\(\int \frac {\sqrt {c^2-d^2 x^2} (A+B x+C x^2+D x^3)}{c+d x} \, dx\) [132]

Optimal result

Integrand size = 39, antiderivative size = 201 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{c+d x} \, dx=\frac {\left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \sqrt {c^2-d^2 x^2}}{d^4}-\frac {\left (4 c C d-4 B d^2-3 c^2 D\right ) x \sqrt {c^2-d^2 x^2}}{8 d^3}+\frac {D x^3 \sqrt {c^2-d^2 x^2}}{4 d}-\frac {(C d-c D) \left (c^2-d^2 x^2\right )^{3/2}}{3 d^4}+\frac {c \left (4 c^2 C d-4 B c d^2+8 A d^3-3 c^3 D\right ) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{8 d^4} \] Output:

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

Mathematica [A] (verified)

Time = 0.92 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.76 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{c+d x} \, dx=\frac {\sqrt {c^2-d^2 x^2} \left (-16 c^3 D+c^2 d (16 C+9 D x)-4 c d^2 (6 B+x (3 C+2 D x))+2 d^3 \left (12 A+x \left (6 B+4 C x+3 D x^2\right )\right )\right )+6 c \left (-4 c^2 C d+4 B c d^2-8 A d^3+3 c^3 D\right ) \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{24 d^4} \] Input:

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

Output:

(Sqrt[c^2 - d^2*x^2]*(-16*c^3*D + c^2*d*(16*C + 9*D*x) - 4*c*d^2*(6*B + x* 
(3*C + 2*D*x)) + 2*d^3*(12*A + x*(6*B + 4*C*x + 3*D*x^2))) + 6*c*(-4*c^2*C 
*d + 4*B*c*d^2 - 8*A*d^3 + 3*c^3*D)*ArcTan[(d*x)/(Sqrt[c^2] - Sqrt[c^2 - d 
^2*x^2])])/(24*d^4)
 

Rubi [A] (verified)

Time = 1.03 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.205, Rules used = {2170, 25, 2170, 27, 667, 676, 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.

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

\(\Big \downarrow \) 2170

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2170

\(\displaystyle \frac {-\frac {\int -\frac {3 d^6 \left (D c^3+4 A d^3-d \left (-5 D c^2+4 C d c-4 B d^2\right ) x\right ) \sqrt {c^2-d^2 x^2}}{c+d x}dx}{3 d^4}-\frac {1}{3} d \left (c^2-d^2 x^2\right )^{3/2} (4 C d-7 c D)}{4 d^5}-\frac {D (c+d x) \left (c^2-d^2 x^2\right )^{3/2}}{4 d^4}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 667

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

\(\Big \downarrow \) 676

\(\displaystyle \frac {d^2 \left (\frac {1}{2} c \left (8 A d^3-4 B c d^2-3 c^3 D+4 c^2 C d\right ) \int \frac {1}{\sqrt {c^2-d^2 x^2}}dx+\frac {4 \sqrt {c^2-d^2 x^2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{d}-\frac {1}{2} x \sqrt {c^2-d^2 x^2} \left (-4 B d^2-5 c^2 D+4 c C d\right )\right )-\frac {1}{3} d \left (c^2-d^2 x^2\right )^{3/2} (4 C d-7 c D)}{4 d^5}-\frac {D (c+d x) \left (c^2-d^2 x^2\right )^{3/2}}{4 d^4}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {d^2 \left (\frac {1}{2} c \left (8 A d^3-4 B c d^2-3 c^3 D+4 c^2 C d\right ) \int \frac {1}{\frac {d^2 x^2}{c^2-d^2 x^2}+1}d\frac {x}{\sqrt {c^2-d^2 x^2}}+\frac {4 \sqrt {c^2-d^2 x^2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{d}-\frac {1}{2} x \sqrt {c^2-d^2 x^2} \left (-4 B d^2-5 c^2 D+4 c C d\right )\right )-\frac {1}{3} d \left (c^2-d^2 x^2\right )^{3/2} (4 C d-7 c D)}{4 d^5}-\frac {D (c+d x) \left (c^2-d^2 x^2\right )^{3/2}}{4 d^4}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {d^2 \left (\frac {c \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right ) \left (8 A d^3-4 B c d^2-3 c^3 D+4 c^2 C d\right )}{2 d}+\frac {4 \sqrt {c^2-d^2 x^2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{d}-\frac {1}{2} x \sqrt {c^2-d^2 x^2} \left (-4 B d^2-5 c^2 D+4 c C d\right )\right )-\frac {1}{3} d \left (c^2-d^2 x^2\right )^{3/2} (4 C d-7 c D)}{4 d^5}-\frac {D (c+d x) \left (c^2-d^2 x^2\right )^{3/2}}{4 d^4}\)

Input:

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

Output:

-1/4*(D*(c + d*x)*(c^2 - d^2*x^2)^(3/2))/d^4 + (-1/3*(d*(4*C*d - 7*c*D)*(c 
^2 - d^2*x^2)^(3/2)) + d^2*((4*(c^2*C*d - B*c*d^2 + A*d^3 - c^3*D)*Sqrt[c^ 
2 - d^2*x^2])/d - ((4*c*C*d - 4*B*d^2 - 5*c^2*D)*x*Sqrt[c^2 - d^2*x^2])/2 
+ (c*(4*c^2*C*d - 4*B*c*d^2 + 8*A*d^3 - 3*c^3*D)*ArcTan[(d*x)/Sqrt[c^2 - d 
^2*x^2]])/(2*d)))/(4*d^5)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 667

Int[(((f_.) + (g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*( 
x_)), x_Symbol] :> Int[(a/d + c*(x/e))*(f + g*x)^n*(a + c*x^2)^(p - 1), x] 
/; FreeQ[{a, c, d, e, f, g, n, p}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[p, 0]
 

rule 676

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x 
_Symbol] :> Simp[(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + (Sim 
p[e*g*x*((a + c*x^2)^(p + 1)/(c*(2*p + 3))), x] - Simp[(a*e*g - c*d*f*(2*p 
+ 3))/(c*(2*p + 3))   Int[(a + c*x^2)^p, x], x]) /; FreeQ[{a, c, d, e, f, g 
, p}, x] &&  !LeQ[p, -1]
 

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. \(384\) vs. \(2(183)=366\).

Time = 0.41 (sec) , antiderivative size = 385, normalized size of antiderivative = 1.92

method	result	size
default	\(\frac {B \,d^{2} \left (\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 \sqrt {d^{2}}}\right )+D c^{2} \left (\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 \sqrt {d^{2}}}\right )-\frac {\left (C d -D c \right ) \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{3 d}+D d^{2} \left (-\frac {x \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{4 d^{2}}+\frac {c^{2} \left (\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 \sqrt {d^{2}}}\right )}{4 d^{2}}\right )-C c d \left (\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 \sqrt {d^{2}}}\right )}{d^{3}}+\frac {\left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right ) \left (\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}+\frac {c d \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{\sqrt {d^{2}}}\right )}{d^{4}}\)	\(385\)

Input:

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

Output:

1/d^3*(B*d^2*(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)))+D*c^2*(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*(C*d-D*c)*(-d 
^2*x^2+c^2)^(3/2)+D*d^2*(-1/4*x*(-d^2*x^2+c^2)^(3/2)/d^2+1/4*c^2/d^2*(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))))-C*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))))+(A*d^3-B*c*d^2+C*c^2*d-D*c^3)/d^4*( 
(-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 = 0.09 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{c+d x} \, dx=\frac {6 \, {\left (3 \, D c^{4} - 4 \, C c^{3} d + 4 \, B c^{2} d^{2} - 8 \, A c d^{3}\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) + {\left (6 \, D d^{3} x^{3} - 16 \, D c^{3} + 16 \, C c^{2} d - 24 \, B c d^{2} + 24 \, A d^{3} - 8 \, {\left (D c d^{2} - C d^{3}\right )} x^{2} + 3 \, {\left (3 \, D c^{2} d - 4 \, C c d^{2} + 4 \, B d^{3}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{24 \, d^{4}} \] Input:

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

Output:

1/24*(6*(3*D*c^4 - 4*C*c^3*d + 4*B*c^2*d^2 - 8*A*c*d^3)*arctan(-(c - sqrt( 
-d^2*x^2 + c^2))/(d*x)) + (6*D*d^3*x^3 - 16*D*c^3 + 16*C*c^2*d - 24*B*c*d^ 
2 + 24*A*d^3 - 8*(D*c*d^2 - C*d^3)*x^2 + 3*(3*D*c^2*d - 4*C*c*d^2 + 4*B*d^ 
3)*x)*sqrt(-d^2*x^2 + c^2))/d^4
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.37 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{c+d x} \, dx=-\frac {3 \, D c^{4} \arcsin \left (\frac {d x}{c}\right )}{8 \, d^{4}} + \frac {C c^{3} \arcsin \left (\frac {d x}{c}\right )}{2 \, d^{3}} - \frac {B c^{2} \arcsin \left (\frac {d x}{c}\right )}{2 \, d^{2}} + \frac {A c \arcsin \left (\frac {d x}{c}\right )}{d} + \frac {5 \, \sqrt {-d^{2} x^{2} + c^{2}} D c^{2} x}{8 \, d^{3}} - \frac {\sqrt {-d^{2} x^{2} + c^{2}} C c x}{2 \, d^{2}} + \frac {\sqrt {-d^{2} x^{2} + c^{2}} B x}{2 \, d} - \frac {\sqrt {-d^{2} x^{2} + c^{2}} D c^{3}}{d^{4}} + \frac {\sqrt {-d^{2} x^{2} + c^{2}} C c^{2}}{d^{3}} - \frac {\sqrt {-d^{2} x^{2} + c^{2}} B c}{d^{2}} + \frac {\sqrt {-d^{2} x^{2} + c^{2}} A}{d} - \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} D x}{4 \, d^{3}} + \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} D c}{3 \, d^{4}} - \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} C}{3 \, d^{3}} \] Input:

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

Output:

-3/8*D*c^4*arcsin(d*x/c)/d^4 + 1/2*C*c^3*arcsin(d*x/c)/d^3 - 1/2*B*c^2*arc 
sin(d*x/c)/d^2 + A*c*arcsin(d*x/c)/d + 5/8*sqrt(-d^2*x^2 + c^2)*D*c^2*x/d^ 
3 - 1/2*sqrt(-d^2*x^2 + c^2)*C*c*x/d^2 + 1/2*sqrt(-d^2*x^2 + c^2)*B*x/d - 
sqrt(-d^2*x^2 + c^2)*D*c^3/d^4 + sqrt(-d^2*x^2 + c^2)*C*c^2/d^3 - sqrt(-d^ 
2*x^2 + c^2)*B*c/d^2 + sqrt(-d^2*x^2 + c^2)*A/d - 1/4*(-d^2*x^2 + c^2)^(3/ 
2)*D*x/d^3 + 1/3*(-d^2*x^2 + c^2)^(3/2)*D*c/d^4 - 1/3*(-d^2*x^2 + c^2)^(3/ 
2)*C/d^3
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.82 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{c+d x} \, dx=\frac {1}{24} \, \sqrt {-d^{2} x^{2} + c^{2}} {\left ({\left (2 \, {\left (\frac {3 \, D x}{d} - \frac {4 \, {\left (D c d^{5} - C d^{6}\right )}}{d^{7}}\right )} x + \frac {3 \, {\left (3 \, D c^{2} d^{4} - 4 \, C c d^{5} + 4 \, B d^{6}\right )}}{d^{7}}\right )} x - \frac {8 \, {\left (2 \, D c^{3} d^{3} - 2 \, C c^{2} d^{4} + 3 \, B c d^{5} - 3 \, A d^{6}\right )}}{d^{7}}\right )} - \frac {{\left (3 \, D c^{4} - 4 \, C c^{3} d + 4 \, B c^{2} d^{2} - 8 \, A c d^{3}\right )} \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{8 \, d^{3} {\left | d \right |}} \] Input:

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

Output:

1/24*sqrt(-d^2*x^2 + c^2)*((2*(3*D*x/d - 4*(D*c*d^5 - C*d^6)/d^7)*x + 3*(3 
*D*c^2*d^4 - 4*C*c*d^5 + 4*B*d^6)/d^7)*x - 8*(2*D*c^3*d^3 - 2*C*c^2*d^4 + 
3*B*c*d^5 - 3*A*d^6)/d^7) - 1/8*(3*D*c^4 - 4*C*c^3*d + 4*B*c^2*d^2 - 8*A*c 
*d^3)*arcsin(d*x/c)*sgn(c)*sgn(d)/(d^3*abs(d))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.78 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{c+d x} \, dx=\frac {8 \mathit {asin} \left (\frac {d x}{c}\right ) a c \,d^{2}-4 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{2} d +\mathit {asin} \left (\frac {d x}{c}\right ) c^{4}+8 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{2}-8 \sqrt {-d^{2} x^{2}+c^{2}}\, b c d +4 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,d^{2} x -\sqrt {-d^{2} x^{2}+c^{2}}\, c^{2} d x +2 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{3} x^{3}-8 a c \,d^{2}+8 b \,c^{2} d}{8 d^{3}} \] Input:

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

Output:

(8*asin((d*x)/c)*a*c*d**2 - 4*asin((d*x)/c)*b*c**2*d + asin((d*x)/c)*c**4 
+ 8*sqrt(c**2 - d**2*x**2)*a*d**2 - 8*sqrt(c**2 - d**2*x**2)*b*c*d + 4*sqr 
t(c**2 - d**2*x**2)*b*d**2*x - sqrt(c**2 - d**2*x**2)*c**2*d*x + 2*sqrt(c* 
*2 - d**2*x**2)*d**3*x**3 - 8*a*c*d**2 + 8*b*c**2*d)/(8*d**3)