\(\int \frac {(c+d x) (A+B x+C x^2+D x^3)}{(c^2-d^2 x^2)^{3/2}} \, dx\) [179]

Optimal result

Integrand size = 37, antiderivative size = 149 \[ \int \frac {(c+d x) \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {\left (c^2 C d+B c d^2+A d^3+c^3 D\right ) (c+d x)}{c d^4 \sqrt {c^2-d^2 x^2}}+\frac {(C d+c D) \sqrt {c^2-d^2 x^2}}{d^4}+\frac {D x \sqrt {c^2-d^2 x^2}}{2 d^3}-\frac {\left (2 c C d+2 B d^2+3 c^2 D\right ) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{2 d^4} \] Output:

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

Mathematica [A] (verified)

Time = 0.79 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.89 \[ \int \frac {(c+d x) \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {\frac {\sqrt {c^2-d^2 x^2} \left (2 A d^3+4 c^3 D+c^2 d (4 C-D x)+c d^2 (2 B-x (2 C+D x))\right )}{c (c-d x)}+2 \left (2 c C d+2 B d^2+3 c^2 D\right ) \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{2 d^4} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.189, Rules used = {2166, 2346, 25, 27, 455, 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.

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

\(\Big \downarrow \) 2166

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

\(\Big \downarrow \) 2346

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

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

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 455

Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( 
a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c   Int[(a + b*x^2)^p, x], x] 
/; FreeQ[{a, b, c, d, p}, x] &&  !LeQ[p, -1]
 

rule 2166

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{Qx = PolynomialQuotient[Pq, a*e + b*d*x, x], R = PolynomialRemainde 
r[Pq, a*e + b*d*x, x]}, Simp[(-d)*R*(d + e*x)^m*((a + b*x^2)^(p + 1)/(2*a*e 
*(p + 1))), x] + Simp[d/(2*a*(p + 1))   Int[(d + e*x)^(m - 1)*(a + b*x^2)^( 
p + 1)*ExpandToSum[2*a*e*(p + 1)*Qx + R*(m + 2*p + 2), x], x], x]] /; FreeQ 
[{a, b, d, e}, x] && PolyQ[Pq, x] && EqQ[b*d^2 + a*e^2, 0] && ILtQ[p + 1/2, 
 0] && GtQ[m, 0]
 

rule 2346

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], 
e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x^2)^(p + 1)/(b*( 
q + 2*p + 1))), x] + Simp[1/(b*(q + 2*p + 1))   Int[(a + b*x^2)^p*ExpandToS 
um[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], 
x], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] &&  !LeQ[p, -1]
 

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.65

method	result	size
default	\(\frac {A x}{c \sqrt {-d^{2} x^{2}+c^{2}}}+\frac {A d +B c}{d^{2} \sqrt {-d^{2} x^{2}+c^{2}}}+\left (B d +C c \right ) \left (\frac {x}{\sqrt {-d^{2} x^{2}+c^{2}}\, d^{2}}-\frac {\arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{d^{2} \sqrt {d^{2}}}\right )+\left (C d +D c \right ) \left (-\frac {x^{2}}{d^{2} \sqrt {-d^{2} x^{2}+c^{2}}}+\frac {2 c^{2}}{d^{4} \sqrt {-d^{2} x^{2}+c^{2}}}\right )+D d \left (-\frac {x^{3}}{2 d^{2} \sqrt {-d^{2} x^{2}+c^{2}}}+\frac {3 c^{2} \left (\frac {x}{\sqrt {-d^{2} x^{2}+c^{2}}\, d^{2}}-\frac {\arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{d^{2} \sqrt {d^{2}}}\right )}{2 d^{2}}\right )\)	\(246\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.52 \[ \int \frac {(c+d x) \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=-\frac {4 \, D c^{4} + 4 \, C c^{3} d + 2 \, B c^{2} d^{2} + 2 \, A c d^{3} - 2 \, {\left (2 \, D c^{3} d + 2 \, C c^{2} d^{2} + B c d^{3} + A d^{4}\right )} x + 2 \, {\left (3 \, D c^{4} + 2 \, C c^{3} d + 2 \, B c^{2} d^{2} - {\left (3 \, D c^{3} d + 2 \, C c^{2} d^{2} + 2 \, B c d^{3}\right )} x\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) - {\left (D c d^{2} x^{2} - 4 \, D c^{3} - 4 \, C c^{2} d - 2 \, B c d^{2} - 2 \, A d^{3} + {\left (D c^{2} d + 2 \, C c d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{2 \, {\left (c d^{5} x - c^{2} d^{4}\right )}} \] Input:

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 9.59 (sec) , antiderivative size = 558, normalized size of antiderivative = 3.74 \[ \int \frac {(c+d x) \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

A*c*Piecewise((-I*x/(c**3*sqrt(-1 + d**2*x**2/c**2)), Abs(d**2*x**2/c**2) 
> 1), (x/(c**3*sqrt(1 - d**2*x**2/c**2)), True)) + A*d*Piecewise((1/(d**2* 
sqrt(c**2 - d**2*x**2)), Ne(d, 0)), (x**2/(2*(c**2)**(3/2)), True)) + B*c* 
Piecewise((1/(d**2*sqrt(c**2 - d**2*x**2)), Ne(d, 0)), (x**2/(2*(c**2)**(3 
/2)), True)) + B*d*Piecewise((I*acosh(d*x/c)/d**3 - I*x/(c*d**2*sqrt(-1 + 
d**2*x**2/c**2)), Abs(d**2*x**2/c**2) > 1), (-asin(d*x/c)/d**3 + x/(c*d**2 
*sqrt(1 - d**2*x**2/c**2)), True)) + C*c*Piecewise((I*acosh(d*x/c)/d**3 - 
I*x/(c*d**2*sqrt(-1 + d**2*x**2/c**2)), Abs(d**2*x**2/c**2) > 1), (-asin(d 
*x/c)/d**3 + x/(c*d**2*sqrt(1 - d**2*x**2/c**2)), True)) + C*d*Piecewise(( 
2*c**2/(d**4*sqrt(c**2 - d**2*x**2)) - x**2/(d**2*sqrt(c**2 - d**2*x**2)), 
 Ne(d, 0)), (x**4/(4*(c**2)**(3/2)), True)) + D*c*Piecewise((2*c**2/(d**4* 
sqrt(c**2 - d**2*x**2)) - x**2/(d**2*sqrt(c**2 - d**2*x**2)), Ne(d, 0)), ( 
x**4/(4*(c**2)**(3/2)), True)) + D*d*Piecewise((3*I*c**2*acosh(d*x/c)/(2*d 
**5) - 3*I*c*x/(2*d**4*sqrt(-1 + d**2*x**2/c**2)) + I*x**3/(2*c*d**2*sqrt( 
-1 + d**2*x**2/c**2)), Abs(d**2*x**2/c**2) > 1), (-3*c**2*asin(d*x/c)/(2*d 
**5) + 3*c*x/(2*d**4*sqrt(1 - d**2*x**2/c**2)) - x**3/(2*c*d**2*sqrt(1 - d 
**2*x**2/c**2)), True))
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.52 \[ \int \frac {(c+d x) \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=-\frac {D x^{3}}{2 \, \sqrt {-d^{2} x^{2} + c^{2}} d} + \frac {A x}{\sqrt {-d^{2} x^{2} + c^{2}} c} + \frac {3 \, D c^{2} x}{2 \, \sqrt {-d^{2} x^{2} + c^{2}} d^{3}} - \frac {{\left (D c + C d\right )} x^{2}}{\sqrt {-d^{2} x^{2} + c^{2}} d^{2}} - \frac {3 \, D c^{2} \arcsin \left (\frac {d x}{c}\right )}{2 \, d^{4}} + \frac {B c}{\sqrt {-d^{2} x^{2} + c^{2}} d^{2}} + \frac {A}{\sqrt {-d^{2} x^{2} + c^{2}} d} + \frac {{\left (C c + B d\right )} x}{\sqrt {-d^{2} x^{2} + c^{2}} d^{2}} - \frac {{\left (C c + B d\right )} \arcsin \left (\frac {d x}{c}\right )}{d^{3}} + \frac {2 \, {\left (D c + C d\right )} c^{2}}{\sqrt {-d^{2} x^{2} + c^{2}} d^{4}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.98 \[ \int \frac {(c+d x) \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {1}{2} \, \sqrt {-d^{2} x^{2} + c^{2}} {\left (\frac {D x}{d^{3}} + \frac {2 \, {\left (D c d^{6} + C d^{7}\right )}}{d^{10}}\right )} - \frac {{\left (3 \, D c^{2} + 2 \, C c d + 2 \, B d^{2}\right )} \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{2 \, d^{3} {\left | d \right |}} + \frac {2 \, {\left (D c^{3} + C c^{2} d + B c d^{2} + A d^{3}\right )}}{c d^{3} {\left (\frac {c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}}{d^{2} x} - 1\right )} {\left | d \right |}} \] Input:

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

Output:

1/2*sqrt(-d^2*x^2 + c^2)*(D*x/d^3 + 2*(D*c*d^6 + C*d^7)/d^10) - 1/2*(3*D*c 
^2 + 2*C*c*d + 2*B*d^2)*arcsin(d*x/c)*sgn(c)*sgn(d)/(d^3*abs(d)) + 2*(D*c^ 
3 + C*c^2*d + B*c*d^2 + A*d^3)/(c*d^3*((c*d + sqrt(-d^2*x^2 + c^2)*abs(d)) 
/(d^2*x) - 1)*abs(d))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.87 \[ \int \frac {(c+d x) \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {-2 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) b c d -5 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) c^{3}+2 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{2} d -2 \mathit {asin} \left (\frac {d x}{c}\right ) b c \,d^{2} x +5 \mathit {asin} \left (\frac {d x}{c}\right ) c^{4}-5 \mathit {asin} \left (\frac {d x}{c}\right ) c^{3} d x -4 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{2}-4 \sqrt {-d^{2} x^{2}+c^{2}}\, b c d -10 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{3}+3 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{2} d x +\sqrt {-d^{2} x^{2}+c^{2}}\, c \,d^{2} x^{2}+4 a c \,d^{2}+4 b \,c^{2} d +10 c^{4}+3 c^{3} d x -4 c^{2} d^{2} x^{2}-c \,d^{3} x^{3}}{2 c \,d^{3} \left (\sqrt {-d^{2} x^{2}+c^{2}}-c +d x \right )} \] Input:

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

Output:

( - 2*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*b*c*d - 5*sqrt(c**2 - d**2*x**2 
)*asin((d*x)/c)*c**3 + 2*asin((d*x)/c)*b*c**2*d - 2*asin((d*x)/c)*b*c*d**2 
*x + 5*asin((d*x)/c)*c**4 - 5*asin((d*x)/c)*c**3*d*x - 4*sqrt(c**2 - d**2* 
x**2)*a*d**2 - 4*sqrt(c**2 - d**2*x**2)*b*c*d - 10*sqrt(c**2 - d**2*x**2)* 
c**3 + 3*sqrt(c**2 - d**2*x**2)*c**2*d*x + sqrt(c**2 - d**2*x**2)*c*d**2*x 
**2 + 4*a*c*d**2 + 4*b*c**2*d + 10*c**4 + 3*c**3*d*x - 4*c**2*d**2*x**2 - 
c*d**3*x**3)/(2*c*d**3*(sqrt(c**2 - d**2*x**2) - c + d*x))