2} \sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {2 \, {\left (15 \, d^{3} x^{3} + 39 \, c d^{2} x^{2} + 17 \, c^{2} d x - 71 \, c^{3}\right )} {\left (d x + c\right )} \sqrt {-d x + c} A}{105 \, {\left (d^{2} x + c d\right )}} + \frac {2 \, {\left (35 \, d^{4} x^{4} + 85 \, c d^{3} x^{3} + 39 \, c^{2} d^{2} x^{2} - 53 \, c^{3} d x - 106 \, c^{4}\right )} {\left (d x + c\right )} \sqrt {-d x + c} B}{315 \, {\left (d^{3} x + c d^{2}\right )}} + \frac {2 \, {\left (105 \, d^{5} x^{5} + 245 \, c d^{4} x^{4} + 115 \, c^{2} d^{3} x^{3} - 93 \, c^{3} d^{2} x^{2} - 124 \, c^{4} d x - 248 \, c^{5}\right )} {\left (d x + c\right )} \sqrt {-d x + c} C}{1155 \, {\left (d^{4} x + c d^{3}\right )}} + \frac {2 \, {\left (3465 \, d^{6} x^{6} + 7875 \, c d^{5} x^{5} + 3745 \, c^{2} d^{4} x^{4} - 2155 \, c^{3} d^{3} x^{3} - 2586 \, c^{4} d^{2} x^{2} - 3448 \, c^{5} d x - 6896 \, c^{6}\right )} {\left (d x + c\right )} \sqrt {-d x + c} D}{45045 \, {\left (d^{5} x + c d^{4}\right )}} \] Input:

integrate((d*x+c)^(3/2)*(-d^2*x^2+c^2)^(1/2)*(D*x^3+C*x^2+B*x+A),x, algori 
thm="maxima")
 

Output:

2/105*(15*d^3*x^3 + 39*c*d^2*x^2 + 17*c^2*d*x - 71*c^3)*(d*x + c)*sqrt(-d* 
x + c)*A/(d^2*x + c*d) + 2/315*(35*d^4*x^4 + 85*c*d^3*x^3 + 39*c^2*d^2*x^2 
 - 53*c^3*d*x - 106*c^4)*(d*x + c)*sqrt(-d*x + c)*B/(d^3*x + c*d^2) + 2/11 
55*(105*d^5*x^5 + 245*c*d^4*x^4 + 115*c^2*d^3*x^3 - 93*c^3*d^2*x^2 - 124*c 
^4*d*x - 248*c^5)*(d*x + c)*sqrt(-d*x + c)*C/(d^4*x + c*d^3) + 2/45045*(34 
65*d^6*x^6 + 7875*c*d^5*x^5 + 3745*c^2*d^4*x^4 - 2155*c^3*d^3*x^3 - 2586*c 
^4*d^2*x^2 - 3448*c^5*d*x - 6896*c^6)*(d*x + c)*sqrt(-d*x + c)*D/(d^5*x + 
c*d^4)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1170 vs. \(2 (270) = 540\).

Time = 0.13 (sec) , antiderivative size = 1170, normalized size of antiderivative = 3.82 \[ \int (c+d x)^{3/2} \sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\text {Too large to display} \] Input:

integrate((d*x+c)^(3/2)*(-d^2*x^2+c^2)^(1/2)*(D*x^3+C*x^2+B*x+A),x, algori 
thm="giac")
 

Output:

-2/45045*(45045*sqrt(-d*x + c)*A*c^3*d^3 - 15015*((-d*x + c)^(3/2) - 3*sqr 
t(-d*x + c)*c)*B*c^3*d^2 - 15015*((-d*x + c)^(3/2) - 3*sqrt(-d*x + c)*c)*A 
*c^2*d^3 + 3003*(3*(d*x - c)^2*sqrt(-d*x + c) - 10*(-d*x + c)^(3/2)*c + 15 
*sqrt(-d*x + c)*c^2)*C*c^3*d + 3003*(3*(d*x - c)^2*sqrt(-d*x + c) - 10*(-d 
*x + c)^(3/2)*c + 15*sqrt(-d*x + c)*c^2)*B*c^2*d^2 - 3003*(3*(d*x - c)^2*s 
qrt(-d*x + c) - 10*(-d*x + c)^(3/2)*c + 15*sqrt(-d*x + c)*c^2)*A*c*d^3 + 1 
287*(5*(d*x - c)^3*sqrt(-d*x + c) + 21*(d*x - c)^2*sqrt(-d*x + c)*c - 35*( 
-d*x + c)^(3/2)*c^2 + 35*sqrt(-d*x + c)*c^3)*D*c^3 + 1287*(5*(d*x - c)^3*s 
qrt(-d*x + c) + 21*(d*x - c)^2*sqrt(-d*x + c)*c - 35*(-d*x + c)^(3/2)*c^2 
+ 35*sqrt(-d*x + c)*c^3)*C*c^2*d - 1287*(5*(d*x - c)^3*sqrt(-d*x + c) + 21 
*(d*x - c)^2*sqrt(-d*x + c)*c - 35*(-d*x + c)^(3/2)*c^2 + 35*sqrt(-d*x + c 
)*c^3)*B*c*d^2 - 1287*(5*(d*x - c)^3*sqrt(-d*x + c) + 21*(d*x - c)^2*sqrt( 
-d*x + c)*c - 35*(-d*x + c)^(3/2)*c^2 + 35*sqrt(-d*x + c)*c^3)*A*d^3 + 143 
*(35*(d*x - c)^4*sqrt(-d*x + c) + 180*(d*x - c)^3*sqrt(-d*x + c)*c + 378*( 
d*x - c)^2*sqrt(-d*x + c)*c^2 - 420*(-d*x + c)^(3/2)*c^3 + 315*sqrt(-d*x + 
 c)*c^4)*D*c^2 - 143*(35*(d*x - c)^4*sqrt(-d*x + c) + 180*(d*x - c)^3*sqrt 
(-d*x + c)*c + 378*(d*x - c)^2*sqrt(-d*x + c)*c^2 - 420*(-d*x + c)^(3/2)*c 
^3 + 315*sqrt(-d*x + c)*c^4)*C*c*d - 143*(35*(d*x - c)^4*sqrt(-d*x + c) + 
180*(d*x - c)^3*sqrt(-d*x + c)*c + 378*(d*x - c)^2*sqrt(-d*x + c)*c^2 - 42 
0*(-d*x + c)^(3/2)*c^3 + 315*sqrt(-d*x + c)*c^4)*B*d^2 - 65*(63*(d*x - ...
 

Mupad [F(-1)]

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

int((c^2 - d^2*x^2)^(1/2)*(c + d*x)^(3/2)*(A + B*x + C*x^2 + x^3*D),x)
 

Output:

int((c^2 - d^2*x^2)^(1/2)*(c + d*x)^(3/2)*(A + B*x + C*x^2 + x^3*D), x)
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.53 \[ \int (c+d x)^{3/2} \sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {2 \sqrt {-d x +c}\, \left (3465 d^{6} x^{6}+11970 c \,d^{5} x^{5}+5005 b \,d^{5} x^{4}+13300 c^{2} d^{4} x^{4}+6435 a \,d^{5} x^{3}+12155 b c \,d^{4} x^{3}+2330 c^{3} d^{3} x^{3}+16731 a c \,d^{4} x^{2}+5577 b \,c^{2} d^{3} x^{2}-6213 c^{4} d^{2} x^{2}+7293 a \,c^{2} d^{3} x -7579 b \,c^{3} d^{2} x -8284 c^{5} d x -30459 a \,c^{3} d^{2}-15158 b \,c^{4} d -16568 c^{6}\right )}{45045 d^{3}} \] Input:

int((d*x+c)^(3/2)*(-d^2*x^2+c^2)^(1/2)*(D*x^3+C*x^2+B*x+A),x)
 

Output:

(2*sqrt(c - d*x)*( - 30459*a*c**3*d**2 + 7293*a*c**2*d**3*x + 16731*a*c*d* 
*4*x**2 + 6435*a*d**5*x**3 - 15158*b*c**4*d - 7579*b*c**3*d**2*x + 5577*b* 
c**2*d**3*x**2 + 12155*b*c*d**4*x**3 + 5005*b*d**5*x**4 - 16568*c**6 - 828 
4*c**5*d*x - 6213*c**4*d**2*x**2 + 2330*c**3*d**3*x**3 + 13300*c**2*d**4*x 
**4 + 11970*c*d**5*x**5 + 3465*d**6*x**6))/(45045*d**3)