\(\int \frac {A+B x+C x^2+D x^3}{(c+d x) \sqrt {c^2-d^2 x^2}} \, dx\) [173]

Optimal result

Integrand size = 39, antiderivative size = 156 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x) \sqrt {c^2-d^2 x^2}} \, dx=-\frac {(C d-c D) \sqrt {c^2-d^2 x^2}}{d^4}-\frac {D x \sqrt {c^2-d^2 x^2}}{2 d^3}-\frac {\left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \sqrt {c^2-d^2 x^2}}{c d^4 (c+d x)}-\frac {\left (2 c C d-2 B d^2-3 c^2 D\right ) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{2 d^4} \] Output:

-(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-(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)-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 = 0.81 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.85 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x) \sqrt {c^2-d^2 x^2}} \, dx=\frac {\frac {\sqrt {c^2-d^2 x^2} \left (-2 A d^3+4 c^3 D+c^2 (-4 C d+d D x)+c d^2 (2 B-x (2 C+D x))\right )}{c (c+d x)}-2 \left (-2 c C d+2 B d^2+3 c^2 D\right ) \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{2 d^4} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.91 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.09, 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, 25, 27, 671, 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.

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

\(\Big \downarrow \) 2170

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2170

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 671

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

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

Time = 0.43 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.67

method	result	size
default	\(\frac {\frac {B \,d^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{\sqrt {d^{2}}}+\frac {D c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{\sqrt {d^{2}}}-\frac {\left (C d -D c \right ) \sqrt {-d^{2} x^{2}+c^{2}}}{d}+d^{2} D \left (-\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2 d^{2}}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 d^{2} \sqrt {d^{2}}}\right )-\frac {C c d \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{\sqrt {d^{2}}}}{d^{3}}-\frac {\left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{d^{5} c \left (x +\frac {c}{d}\right )}\)	\(261\)

Input:

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

Output:

1/d^3*(B*d^2/(d^2)^(1/2)*arctan((d^2)^(1/2)*x/(-d^2*x^2+c^2)^(1/2))+D*c^2/ 
(d^2)^(1/2)*arctan((d^2)^(1/2)*x/(-d^2*x^2+c^2)^(1/2))-1/d*(C*d-D*c)*(-d^2 
*x^2+c^2)^(1/2)+d^2*D*(-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*c*d/(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^5/c/( 
x+c/d)*(-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 = 227, normalized size of antiderivative = 1.46 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x) \sqrt {c^2-d^2 x^2}} \, dx=\frac {4 \, D c^{4} - 4 \, C c^{3} d + 2 \, B c^{2} d^{2} - 2 \, A c d^{3} + 2 \, {\left (2 \, D c^{3} d - 2 \, C c^{2} d^{2} + B c d^{3} - A d^{4}\right )} x - 2 \, {\left (3 \, D c^{4} - 2 \, C c^{3} d + 2 \, B c^{2} d^{2} + {\left (3 \, D c^{3} d - 2 \, C c^{2} d^{2} + 2 \, B c d^{3}\right )} x\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) - {\left (D c d^{2} x^{2} - 4 \, D c^{3} + 4 \, C c^{2} d - 2 \, B c d^{2} + 2 \, A d^{3} - {\left (D c^{2} d - 2 \, C c d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{2 \, {\left (c d^{5} x + c^{2} d^{4}\right )}} \] Input:

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

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

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

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.46 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x) \sqrt {c^2-d^2 x^2}} \, dx=\frac {\sqrt {-d^{2} x^{2} + c^{2}} D c^{2}}{d^{5} x + c d^{4}} - \frac {\sqrt {-d^{2} x^{2} + c^{2}} C c}{d^{4} x + c d^{3}} - \frac {\sqrt {-d^{2} x^{2} + c^{2}} A}{c d^{2} x + c^{2} d} + \frac {\sqrt {-d^{2} x^{2} + c^{2}} B}{d^{3} x + c d^{2}} + \frac {3 \, D c^{2} \arcsin \left (\frac {d x}{c}\right )}{2 \, d^{4}} - \frac {C c \arcsin \left (\frac {d x}{c}\right )}{d^{3}} + \frac {B \arcsin \left (\frac {d x}{c}\right )}{d^{2}} - \frac {\sqrt {-d^{2} x^{2} + c^{2}} D x}{2 \, d^{3}} + \frac {\sqrt {-d^{2} x^{2} + c^{2}} D c}{d^{4}} - \frac {\sqrt {-d^{2} x^{2} + c^{2}} C}{d^{3}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.96 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x) \sqrt {c^2-d^2 x^2}} \, dx=-\frac {1}{2} \, \sqrt {-d^{2} x^{2} + c^{2}} {\left (\frac {D x}{d^{3}} - \frac {2 \, {\left (D c d^{6} - C d^{7}\right )}}{d^{10}}\right )} + \frac {{\left (3 \, D c^{2} - 2 \, C c d + 2 \, B d^{2}\right )} \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{2 \, d^{3} {\left | d \right |}} - \frac {2 \, {\left (D c^{3} - C c^{2} d + B c d^{2} - A d^{3}\right )}}{c d^{3} {\left (\frac {c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}}{d^{2} x} + 1\right )} {\left | d \right |}} \] Input:

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.69 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x) \sqrt {c^2-d^2 x^2}} \, dx=\frac {2 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) b c d +\sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) c^{3}-2 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{2} d -2 \mathit {asin} \left (\frac {d x}{c}\right ) b c \,d^{2} x -\mathit {asin} \left (\frac {d x}{c}\right ) c^{4}-\mathit {asin} \left (\frac {d x}{c}\right ) c^{3} d x +4 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{2}-4 \sqrt {-d^{2} x^{2}+c^{2}}\, b c d -2 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{3}+\sqrt {-d^{2} x^{2}+c^{2}}\, c^{2} d x +\sqrt {-d^{2} x^{2}+c^{2}}\, c \,d^{2} x^{2}-4 a c \,d^{2}+4 b \,c^{2} d +2 c^{4}+c^{3} d x +c \,d^{3} x^{3}}{2 c \,d^{3} \left (\sqrt {-d^{2} x^{2}+c^{2}}-c -d x \right )} \] Input:

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

Output:

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