\(\int \frac {(c^2-d^2 x^2)^{3/2} (A+B x+C x^2+D x^3)}{(c+d x)^3} \, dx\) [146]

Optimal result

Integrand size = 39, antiderivative size = 257 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^3} \, dx=-\frac {\left (7 c^2 C d-5 B c d^2+3 A d^3-9 c^3 D\right ) \sqrt {c^2-d^2 x^2}}{d^4}+\frac {\left (12 c C d-4 B d^2-19 c^2 D\right ) x \sqrt {c^2-d^2 x^2}}{8 d^3}-\frac {D x^3 \sqrt {c^2-d^2 x^2}}{4 d}+\frac {(C d-3 c D) \left (c^2-d^2 x^2\right )^{3/2}}{3 d^4}-\frac {2 \left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \left (c^2-d^2 x^2\right )^{3/2}}{d^4 (c+d x)^2}-\frac {c \left (44 c^2 C d-36 B c d^2+24 A d^3-51 c^3 D\right ) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{8 d^4} \] Output:

-(3*A*d^3-5*B*c*d^2+7*C*c^2*d-9*D*c^3)*(-d^2*x^2+c^2)^(1/2)/d^4+1/8*(-4*B* 
d^2+12*C*c*d-19*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-3*D*c)*(-d^2*x^2+c^2)^(3/2)/d^4-2*(A*d^3-B*c*d^2+C*c^2*d 
-D*c^3)*(-d^2*x^2+c^2)^(3/2)/d^4/(d*x+c)^2-1/8*c*(24*A*d^3-36*B*c*d^2+44*C 
*c^2*d-51*D*c^3)*arctan(d*x/(-d^2*x^2+c^2)^(1/2))/d^4
 

Mathematica [A] (verified)

Time = 1.38 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.74 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^3} \, dx=\frac {\frac {\sqrt {c^2-d^2 x^2} \left (240 c^4 D+c^3 (-208 C d+87 d D x)+c^2 d^2 (168 B-x (76 C+33 D x))-2 d^4 x \left (12 A+x \left (6 B+4 C x+3 D x^2\right )\right )+2 c d^3 \left (-60 A+x \left (30 B+14 C x+9 D x^2\right )\right )\right )}{c+d x}-6 c \left (-44 c^2 C d+36 B c d^2-24 A d^3+51 c^3 D\right ) \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{24 d^4} \] Input:

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

Output:

((Sqrt[c^2 - d^2*x^2]*(240*c^4*D + c^3*(-208*C*d + 87*d*D*x) + c^2*d^2*(16 
8*B - x*(76*C + 33*D*x)) - 2*d^4*x*(12*A + x*(6*B + 4*C*x + 3*D*x^2)) + 2* 
c*d^3*(-60*A + x*(30*B + 14*C*x + 9*D*x^2))))/(c + d*x) - 6*c*(-44*c^2*C*d 
 + 36*B*c*d^2 - 24*A*d^3 + 51*c^3*D)*ArcTan[(d*x)/(Sqrt[c^2] - Sqrt[c^2 - 
d^2*x^2])])/(24*d^4)
 

Rubi [A] (verified)

Time = 1.07 (sec) , antiderivative size = 254, normalized size of antiderivative = 0.99, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2170, 25, 2170, 27, 671, 466, 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.

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

\(\Big \downarrow \) 2170

\(\displaystyle -\frac {\int -\frac {\left (c^2-d^2 x^2\right )^{3/2} \left ((4 C d-9 c D) x^2 d^4+2 \left (2 B d^2-3 c^2 D\right ) x d^3+\left (4 A d^3-c^3 D\right ) d^2\right )}{(c+d x)^3}dx}{4 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{5/2}}{4 d^4 (c+d x)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left ((4 C d-9 c D) x^2 d^4+2 \left (2 B d^2-3 c^2 D\right ) x d^3+\left (4 A d^3-c^3 D\right ) d^2\right )}{(c+d x)^3}dx}{4 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{5/2}}{4 d^4 (c+d x)}\)

\(\Big \downarrow \) 2170

\(\displaystyle \frac {-\frac {\int \frac {d^6 \left (-15 D c^3+8 C d c^2-12 A d^3+d \left (-27 D c^2+20 C d c-12 B d^2\right ) x\right ) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^3}dx}{3 d^4}-\frac {d \left (c^2-d^2 x^2\right )^{5/2} (4 C d-9 c D)}{3 (c+d x)^2}}{4 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{5/2}}{4 d^4 (c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {1}{3} d^2 \int \frac {\left (-15 D c^3+8 C d c^2-12 A d^3+d \left (-27 D c^2+20 C d c-12 B d^2\right ) x\right ) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^3}dx-\frac {d \left (c^2-d^2 x^2\right )^{5/2} (4 C d-9 c D)}{3 (c+d x)^2}}{4 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{5/2}}{4 d^4 (c+d x)}\)

\(\Big \downarrow \) 671

\(\displaystyle \frac {-\frac {1}{3} d^2 \left (\frac {\left (24 A d^3-36 B c d^2-51 c^3 D+44 c^2 C d\right ) \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^2}dx}{c}+\frac {12 \left (c^2-d^2 x^2\right )^{5/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (c+d x)^3}\right )-\frac {d \left (c^2-d^2 x^2\right )^{5/2} (4 C d-9 c D)}{3 (c+d x)^2}}{4 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{5/2}}{4 d^4 (c+d x)}\)

\(\Big \downarrow \) 466

\(\displaystyle \frac {-\frac {1}{3} d^2 \left (\frac {\left (24 A d^3-36 B c d^2-51 c^3 D+44 c^2 C d\right ) \left (\frac {3}{2} c \int \frac {\sqrt {c^2-d^2 x^2}}{c+d x}dx+\frac {\left (c^2-d^2 x^2\right )^{3/2}}{2 d (c+d x)}\right )}{c}+\frac {12 \left (c^2-d^2 x^2\right )^{5/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (c+d x)^3}\right )-\frac {d \left (c^2-d^2 x^2\right )^{5/2} (4 C d-9 c D)}{3 (c+d x)^2}}{4 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{5/2}}{4 d^4 (c+d x)}\)

\(\Big \downarrow \) 466

\(\displaystyle \frac {-\frac {1}{3} d^2 \left (\frac {\left (24 A d^3-36 B c d^2-51 c^3 D+44 c^2 C d\right ) \left (\frac {3}{2} c \left (c \int \frac {1}{\sqrt {c^2-d^2 x^2}}dx+\frac {\sqrt {c^2-d^2 x^2}}{d}\right )+\frac {\left (c^2-d^2 x^2\right )^{3/2}}{2 d (c+d x)}\right )}{c}+\frac {12 \left (c^2-d^2 x^2\right )^{5/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (c+d x)^3}\right )-\frac {d \left (c^2-d^2 x^2\right )^{5/2} (4 C d-9 c D)}{3 (c+d x)^2}}{4 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{5/2}}{4 d^4 (c+d x)}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {-\frac {1}{3} d^2 \left (\frac {\left (24 A d^3-36 B c d^2-51 c^3 D+44 c^2 C d\right ) \left (\frac {3}{2} c \left (c \int \frac {1}{\frac {d^2 x^2}{c^2-d^2 x^2}+1}d\frac {x}{\sqrt {c^2-d^2 x^2}}+\frac {\sqrt {c^2-d^2 x^2}}{d}\right )+\frac {\left (c^2-d^2 x^2\right )^{3/2}}{2 d (c+d x)}\right )}{c}+\frac {12 \left (c^2-d^2 x^2\right )^{5/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (c+d x)^3}\right )-\frac {d \left (c^2-d^2 x^2\right )^{5/2} (4 C d-9 c D)}{3 (c+d x)^2}}{4 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{5/2}}{4 d^4 (c+d x)}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {-\frac {1}{3} d^2 \left (\frac {\left (\frac {3}{2} c \left (\frac {c \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{d}+\frac {\sqrt {c^2-d^2 x^2}}{d}\right )+\frac {\left (c^2-d^2 x^2\right )^{3/2}}{2 d (c+d x)}\right ) \left (24 A d^3-36 B c d^2-51 c^3 D+44 c^2 C d\right )}{c}+\frac {12 \left (c^2-d^2 x^2\right )^{5/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (c+d x)^3}\right )-\frac {d \left (c^2-d^2 x^2\right )^{5/2} (4 C d-9 c D)}{3 (c+d x)^2}}{4 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{5/2}}{4 d^4 (c+d x)}\)

Input:

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

Output:

-1/4*(D*(c^2 - d^2*x^2)^(5/2))/(d^4*(c + d*x)) + (-1/3*(d*(4*C*d - 9*c*D)* 
(c^2 - d^2*x^2)^(5/2))/(c + d*x)^2 - (d^2*((12*(c^2*C*d - B*c*d^2 + A*d^3 
- c^3*D)*(c^2 - d^2*x^2)^(5/2))/(c*d*(c + d*x)^3) + ((44*c^2*C*d - 36*B*c* 
d^2 + 24*A*d^3 - 51*c^3*D)*((c^2 - d^2*x^2)^(3/2)/(2*d*(c + d*x)) + (3*c*( 
Sqrt[c^2 - d^2*x^2]/d + (c*ArcTan[(d*x)/Sqrt[c^2 - d^2*x^2]])/d))/2))/c))/ 
3)/(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 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. \(688\) vs. \(2(237)=474\).

Time = 0.42 (sec) , antiderivative size = 689, normalized size of antiderivative = 2.68

method	result	size
default	\(\frac {D \left (\frac {x \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{4}+\frac {3 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}\right )}{d^{3}}+\frac {\left (C d -3 D c \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}}}{3}+c d \left (-\frac {\left (-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d \right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{4 d^{2}}+\frac {c^{2} \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 )}{2 \sqrt {d^{2}}}\right )\right )}{d^{4}}+\frac {\left (B \,d^{2}-2 C c d +3 D c^{2}\right ) \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {5}{2}}}{c d \left (x +\frac {c}{d}\right )^{2}}+\frac {3 d \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}}}{3}+c d \left (-\frac {\left (-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d \right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{4 d^{2}}+\frac {c^{2} \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 )}{2 \sqrt {d^{2}}}\right )\right )}{c}\right )}{d^{5}}+\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 {5}{2}}}{c d \left (x +\frac {c}{d}\right )^{3}}-\frac {2 d \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {5}{2}}}{c d \left (x +\frac {c}{d}\right )^{2}}+\frac {3 d \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}}}{3}+c d \left (-\frac {\left (-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d \right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{4 d^{2}}+\frac {c^{2} \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 )}{2 \sqrt {d^{2}}}\right )\right )}{c}\right )}{c}\right )}{d^{6}}\)	\(689\)

Input:

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

Output:

D/d^3*(1/4*x*(-d^2*x^2+c^2)^(3/2)+3/4*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/d^4*(C*d-3* 
D*c)*(1/3*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(3/2)+c*d*(-1/4*(-2*d^2*(x+c/d)+2 
*c*d)/d^2*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(1/2)+1/2*c^2/(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*(B*d^2-2*C*c*d 
+3*D*c^2)*(1/c/d/(x+c/d)^2*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(5/2)+3*d/c*(1/3 
*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(3/2)+c*d*(-1/4*(-2*d^2*(x+c/d)+2*c*d)/d^2 
*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(1/2)+1/2*c^2/(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^6*(A*d^3-B*c*d^2+C*c^2*d 
-D*c^3)*(-1/c/d/(x+c/d)^3*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(5/2)-2*d/c*(1/c/ 
d/(x+c/d)^2*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(5/2)+3*d/c*(1/3*(-d^2*(x+c/d)^ 
2+2*c*d*(x+c/d))^(3/2)+c*d*(-1/4*(-2*d^2*(x+c/d)+2*c*d)/d^2*(-d^2*(x+c/d)^ 
2+2*c*d*(x+c/d))^(1/2)+1/2*c^2/(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 = 313, normalized size of antiderivative = 1.22 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^3} \, dx=\frac {240 \, D c^{5} - 208 \, C c^{4} d + 168 \, B c^{3} d^{2} - 120 \, A c^{2} d^{3} + 8 \, {\left (30 \, D c^{4} d - 26 \, C c^{3} d^{2} + 21 \, B c^{2} d^{3} - 15 \, A c d^{4}\right )} x - 6 \, {\left (51 \, D c^{5} - 44 \, C c^{4} d + 36 \, B c^{3} d^{2} - 24 \, A c^{2} d^{3} + {\left (51 \, D c^{4} d - 44 \, C c^{3} d^{2} + 36 \, B c^{2} d^{3} - 24 \, A c d^{4}\right )} x\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) - {\left (6 \, D d^{4} x^{4} - 240 \, D c^{4} + 208 \, C c^{3} d - 168 \, B c^{2} d^{2} + 120 \, A c d^{3} - 2 \, {\left (9 \, D c d^{3} - 4 \, C d^{4}\right )} x^{3} + {\left (33 \, D c^{2} d^{2} - 28 \, C c d^{3} + 12 \, B d^{4}\right )} x^{2} - {\left (87 \, D c^{3} d - 76 \, C c^{2} d^{2} + 60 \, B c d^{3} - 24 \, A d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{24 \, {\left (d^{5} x + c d^{4}\right )}} \] Input:

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

Output:

1/24*(240*D*c^5 - 208*C*c^4*d + 168*B*c^3*d^2 - 120*A*c^2*d^3 + 8*(30*D*c^ 
4*d - 26*C*c^3*d^2 + 21*B*c^2*d^3 - 15*A*c*d^4)*x - 6*(51*D*c^5 - 44*C*c^4 
*d + 36*B*c^3*d^2 - 24*A*c^2*d^3 + (51*D*c^4*d - 44*C*c^3*d^2 + 36*B*c^2*d 
^3 - 24*A*c*d^4)*x)*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d*x)) - (6*D*d^4*x 
^4 - 240*D*c^4 + 208*C*c^3*d - 168*B*c^2*d^2 + 120*A*c*d^3 - 2*(9*D*c*d^3 
- 4*C*d^4)*x^3 + (33*D*c^2*d^2 - 28*C*c*d^3 + 12*B*d^4)*x^2 - (87*D*c^3*d 
- 76*C*c^2*d^2 + 60*B*c*d^3 - 24*A*d^4)*x)*sqrt(-d^2*x^2 + c^2))/(d^5*x + 
c*d^4)
 

Sympy [F]

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

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

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.15 (sec) , antiderivative size = 756, normalized size of antiderivative = 2.94 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^3} \, dx =\text {Too large to display} \] Input:

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

Output:

-(-d^2*x^2 + c^2)^(3/2)*D*c^3/(d^6*x^2 + 2*c*d^5*x + c^2*d^4) + 6*sqrt(-d^ 
2*x^2 + c^2)*D*c^4/(d^5*x + c*d^4) + (-d^2*x^2 + c^2)^(3/2)*C*c^2/(d^5*x^2 
 + 2*c*d^4*x + c^2*d^3) + 3/2*(-d^2*x^2 + c^2)^(3/2)*D*c^2/(d^5*x + c*d^4) 
 - 6*sqrt(-d^2*x^2 + c^2)*C*c^3/(d^4*x + c*d^3) - (-d^2*x^2 + c^2)^(3/2)*B 
*c/(d^4*x^2 + 2*c*d^3*x + c^2*d^2) - (-d^2*x^2 + c^2)^(3/2)*C*c/(d^4*x + c 
*d^3) + 6*sqrt(-d^2*x^2 + c^2)*B*c^2/(d^3*x + c*d^2) + 3/2*I*D*c^4*arcsin( 
d*x/c + 2)/d^4 - 1/2*I*C*c^3*arcsin(d*x/c + 2)/d^3 + 63/8*D*c^4*arcsin(d*x 
/c)/d^4 - 6*C*c^3*arcsin(d*x/c)/d^3 + 9/2*B*c^2*arcsin(d*x/c)/d^2 - 3*A*c* 
arcsin(d*x/c)/d + (-d^2*x^2 + c^2)^(3/2)*A/(d^3*x^2 + 2*c*d^2*x + c^2*d) + 
 1/2*(-d^2*x^2 + c^2)^(3/2)*B/(d^3*x + c*d^2) - 6*sqrt(-d^2*x^2 + c^2)*A*c 
/(d^2*x + c*d) - 3/2*sqrt(d^2*x^2 + 4*c*d*x + 3*c^2)*D*c^2*x/d^3 + 3/8*sqr 
t(-d^2*x^2 + c^2)*D*c^2*x/d^3 + 1/2*sqrt(d^2*x^2 + 4*c*d*x + 3*c^2)*C*c*x/ 
d^2 - 3*sqrt(d^2*x^2 + 4*c*d*x + 3*c^2)*D*c^3/d^4 + 9/2*sqrt(-d^2*x^2 + c^ 
2)*D*c^3/d^4 + sqrt(d^2*x^2 + 4*c*d*x + 3*c^2)*C*c^2/d^3 - 3*sqrt(-d^2*x^2 
 + c^2)*C*c^2/d^3 + 3/2*sqrt(-d^2*x^2 + c^2)*B*c/d^2 + 1/4*(-d^2*x^2 + c^2 
)^(3/2)*D*x/d^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 = 235, normalized size of antiderivative = 0.91 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^3} \, dx=-\frac {1}{24} \, \sqrt {-d^{2} x^{2} + c^{2}} {\left ({\left (2 \, {\left (\frac {3 \, D x}{d} - \frac {4 \, {\left (3 \, D c d^{11} - C d^{12}\right )}}{d^{13}}\right )} x + \frac {3 \, {\left (19 \, D c^{2} d^{10} - 12 \, C c d^{11} + 4 \, B d^{12}\right )}}{d^{13}}\right )} x - \frac {8 \, {\left (18 \, D c^{3} d^{9} - 14 \, C c^{2} d^{10} + 9 \, B c d^{11} - 3 \, A d^{12}\right )}}{d^{13}}\right )} + \frac {{\left (51 \, D c^{4} - 44 \, C c^{3} d + 36 \, B c^{2} d^{2} - 24 \, A c d^{3}\right )} \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{8 \, d^{3} {\left | d \right |}} - \frac {8 \, {\left (D c^{4} - C c^{3} d + B c^{2} d^{2} - A c d^{3}\right )}}{d^{3} {\left (\frac {c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}}{d^{2} x} + 1\right )} {\left | d \right |}} \] Input:

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

Output:

-1/24*sqrt(-d^2*x^2 + c^2)*((2*(3*D*x/d - 4*(3*D*c*d^11 - C*d^12)/d^13)*x 
+ 3*(19*D*c^2*d^10 - 12*C*c*d^11 + 4*B*d^12)/d^13)*x - 8*(18*D*c^3*d^9 - 1 
4*C*c^2*d^10 + 9*B*c*d^11 - 3*A*d^12)/d^13) + 1/8*(51*D*c^4 - 44*C*c^3*d + 
 36*B*c^2*d^2 - 24*A*c*d^3)*arcsin(d*x/c)*sgn(c)*sgn(d)/(d^3*abs(d)) - 8*( 
D*c^4 - C*c^3*d + B*c^2*d^2 - A*c*d^3)/(d^3*((c*d + sqrt(-d^2*x^2 + c^2)*a 
bs(d))/(d^2*x) + 1)*abs(d))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 517, normalized size of antiderivative = 2.01 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^3} \, dx=\frac {42 c^{5}+5 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{2} d^{2} x^{2}-42 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{4}+6 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{4} x^{4}+6 d^{5} x^{5}+12 b \,d^{4} x^{3}-72 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) a c \,d^{2}+72 \mathit {asin} \left (\frac {d x}{c}\right ) a \,c^{2} d^{2}+24 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{3} x -216 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2} d +12 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,d^{3} x^{2}-108 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{3} d -21 \mathit {asin} \left (\frac {d x}{c}\right ) c^{4} d x +192 \sqrt {-d^{2} x^{2}+c^{2}}\, a c \,d^{2}-60 b \,c^{2} d^{2} x -72 b c \,d^{3} x^{2}+24 a c \,d^{3} x +21 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) c^{4}-192 a \,c^{2} d^{2}+24 a \,d^{4} x^{2}+108 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{2} d -108 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{2} d^{2} x -60 \sqrt {-d^{2} x^{2}+c^{2}}\, b c \,d^{2} x -11 c^{4} d x -16 c^{3} d^{2} x^{2}+15 c^{2} d^{3} x^{3}-16 c \,d^{4} x^{4}-11 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{3} d x -10 \sqrt {-d^{2} x^{2}+c^{2}}\, c \,d^{3} x^{3}-21 \mathit {asin} \left (\frac {d x}{c}\right ) c^{5}+216 b \,c^{3} d +72 \mathit {asin} \left (\frac {d x}{c}\right ) a c \,d^{3} x}{24 d^{3} \left (\sqrt {-d^{2} x^{2}+c^{2}}-c -d x \right )} \] Input:

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

Output:

( - 72*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*a*c*d**2 + 108*sqrt(c**2 - d** 
2*x**2)*asin((d*x)/c)*b*c**2*d + 21*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*c 
**4 + 72*asin((d*x)/c)*a*c**2*d**2 + 72*asin((d*x)/c)*a*c*d**3*x - 108*asi 
n((d*x)/c)*b*c**3*d - 108*asin((d*x)/c)*b*c**2*d**2*x - 21*asin((d*x)/c)*c 
**5 - 21*asin((d*x)/c)*c**4*d*x + 192*sqrt(c**2 - d**2*x**2)*a*c*d**2 + 24 
*sqrt(c**2 - d**2*x**2)*a*d**3*x - 216*sqrt(c**2 - d**2*x**2)*b*c**2*d - 6 
0*sqrt(c**2 - d**2*x**2)*b*c*d**2*x + 12*sqrt(c**2 - d**2*x**2)*b*d**3*x** 
2 - 42*sqrt(c**2 - d**2*x**2)*c**4 - 11*sqrt(c**2 - d**2*x**2)*c**3*d*x + 
5*sqrt(c**2 - d**2*x**2)*c**2*d**2*x**2 - 10*sqrt(c**2 - d**2*x**2)*c*d**3 
*x**3 + 6*sqrt(c**2 - d**2*x**2)*d**4*x**4 - 192*a*c**2*d**2 + 24*a*c*d**3 
*x + 24*a*d**4*x**2 + 216*b*c**3*d - 60*b*c**2*d**2*x - 72*b*c*d**3*x**2 + 
 12*b*d**4*x**3 + 42*c**5 - 11*c**4*d*x - 16*c**3*d**2*x**2 + 15*c**2*d**3 
*x**3 - 16*c*d**4*x**4 + 6*d**5*x**5)/(24*d**3*(sqrt(c**2 - d**2*x**2) - c 
 - d*x))