\(\int \frac {(c+d x) (A+B x+C x^2+D x^3)}{\sqrt {c^2-d^2 x^2}} \, dx\) [171]

Optimal result

Integrand size = 37, antiderivative size = 199 \[ \int \frac {(c+d x) \left (A+B x+C x^2+D x^3\right )}{\sqrt {c^2-d^2 x^2}} \, dx=-\frac {\left (c^2 C d+B c d^2+A d^3+c^3 D\right ) \sqrt {c^2-d^2 x^2}}{d^4}-\frac {\left (4 c C d+4 B d^2+3 c^2 D\right ) x \sqrt {c^2-d^2 x^2}}{8 d^3}-\frac {D x^3 \sqrt {c^2-d^2 x^2}}{4 d}+\frac {(C d+c D) \left (c^2-d^2 x^2\right )^{3/2}}{3 d^4}+\frac {c \left (4 c^2 C d+4 B c d^2+8 A d^3+3 c^3 D\right ) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{8 d^4} \] 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 = 1.03 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.76 \[ \int \frac {(c+d x) \left (A+B x+C x^2+D x^3\right )}{\sqrt {c^2-d^2 x^2}} \, dx=-\frac {\sqrt {c^2-d^2 x^2} \left (16 c^3 D+c^2 d (16 C+9 D x)+4 c d^2 (6 B+x (3 C+2 D x))+2 d^3 \left (12 A+x \left (6 B+4 C x+3 D x^2\right )\right )\right )+6 c \left (4 c^2 C d+4 B c d^2+8 A d^3+3 c^3 D\right ) \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{24 d^4} \] Input:

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

Output:

-1/24*(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])])/d^4
 

Rubi [A] (verified)

Time = 1.11 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.11, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.270, Rules used = {2346, 25, 2346, 25, 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.

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

\(\Big \downarrow \) 2346

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2346

\(\displaystyle \frac {-\frac {\int -\frac {12 A c d^4+3 \left (3 D c^2+4 C d c+4 B d^2\right ) x^2 d^3+4 \left (2 D c^3+2 C d c^2+3 B d^2 c+3 A d^3\right ) x d^2}{\sqrt {c^2-d^2 x^2}}dx}{3 d^2}-\frac {4}{3} x^2 \sqrt {c^2-d^2 x^2} (c D+C d)}{4 d^2}-\frac {D x^3 \sqrt {c^2-d^2 x^2}}{4 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {12 A c d^4+3 \left (3 D c^2+4 C d c+4 B d^2\right ) x^2 d^3+4 \left (2 D c^3+2 C d c^2+3 B d^2 c+3 A d^3\right ) x d^2}{\sqrt {c^2-d^2 x^2}}dx}{3 d^2}-\frac {4}{3} x^2 \sqrt {c^2-d^2 x^2} (c D+C d)}{4 d^2}-\frac {D x^3 \sqrt {c^2-d^2 x^2}}{4 d}\)

\(\Big \downarrow \) 2346

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

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

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 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 = 0.39 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.33

method	result	size
default	\(\frac {A c \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{\sqrt {d^{2}}}-\frac {\left (A d +B c \right ) \sqrt {-d^{2} x^{2}+c^{2}}}{d^{2}}+\left (B d +C c \right ) \left (-\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2 d^{2}}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 d^{2} \sqrt {d^{2}}}\right )+\left (C d +D c \right ) \left (-\frac {x^{2} \sqrt {-d^{2} x^{2}+c^{2}}}{3 d^{2}}-\frac {2 c^{2} \sqrt {-d^{2} x^{2}+c^{2}}}{3 d^{4}}\right )+D d \left (-\frac {x^{3} \sqrt {-d^{2} x^{2}+c^{2}}}{4 d^{2}}+\frac {3 c^{2} \left (-\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2 d^{2}}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 d^{2} \sqrt {d^{2}}}\right )}{4 d^{2}}\right )\)	\(265\)

Input:

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

Output:

A*c/(d^2)^(1/2)*arctan((d^2)^(1/2)*x/(-d^2*x^2+c^2)^(1/2))-(A*d+B*c)*(-d^2 
*x^2+c^2)^(1/2)/d^2+(B*d+C*c)*(-1/2*x*(-d^2*x^2+c^2)^(1/2)/d^2+1/2*c^2/d^2 
/(d^2)^(1/2)*arctan((d^2)^(1/2)*x/(-d^2*x^2+c^2)^(1/2)))+(C*d+D*c)*(-1/3*x 
^2/d^2*(-d^2*x^2+c^2)^(1/2)-2/3*c^2*(-d^2*x^2+c^2)^(1/2)/d^4)+D*d*(-1/4*x^ 
3*(-d^2*x^2+c^2)^(1/2)/d^2+3/4*c^2/d^2*(-1/2*x*(-d^2*x^2+c^2)^(1/2)/d^2+1/ 
2*c^2/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 = 0.09 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.79 \[ \int \frac {(c+d x) \left (A+B x+C x^2+D x^3\right )}{\sqrt {c^2-d^2 x^2}} \, dx=-\frac {6 \, {\left (3 \, D c^{4} + 4 \, C c^{3} d + 4 \, B c^{2} d^{2} + 8 \, A c d^{3}\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) + {\left (6 \, D d^{3} x^{3} + 16 \, D c^{3} + 16 \, C c^{2} d + 24 \, B c d^{2} + 24 \, A d^{3} + 8 \, {\left (D c d^{2} + C d^{3}\right )} x^{2} + 3 \, {\left (3 \, D c^{2} d + 4 \, C c d^{2} + 4 \, B d^{3}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{24 \, d^{4}} \] Input:

integrate((d*x+c)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(1/2),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 [A] (verification not implemented)

Time = 0.71 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.18 \[ \int \frac {(c+d x) \left (A+B x+C x^2+D x^3\right )}{\sqrt {c^2-d^2 x^2}} \, dx=\begin {cases} \sqrt {c^{2} - d^{2} x^{2}} \left (- \frac {D x^{3}}{4 d} - \frac {x^{2} \left (C d + D c\right )}{3 d^{2}} - \frac {x \left (B d + C c + \frac {3 D c^{2}}{4 d}\right )}{2 d^{2}} - \frac {A d + B c + \frac {2 c^{2} \left (C d + D c\right )}{3 d^{2}}}{d^{2}}\right ) + \left (A c + \frac {c^{2} \left (B d + C c + \frac {3 D c^{2}}{4 d}\right )}{2 d^{2}}\right ) \left (\begin {cases} \frac {\log {\left (- 2 d^{2} x + 2 \sqrt {- d^{2}} \sqrt {c^{2} - d^{2} x^{2}} \right )}}{\sqrt {- d^{2}}} & \text {for}\: c^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- d^{2} x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: d^{2} \neq 0 \\\frac {A c x + \frac {D d x^{5}}{5} + \frac {x^{4} \left (C d + D c\right )}{4} + \frac {x^{3} \left (B d + C c\right )}{3} + \frac {x^{2} \left (A d + B c\right )}{2}}{\sqrt {c^{2}}} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((sqrt(c**2 - d**2*x**2)*(-D*x**3/(4*d) - x**2*(C*d + D*c)/(3*d** 
2) - x*(B*d + C*c + 3*D*c**2/(4*d))/(2*d**2) - (A*d + B*c + 2*c**2*(C*d + 
D*c)/(3*d**2))/d**2) + (A*c + c**2*(B*d + C*c + 3*D*c**2/(4*d))/(2*d**2))* 
Piecewise((log(-2*d**2*x + 2*sqrt(-d**2)*sqrt(c**2 - d**2*x**2))/sqrt(-d** 
2), Ne(c**2, 0)), (x*log(x)/sqrt(-d**2*x**2), True)), Ne(d**2, 0)), ((A*c* 
x + D*d*x**5/5 + x**4*(C*d + D*c)/4 + x**3*(B*d + C*c)/3 + x**2*(A*d + B*c 
)/2)/sqrt(c**2), True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.13 \[ \int \frac {(c+d x) \left (A+B x+C x^2+D x^3\right )}{\sqrt {c^2-d^2 x^2}} \, dx=-\frac {\sqrt {-d^{2} x^{2} + c^{2}} D x^{3}}{4 \, d} + \frac {3 \, D c^{4} \arcsin \left (\frac {d x}{c}\right )}{8 \, d^{4}} + \frac {A c \arcsin \left (\frac {d x}{c}\right )}{d} - \frac {3 \, \sqrt {-d^{2} x^{2} + c^{2}} D c^{2} x}{8 \, d^{3}} - \frac {\sqrt {-d^{2} x^{2} + c^{2}} {\left (D c + C d\right )} x^{2}}{3 \, d^{2}} + \frac {{\left (C c + B d\right )} c^{2} \arcsin \left (\frac {d x}{c}\right )}{2 \, d^{3}} - \frac {\sqrt {-d^{2} x^{2} + c^{2}} B c}{d^{2}} - \frac {\sqrt {-d^{2} x^{2} + c^{2}} A}{d} - \frac {\sqrt {-d^{2} x^{2} + c^{2}} {\left (C c + B d\right )} x}{2 \, d^{2}} - \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} {\left (D c + C d\right )} c^{2}}{3 \, d^{4}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.82 \[ \int \frac {(c+d x) \left (A+B x+C x^2+D x^3\right )}{\sqrt {c^2-d^2 x^2}} \, dx=-\frac {1}{24} \, \sqrt {-d^{2} x^{2} + c^{2}} {\left ({\left (2 \, {\left (\frac {3 \, D x}{d} + \frac {4 \, {\left (D c d^{5} + C d^{6}\right )}}{d^{7}}\right )} x + \frac {3 \, {\left (3 \, D c^{2} d^{4} + 4 \, C c d^{5} + 4 \, B d^{6}\right )}}{d^{7}}\right )} x + \frac {8 \, {\left (2 \, D c^{3} d^{3} + 2 \, C c^{2} d^{4} + 3 \, B c d^{5} + 3 \, A d^{6}\right )}}{d^{7}}\right )} + \frac {{\left (3 \, D c^{4} + 4 \, C c^{3} d + 4 \, B c^{2} d^{2} + 8 \, A c d^{3}\right )} \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{8 \, d^{3} {\left | d \right |}} \] Input:

integrate((d*x+c)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(1/2),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. \[ \int \frac {(c+d x) \left (A+B x+C x^2+D x^3\right )}{\sqrt {c^2-d^2 x^2}} \, dx=\int \frac {\left (c+d\,x\right )\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{\sqrt {c^2-d^2\,x^2}} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.02 \[ \int \frac {(c+d x) \left (A+B x+C x^2+D x^3\right )}{\sqrt {c^2-d^2 x^2}} \, dx=\frac {24 \mathit {asin} \left (\frac {d x}{c}\right ) a c \,d^{2}+12 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{2} d +21 \mathit {asin} \left (\frac {d x}{c}\right ) c^{4}-24 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{2}-24 \sqrt {-d^{2} x^{2}+c^{2}}\, b c d -12 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,d^{2} x -32 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{3}-21 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{2} d x -16 \sqrt {-d^{2} x^{2}+c^{2}}\, c \,d^{2} x^{2}-6 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{3} x^{3}+24 a c \,d^{2}+24 b \,c^{2} d +32 c^{4}}{24 d^{3}} \] Input:

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

Output:

(24*asin((d*x)/c)*a*c*d**2 + 12*asin((d*x)/c)*b*c**2*d + 21*asin((d*x)/c)* 
c**4 - 24*sqrt(c**2 - d**2*x**2)*a*d**2 - 24*sqrt(c**2 - d**2*x**2)*b*c*d 
- 12*sqrt(c**2 - d**2*x**2)*b*d**2*x - 32*sqrt(c**2 - d**2*x**2)*c**3 - 21 
*sqrt(c**2 - d**2*x**2)*c**2*d*x - 16*sqrt(c**2 - d**2*x**2)*c*d**2*x**2 - 
 6*sqrt(c**2 - d**2*x**2)*d**3*x**3 + 24*a*c*d**2 + 24*b*c**2*d + 32*c**4) 
/(24*d**3)