\(\int \frac {\sqrt {c^2-d^2 x^2} (A+B x+C x^2+D x^3)}{(c+d x)^2} \, dx\) [133]

Optimal result

Integrand size = 39, antiderivative size = 203 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^2} \, dx=-\frac {\left (2 c C d-B d^2-3 c^2 D\right ) \sqrt {c^2-d^2 x^2}}{d^4}+\frac {(C d-2 c D) x \sqrt {c^2-d^2 x^2}}{2 d^3}-\frac {2 \left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \sqrt {c^2-d^2 x^2}}{d^4 (c+d x)}-\frac {D \left (c^2-d^2 x^2\right )^{3/2}}{3 d^4}-\frac {\left (5 c^2 C d-4 B c d^2+2 A d^3-6 c^3 D\right ) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{2 d^4} \] Output:

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

Mathematica [A] (verified)

Time = 1.07 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.78 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^2} \, dx=\frac {\frac {\sqrt {c^2-d^2 x^2} \left (28 c^3 D+c^2 (-24 C d+10 d D x)+c d^2 (18 B-x (9 C+4 D x))+d^3 \left (-12 A+x \left (6 B+3 C x+2 D x^2\right )\right )\right )}{c+d x}+6 \left (5 c^2 C d-4 B c d^2+2 A d^3-6 c^3 D\right ) \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{6 d^4} \] Input:

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

Output:

((Sqrt[c^2 - d^2*x^2]*(28*c^3*D + c^2*(-24*C*d + 10*d*D*x) + c*d^2*(18*B - 
 x*(9*C + 4*D*x)) + d^3*(-12*A + x*(6*B + 3*C*x + 2*D*x^2))))/(c + d*x) + 
6*(5*c^2*C*d - 4*B*c*d^2 + 2*A*d^3 - 6*c^3*D)*ArcTan[(d*x)/(Sqrt[c^2] - Sq 
rt[c^2 - d^2*x^2])])/(6*d^4)
 

Rubi [A] (verified)

Time = 0.97 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.205, Rules used = {2170, 27, 2170, 27, 671, 466, 224, 216}

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.

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

\(\Big \downarrow \) 2170

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2170

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 671

\(\displaystyle \frac {-\frac {1}{2} d^2 \left (\frac {\left (2 A d^3-4 B c d^2-6 c^3 D+5 c^2 C d\right ) \int \frac {\sqrt {c^2-d^2 x^2}}{c+d x}dx}{c}+\frac {2 \left (c^2-d^2 x^2\right )^{3/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (c+d x)^2}\right )-\frac {d \left (c^2-d^2 x^2\right )^{3/2} (C d-2 c D)}{2 (c+d x)}}{d^5}-\frac {D \left (c^2-d^2 x^2\right )^{3/2}}{3 d^4}\)

\(\Big \downarrow \) 466

\(\displaystyle \frac {-\frac {1}{2} d^2 \left (\frac {\left (2 A d^3-4 B c d^2-6 c^3 D+5 c^2 C d\right ) \left (c \int \frac {1}{\sqrt {c^2-d^2 x^2}}dx+\frac {\sqrt {c^2-d^2 x^2}}{d}\right )}{c}+\frac {2 \left (c^2-d^2 x^2\right )^{3/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (c+d x)^2}\right )-\frac {d \left (c^2-d^2 x^2\right )^{3/2} (C d-2 c D)}{2 (c+d x)}}{d^5}-\frac {D \left (c^2-d^2 x^2\right )^{3/2}}{3 d^4}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {-\frac {1}{2} d^2 \left (\frac {\left (2 A d^3-4 B c d^2-6 c^3 D+5 c^2 C d\right ) \left (c \int \frac {1}{\frac {d^2 x^2}{c^2-d^2 x^2}+1}d\frac {x}{\sqrt {c^2-d^2 x^2}}+\frac {\sqrt {c^2-d^2 x^2}}{d}\right )}{c}+\frac {2 \left (c^2-d^2 x^2\right )^{3/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (c+d x)^2}\right )-\frac {d \left (c^2-d^2 x^2\right )^{3/2} (C d-2 c D)}{2 (c+d x)}}{d^5}-\frac {D \left (c^2-d^2 x^2\right )^{3/2}}{3 d^4}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {-\frac {1}{2} d^2 \left (\frac {\left (\frac {c \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{d}+\frac {\sqrt {c^2-d^2 x^2}}{d}\right ) \left (2 A d^3-4 B c d^2-6 c^3 D+5 c^2 C d\right )}{c}+\frac {2 \left (c^2-d^2 x^2\right )^{3/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (c+d x)^2}\right )-\frac {d \left (c^2-d^2 x^2\right )^{3/2} (C d-2 c D)}{2 (c+d x)}}{d^5}-\frac {D \left (c^2-d^2 x^2\right )^{3/2}}{3 d^4}\)

Input:

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

Output:

-1/3*(D*(c^2 - d^2*x^2)^(3/2))/d^4 + (-1/2*(d*(C*d - 2*c*D)*(c^2 - d^2*x^2 
)^(3/2))/(c + d*x) - (d^2*((2*(c^2*C*d - B*c*d^2 + A*d^3 - c^3*D)*(c^2 - d 
^2*x^2)^(3/2))/(c*d*(c + d*x)^2) + ((5*c^2*C*d - 4*B*c*d^2 + 2*A*d^3 - 6*c 
^3*D)*(Sqrt[c^2 - d^2*x^2]/d + (c*ArcTan[(d*x)/Sqrt[c^2 - d^2*x^2]])/d))/c 
))/2)/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 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

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 671

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_ 
), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + c*x^2)^(p + 1)/(2*c*d*(m 
 + p + 1))), x] + Simp[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d)*(m 
+ p + 1))   Int[(d + e*x)^(m + 1)*(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] && ((LtQ[m, -1] &&  !IGtQ[m + p 
 + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p 
 + 1, 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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(381\) vs. \(2(187)=374\).

Time = 0.43 (sec) , antiderivative size = 382, normalized size of antiderivative = 1.88

method	result	size
default	\(\frac {C d \left (\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 \sqrt {d^{2}}}\right )-\frac {D \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{3 d}-2 D c \left (\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 \sqrt {d^{2}}}\right )}{d^{3}}+\frac {\left (B \,d^{2}-2 C c d +3 D c^{2}\right ) \left (\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}+\frac {c d \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{\sqrt {d^{2}}}\right )}{d^{4}}+\frac {\left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right ) \left (-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{c d \left (x +\frac {c}{d}\right )^{2}}-\frac {d \left (\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}+\frac {c d \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{\sqrt {d^{2}}}\right )}{c}\right )}{d^{5}}\)	\(382\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.29 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^2} \, dx=\frac {28 \, D c^{4} - 24 \, C c^{3} d + 18 \, B c^{2} d^{2} - 12 \, A c d^{3} + 2 \, {\left (14 \, D c^{3} d - 12 \, C c^{2} d^{2} + 9 \, B c d^{3} - 6 \, A d^{4}\right )} x - 6 \, {\left (6 \, D c^{4} - 5 \, C c^{3} d + 4 \, B c^{2} d^{2} - 2 \, A c d^{3} + {\left (6 \, D c^{3} d - 5 \, C c^{2} d^{2} + 4 \, B c d^{3} - 2 \, A d^{4}\right )} x\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) + {\left (2 \, D d^{3} x^{3} + 28 \, D c^{3} - 24 \, C c^{2} d + 18 \, B c d^{2} - 12 \, A d^{3} - {\left (4 \, D c d^{2} - 3 \, C d^{3}\right )} x^{2} + {\left (10 \, D c^{2} d - 9 \, C c d^{2} + 6 \, B d^{3}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{6 \, {\left (d^{5} x + c d^{4}\right )}} \] Input:

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

Output:

1/6*(28*D*c^4 - 24*C*c^3*d + 18*B*c^2*d^2 - 12*A*c*d^3 + 2*(14*D*c^3*d - 1 
2*C*c^2*d^2 + 9*B*c*d^3 - 6*A*d^4)*x - 6*(6*D*c^4 - 5*C*c^3*d + 4*B*c^2*d^ 
2 - 2*A*c*d^3 + (6*D*c^3*d - 5*C*c^2*d^2 + 4*B*c*d^3 - 2*A*d^4)*x)*arctan( 
-(c - sqrt(-d^2*x^2 + c^2))/(d*x)) + (2*D*d^3*x^3 + 28*D*c^3 - 24*C*c^2*d 
+ 18*B*c*d^2 - 12*A*d^3 - (4*D*c*d^2 - 3*C*d^3)*x^2 + (10*D*c^2*d - 9*C*c* 
d^2 + 6*B*d^3)*x)*sqrt(-d^2*x^2 + c^2))/(d^5*x + c*d^4)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.53 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^2} \, dx=\frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} D c^{3}}{d^{5} x + c d^{4}} - \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} C c^{2}}{d^{4} x + c d^{3}} + \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} B c}{d^{3} x + c d^{2}} + \frac {3 \, D c^{3} \arcsin \left (\frac {d x}{c}\right )}{d^{4}} - \frac {5 \, C c^{2} \arcsin \left (\frac {d x}{c}\right )}{2 \, d^{3}} + \frac {2 \, B c \arcsin \left (\frac {d x}{c}\right )}{d^{2}} - \frac {A \arcsin \left (\frac {d x}{c}\right )}{d} - \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} A}{d^{2} x + c d} - \frac {\sqrt {-d^{2} x^{2} + c^{2}} D c x}{d^{3}} + \frac {\sqrt {-d^{2} x^{2} + c^{2}} C x}{2 \, d^{2}} + \frac {3 \, \sqrt {-d^{2} x^{2} + c^{2}} D c^{2}}{d^{4}} - \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} C c}{d^{3}} + \frac {\sqrt {-d^{2} x^{2} + c^{2}} B}{d^{2}} - \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} D}{3 \, d^{4}} \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 542 vs. \(2 (186) = 372\).

Time = 0.17 (sec) , antiderivative size = 542, normalized size of antiderivative = 2.67 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^2} \, dx =\text {Too large to display} \] Input:

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

Output:

1/24*(48*D*c^4*d^4*sqrt(2*c/(d*x + c) - 1)*sgn(1/(d*x + c))*sgn(d) - 48*C* 
c^3*d^5*sqrt(2*c/(d*x + c) - 1)*sgn(1/(d*x + c))*sgn(d) + 48*B*c^2*d^6*sqr 
t(2*c/(d*x + c) - 1)*sgn(1/(d*x + c))*sgn(d) - 48*A*c*d^7*sqrt(2*c/(d*x + 
c) - 1)*sgn(1/(d*x + c))*sgn(d) - 24*(6*D*c^4*d^4*sgn(1/(d*x + c))*sgn(d) 
- 5*C*c^3*d^5*sgn(1/(d*x + c))*sgn(d) + 4*B*c^2*d^6*sgn(1/(d*x + c))*sgn(d 
) - 2*A*c*d^7*sgn(1/(d*x + c))*sgn(d))*arctan(sqrt(2*c/(d*x + c) - 1)) + ( 
24*D*c^4*d^4*(2*c/(d*x + c) - 1)^(5/2)*sgn(1/(d*x + c))*sgn(d) - 15*C*c^3* 
d^5*(2*c/(d*x + c) - 1)^(5/2)*sgn(1/(d*x + c))*sgn(d) + 6*B*c^2*d^6*(2*c/( 
d*x + c) - 1)^(5/2)*sgn(1/(d*x + c))*sgn(d) + 28*D*c^4*d^4*(2*c/(d*x + c) 
- 1)^(3/2)*sgn(1/(d*x + c))*sgn(d) - 24*C*c^3*d^5*(2*c/(d*x + c) - 1)^(3/2 
)*sgn(1/(d*x + c))*sgn(d) + 12*B*c^2*d^6*(2*c/(d*x + c) - 1)^(3/2)*sgn(1/( 
d*x + c))*sgn(d) + 12*D*c^4*d^4*sqrt(2*c/(d*x + c) - 1)*sgn(1/(d*x + c))*s 
gn(d) - 9*C*c^3*d^5*sqrt(2*c/(d*x + c) - 1)*sgn(1/(d*x + c))*sgn(d) + 6*B* 
c^2*d^6*sqrt(2*c/(d*x + c) - 1)*sgn(1/(d*x + c))*sgn(d))*(d*x + c)^3/c^3)* 
abs(d)/(c*d^9)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 396, normalized size of antiderivative = 1.95 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^2} \, dx=\frac {-c^{3} d x +3 c \,d^{3} x^{3}-2 d^{4} x^{4}-6 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{3}-\sqrt {-d^{2} x^{2}+c^{2}}\, c^{2} d x +\sqrt {-d^{2} x^{2}+c^{2}}\, c \,d^{2} x^{2}-6 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) a \,d^{2}+6 \mathit {asin} \left (\frac {d x}{c}\right ) a \,d^{3} x -3 \mathit {asin} \left (\frac {d x}{c}\right ) c^{3} d x -6 b c \,d^{2} x -2 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{3} x^{3}-3 \mathit {asin} \left (\frac {d x}{c}\right ) c^{4}+6 c^{4}+6 \mathit {asin} \left (\frac {d x}{c}\right ) a c \,d^{2}-12 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{2} d -24 \sqrt {-d^{2} x^{2}+c^{2}}\, b c d -6 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,d^{2} x +3 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) c^{3}-6 b \,d^{3} x^{2}-2 c^{2} d^{2} x^{2}+24 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{2}-24 a c \,d^{2}+24 b \,c^{2} d +12 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) b c d -12 \mathit {asin} \left (\frac {d x}{c}\right ) b c \,d^{2} x}{6 d^{3} \left (\sqrt {-d^{2} x^{2}+c^{2}}-c -d x \right )} \] Input:

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

Output:

( - 6*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*a*d**2 + 12*sqrt(c**2 - d**2*x* 
*2)*asin((d*x)/c)*b*c*d + 3*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*c**3 + 6* 
asin((d*x)/c)*a*c*d**2 + 6*asin((d*x)/c)*a*d**3*x - 12*asin((d*x)/c)*b*c** 
2*d - 12*asin((d*x)/c)*b*c*d**2*x - 3*asin((d*x)/c)*c**4 - 3*asin((d*x)/c) 
*c**3*d*x + 24*sqrt(c**2 - d**2*x**2)*a*d**2 - 24*sqrt(c**2 - d**2*x**2)*b 
*c*d - 6*sqrt(c**2 - d**2*x**2)*b*d**2*x - 6*sqrt(c**2 - d**2*x**2)*c**3 - 
 sqrt(c**2 - d**2*x**2)*c**2*d*x + sqrt(c**2 - d**2*x**2)*c*d**2*x**2 - 2* 
sqrt(c**2 - d**2*x**2)*d**3*x**3 - 24*a*c*d**2 + 24*b*c**2*d - 6*b*c*d**2* 
x - 6*b*d**3*x**2 + 6*c**4 - c**3*d*x - 2*c**2*d**2*x**2 + 3*c*d**3*x**3 - 
 2*d**4*x**4)/(6*d**3*(sqrt(c**2 - d**2*x**2) - c - d*x))