\(\int \frac {(c^2-d^2 x^2)^{3/2} (A+B x+C x^2+D x^3)}{\sqrt {c+d x}} \, dx\) [205]

Optimal result

Integrand size = 41, antiderivative size = 244 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c+d x}} \, dx=-\frac {8 c \left (247 c^2 C d-143 B c d^2+1287 A d^3-63 c^3 D\right ) \left (c^2-d^2 x^2\right )^{5/2}}{45045 d^4 (c+d x)^{5/2}}-\frac {2 \left (247 c^2 C d-143 B c d^2+1287 A d^3-63 c^3 D\right ) \left (c^2-d^2 x^2\right )^{5/2}}{9009 d^4 (c+d x)^{3/2}}+\frac {2 \left (130 c C d-143 B d^2-153 c^2 D\right ) \left (c^2-d^2 x^2\right )^{5/2}}{1287 d^4 \sqrt {c+d x}}-\frac {2 (13 C d-23 c D) \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2}}{143 d^4}-\frac {2 D (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{5/2}}{13 d^4} \] Output:

-8/45045*c*(1287*A*d^3-143*B*c*d^2+247*C*c^2*d-63*D*c^3)*(-d^2*x^2+c^2)^(5 
/2)/d^4/(d*x+c)^(5/2)-2/9009*(1287*A*d^3-143*B*c*d^2+247*C*c^2*d-63*D*c^3) 
*(-d^2*x^2+c^2)^(5/2)/d^4/(d*x+c)^(3/2)+2/1287*(-143*B*d^2+130*C*c*d-153*D 
*c^2)*(-d^2*x^2+c^2)^(5/2)/d^4/(d*x+c)^(1/2)-2/143*(13*C*d-23*D*c)*(d*x+c) 
^(1/2)*(-d^2*x^2+c^2)^(5/2)/d^4-2/13*D*(d*x+c)^(3/2)*(-d^2*x^2+c^2)^(5/2)/ 
d^4
 

Mathematica [A] (verified)

Time = 2.42 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.56 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c+d x}} \, dx=-\frac {2 (c-d x)^2 \sqrt {c^2-d^2 x^2} \left (1008 c^4 D+8 c^3 d (221 C+315 D x)+c d^3 \left (11583 A+9295 B x+7735 C x^2+6615 D x^3\right )+2 c^2 d^2 (1859 B+5 x (442 C+441 D x))+5 d^4 x \left (1287 A+7 x \left (143 B+117 C x+99 D x^2\right )\right )\right )}{45045 d^4 \sqrt {c+d x}} \] Input:

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

Output:

(-2*(c - d*x)^2*Sqrt[c^2 - d^2*x^2]*(1008*c^4*D + 8*c^3*d*(221*C + 315*D*x 
) + c*d^3*(11583*A + 9295*B*x + 7735*C*x^2 + 6615*D*x^3) + 2*c^2*d^2*(1859 
*B + 5*x*(442*C + 441*D*x)) + 5*d^4*x*(1287*A + 7*x*(143*B + 117*C*x + 99* 
D*x^2))))/(45045*d^4*Sqrt[c + d*x])
 

Rubi [A] (verified)

Time = 1.05 (sec) , antiderivative size = 236, normalized size of antiderivative = 0.97, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {2170, 27, 2170, 27, 672, 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.

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

\(\Big \downarrow \) 2170

\(\displaystyle -\frac {2 \int -\frac {\left (c^2-d^2 x^2\right )^{3/2} \left ((13 C d-23 c D) x^2 d^4+\left (13 B d^2-7 c^2 D\right ) x d^3+\left (3 D c^3+13 A d^3\right ) d^2\right )}{2 \sqrt {c+d x}}dx}{13 d^5}-\frac {2 D (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{5/2}}{13 d^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left ((13 C d-23 c D) x^2 d^4+\left (13 B d^2-7 c^2 D\right ) x d^3+\left (3 D c^3+13 A d^3\right ) d^2\right )}{\sqrt {c+d x}}dx}{13 d^5}-\frac {2 D (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{5/2}}{13 d^4}\)

\(\Big \downarrow \) 2170

\(\displaystyle \frac {-\frac {2 \int -\frac {d^6 \left (10 D c^3+13 C d c^2+143 A d^3-d \left (-153 D c^2+130 C d c-143 B d^2\right ) x\right ) \left (c^2-d^2 x^2\right )^{3/2}}{2 \sqrt {c+d x}}dx}{11 d^4}-\frac {2}{11} d \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2} (13 C d-23 c D)}{13 d^5}-\frac {2 D (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{5/2}}{13 d^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{11} d^2 \int \frac {\left (10 D c^3+13 C d c^2+143 A d^3-d \left (-153 D c^2+130 C d c-143 B d^2\right ) x\right ) \left (c^2-d^2 x^2\right )^{3/2}}{\sqrt {c+d x}}dx-\frac {2}{11} d \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2} (13 C d-23 c D)}{13 d^5}-\frac {2 D (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{5/2}}{13 d^4}\)

\(\Big \downarrow \) 672

\(\displaystyle \frac {\frac {1}{11} d^2 \left (\frac {1}{9} \left (1287 A d^3-143 B c d^2-63 c^3 D+247 c^2 C d\right ) \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{\sqrt {c+d x}}dx+\frac {2 \left (c^2-d^2 x^2\right )^{5/2} \left (-143 B d^2-153 c^2 D+130 c C d\right )}{9 d \sqrt {c+d x}}\right )-\frac {2}{11} d \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2} (13 C d-23 c D)}{13 d^5}-\frac {2 D (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{5/2}}{13 d^4}\)

\(\Big \downarrow \) 459

\(\displaystyle \frac {\frac {1}{11} d^2 \left (\frac {1}{9} \left (1287 A d^3-143 B c d^2-63 c^3 D+247 c^2 C d\right ) \left (\frac {4}{7} c \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{3/2}}dx-\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{7 d (c+d x)^{3/2}}\right )+\frac {2 \left (c^2-d^2 x^2\right )^{5/2} \left (-143 B d^2-153 c^2 D+130 c C d\right )}{9 d \sqrt {c+d x}}\right )-\frac {2}{11} d \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2} (13 C d-23 c D)}{13 d^5}-\frac {2 D (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{5/2}}{13 d^4}\)

\(\Big \downarrow \) 458

\(\displaystyle \frac {\frac {1}{11} d^2 \left (\frac {1}{9} \left (-\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{7 d (c+d x)^{3/2}}-\frac {8 c \left (c^2-d^2 x^2\right )^{5/2}}{35 d (c+d x)^{5/2}}\right ) \left (1287 A d^3-143 B c d^2-63 c^3 D+247 c^2 C d\right )+\frac {2 \left (c^2-d^2 x^2\right )^{5/2} \left (-143 B d^2-153 c^2 D+130 c C d\right )}{9 d \sqrt {c+d x}}\right )-\frac {2}{11} d \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2} (13 C d-23 c D)}{13 d^5}-\frac {2 D (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{5/2}}{13 d^4}\)

Input:

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

Output:

(-2*D*(c + d*x)^(3/2)*(c^2 - d^2*x^2)^(5/2))/(13*d^4) + ((-2*d*(13*C*d - 2 
3*c*D)*Sqrt[c + d*x]*(c^2 - d^2*x^2)^(5/2))/11 + (d^2*((2*(130*c*C*d - 143 
*B*d^2 - 153*c^2*D)*(c^2 - d^2*x^2)^(5/2))/(9*d*Sqrt[c + d*x]) + ((247*c^2 
*C*d - 143*B*c*d^2 + 1287*A*d^3 - 63*c^3*D)*((-8*c*(c^2 - d^2*x^2)^(5/2))/ 
(35*d*(c + d*x)^(5/2)) - (2*(c^2 - d^2*x^2)^(5/2))/(7*d*(c + d*x)^(3/2)))) 
/9))/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 = 0.37 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.64

method	result	size
gosper	\(-\frac {2 \left (-d x +c \right ) \left (3465 D x^{4} d^{4}+4095 C \,d^{4} x^{3}+6615 D c \,d^{3} x^{3}+5005 B \,d^{4} x^{2}+7735 C c \,d^{3} x^{2}+4410 D c^{2} d^{2} x^{2}+6435 A \,d^{4} x +9295 B c \,d^{3} x +4420 C \,c^{2} d^{2} x +2520 D c^{3} d x +11583 A c \,d^{3}+3718 B \,c^{2} d^{2}+1768 C \,c^{3} d +1008 c^{4} D\right ) \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{45045 d^{4} \left (d x +c \right )^{\frac {3}{2}}}\)	\(155\)
orering	\(-\frac {2 \left (-d x +c \right ) \left (3465 D x^{4} d^{4}+4095 C \,d^{4} x^{3}+6615 D c \,d^{3} x^{3}+5005 B \,d^{4} x^{2}+7735 C c \,d^{3} x^{2}+4410 D c^{2} d^{2} x^{2}+6435 A \,d^{4} x +9295 B c \,d^{3} x +4420 C \,c^{2} d^{2} x +2520 D c^{3} d x +11583 A c \,d^{3}+3718 B \,c^{2} d^{2}+1768 C \,c^{3} d +1008 c^{4} D\right ) \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{45045 d^{4} \left (d x +c \right )^{\frac {3}{2}}}\)	\(155\)
default	\(-\frac {2 \sqrt {-d^{2} x^{2}+c^{2}}\, \left (-d x +c \right )^{2} \left (3465 D x^{4} d^{4}+4095 C \,d^{4} x^{3}+6615 D c \,d^{3} x^{3}+5005 B \,d^{4} x^{2}+7735 C c \,d^{3} x^{2}+4410 D c^{2} d^{2} x^{2}+6435 A \,d^{4} x +9295 B c \,d^{3} x +4420 C \,c^{2} d^{2} x +2520 D c^{3} d x +11583 A c \,d^{3}+3718 B \,c^{2} d^{2}+1768 C \,c^{3} d +1008 c^{4} D\right )}{45045 \sqrt {d x +c}\, d^{4}}\)	\(157\)

Input:

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

Output:

-2/45045*(-d*x+c)*(3465*D*d^4*x^4+4095*C*d^4*x^3+6615*D*c*d^3*x^3+5005*B*d 
^4*x^2+7735*C*c*d^3*x^2+4410*D*c^2*d^2*x^2+6435*A*d^4*x+9295*B*c*d^3*x+442 
0*C*c^2*d^2*x+2520*D*c^3*d*x+11583*A*c*d^3+3718*B*c^2*d^2+1768*C*c^3*d+100 
8*D*c^4)*(-d^2*x^2+c^2)^(3/2)/d^4/(d*x+c)^(3/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 237, normalized size of antiderivative = 0.97 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c+d x}} \, dx=-\frac {2 \, {\left (3465 \, D d^{6} x^{6} + 1008 \, D c^{6} + 1768 \, C c^{5} d + 3718 \, B c^{4} d^{2} + 11583 \, A c^{3} d^{3} - 315 \, {\left (D c d^{5} - 13 \, C d^{6}\right )} x^{5} - 35 \, {\left (153 \, D c^{2} d^{4} + 13 \, C c d^{5} - 143 \, B d^{6}\right )} x^{4} + 5 \, {\left (63 \, D c^{3} d^{3} - 1391 \, C c^{2} d^{4} - 143 \, B c d^{5} + 1287 \, A d^{6}\right )} x^{3} + 3 \, {\left (126 \, D c^{4} d^{2} + 221 \, C c^{3} d^{3} - 3289 \, B c^{2} d^{4} - 429 \, A c d^{5}\right )} x^{2} + {\left (504 \, D c^{5} d + 884 \, C c^{4} d^{2} + 1859 \, B c^{3} d^{3} - 16731 \, A c^{2} d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{45045 \, {\left (d^{5} x + c d^{4}\right )}} \] Input:

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

Output:

-2/45045*(3465*D*d^6*x^6 + 1008*D*c^6 + 1768*C*c^5*d + 3718*B*c^4*d^2 + 11 
583*A*c^3*d^3 - 315*(D*c*d^5 - 13*C*d^6)*x^5 - 35*(153*D*c^2*d^4 + 13*C*c* 
d^5 - 143*B*d^6)*x^4 + 5*(63*D*c^3*d^3 - 1391*C*c^2*d^4 - 143*B*c*d^5 + 12 
87*A*d^6)*x^3 + 3*(126*D*c^4*d^2 + 221*C*c^3*d^3 - 3289*B*c^2*d^4 - 429*A* 
c*d^5)*x^2 + (504*D*c^5*d + 884*C*c^4*d^2 + 1859*B*c^3*d^3 - 16731*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]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.00 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c+d x}} \, dx=-\frac {2 \, {\left (5 \, d^{3} x^{3} - c d^{2} x^{2} - 13 \, c^{2} d x + 9 \, c^{3}\right )} \sqrt {-d x + c} A}{35 \, d} - \frac {2 \, {\left (35 \, d^{4} x^{4} - 5 \, c d^{3} x^{3} - 69 \, c^{2} d^{2} x^{2} + 13 \, c^{3} d x + 26 \, c^{4}\right )} \sqrt {-d x + c} B}{315 \, d^{2}} - \frac {2 \, {\left (315 \, d^{5} x^{5} - 35 \, c d^{4} x^{4} - 535 \, c^{2} d^{3} x^{3} + 51 \, c^{3} d^{2} x^{2} + 68 \, c^{4} d x + 136 \, c^{5}\right )} \sqrt {-d x + c} C}{3465 \, d^{3}} - \frac {2 \, {\left (55 \, d^{6} x^{6} - 5 \, c d^{5} x^{5} - 85 \, c^{2} d^{4} x^{4} + 5 \, c^{3} d^{3} x^{3} + 6 \, c^{4} d^{2} x^{2} + 8 \, c^{5} d x + 16 \, c^{6}\right )} \sqrt {-d x + c} D}{715 \, d^{4}} \] Input:

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

Output:

-2/35*(5*d^3*x^3 - c*d^2*x^2 - 13*c^2*d*x + 9*c^3)*sqrt(-d*x + c)*A/d - 2/ 
315*(35*d^4*x^4 - 5*c*d^3*x^3 - 69*c^2*d^2*x^2 + 13*c^3*d*x + 26*c^4)*sqrt 
(-d*x + c)*B/d^2 - 2/3465*(315*d^5*x^5 - 35*c*d^4*x^4 - 535*c^2*d^3*x^3 + 
51*c^3*d^2*x^2 + 68*c^4*d*x + 136*c^5)*sqrt(-d*x + c)*C/d^3 - 2/715*(55*d^ 
6*x^6 - 5*c*d^5*x^5 - 85*c^2*d^4*x^4 + 5*c^3*d^3*x^3 + 6*c^4*d^2*x^2 + 8*c 
^5*d*x + 16*c^6)*sqrt(-d*x + c)*D/d^4
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 567 vs. \(2 (214) = 428\).

Time = 0.13 (sec) , antiderivative size = 567, normalized size of antiderivative = 2.32 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c+d x}} \, dx=-\frac {2 \, {\left (15015 \, {\left (-d x + c\right )}^{\frac {3}{2}} A c^{2} d^{3} - 3003 \, {\left (3 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} - 5 \, {\left (-d x + c\right )}^{\frac {3}{2}} c\right )} B c^{2} d^{2} - 429 \, {\left (15 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} + 42 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} c - 35 \, {\left (-d x + c\right )}^{\frac {3}{2}} c^{2}\right )} C c^{2} d + 429 \, {\left (15 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} + 42 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} c - 35 \, {\left (-d x + c\right )}^{\frac {3}{2}} c^{2}\right )} A d^{3} - 143 \, {\left (35 \, {\left (d x - c\right )}^{4} \sqrt {-d x + c} + 135 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} c + 189 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} c^{2} - 105 \, {\left (-d x + c\right )}^{\frac {3}{2}} c^{3}\right )} D c^{2} + 143 \, {\left (35 \, {\left (d x - c\right )}^{4} \sqrt {-d x + c} + 135 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} c + 189 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} c^{2} - 105 \, {\left (-d x + c\right )}^{\frac {3}{2}} c^{3}\right )} B d^{2} + 13 \, {\left (315 \, {\left (d x - c\right )}^{5} \sqrt {-d x + c} + 1540 \, {\left (d x - c\right )}^{4} \sqrt {-d x + c} c + 2970 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} c^{2} + 2772 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} c^{3} - 1155 \, {\left (-d x + c\right )}^{\frac {3}{2}} c^{4}\right )} C d + 5 \, {\left (693 \, {\left (d x - c\right )}^{6} \sqrt {-d x + c} + 4095 \, {\left (d x - c\right )}^{5} \sqrt {-d x + c} c + 10010 \, {\left (d x - c\right )}^{4} \sqrt {-d x + c} c^{2} + 12870 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} c^{3} + 9009 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} c^{4} - 3003 \, {\left (-d x + c\right )}^{\frac {3}{2}} c^{5}\right )} D\right )}}{45045 \, d^{4}} \] Input:

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

Output:

-2/45045*(15015*(-d*x + c)^(3/2)*A*c^2*d^3 - 3003*(3*(d*x - c)^2*sqrt(-d*x 
 + c) - 5*(-d*x + c)^(3/2)*c)*B*c^2*d^2 - 429*(15*(d*x - c)^3*sqrt(-d*x + 
c) + 42*(d*x - c)^2*sqrt(-d*x + c)*c - 35*(-d*x + c)^(3/2)*c^2)*C*c^2*d + 
429*(15*(d*x - c)^3*sqrt(-d*x + c) + 42*(d*x - c)^2*sqrt(-d*x + c)*c - 35* 
(-d*x + c)^(3/2)*c^2)*A*d^3 - 143*(35*(d*x - c)^4*sqrt(-d*x + c) + 135*(d* 
x - c)^3*sqrt(-d*x + c)*c + 189*(d*x - c)^2*sqrt(-d*x + c)*c^2 - 105*(-d*x 
 + c)^(3/2)*c^3)*D*c^2 + 143*(35*(d*x - c)^4*sqrt(-d*x + c) + 135*(d*x - c 
)^3*sqrt(-d*x + c)*c + 189*(d*x - c)^2*sqrt(-d*x + c)*c^2 - 105*(-d*x + c) 
^(3/2)*c^3)*B*d^2 + 13*(315*(d*x - c)^5*sqrt(-d*x + c) + 1540*(d*x - c)^4* 
sqrt(-d*x + c)*c + 2970*(d*x - c)^3*sqrt(-d*x + c)*c^2 + 2772*(d*x - c)^2* 
sqrt(-d*x + c)*c^3 - 1155*(-d*x + c)^(3/2)*c^4)*C*d + 5*(693*(d*x - c)^6*s 
qrt(-d*x + c) + 4095*(d*x - c)^5*sqrt(-d*x + c)*c + 10010*(d*x - c)^4*sqrt 
(-d*x + c)*c^2 + 12870*(d*x - c)^3*sqrt(-d*x + c)*c^3 + 9009*(d*x - c)^2*s 
qrt(-d*x + c)*c^4 - 3003*(-d*x + c)^(3/2)*c^5)*D)/d^4
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.66 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c+d x}} \, dx=\frac {2 \sqrt {-d x +c}\, \left (-3465 d^{6} x^{6}-3780 c \,d^{5} x^{5}-5005 b \,d^{5} x^{4}+5810 c^{2} d^{4} x^{4}-6435 a \,d^{5} x^{3}+715 b c \,d^{4} x^{3}+6640 c^{3} d^{3} x^{3}+1287 a c \,d^{4} x^{2}+9867 b \,c^{2} d^{3} x^{2}-1041 c^{4} d^{2} x^{2}+16731 a \,c^{2} d^{3} x -1859 b \,c^{3} d^{2} x -1388 c^{5} d x -11583 a \,c^{3} d^{2}-3718 b \,c^{4} d -2776 c^{6}\right )}{45045 d^{3}} \] Input:

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

Output:

(2*sqrt(c - d*x)*( - 11583*a*c**3*d**2 + 16731*a*c**2*d**3*x + 1287*a*c*d* 
*4*x**2 - 6435*a*d**5*x**3 - 3718*b*c**4*d - 1859*b*c**3*d**2*x + 9867*b*c 
**2*d**3*x**2 + 715*b*c*d**4*x**3 - 5005*b*d**5*x**4 - 2776*c**6 - 1388*c* 
*5*d*x - 1041*c**4*d**2*x**2 + 6640*c**3*d**3*x**3 + 5810*c**2*d**4*x**4 - 
 3780*c*d**5*x**5 - 3465*d**6*x**6))/(45045*d**3)