\(\int \frac {\sqrt {c^2-d^2 x^2} (A+B x+C x^2+D x^3)}{(c+d x)^{3/2}} \, dx\) [197]

Optimal result

Integrand size = 41, antiderivative size = 257 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{3/2}} \, dx=\frac {2 \left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \sqrt {c^2-d^2 x^2}}{d^4 \sqrt {c+d x}}+\frac {2 \left (42 c C d-35 B d^2-53 c^2 D\right ) \left (c^2-d^2 x^2\right )^{3/2}}{105 d^4 (c+d x)^{3/2}}-\frac {2 (7 C d-13 c D) \left (c^2-d^2 x^2\right )^{3/2}}{35 d^4 \sqrt {c+d x}}-\frac {2 D \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}{7 d^4}-\frac {2 \sqrt {2} \sqrt {c} \left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{d^4} \] Output:

2*(A*d^3-B*c*d^2+C*c^2*d-D*c^3)*(-d^2*x^2+c^2)^(1/2)/d^4/(d*x+c)^(1/2)+2/1 
05*(-35*B*d^2+42*C*c*d-53*D*c^2)*(-d^2*x^2+c^2)^(3/2)/d^4/(d*x+c)^(3/2)-2/ 
35*(7*C*d-13*D*c)*(-d^2*x^2+c^2)^(3/2)/d^4/(d*x+c)^(1/2)-2/7*D*(d*x+c)^(1/ 
2)*(-d^2*x^2+c^2)^(3/2)/d^4-2*2^(1/2)*c^(1/2)*(A*d^3-B*c*d^2+C*c^2*d-D*c^3 
)*arctanh(2^(1/2)*c^(1/2)*(d*x+c)^(1/2)/(-d^2*x^2+c^2)^(1/2))/d^4
 

Mathematica [A] (verified)

Time = 0.73 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.68 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{3/2}} \, dx=\frac {2 \left (\frac {\sqrt {c^2-d^2 x^2} \left (-134 c^3 D+2 c^2 d (63 C+19 D x)-2 c d^2 (70 B+3 x (7 C+4 D x))+d^3 (105 A+x (35 B+3 x (7 C+5 D x)))\right )}{\sqrt {c+d x}}+105 \sqrt {2} \sqrt {c} \left (-c^2 C d+B c d^2-A d^3+c^3 D\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )\right )}{105 d^4} \] Input:

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

Output:

(2*((Sqrt[c^2 - d^2*x^2]*(-134*c^3*D + 2*c^2*d*(63*C + 19*D*x) - 2*c*d^2*( 
70*B + 3*x*(7*C + 4*D*x)) + d^3*(105*A + x*(35*B + 3*x*(7*C + 5*D*x)))))/S 
qrt[c + d*x] + 105*Sqrt[2]*Sqrt[c]*(-(c^2*C*d) + B*c*d^2 - A*d^3 + c^3*D)* 
ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[c + d*x])/Sqrt[c^2 - d^2*x^2]]))/(105*d^4)
 

Rubi [A] (verified)

Time = 1.12 (sec) , antiderivative size = 249, normalized size of antiderivative = 0.97, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {2170, 27, 2170, 27, 672, 466, 471, 221}

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.

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

\(\Big \downarrow \) 2170

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2170

\(\displaystyle \frac {-\frac {2 \int \frac {d^6 \left (-18 D c^3+7 C d c^2-35 A d^3+d \left (-53 D c^2+42 C d c-35 B d^2\right ) x\right ) \sqrt {c^2-d^2 x^2}}{2 (c+d x)^{3/2}}dx}{5 d^4}-\frac {2 d \left (c^2-d^2 x^2\right )^{3/2} (7 C d-13 c D)}{5 \sqrt {c+d x}}}{7 d^5}-\frac {2 D \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}{7 d^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {1}{5} d^2 \int \frac {\left (-18 D c^3+7 C d c^2-35 A d^3+d \left (-53 D c^2+42 C d c-35 B d^2\right ) x\right ) \sqrt {c^2-d^2 x^2}}{(c+d x)^{3/2}}dx-\frac {2 d \left (c^2-d^2 x^2\right )^{3/2} (7 C d-13 c D)}{5 \sqrt {c+d x}}}{7 d^5}-\frac {2 D \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}{7 d^4}\)

\(\Big \downarrow \) 672

\(\displaystyle \frac {-\frac {1}{5} d^2 \left (-35 \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right ) \int \frac {\sqrt {c^2-d^2 x^2}}{(c+d x)^{3/2}}dx-\frac {2 \left (c^2-d^2 x^2\right )^{3/2} \left (-35 B d^2-53 c^2 D+42 c C d\right )}{3 d (c+d x)^{3/2}}\right )-\frac {2 d \left (c^2-d^2 x^2\right )^{3/2} (7 C d-13 c D)}{5 \sqrt {c+d x}}}{7 d^5}-\frac {2 D \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}{7 d^4}\)

\(\Big \downarrow \) 466

\(\displaystyle \frac {-\frac {1}{5} d^2 \left (-35 \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right ) \left (2 c \int \frac {1}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx+\frac {2 \sqrt {c^2-d^2 x^2}}{d \sqrt {c+d x}}\right )-\frac {2 \left (c^2-d^2 x^2\right )^{3/2} \left (-35 B d^2-53 c^2 D+42 c C d\right )}{3 d (c+d x)^{3/2}}\right )-\frac {2 d \left (c^2-d^2 x^2\right )^{3/2} (7 C d-13 c D)}{5 \sqrt {c+d x}}}{7 d^5}-\frac {2 D \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}{7 d^4}\)

\(\Big \downarrow \) 471

\(\displaystyle \frac {-\frac {1}{5} d^2 \left (-35 \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right ) \left (4 c d \int \frac {1}{\frac {d^2 \left (c^2-d^2 x^2\right )}{c+d x}-2 c d^2}d\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}+\frac {2 \sqrt {c^2-d^2 x^2}}{d \sqrt {c+d x}}\right )-\frac {2 \left (c^2-d^2 x^2\right )^{3/2} \left (-35 B d^2-53 c^2 D+42 c C d\right )}{3 d (c+d x)^{3/2}}\right )-\frac {2 d \left (c^2-d^2 x^2\right )^{3/2} (7 C d-13 c D)}{5 \sqrt {c+d x}}}{7 d^5}-\frac {2 D \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}{7 d^4}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {-\frac {1}{5} d^2 \left (-35 \left (\frac {2 \sqrt {c^2-d^2 x^2}}{d \sqrt {c+d x}}-\frac {2 \sqrt {2} \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {c+d x}}\right )}{d}\right ) \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )-\frac {2 \left (c^2-d^2 x^2\right )^{3/2} \left (-35 B d^2-53 c^2 D+42 c C d\right )}{3 d (c+d x)^{3/2}}\right )-\frac {2 d \left (c^2-d^2 x^2\right )^{3/2} (7 C d-13 c D)}{5 \sqrt {c+d x}}}{7 d^5}-\frac {2 D \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}{7 d^4}\)

Input:

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

Output:

(-2*D*Sqrt[c + d*x]*(c^2 - d^2*x^2)^(3/2))/(7*d^4) + ((-2*d*(7*C*d - 13*c* 
D)*(c^2 - d^2*x^2)^(3/2))/(5*Sqrt[c + d*x]) - (d^2*((-2*(42*c*C*d - 35*B*d 
^2 - 53*c^2*D)*(c^2 - d^2*x^2)^(3/2))/(3*d*(c + d*x)^(3/2)) - 35*(c^2*C*d 
- B*c*d^2 + A*d^3 - c^3*D)*((2*Sqrt[c^2 - d^2*x^2])/(d*Sqrt[c + d*x]) - (2 
*Sqrt[2]*Sqrt[c]*ArcTanh[Sqrt[c^2 - d^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[c + d*x 
])])/d)))/5)/(7*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 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 466

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

rule 471

Int[1/(Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Sim 
p[2*d   Subst[Int[1/(2*b*c + d^2*x^2), x], x, Sqrt[a + b*x^2]/Sqrt[c + d*x] 
], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 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 = 304, normalized size of antiderivative = 1.18

method	result	size
default	\(-\frac {2 \sqrt {-d^{2} x^{2}+c^{2}}\, \left (-15 D d^{3} x^{3} \sqrt {-d x +c}+105 A \sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) d^{3}-105 B \,c^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) d^{2}+105 C \,c^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) d -21 C \,d^{3} x^{2} \sqrt {-d x +c}-105 D c^{\frac {7}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right )+24 D c \,d^{2} x^{2} \sqrt {-d x +c}-35 B \,d^{3} x \sqrt {-d x +c}+42 C c \,d^{2} x \sqrt {-d x +c}-38 D c^{2} d x \sqrt {-d x +c}-105 A \,d^{3} \sqrt {-d x +c}+140 B c \,d^{2} \sqrt {-d x +c}-126 C \,c^{2} d \sqrt {-d x +c}+134 D c^{3} \sqrt {-d x +c}\right )}{105 \sqrt {d x +c}\, \sqrt {-d x +c}\, d^{4}}\)	\(304\)

Input:

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

Output:

-2/105*(-d^2*x^2+c^2)^(1/2)*(-15*D*d^3*x^3*(-d*x+c)^(1/2)+105*A*c^(1/2)*2^ 
(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*d^3-105*B*c^(3/2)*2^(1/2 
)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*d^2+105*C*c^(5/2)*2^(1/2)*ar 
ctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*d-21*C*d^3*x^2*(-d*x+c)^(1/2)-10 
5*D*c^(7/2)*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))+24*D*c*d^2 
*x^2*(-d*x+c)^(1/2)-35*B*d^3*x*(-d*x+c)^(1/2)+42*C*c*d^2*x*(-d*x+c)^(1/2)- 
38*D*c^2*d*x*(-d*x+c)^(1/2)-105*A*d^3*(-d*x+c)^(1/2)+140*B*c*d^2*(-d*x+c)^ 
(1/2)-126*C*c^2*d*(-d*x+c)^(1/2)+134*D*c^3*(-d*x+c)^(1/2))/(d*x+c)^(1/2)/( 
-d*x+c)^(1/2)/d^4
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 479, normalized size of antiderivative = 1.86 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{3/2}} \, dx=\left [\frac {105 \, \sqrt {2} {\left (D c^{4} - C c^{3} d + B c^{2} d^{2} - A c d^{3} + {\left (D c^{3} d - C c^{2} d^{2} + B c d^{3} - A d^{4}\right )} x\right )} \sqrt {c} \log \left (-\frac {d^{2} x^{2} - 2 \, c d x - 2 \, \sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {c} - 3 \, c^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + 2 \, {\left (15 \, D d^{3} x^{3} - 134 \, D c^{3} + 126 \, C c^{2} d - 140 \, B c d^{2} + 105 \, A d^{3} - 3 \, {\left (8 \, D c d^{2} - 7 \, C d^{3}\right )} x^{2} + {\left (38 \, D c^{2} d - 42 \, C c d^{2} + 35 \, B d^{3}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{105 \, {\left (d^{5} x + c d^{4}\right )}}, -\frac {2 \, {\left (105 \, \sqrt {2} {\left (D c^{4} - C c^{3} d + B c^{2} d^{2} - A c d^{3} + {\left (D c^{3} d - C c^{2} d^{2} + B c d^{3} - A d^{4}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {-c}}{2 \, {\left (c d x + c^{2}\right )}}\right ) - {\left (15 \, D d^{3} x^{3} - 134 \, D c^{3} + 126 \, C c^{2} d - 140 \, B c d^{2} + 105 \, A d^{3} - 3 \, {\left (8 \, D c d^{2} - 7 \, C d^{3}\right )} x^{2} + {\left (38 \, D c^{2} d - 42 \, C c d^{2} + 35 \, B d^{3}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}\right )}}{105 \, {\left (d^{5} x + c d^{4}\right )}}\right ] \] Input:

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

Output:

[1/105*(105*sqrt(2)*(D*c^4 - C*c^3*d + B*c^2*d^2 - A*c*d^3 + (D*c^3*d - C* 
c^2*d^2 + B*c*d^3 - A*d^4)*x)*sqrt(c)*log(-(d^2*x^2 - 2*c*d*x - 2*sqrt(2)* 
sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)*sqrt(c) - 3*c^2)/(d^2*x^2 + 2*c*d*x + c 
^2)) + 2*(15*D*d^3*x^3 - 134*D*c^3 + 126*C*c^2*d - 140*B*c*d^2 + 105*A*d^3 
 - 3*(8*D*c*d^2 - 7*C*d^3)*x^2 + (38*D*c^2*d - 42*C*c*d^2 + 35*B*d^3)*x)*s 
qrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(d^5*x + c*d^4), -2/105*(105*sqrt(2)*(D 
*c^4 - C*c^3*d + B*c^2*d^2 - A*c*d^3 + (D*c^3*d - C*c^2*d^2 + B*c*d^3 - A* 
d^4)*x)*sqrt(-c)*arctan(1/2*sqrt(2)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)*sqr 
t(-c)/(c*d*x + c^2)) - (15*D*d^3*x^3 - 134*D*c^3 + 126*C*c^2*d - 140*B*c*d 
^2 + 105*A*d^3 - 3*(8*D*c*d^2 - 7*C*d^3)*x^2 + (38*D*c^2*d - 42*C*c*d^2 + 
35*B*d^3)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(d^5*x + c*d^4)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.82 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{3/2}} \, dx=\frac {2 \, {\left (15 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} D + 21 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} D c - 35 \, {\left (-d x + c\right )}^{\frac {3}{2}} D c^{2} - 105 \, \sqrt {-d x + c} D c^{3} + 21 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} C d + 105 \, \sqrt {-d x + c} C c^{2} d - 35 \, {\left (-d x + c\right )}^{\frac {3}{2}} B d^{2} - 105 \, \sqrt {-d x + c} B c d^{2} + 105 \, \sqrt {-d x + c} A d^{3} - \frac {105 \, \sqrt {2} {\left (D c^{4} - C c^{3} d + B c^{2} d^{2} - A c d^{3}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-d x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c}}\right )}}{105 \, d^{4}} \] Input:

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

Output:

2/105*(15*(d*x - c)^3*sqrt(-d*x + c)*D + 21*(d*x - c)^2*sqrt(-d*x + c)*D*c 
 - 35*(-d*x + c)^(3/2)*D*c^2 - 105*sqrt(-d*x + c)*D*c^3 + 21*(d*x - c)^2*s 
qrt(-d*x + c)*C*d + 105*sqrt(-d*x + c)*C*c^2*d - 35*(-d*x + c)^(3/2)*B*d^2 
 - 105*sqrt(-d*x + c)*B*c*d^2 + 105*sqrt(-d*x + c)*A*d^3 - 105*sqrt(2)*(D* 
c^4 - C*c^3*d + B*c^2*d^2 - A*c*d^3)*arctan(1/2*sqrt(2)*sqrt(-d*x + c)/sqr 
t(-c))/sqrt(-c))/d^4
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.74 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{3/2}} \, dx=\frac {2 \sqrt {-d x +c}\, a \,d^{2}-\frac {8 \sqrt {-d x +c}\, b c d}{3}+\frac {2 \sqrt {-d x +c}\, b \,d^{2} x}{3}-\frac {16 \sqrt {-d x +c}\, c^{3}}{105}-\frac {8 \sqrt {-d x +c}\, c^{2} d x}{105}-\frac {2 \sqrt {-d x +c}\, c \,d^{2} x^{2}}{35}+\frac {2 \sqrt {-d x +c}\, d^{3} x^{3}}{7}+2 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) a \,d^{2}-2 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) b c d -2 \sqrt {c}\, \sqrt {2}\, a \,d^{2}+\frac {10 \sqrt {c}\, \sqrt {2}\, b c d}{3}+\frac {44 \sqrt {c}\, \sqrt {2}\, c^{3}}{105}}{d^{3}} \] Input:

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

Output:

(2*(105*sqrt(c - d*x)*a*d**2 - 140*sqrt(c - d*x)*b*c*d + 35*sqrt(c - d*x)* 
b*d**2*x - 8*sqrt(c - d*x)*c**3 - 4*sqrt(c - d*x)*c**2*d*x - 3*sqrt(c - d* 
x)*c*d**2*x**2 + 15*sqrt(c - d*x)*d**3*x**3 + 105*sqrt(c)*sqrt(2)*log(tan( 
asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*d**2 - 105*sqrt(c)*sqrt(2)*log 
(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*b*c*d - 105*sqrt(c)*sqrt(2) 
*a*d**2 + 175*sqrt(c)*sqrt(2)*b*c*d + 22*sqrt(c)*sqrt(2)*c**3))/(105*d**3)