\(\int \frac {(c^2-d^2 x^2)^{3/2} (A+B x+C x^2+D x^3)}{(c+d x)^{3/2}} \, dx\) [206]

Optimal result

Integrand size = 41, antiderivative size = 256 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{3/2}} \, dx=-\frac {2 \left (253 c^2 C d-297 B c d^2+693 A d^3-177 c^3 D\right ) \left (c^2-d^2 x^2\right )^{5/2}}{3465 d^4 (c+d x)^{5/2}}-\frac {\left (253 c^2 C d-297 B c d^2+693 A d^3-177 c^3 D\right ) \left (c^2-d^2 x^2\right )^{5/2}}{1386 c d^4 (c+d x)^{3/2}}+\frac {\left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \left (c^2-d^2 x^2\right )^{5/2}}{2 c d^4 (c+d x)^{3/2}}-\frac {2 (11 C d-21 c D) \left (c^2-d^2 x^2\right )^{5/2}}{99 d^4 \sqrt {c+d x}}-\frac {2 D \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2}}{11 d^4} \] Output:

-2/3465*(693*A*d^3-297*B*c*d^2+253*C*c^2*d-177*D*c^3)*(-d^2*x^2+c^2)^(5/2) 
/d^4/(d*x+c)^(5/2)-1/1386*(693*A*d^3-297*B*c*d^2+253*C*c^2*d-177*D*c^3)*(- 
d^2*x^2+c^2)^(5/2)/c/d^4/(d*x+c)^(3/2)+1/2*(A*d^3-B*c*d^2+C*c^2*d-D*c^3)*( 
-d^2*x^2+c^2)^(5/2)/c/d^4/(d*x+c)^(3/2)-2/99*(11*C*d-21*D*c)*(-d^2*x^2+c^2 
)^(5/2)/d^4/(d*x+c)^(1/2)-2/11*D*(d*x+c)^(1/2)*(-d^2*x^2+c^2)^(5/2)/d^4
 

Mathematica [A] (verified)

Time = 2.37 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.42 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{3/2}} \, dx=-\frac {2 (c-d x)^2 \sqrt {c^2-d^2 x^2} \left (48 c^3 D+8 c^2 d (11 C+15 D x)+2 c d^2 (99 B+5 x (22 C+21 D x))+d^3 \left (693 A+5 x \left (99 B+77 C x+63 D x^2\right )\right )\right )}{3465 d^4 \sqrt {c+d x}} \] Input:

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

Output:

(-2*(c - d*x)^2*Sqrt[c^2 - d^2*x^2]*(48*c^3*D + 8*c^2*d*(11*C + 15*D*x) + 
2*c*d^2*(99*B + 5*x*(22*C + 21*D*x)) + d^3*(693*A + 5*x*(99*B + 77*C*x + 6 
3*D*x^2))))/(3465*d^4*Sqrt[c + d*x])
 

Rubi [A] (verified)

Time = 1.02 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.77, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {2170, 27, 2170, 27, 672, 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.

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

\(\Big \downarrow \) 2170

\(\displaystyle -\frac {2 \int -\frac {\left (c^2-d^2 x^2\right )^{3/2} \left ((11 C d-21 c D) x^2 d^4+\left (11 B d^2-9 c^2 D\right ) x d^3+\left (D c^3+11 A d^3\right ) d^2\right )}{2 (c+d x)^{3/2}}dx}{11 d^5}-\frac {2 D \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2}}{11 d^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left ((11 C d-21 c D) x^2 d^4+\left (11 B d^2-9 c^2 D\right ) x d^3+\left (D c^3+11 A d^3\right ) d^2\right )}{(c+d x)^{3/2}}dx}{11 d^5}-\frac {2 D \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2}}{11 d^4}\)

\(\Big \downarrow \) 2170

\(\displaystyle \frac {-\frac {2 \int \frac {d^6 \left (-30 D c^3+11 C d c^2-99 A d^3+d \left (-129 D c^2+110 C d c-99 B d^2\right ) x\right ) \left (c^2-d^2 x^2\right )^{3/2}}{2 (c+d x)^{3/2}}dx}{9 d^4}-\frac {2 d \left (c^2-d^2 x^2\right )^{5/2} (11 C d-21 c D)}{9 \sqrt {c+d x}}}{11 d^5}-\frac {2 D \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2}}{11 d^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {1}{9} d^2 \int \frac {\left (-30 D c^3+11 C d c^2-99 A d^3+d \left (-129 D c^2+110 C d c-99 B d^2\right ) x\right ) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{3/2}}dx-\frac {2 d \left (c^2-d^2 x^2\right )^{5/2} (11 C d-21 c D)}{9 \sqrt {c+d x}}}{11 d^5}-\frac {2 D \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2}}{11 d^4}\)

\(\Big \downarrow \) 672

\(\displaystyle \frac {-\frac {1}{9} d^2 \left (-\frac {1}{7} \left (693 A d^3-297 B c d^2-177 c^3 D+253 c^2 C d\right ) \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{3/2}}dx-\frac {2 \left (c^2-d^2 x^2\right )^{5/2} \left (-99 B d^2-129 c^2 D+110 c C d\right )}{7 d (c+d x)^{3/2}}\right )-\frac {2 d \left (c^2-d^2 x^2\right )^{5/2} (11 C d-21 c D)}{9 \sqrt {c+d x}}}{11 d^5}-\frac {2 D \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2}}{11 d^4}\)

\(\Big \downarrow \) 458

\(\displaystyle \frac {-\frac {1}{9} d^2 \left (\frac {2 \left (c^2-d^2 x^2\right )^{5/2} \left (693 A d^3-297 B c d^2-177 c^3 D+253 c^2 C d\right )}{35 d (c+d x)^{5/2}}-\frac {2 \left (c^2-d^2 x^2\right )^{5/2} \left (-99 B d^2-129 c^2 D+110 c C d\right )}{7 d (c+d x)^{3/2}}\right )-\frac {2 d \left (c^2-d^2 x^2\right )^{5/2} (11 C d-21 c D)}{9 \sqrt {c+d x}}}{11 d^5}-\frac {2 D \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{5/2}}{11 d^4}\)

Input:

Int[((c^2 - d^2*x^2)^(3/2)*(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)^(5/2))/(11*d^4) + ((-2*d*(11*C*d - 21* 
c*D)*(c^2 - d^2*x^2)^(5/2))/(9*Sqrt[c + d*x]) - (d^2*((2*(253*c^2*C*d - 29 
7*B*c*d^2 + 693*A*d^3 - 177*c^3*D)*(c^2 - d^2*x^2)^(5/2))/(35*d*(c + d*x)^ 
(5/2)) - (2*(110*c*C*d - 99*B*d^2 - 129*c^2*D)*(c^2 - d^2*x^2)^(5/2))/(7*d 
*(c + d*x)^(3/2))))/9)/(11*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 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 = 111, normalized size of antiderivative = 0.43

method	result	size
gosper	\(-\frac {2 \left (-d x +c \right ) \left (315 D x^{3} d^{3}+385 C \,d^{3} x^{2}+210 D c \,d^{2} x^{2}+495 B \,d^{3} x +220 C c \,d^{2} x +120 D c^{2} d x +693 A \,d^{3}+198 B c \,d^{2}+88 C \,c^{2} d +48 D c^{3}\right ) \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{3465 d^{4} \left (d x +c \right )^{\frac {3}{2}}}\)	\(111\)
orering	\(-\frac {2 \left (-d x +c \right ) \left (315 D x^{3} d^{3}+385 C \,d^{3} x^{2}+210 D c \,d^{2} x^{2}+495 B \,d^{3} x +220 C c \,d^{2} x +120 D c^{2} d x +693 A \,d^{3}+198 B c \,d^{2}+88 C \,c^{2} d +48 D c^{3}\right ) \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{3465 d^{4} \left (d x +c \right )^{\frac {3}{2}}}\)	\(111\)
default	\(-\frac {2 \sqrt {-d^{2} x^{2}+c^{2}}\, \left (-d x +c \right )^{2} \left (315 D x^{3} d^{3}+385 C \,d^{3} x^{2}+210 D c \,d^{2} x^{2}+495 B \,d^{3} x +220 C c \,d^{2} x +120 D c^{2} d x +693 A \,d^{3}+198 B c \,d^{2}+88 C \,c^{2} d +48 D c^{3}\right )}{3465 \sqrt {d x +c}\, d^{4}}\)	\(113\)

Input:

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

Output:

-2/3465*(-d*x+c)*(315*D*d^3*x^3+385*C*d^3*x^2+210*D*c*d^2*x^2+495*B*d^3*x+ 
220*C*c*d^2*x+120*D*c^2*d*x+693*A*d^3+198*B*c*d^2+88*C*c^2*d+48*D*c^3)*(-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 = 196, normalized size of antiderivative = 0.77 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{3/2}} \, dx=-\frac {2 \, {\left (315 \, D d^{5} x^{5} + 48 \, D c^{5} + 88 \, C c^{4} d + 198 \, B c^{3} d^{2} + 693 \, A c^{2} d^{3} - 35 \, {\left (12 \, D c d^{4} - 11 \, C d^{5}\right )} x^{4} + 5 \, {\left (3 \, D c^{2} d^{3} - 110 \, C c d^{4} + 99 \, B d^{5}\right )} x^{3} + 3 \, {\left (6 \, D c^{3} d^{2} + 11 \, C c^{2} d^{3} - 264 \, B c d^{4} + 231 \, A d^{5}\right )} x^{2} + {\left (24 \, D c^{4} d + 44 \, C c^{3} d^{2} + 99 \, B c^{2} d^{3} - 1386 \, A c d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{3465 \, {\left (d^{5} x + c d^{4}\right )}} \] Input:

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

Output:

-2/3465*(315*D*d^5*x^5 + 48*D*c^5 + 88*C*c^4*d + 198*B*c^3*d^2 + 693*A*c^2 
*d^3 - 35*(12*D*c*d^4 - 11*C*d^5)*x^4 + 5*(3*D*c^2*d^3 - 110*C*c*d^4 + 99* 
B*d^5)*x^3 + 3*(6*D*c^3*d^2 + 11*C*c^2*d^3 - 264*B*c*d^4 + 231*A*d^5)*x^2 
+ (24*D*c^4*d + 44*C*c^3*d^2 + 99*B*c^2*d^3 - 1386*A*c*d^4)*x)*sqrt(-d^2*x 
^2 + c^2)*sqrt(d*x + c)/(d^5*x + c*d^4)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.76 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{3/2}} \, dx=-\frac {2 \, {\left (d^{2} x^{2} - 2 \, c d x + c^{2}\right )} \sqrt {-d x + c} A}{5 \, d} - \frac {2 \, {\left (5 \, d^{3} x^{3} - 8 \, c d^{2} x^{2} + c^{2} d x + 2 \, c^{3}\right )} \sqrt {-d x + c} B}{35 \, d^{2}} - \frac {2 \, {\left (35 \, d^{4} x^{4} - 50 \, c d^{3} x^{3} + 3 \, c^{2} d^{2} x^{2} + 4 \, c^{3} d x + 8 \, c^{4}\right )} \sqrt {-d x + c} C}{315 \, d^{3}} - \frac {2 \, {\left (105 \, d^{5} x^{5} - 140 \, c d^{4} x^{4} + 5 \, c^{2} d^{3} x^{3} + 6 \, c^{3} d^{2} x^{2} + 8 \, c^{4} d x + 16 \, c^{5}\right )} \sqrt {-d x + c} D}{1155 \, d^{4}} \] Input:

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

Output:

-2/5*(d^2*x^2 - 2*c*d*x + c^2)*sqrt(-d*x + c)*A/d - 2/35*(5*d^3*x^3 - 8*c* 
d^2*x^2 + c^2*d*x + 2*c^3)*sqrt(-d*x + c)*B/d^2 - 2/315*(35*d^4*x^4 - 50*c 
*d^3*x^3 + 3*c^2*d^2*x^2 + 4*c^3*d*x + 8*c^4)*sqrt(-d*x + c)*C/d^3 - 2/115 
5*(105*d^5*x^5 - 140*c*d^4*x^4 + 5*c^2*d^3*x^3 + 6*c^3*d^2*x^2 + 8*c^4*d*x 
 + 16*c^5)*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. 470 vs. \(2 (226) = 452\).

Time = 0.13 (sec) , antiderivative size = 470, normalized size of antiderivative = 1.84 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{3/2}} \, dx=-\frac {2 \, {\left (1155 \, {\left (-d x + c\right )}^{\frac {3}{2}} A c d^{3} - 231 \, {\left (3 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} - 5 \, {\left (-d x + c\right )}^{\frac {3}{2}} c\right )} B c d^{2} + 231 \, {\left (3 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} - 5 \, {\left (-d x + c\right )}^{\frac {3}{2}} c\right )} A d^{3} - 33 \, {\left (15 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} + 42 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} c - 35 \, {\left (-d x + c\right )}^{\frac {3}{2}} c^{2}\right )} C c d + 33 \, {\left (15 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} + 42 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} c - 35 \, {\left (-d x + c\right )}^{\frac {3}{2}} c^{2}\right )} B d^{2} - 11 \, {\left (35 \, {\left (d x - c\right )}^{4} \sqrt {-d x + c} + 135 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} c + 189 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} c^{2} - 105 \, {\left (-d x + c\right )}^{\frac {3}{2}} c^{3}\right )} D c + 11 \, {\left (35 \, {\left (d x - c\right )}^{4} \sqrt {-d x + c} + 135 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} c + 189 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} c^{2} - 105 \, {\left (-d x + c\right )}^{\frac {3}{2}} c^{3}\right )} C d + {\left (315 \, {\left (d x - c\right )}^{5} \sqrt {-d x + c} + 1540 \, {\left (d x - c\right )}^{4} \sqrt {-d x + c} c + 2970 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} c^{2} + 2772 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} c^{3} - 1155 \, {\left (-d x + c\right )}^{\frac {3}{2}} c^{4}\right )} D\right )}}{3465 \, d^{4}} \] Input:

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

Output:

-2/3465*(1155*(-d*x + c)^(3/2)*A*c*d^3 - 231*(3*(d*x - c)^2*sqrt(-d*x + c) 
 - 5*(-d*x + c)^(3/2)*c)*B*c*d^2 + 231*(3*(d*x - c)^2*sqrt(-d*x + c) - 5*( 
-d*x + c)^(3/2)*c)*A*d^3 - 33*(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*d + 33*(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)*B*d^2 - 11*(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 
 + 11*(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)*C*d + (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)*D)/d^4
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.49 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{3/2}} \, dx=\frac {2 \sqrt {-d x +c}\, \left (-315 d^{5} x^{5}+35 c \,d^{4} x^{4}-495 b \,d^{4} x^{3}+535 c^{2} d^{3} x^{3}-693 a \,d^{4} x^{2}+792 b c \,d^{3} x^{2}-51 c^{3} d^{2} x^{2}+1386 a c \,d^{3} x -99 b \,c^{2} d^{2} x -68 c^{4} d x -693 a \,c^{2} d^{2}-198 b \,c^{3} d -136 c^{5}\right )}{3465 d^{3}} \] Input:

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

Output:

(2*sqrt(c - d*x)*( - 693*a*c**2*d**2 + 1386*a*c*d**3*x - 693*a*d**4*x**2 - 
 198*b*c**3*d - 99*b*c**2*d**2*x + 792*b*c*d**3*x**2 - 495*b*d**4*x**3 - 1 
36*c**5 - 68*c**4*d*x - 51*c**3*d**2*x**2 + 535*c**2*d**3*x**3 + 35*c*d**4 
*x**4 - 315*d**5*x**5))/(3465*d**3)