\(\int \frac {(c^2-d^2 x^2)^{3/2} (A+B x+C x^2+D x^3)}{(c+d x)^2} \, dx\) [145]

Optimal result

Integrand size = 39, antiderivative size = 255 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^2} \, dx=\frac {2 c \left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \sqrt {c^2-d^2 x^2}}{d^4}-\frac {\left (7 c^2 C d-8 B c d^2+4 A d^3-6 c^3 D\right ) x \sqrt {c^2-d^2 x^2}}{8 d^3}-\frac {(C d-2 c D) x^3 \sqrt {c^2-d^2 x^2}}{4 d}-\frac {\left (2 c C d-B d^2-3 c^2 D\right ) \left (c^2-d^2 x^2\right )^{3/2}}{3 d^4}-\frac {D \left (c^2-d^2 x^2\right )^{5/2}}{5 d^4}+\frac {c^2 \left (7 c^2 C d-8 B c d^2+12 A d^3-6 c^3 D\right ) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{8 d^4} \] Output:

2*c*(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*A*d^3-8* 
B*c*d^2+7*C*c^2*d-6*D*c^3)*x*(-d^2*x^2+c^2)^(1/2)/d^3-1/4*(C*d-2*D*c)*x^3* 
(-d^2*x^2+c^2)^(1/2)/d-1/3*(-B*d^2+2*C*c*d-3*D*c^2)*(-d^2*x^2+c^2)^(3/2)/d 
^4-1/5*D*(-d^2*x^2+c^2)^(5/2)/d^4+1/8*c^2*(12*A*d^3-8*B*c*d^2+7*C*c^2*d-6* 
D*c^3)*arctan(d*x/(-d^2*x^2+c^2)^(1/2))/d^4
 

Mathematica [A] (verified)

Time = 1.10 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.73 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^2} \, dx=\frac {-\sqrt {c^2-d^2 x^2} \left (144 c^4 D-10 c^3 d (16 C+9 D x)+c^2 d^2 (200 B+3 x (35 C+24 D x))-20 c d^3 (12 A+x (6 B+x (4 C+3 D x)))+2 d^4 x (30 A+x (20 B+3 x (5 C+4 D x)))\right )+30 c^2 \left (-7 c^2 C d+8 B c d^2-12 A d^3+6 c^3 D\right ) \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{120 d^4} \] Input:

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

Output:

(-(Sqrt[c^2 - d^2*x^2]*(144*c^4*D - 10*c^3*d*(16*C + 9*D*x) + c^2*d^2*(200 
*B + 3*x*(35*C + 24*D*x)) - 20*c*d^3*(12*A + x*(6*B + x*(4*C + 3*D*x))) + 
2*d^4*x*(30*A + x*(20*B + 3*x*(5*C + 4*D*x))))) + 30*c^2*(-7*c^2*C*d + 8*B 
*c*d^2 - 12*A*d^3 + 6*c^3*D)*ArcTan[(d*x)/(Sqrt[c^2] - Sqrt[c^2 - d^2*x^2] 
)])/(120*d^4)
 

Rubi [A] (verified)

Time = 1.05 (sec) , antiderivative size = 240, normalized size of antiderivative = 0.94, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2170, 27, 2170, 27, 671, 466, 211, 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.

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

\(\Big \downarrow \) 2170

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2170

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 671

\(\displaystyle \frac {-\frac {1}{4} d^2 \left (-\frac {\left (12 A d^3-8 B c d^2-6 c^3 D+7 c^2 C d\right ) \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{c+d x}dx}{c}-\frac {4 \left (c^2-d^2 x^2\right )^{5/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (c+d x)^2}\right )-\frac {d \left (c^2-d^2 x^2\right )^{5/2} (C d-2 c D)}{4 (c+d x)}}{d^5}-\frac {D \left (c^2-d^2 x^2\right )^{5/2}}{5 d^4}\)

\(\Big \downarrow \) 466

\(\displaystyle \frac {-\frac {1}{4} d^2 \left (-\frac {\left (12 A d^3-8 B c d^2-6 c^3 D+7 c^2 C d\right ) \left (c \int \sqrt {c^2-d^2 x^2}dx+\frac {\left (c^2-d^2 x^2\right )^{3/2}}{3 d}\right )}{c}-\frac {4 \left (c^2-d^2 x^2\right )^{5/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (c+d x)^2}\right )-\frac {d \left (c^2-d^2 x^2\right )^{5/2} (C d-2 c D)}{4 (c+d x)}}{d^5}-\frac {D \left (c^2-d^2 x^2\right )^{5/2}}{5 d^4}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {-\frac {1}{4} d^2 \left (-\frac {\left (12 A d^3-8 B c d^2-6 c^3 D+7 c^2 C d\right ) \left (c \left (\frac {1}{2} c^2 \int \frac {1}{\sqrt {c^2-d^2 x^2}}dx+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )+\frac {\left (c^2-d^2 x^2\right )^{3/2}}{3 d}\right )}{c}-\frac {4 \left (c^2-d^2 x^2\right )^{5/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (c+d x)^2}\right )-\frac {d \left (c^2-d^2 x^2\right )^{5/2} (C d-2 c D)}{4 (c+d x)}}{d^5}-\frac {D \left (c^2-d^2 x^2\right )^{5/2}}{5 d^4}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {-\frac {1}{4} d^2 \left (-\frac {\left (12 A d^3-8 B c d^2-6 c^3 D+7 c^2 C d\right ) \left (c \left (\frac {1}{2} c^2 \int \frac {1}{\frac {d^2 x^2}{c^2-d^2 x^2}+1}d\frac {x}{\sqrt {c^2-d^2 x^2}}+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )+\frac {\left (c^2-d^2 x^2\right )^{3/2}}{3 d}\right )}{c}-\frac {4 \left (c^2-d^2 x^2\right )^{5/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (c+d x)^2}\right )-\frac {d \left (c^2-d^2 x^2\right )^{5/2} (C d-2 c D)}{4 (c+d x)}}{d^5}-\frac {D \left (c^2-d^2 x^2\right )^{5/2}}{5 d^4}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {-\frac {1}{4} d^2 \left (-\frac {\left (c \left (\frac {c^2 \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{2 d}+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )+\frac {\left (c^2-d^2 x^2\right )^{3/2}}{3 d}\right ) \left (12 A d^3-8 B c d^2-6 c^3 D+7 c^2 C d\right )}{c}-\frac {4 \left (c^2-d^2 x^2\right )^{5/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (c+d x)^2}\right )-\frac {d \left (c^2-d^2 x^2\right )^{5/2} (C d-2 c D)}{4 (c+d x)}}{d^5}-\frac {D \left (c^2-d^2 x^2\right )^{5/2}}{5 d^4}\)

Input:

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

Output:

-1/5*(D*(c^2 - d^2*x^2)^(5/2))/d^4 + (-1/4*(d*(C*d - 2*c*D)*(c^2 - d^2*x^2 
)^(5/2))/(c + d*x) - (d^2*((-4*(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)^2) - ((7*c^2*C*d - 8*B*c*d^2 + 12*A*d^3 - 6 
*c^3*D)*((c^2 - d^2*x^2)^(3/2)/(3*d) + c*((x*Sqrt[c^2 - d^2*x^2])/2 + (c^2 
*ArcTan[(d*x)/Sqrt[c^2 - d^2*x^2]])/(2*d))))/c))/4)/d^5
 

Defintions of rubi rules used

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 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

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. \(542\) vs. \(2(233)=466\).

Time = 0.42 (sec) , antiderivative size = 543, normalized size of antiderivative = 2.13

method	result	size
default	\(\frac {C d \left (\frac {x \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{4}+\frac {3 c^{2} \left (\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 \sqrt {d^{2}}}\right )}{4}\right )-\frac {D \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}{5 d}-2 D c \left (\frac {x \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{4}+\frac {3 c^{2} \left (\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 \sqrt {d^{2}}}\right )}{4}\right )}{d^{3}}+\frac {\left (B \,d^{2}-2 C c d +3 D c^{2}\right ) \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{3}+c d \left (-\frac {\left (-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d \right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{4 d^{2}}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{2 \sqrt {d^{2}}}\right )\right )}{d^{4}}+\frac {\left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right ) \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {5}{2}}}{c d \left (x +\frac {c}{d}\right )^{2}}+\frac {3 d \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{3}+c d \left (-\frac {\left (-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d \right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{4 d^{2}}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{2 \sqrt {d^{2}}}\right )\right )}{c}\right )}{d^{5}}\)	\(543\)

Input:

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

Output:

1/d^3*(C*d*(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))))-1/5*D/d* 
(-d^2*x^2+c^2)^(5/2)-2*D*c*(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*(B*d^2-2*C*c*d+3*D*c^2)*(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*(A*d^3-B*c*d^2+C*c^2*d-D*c^3)*(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 = 202, normalized size of antiderivative = 0.79 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^2} \, dx=\frac {30 \, {\left (6 \, D c^{5} - 7 \, C c^{4} d + 8 \, B c^{3} d^{2} - 12 \, A c^{2} d^{3}\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) - {\left (24 \, D d^{4} x^{4} + 144 \, D c^{4} - 160 \, C c^{3} d + 200 \, B c^{2} d^{2} - 240 \, A c d^{3} - 30 \, {\left (2 \, D c d^{3} - C d^{4}\right )} x^{3} + 8 \, {\left (9 \, D c^{2} d^{2} - 10 \, C c d^{3} + 5 \, B d^{4}\right )} x^{2} - 15 \, {\left (6 \, D c^{3} d - 7 \, C c^{2} d^{2} + 8 \, B c d^{3} - 4 \, A d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{120 \, d^{4}} \] Input:

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

Output:

1/120*(30*(6*D*c^5 - 7*C*c^4*d + 8*B*c^3*d^2 - 12*A*c^2*d^3)*arctan(-(c - 
sqrt(-d^2*x^2 + c^2))/(d*x)) - (24*D*d^4*x^4 + 144*D*c^4 - 160*C*c^3*d + 2 
00*B*c^2*d^2 - 240*A*c*d^3 - 30*(2*D*c*d^3 - C*d^4)*x^3 + 8*(9*D*c^2*d^2 - 
 10*C*c*d^3 + 5*B*d^4)*x^2 - 15*(6*D*c^3*d - 7*C*c^2*d^2 + 8*B*c*d^3 - 4*A 
*d^4)*x)*sqrt(-d^2*x^2 + c^2))/d^4
 

Sympy [F]

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

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

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.14 (sec) , antiderivative size = 682, normalized size of antiderivative = 2.67 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^2} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/2*(-d^2*x^2 + c^2)^(3/2)*D*c^3/(d^5*x + c*d^4) + 1/2*(-d^2*x^2 + c^2)^( 
3/2)*C*c^2/(d^4*x + c*d^3) - 3/2*I*D*c^5*arcsin(d*x/c + 2)/d^4 + I*C*c^4*a 
rcsin(d*x/c + 2)/d^3 - 1/2*I*B*c^3*arcsin(d*x/c + 2)/d^2 - 9/4*D*c^5*arcsi 
n(d*x/c)/d^4 + 15/8*C*c^4*arcsin(d*x/c)/d^3 - 3/2*B*c^3*arcsin(d*x/c)/d^2 
+ 3/2*A*c^2*arcsin(d*x/c)/d - 1/2*(-d^2*x^2 + c^2)^(3/2)*B*c/(d^3*x + c*d^ 
2) + 3/2*sqrt(d^2*x^2 + 4*c*d*x + 3*c^2)*D*c^3*x/d^3 - 3/4*sqrt(-d^2*x^2 + 
 c^2)*D*c^3*x/d^3 - sqrt(d^2*x^2 + 4*c*d*x + 3*c^2)*C*c^2*x/d^2 + 3/8*sqrt 
(-d^2*x^2 + c^2)*C*c^2*x/d^2 + 1/2*sqrt(d^2*x^2 + 4*c*d*x + 3*c^2)*B*c*x/d 
 + 1/2*(-d^2*x^2 + c^2)^(3/2)*A/(d^2*x + c*d) + 3*sqrt(d^2*x^2 + 4*c*d*x + 
 3*c^2)*D*c^4/d^4 - 3/2*sqrt(-d^2*x^2 + c^2)*D*c^4/d^4 - 2*sqrt(d^2*x^2 + 
4*c*d*x + 3*c^2)*C*c^3/d^3 + 3/2*sqrt(-d^2*x^2 + c^2)*C*c^3/d^3 + sqrt(d^2 
*x^2 + 4*c*d*x + 3*c^2)*B*c^2/d^2 - 3/2*sqrt(-d^2*x^2 + c^2)*B*c^2/d^2 + 3 
/2*sqrt(-d^2*x^2 + c^2)*A*c/d - 1/2*(-d^2*x^2 + c^2)^(3/2)*D*c*x/d^3 + 1/4 
*(-d^2*x^2 + c^2)^(3/2)*C*x/d^2 + (-d^2*x^2 + c^2)^(3/2)*D*c^2/d^4 - 2/3*( 
-d^2*x^2 + c^2)^(3/2)*C*c/d^3 + 1/3*(-d^2*x^2 + c^2)^(3/2)*B/d^2 - 1/5*(-d 
^2*x^2 + c^2)^(5/2)*D/d^4
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 778 vs. \(2 (233) = 466\).

Time = 0.20 (sec) , antiderivative size = 778, normalized size of antiderivative = 3.05 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^2} \, dx=\text {Too large to display} \] Input:

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

Output:

1/1920*(480*(6*D*c^6*d^6*sgn(1/(d*x + c))*sgn(d) - 7*C*c^5*d^7*sgn(1/(d*x 
+ c))*sgn(d) + 8*B*c^4*d^8*sgn(1/(d*x + c))*sgn(d) - 12*A*c^3*d^9*sgn(1/(d 
*x + c))*sgn(d))*arctan(sqrt(2*c/(d*x + c) - 1)) - (390*D*c^6*d^6*(2*c/(d* 
x + c) - 1)^(9/2)*sgn(1/(d*x + c))*sgn(d) - 375*C*c^5*d^7*(2*c/(d*x + c) - 
 1)^(9/2)*sgn(1/(d*x + c))*sgn(d) + 360*B*c^4*d^8*(2*c/(d*x + c) - 1)^(9/2 
)*sgn(1/(d*x + c))*sgn(d) - 300*A*c^3*d^9*(2*c/(d*x + c) - 1)^(9/2)*sgn(1/ 
(d*x + c))*sgn(d) + 540*D*c^6*d^6*(2*c/(d*x + c) - 1)^(7/2)*sgn(1/(d*x + c 
))*sgn(d) - 790*C*c^5*d^7*(2*c/(d*x + c) - 1)^(7/2)*sgn(1/(d*x + c))*sgn(d 
) + 1040*B*c^4*d^8*(2*c/(d*x + c) - 1)^(7/2)*sgn(1/(d*x + c))*sgn(d) - 108 
0*A*c^3*d^9*(2*c/(d*x + c) - 1)^(7/2)*sgn(1/(d*x + c))*sgn(d) + 864*D*c^6* 
d^6*(2*c/(d*x + c) - 1)^(5/2)*sgn(1/(d*x + c))*sgn(d) - 800*C*c^5*d^7*(2*c 
/(d*x + c) - 1)^(5/2)*sgn(1/(d*x + c))*sgn(d) + 1120*B*c^4*d^8*(2*c/(d*x + 
 c) - 1)^(5/2)*sgn(1/(d*x + c))*sgn(d) - 1440*A*c^3*d^9*(2*c/(d*x + c) - 1 
)^(5/2)*sgn(1/(d*x + c))*sgn(d) + 420*D*c^6*d^6*(2*c/(d*x + c) - 1)^(3/2)* 
sgn(1/(d*x + c))*sgn(d) - 490*C*c^5*d^7*(2*c/(d*x + c) - 1)^(3/2)*sgn(1/(d 
*x + c))*sgn(d) + 560*B*c^4*d^8*(2*c/(d*x + c) - 1)^(3/2)*sgn(1/(d*x + c)) 
*sgn(d) - 840*A*c^3*d^9*(2*c/(d*x + c) - 1)^(3/2)*sgn(1/(d*x + c))*sgn(d) 
+ 90*D*c^6*d^6*sqrt(2*c/(d*x + c) - 1)*sgn(1/(d*x + c))*sgn(d) - 105*C*c^5 
*d^7*sqrt(2*c/(d*x + c) - 1)*sgn(1/(d*x + c))*sgn(d) + 120*B*c^4*d^8*sqrt( 
2*c/(d*x + c) - 1)*sgn(1/(d*x + c))*sgn(d) - 180*A*c^3*d^9*sqrt(2*c/(d*...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.09 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^2} \, dx=\frac {180 \mathit {asin} \left (\frac {d x}{c}\right ) a \,c^{2} d^{2}-120 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{3} d +15 \mathit {asin} \left (\frac {d x}{c}\right ) c^{5}+240 \sqrt {-d^{2} x^{2}+c^{2}}\, a c \,d^{2}-60 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{3} x -200 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2} d +120 \sqrt {-d^{2} x^{2}+c^{2}}\, b c \,d^{2} x -40 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,d^{3} x^{2}+16 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{4}-15 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{3} d x +8 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{2} d^{2} x^{2}+30 \sqrt {-d^{2} x^{2}+c^{2}}\, c \,d^{3} x^{3}-24 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{4} x^{4}-240 a \,c^{2} d^{2}+200 b \,c^{3} d -16 c^{5}}{120 d^{3}} \] Input:

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

Output:

(180*asin((d*x)/c)*a*c**2*d**2 - 120*asin((d*x)/c)*b*c**3*d + 15*asin((d*x 
)/c)*c**5 + 240*sqrt(c**2 - d**2*x**2)*a*c*d**2 - 60*sqrt(c**2 - d**2*x**2 
)*a*d**3*x - 200*sqrt(c**2 - d**2*x**2)*b*c**2*d + 120*sqrt(c**2 - d**2*x* 
*2)*b*c*d**2*x - 40*sqrt(c**2 - d**2*x**2)*b*d**3*x**2 + 16*sqrt(c**2 - d* 
*2*x**2)*c**4 - 15*sqrt(c**2 - d**2*x**2)*c**3*d*x + 8*sqrt(c**2 - d**2*x* 
*2)*c**2*d**2*x**2 + 30*sqrt(c**2 - d**2*x**2)*c*d**3*x**3 - 24*sqrt(c**2 
- d**2*x**2)*d**4*x**4 - 240*a*c**2*d**2 + 200*b*c**3*d - 16*c**5)/(120*d* 
*3)