\(\int \frac {A+B x+C x^2+D x^3}{(c+d x)^3 \sqrt {c^2-d^2 x^2}} \, dx\) [175]

Optimal result

Integrand size = 39, antiderivative size = 204 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^3 \sqrt {c^2-d^2 x^2}} \, dx=-\frac {\left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \sqrt {c^2-d^2 x^2}}{5 c d^4 (c+d x)^3}+\frac {\left (8 c^2 C d-3 B c d^2-2 A d^3-13 c^3 D\right ) \sqrt {c^2-d^2 x^2}}{15 c^2 d^4 (c+d x)^2}-\frac {\left (7 c^2 C d+3 B c d^2+2 A d^3-32 c^3 D\right ) \sqrt {c^2-d^2 x^2}}{15 c^3 d^4 (c+d x)}+\frac {D \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{d^4} \] Output:

-1/5*(A*d^3-B*c*d^2+C*c^2*d-D*c^3)*(-d^2*x^2+c^2)^(1/2)/c/d^4/(d*x+c)^3+1/ 
15*(-2*A*d^3-3*B*c*d^2+8*C*c^2*d-13*D*c^3)*(-d^2*x^2+c^2)^(1/2)/c^2/d^4/(d 
*x+c)^2-1/15*(2*A*d^3+3*B*c*d^2+7*C*c^2*d-32*D*c^3)*(-d^2*x^2+c^2)^(1/2)/c 
^3/d^4/(d*x+c)+D*arctan(d*x/(-d^2*x^2+c^2)^(1/2))/d^4
 

Mathematica [A] (verified)

Time = 1.18 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.77 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^3 \sqrt {c^2-d^2 x^2}} \, dx=\frac {\frac {\sqrt {c^2-d^2 x^2} \left (22 c^5 D-2 A d^5 x^2-3 c d^4 x (2 A+B x)+c^4 (-2 C d+51 d D x)+c^3 d^2 \left (-3 B-6 C x+32 D x^2\right )-c^2 d^3 (7 A+x (9 B+7 C x))\right )}{c^3 (c+d x)^3}-30 D \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{15 d^4} \] Input:

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

Output:

((Sqrt[c^2 - d^2*x^2]*(22*c^5*D - 2*A*d^5*x^2 - 3*c*d^4*x*(2*A + B*x) + c^ 
4*(-2*C*d + 51*d*D*x) + c^3*d^2*(-3*B - 6*C*x + 32*D*x^2) - c^2*d^3*(7*A + 
 x*(9*B + 7*C*x))))/(c^3*(c + d*x)^3) - 30*D*ArcTan[(d*x)/(Sqrt[c^2] - Sqr 
t[c^2 - d^2*x^2])])/(15*d^4)
 

Rubi [A] (verified)

Time = 0.94 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.67, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {2168, 2009}

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.

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

\(\Big \downarrow \) 2168

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 \sqrt {c^2-d^2 x^2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{15 c^3 d^4 (c+d x)}-\frac {2 \sqrt {c^2-d^2 x^2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{15 c^2 d^4 (c+d x)^2}-\frac {\sqrt {c^2-d^2 x^2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{5 c d^4 (c+d x)^3}+\frac {D \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{d^4}+\frac {\sqrt {c^2-d^2 x^2} \left (-B d^2-3 c^2 D+2 c C d\right )}{3 c^2 d^4 (c+d x)}+\frac {\sqrt {c^2-d^2 x^2} \left (-B d^2-3 c^2 D+2 c C d\right )}{3 c d^4 (c+d x)^2}-\frac {\sqrt {c^2-d^2 x^2} (C d-3 c D)}{c d^4 (c+d x)}\)

Input:

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

Output:

-1/5*((c^2*C*d - B*c*d^2 + A*d^3 - c^3*D)*Sqrt[c^2 - d^2*x^2])/(c*d^4*(c + 
 d*x)^3) + ((2*c*C*d - B*d^2 - 3*c^2*D)*Sqrt[c^2 - d^2*x^2])/(3*c*d^4*(c + 
 d*x)^2) - (2*(c^2*C*d - B*c*d^2 + A*d^3 - c^3*D)*Sqrt[c^2 - d^2*x^2])/(15 
*c^2*d^4*(c + d*x)^2) - ((C*d - 3*c*D)*Sqrt[c^2 - d^2*x^2])/(c*d^4*(c + d* 
x)) + ((2*c*C*d - B*d^2 - 3*c^2*D)*Sqrt[c^2 - d^2*x^2])/(3*c^2*d^4*(c + d* 
x)) - (2*(c^2*C*d - B*c*d^2 + A*d^3 - c^3*D)*Sqrt[c^2 - d^2*x^2])/(15*c^3* 
d^4*(c + d*x)) + (D*ArcTan[(d*x)/Sqrt[c^2 - d^2*x^2]])/d^4
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2168

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
 Int[ExpandIntegrand[(a + b*x^2)^p, (d + e*x)^m*Pq, x], x] /; FreeQ[{a, b, 
d, e}, x] && PolyQ[Pq, x] && EqQ[b*d^2 + a*e^2, 0] && EqQ[m + Expon[Pq, x] 
+ 2*p + 1, 0] && ILtQ[m, 0]
 

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 365, normalized size of antiderivative = 1.79

method	result	size
default	\(\frac {D \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{d^{3} \sqrt {d^{2}}}-\frac {\left (C d -3 D c \right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{d^{5} c \left (x +\frac {c}{d}\right )}+\frac {\left (B \,d^{2}-2 C c d +3 D c^{2}\right ) \left (-\frac {\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{3 c d \left (x +\frac {c}{d}\right )^{2}}-\frac {\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{3 c^{2} \left (x +\frac {c}{d}\right )}\right )}{d^{5}}+\frac {\left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right ) \left (-\frac {\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{5 c d \left (x +\frac {c}{d}\right )^{3}}+\frac {2 d \left (-\frac {\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{3 c d \left (x +\frac {c}{d}\right )^{2}}-\frac {\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{3 c^{2} \left (x +\frac {c}{d}\right )}\right )}{5 c}\right )}{d^{6}}\)	\(365\)

Input:

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

Output:

D/d^3/(d^2)^(1/2)*arctan((d^2)^(1/2)*x/(-d^2*x^2+c^2)^(1/2))-1/d^5*(C*d-3* 
D*c)/c/(x+c/d)*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(1/2)+1/d^5*(B*d^2-2*C*c*d+3 
*D*c^2)*(-1/3/c/d/(x+c/d)^2*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(1/2)-1/3/c^2/( 
x+c/d)*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(1/2))+1/d^6*(A*d^3-B*c*d^2+C*c^2*d- 
D*c^3)*(-1/5/c/d/(x+c/d)^3*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(1/2)+2/5*d/c*(- 
1/3/c/d/(x+c/d)^2*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(1/2)-1/3/c^2/(x+c/d)*(-d 
^2*(x+c/d)^2+2*c*d*(x+c/d))^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 373, normalized size of antiderivative = 1.83 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^3 \sqrt {c^2-d^2 x^2}} \, dx=\frac {22 \, D c^{6} - 2 \, C c^{5} d - 3 \, B c^{4} d^{2} - 7 \, A c^{3} d^{3} + {\left (22 \, D c^{3} d^{3} - 2 \, C c^{2} d^{4} - 3 \, B c d^{5} - 7 \, A d^{6}\right )} x^{3} + 3 \, {\left (22 \, D c^{4} d^{2} - 2 \, C c^{3} d^{3} - 3 \, B c^{2} d^{4} - 7 \, A c d^{5}\right )} x^{2} + 3 \, {\left (22 \, D c^{5} d - 2 \, C c^{4} d^{2} - 3 \, B c^{3} d^{3} - 7 \, A c^{2} d^{4}\right )} x - 30 \, {\left (D c^{3} d^{3} x^{3} + 3 \, D c^{4} d^{2} x^{2} + 3 \, D c^{5} d x + D c^{6}\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) + {\left (22 \, D c^{5} - 2 \, C c^{4} d - 3 \, B c^{3} d^{2} - 7 \, A c^{2} d^{3} + {\left (32 \, D c^{3} d^{2} - 7 \, C c^{2} d^{3} - 3 \, B c d^{4} - 2 \, A d^{5}\right )} x^{2} + 3 \, {\left (17 \, D c^{4} d - 2 \, C c^{3} d^{2} - 3 \, B c^{2} d^{3} - 2 \, A c d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{15 \, {\left (c^{3} d^{7} x^{3} + 3 \, c^{4} d^{6} x^{2} + 3 \, c^{5} d^{5} x + c^{6} d^{4}\right )}} \] Input:

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

Output:

1/15*(22*D*c^6 - 2*C*c^5*d - 3*B*c^4*d^2 - 7*A*c^3*d^3 + (22*D*c^3*d^3 - 2 
*C*c^2*d^4 - 3*B*c*d^5 - 7*A*d^6)*x^3 + 3*(22*D*c^4*d^2 - 2*C*c^3*d^3 - 3* 
B*c^2*d^4 - 7*A*c*d^5)*x^2 + 3*(22*D*c^5*d - 2*C*c^4*d^2 - 3*B*c^3*d^3 - 7 
*A*c^2*d^4)*x - 30*(D*c^3*d^3*x^3 + 3*D*c^4*d^2*x^2 + 3*D*c^5*d*x + D*c^6) 
*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d*x)) + (22*D*c^5 - 2*C*c^4*d - 3*B*c 
^3*d^2 - 7*A*c^2*d^3 + (32*D*c^3*d^2 - 7*C*c^2*d^3 - 3*B*c*d^4 - 2*A*d^5)* 
x^2 + 3*(17*D*c^4*d - 2*C*c^3*d^2 - 3*B*c^2*d^3 - 2*A*c*d^4)*x)*sqrt(-d^2* 
x^2 + c^2))/(c^3*d^7*x^3 + 3*c^4*d^6*x^2 + 3*c^5*d^5*x + c^6*d^4)
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 880 vs. \(2 (190) = 380\).

Time = 0.13 (sec) , antiderivative size = 880, normalized size of antiderivative = 4.31 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^3 \sqrt {c^2-d^2 x^2}} \, dx =\text {Too large to display} \] Input:

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

Output:

1/5*sqrt(-d^2*x^2 + c^2)*D*c^3/(c*d^7*x^3 + 3*c^2*d^6*x^2 + 3*c^3*d^5*x + 
c^4*d^4) + 2/15*sqrt(-d^2*x^2 + c^2)*D*c^3/(c^2*d^6*x^2 + 2*c^3*d^5*x + c^ 
4*d^4) + 2/15*sqrt(-d^2*x^2 + c^2)*D*c^3/(c^3*d^5*x + c^4*d^4) - 1/5*sqrt( 
-d^2*x^2 + c^2)*C*c^2/(c*d^6*x^3 + 3*c^2*d^5*x^2 + 3*c^3*d^4*x + c^4*d^3) 
- 2/15*sqrt(-d^2*x^2 + c^2)*C*c^2/(c^2*d^5*x^2 + 2*c^3*d^4*x + c^4*d^3) - 
2/15*sqrt(-d^2*x^2 + c^2)*C*c^2/(c^3*d^4*x + c^4*d^3) - sqrt(-d^2*x^2 + c^ 
2)*D*c^2/(c*d^6*x^2 + 2*c^2*d^5*x + c^3*d^4) - sqrt(-d^2*x^2 + c^2)*D*c^2/ 
(c^2*d^5*x + c^3*d^4) + 1/5*sqrt(-d^2*x^2 + c^2)*B*c/(c*d^5*x^3 + 3*c^2*d^ 
4*x^2 + 3*c^3*d^3*x + c^4*d^2) + 2/15*sqrt(-d^2*x^2 + c^2)*B*c/(c^2*d^4*x^ 
2 + 2*c^3*d^3*x + c^4*d^2) + 2/15*sqrt(-d^2*x^2 + c^2)*B*c/(c^3*d^3*x + c^ 
4*d^2) + 2/3*sqrt(-d^2*x^2 + c^2)*C*c/(c*d^5*x^2 + 2*c^2*d^4*x + c^3*d^3) 
+ 2/3*sqrt(-d^2*x^2 + c^2)*C*c/(c^2*d^4*x + c^3*d^3) - 1/5*sqrt(-d^2*x^2 + 
 c^2)*A/(c*d^4*x^3 + 3*c^2*d^3*x^2 + 3*c^3*d^2*x + c^4*d) - 2/15*sqrt(-d^2 
*x^2 + c^2)*A/(c^2*d^3*x^2 + 2*c^3*d^2*x + c^4*d) - 2/15*sqrt(-d^2*x^2 + c 
^2)*A/(c^3*d^2*x + c^4*d) - 1/3*sqrt(-d^2*x^2 + c^2)*B/(c*d^4*x^2 + 2*c^2* 
d^3*x + c^3*d^2) - 1/3*sqrt(-d^2*x^2 + c^2)*B/(c^2*d^3*x + c^3*d^2) - sqrt 
(-d^2*x^2 + c^2)*C/(c*d^4*x + c^2*d^3) + 3*sqrt(-d^2*x^2 + c^2)*D/(d^5*x + 
 c*d^4) + D*arcsin(d*x/c)/d^4
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 516 vs. \(2 (190) = 380\).

Time = 0.15 (sec) , antiderivative size = 516, normalized size of antiderivative = 2.53 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^3 \sqrt {c^2-d^2 x^2}} \, dx=\frac {D \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{d^{3} {\left | d \right |}} - \frac {2 \, {\left (22 \, D c^{3} - 2 \, C c^{2} d - 3 \, B c d^{2} - 7 \, A d^{3} - \frac {15 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} B c}{x} + \frac {95 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} D c^{3}}{d^{2} x} - \frac {10 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} C c^{2}}{d x} - \frac {20 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} A d}{x} + \frac {145 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} D c^{3}}{d^{4} x^{2}} - \frac {20 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} C c^{2}}{d^{3} x^{2}} - \frac {15 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} B c}{d^{2} x^{2}} - \frac {40 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} A}{d x^{2}} + \frac {75 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{3} D c^{3}}{d^{6} x^{3}} - \frac {15 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{3} B c}{d^{4} x^{3}} - \frac {30 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{3} A}{d^{3} x^{3}} + \frac {15 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{4} D c^{3}}{d^{8} x^{4}} - \frac {15 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{4} A}{d^{5} x^{4}}\right )}}{15 \, c^{3} d^{3} {\left (\frac {c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}}{d^{2} x} + 1\right )}^{5} {\left | d \right |}} \] Input:

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

Output:

D*arcsin(d*x/c)*sgn(c)*sgn(d)/(d^3*abs(d)) - 2/15*(22*D*c^3 - 2*C*c^2*d - 
3*B*c*d^2 - 7*A*d^3 - 15*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))*B*c/x + 95*(c 
*d + sqrt(-d^2*x^2 + c^2)*abs(d))*D*c^3/(d^2*x) - 10*(c*d + sqrt(-d^2*x^2 
+ c^2)*abs(d))*C*c^2/(d*x) - 20*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))*A*d/x 
+ 145*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^2*D*c^3/(d^4*x^2) - 20*(c*d + sq 
rt(-d^2*x^2 + c^2)*abs(d))^2*C*c^2/(d^3*x^2) - 15*(c*d + sqrt(-d^2*x^2 + c 
^2)*abs(d))^2*B*c/(d^2*x^2) - 40*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^2*A/( 
d*x^2) + 75*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^3*D*c^3/(d^6*x^3) - 15*(c* 
d + sqrt(-d^2*x^2 + c^2)*abs(d))^3*B*c/(d^4*x^3) - 30*(c*d + sqrt(-d^2*x^2 
 + c^2)*abs(d))^3*A/(d^3*x^3) + 15*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^4*D 
*c^3/(d^8*x^4) - 15*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^4*A/(d^5*x^4))/(c^ 
3*d^3*((c*d + sqrt(-d^2*x^2 + c^2)*abs(d))/(d^2*x) + 1)^5*abs(d))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 392, normalized size of antiderivative = 1.92 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^3 \sqrt {c^2-d^2 x^2}} \, dx=\frac {15 \mathit {asin} \left (\frac {d x}{c}\right ) \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{5} c^{3}+75 \mathit {asin} \left (\frac {d x}{c}\right ) \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{4} c^{3}+150 \mathit {asin} \left (\frac {d x}{c}\right ) \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{3} c^{3}+150 \mathit {asin} \left (\frac {d x}{c}\right ) \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{2} c^{3}+75 \mathit {asin} \left (\frac {d x}{c}\right ) \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right ) c^{3}+15 \mathit {asin} \left (\frac {d x}{c}\right ) c^{3}+6 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{5} a \,d^{2}-6 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{5} c^{3}-30 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{3} b c d +90 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{3} c^{3}-20 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{2} a \,d^{2}-30 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{2} b c d +190 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{2} c^{3}-10 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right ) a \,d^{2}-30 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right ) b c d +140 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right ) c^{3}-8 a \,d^{2}-6 b c d +34 c^{3}}{15 c^{3} d^{3} \left (\tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{5}+5 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{4}+10 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{3}+10 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{2}+5 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )+1\right )} \] Input:

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

Output:

(15*asin((d*x)/c)*tan(asin((d*x)/c)/2)**5*c**3 + 75*asin((d*x)/c)*tan(asin 
((d*x)/c)/2)**4*c**3 + 150*asin((d*x)/c)*tan(asin((d*x)/c)/2)**3*c**3 + 15 
0*asin((d*x)/c)*tan(asin((d*x)/c)/2)**2*c**3 + 75*asin((d*x)/c)*tan(asin(( 
d*x)/c)/2)*c**3 + 15*asin((d*x)/c)*c**3 + 6*tan(asin((d*x)/c)/2)**5*a*d**2 
 - 6*tan(asin((d*x)/c)/2)**5*c**3 - 30*tan(asin((d*x)/c)/2)**3*b*c*d + 90* 
tan(asin((d*x)/c)/2)**3*c**3 - 20*tan(asin((d*x)/c)/2)**2*a*d**2 - 30*tan( 
asin((d*x)/c)/2)**2*b*c*d + 190*tan(asin((d*x)/c)/2)**2*c**3 - 10*tan(asin 
((d*x)/c)/2)*a*d**2 - 30*tan(asin((d*x)/c)/2)*b*c*d + 140*tan(asin((d*x)/c 
)/2)*c**3 - 8*a*d**2 - 6*b*c*d + 34*c**3)/(15*c**3*d**3*(tan(asin((d*x)/c) 
/2)**5 + 5*tan(asin((d*x)/c)/2)**4 + 10*tan(asin((d*x)/c)/2)**3 + 10*tan(a 
sin((d*x)/c)/2)**2 + 5*tan(asin((d*x)/c)/2) + 1))