\(\int \frac {(c^2-d^2 x^2)^{3/2} (A+B x+C x^2+D x^3)}{(c+d x)^4} \, dx\) [147]

Optimal result

Integrand size = 39, antiderivative size = 258 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^4} \, dx=\frac {\left (4 c C d-B d^2-9 c^2 D\right ) \sqrt {c^2-d^2 x^2}}{d^4}-\frac {(C d-4 c D) x \sqrt {c^2-d^2 x^2}}{2 d^3}+\frac {2 \left (5 c^2 C d-3 B c d^2+A d^3-7 c^3 D\right ) \sqrt {c^2-d^2 x^2}}{d^4 (c+d x)}+\frac {D \left (c^2-d^2 x^2\right )^{3/2}}{3 d^4}-\frac {2 \left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \left (c^2-d^2 x^2\right )^{3/2}}{3 d^4 (c+d x)^3}+\frac {\left (17 c^2 C d-8 B c d^2+2 A d^3-28 c^3 D\right ) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{2 d^4} \] Output:

(-B*d^2+4*C*c*d-9*D*c^2)*(-d^2*x^2+c^2)^(1/2)/d^4-1/2*(C*d-4*D*c)*x*(-d^2* 
x^2+c^2)^(1/2)/d^3+2*(A*d^3-3*B*c*d^2+5*C*c^2*d-7*D*c^3)*(-d^2*x^2+c^2)^(1 
/2)/d^4/(d*x+c)+1/3*D*(-d^2*x^2+c^2)^(3/2)/d^4-2/3*(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)^3+1/2*(2*A*d^3-8*B*c*d^2+17*C*c^2* 
d-28*D*c^3)*arctan(d*x/(-d^2*x^2+c^2)^(1/2))/d^4
 

Mathematica [A] (verified)

Time = 1.63 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.72 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^4} \, dx=\frac {-\frac {\sqrt {c^2-d^2 x^2} \left (132 c^4 D-20 c^3 d (4 C-9 D x)+c^2 d^2 (38 B+x (-109 C+30 D x))+d^4 x \left (-16 A+x \left (6 B+3 C x+2 D x^2\right )\right )-2 c d^3 \left (4 A+x \left (-26 B+9 C x+4 D x^2\right )\right )\right )}{(c+d x)^2}+6 \left (-17 c^2 C d+8 B c d^2-2 A d^3+28 c^3 D\right ) \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{6 d^4} \] Input:

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

Output:

(-((Sqrt[c^2 - d^2*x^2]*(132*c^4*D - 20*c^3*d*(4*C - 9*D*x) + c^2*d^2*(38* 
B + x*(-109*C + 30*D*x)) + d^4*x*(-16*A + x*(6*B + 3*C*x + 2*D*x^2)) - 2*c 
*d^3*(4*A + x*(-26*B + 9*C*x + 4*D*x^2))))/(c + d*x)^2) + 6*(-17*c^2*C*d + 
 8*B*c*d^2 - 2*A*d^3 + 28*c^3*D)*ArcTan[(d*x)/(Sqrt[c^2] - Sqrt[c^2 - d^2* 
x^2])])/(6*d^4)
 

Rubi [A] (verified)

Time = 1.15 (sec) , antiderivative size = 249, normalized size of antiderivative = 0.97, 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, 463, 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.

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

\(\Big \downarrow \) 2170

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2170

\(\displaystyle \frac {-\frac {\int \frac {d^6 \left (-20 D c^3+9 C d c^2-6 A d^3+d \left (-26 D c^2+15 C d c-6 B d^2\right ) x\right ) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^4}dx}{2 d^4}-\frac {d \left (c^2-d^2 x^2\right )^{5/2} (3 C d-8 c D)}{2 (c+d x)^3}}{3 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{5/2}}{3 d^4 (c+d x)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {1}{2} d^2 \int \frac {\left (-20 D c^3+9 C d c^2-6 A d^3+d \left (-26 D c^2+15 C d c-6 B d^2\right ) x\right ) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^4}dx-\frac {d \left (c^2-d^2 x^2\right )^{5/2} (3 C d-8 c D)}{2 (c+d x)^3}}{3 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{5/2}}{3 d^4 (c+d x)^2}\)

\(\Big \downarrow \) 671

\(\displaystyle \frac {-\frac {1}{2} d^2 \left (\frac {\left (2 A d^3-8 B c d^2-28 c^3 D+17 c^2 C d\right ) \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^3}dx}{c}+\frac {2 \left (c^2-d^2 x^2\right )^{5/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (c+d x)^4}\right )-\frac {d \left (c^2-d^2 x^2\right )^{5/2} (3 C d-8 c D)}{2 (c+d x)^3}}{3 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{5/2}}{3 d^4 (c+d x)^2}\)

\(\Big \downarrow \) 463

\(\displaystyle \frac {-\frac {1}{2} d^2 \left (\frac {\left (2 A d^3-8 B c d^2-28 c^3 D+17 c^2 C d\right ) \left (-\int \frac {3 c-d x}{\sqrt {c^2-d^2 x^2}}dx-\frac {4 c \sqrt {c^2-d^2 x^2}}{d (c+d x)}\right )}{c}+\frac {2 \left (c^2-d^2 x^2\right )^{5/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (c+d x)^4}\right )-\frac {d \left (c^2-d^2 x^2\right )^{5/2} (3 C d-8 c D)}{2 (c+d x)^3}}{3 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{5/2}}{3 d^4 (c+d x)^2}\)

\(\Big \downarrow \) 455

\(\displaystyle \frac {-\frac {1}{2} d^2 \left (\frac {\left (2 A d^3-8 B c d^2-28 c^3 D+17 c^2 C d\right ) \left (-3 c \int \frac {1}{\sqrt {c^2-d^2 x^2}}dx-\frac {4 c \sqrt {c^2-d^2 x^2}}{d (c+d x)}-\frac {\sqrt {c^2-d^2 x^2}}{d}\right )}{c}+\frac {2 \left (c^2-d^2 x^2\right )^{5/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (c+d x)^4}\right )-\frac {d \left (c^2-d^2 x^2\right )^{5/2} (3 C d-8 c D)}{2 (c+d x)^3}}{3 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{5/2}}{3 d^4 (c+d x)^2}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {-\frac {1}{2} d^2 \left (\frac {\left (2 A d^3-8 B c d^2-28 c^3 D+17 c^2 C d\right ) \left (-3 c \int \frac {1}{\frac {d^2 x^2}{c^2-d^2 x^2}+1}d\frac {x}{\sqrt {c^2-d^2 x^2}}-\frac {4 c \sqrt {c^2-d^2 x^2}}{d (c+d x)}-\frac {\sqrt {c^2-d^2 x^2}}{d}\right )}{c}+\frac {2 \left (c^2-d^2 x^2\right )^{5/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (c+d x)^4}\right )-\frac {d \left (c^2-d^2 x^2\right )^{5/2} (3 C d-8 c D)}{2 (c+d x)^3}}{3 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{5/2}}{3 d^4 (c+d x)^2}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {-\frac {1}{2} d^2 \left (\frac {\left (-\frac {3 c \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{d}-\frac {4 c \sqrt {c^2-d^2 x^2}}{d (c+d x)}-\frac {\sqrt {c^2-d^2 x^2}}{d}\right ) \left (2 A d^3-8 B c d^2-28 c^3 D+17 c^2 C d\right )}{c}+\frac {2 \left (c^2-d^2 x^2\right )^{5/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (c+d x)^4}\right )-\frac {d \left (c^2-d^2 x^2\right )^{5/2} (3 C d-8 c D)}{2 (c+d x)^3}}{3 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{5/2}}{3 d^4 (c+d x)^2}\)

Input:

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

Output:

-1/3*(D*(c^2 - d^2*x^2)^(5/2))/(d^4*(c + d*x)^2) + (-1/2*(d*(3*C*d - 8*c*D 
)*(c^2 - d^2*x^2)^(5/2))/(c + d*x)^3 - (d^2*((2*(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)^4) + ((17*c^2*C*d - 8*B*c* 
d^2 + 2*A*d^3 - 28*c^3*D)*(-(Sqrt[c^2 - d^2*x^2]/d) - (4*c*Sqrt[c^2 - d^2* 
x^2])/(d*(c + d*x)) - (3*c*ArcTan[(d*x)/Sqrt[c^2 - d^2*x^2]])/d))/c))/2)/( 
3*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 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 463

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(-(-c)^(-n - 2))*d^(2*n + 3)*(Sqrt[a + b*x^2]/(2^(n + 1)*b^(n + 2)*(c + d*x 
))), x] - Simp[d^(2*n + 2)/b^(n + 1)   Int[(1/Sqrt[a + b*x^2])*ExpandToSum[ 
(2^(-n - 1)*(-c)^(-n - 1) - (-c + d*x)^(-n - 1))/(c + d*x), x], x], x] /; F 
reeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && ILtQ[n, 0] && EqQ[n + p, 
-3/2]
 

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. \(901\) vs. \(2(238)=476\).

Time = 0.48 (sec) , antiderivative size = 902, normalized size of antiderivative = 3.50

method	result	size
default	\(\text {Expression too large to display}\)	\(902\)

Input:

int((-d^2*x^2+c^2)^(3/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^4,x,method=_RETURNVER 
BOSE)
 

Output:

D/d^4*(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))))+(C*d-3*D*c)/d^5*(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)))))+(B*d^2-2*C*c*d+3*D*c^2)/d^6*(-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))))))+(A*d^3-B*c*d^2+C*c^2*d-D*c^3)/d^7*(-1/3/c/d/(x+c/d)^4*(-d^2*(x+ 
c/d)^2+2*c*d*(x+c/d))^(5/2)-1/3*d/c*(-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)*arc 
tan((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 = 0.13 (sec) , antiderivative size = 396, normalized size of antiderivative = 1.53 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^4} \, dx=-\frac {132 \, D c^{5} - 80 \, C c^{4} d + 38 \, B c^{3} d^{2} - 8 \, A c^{2} d^{3} + 2 \, {\left (66 \, D c^{3} d^{2} - 40 \, C c^{2} d^{3} + 19 \, B c d^{4} - 4 \, A d^{5}\right )} x^{2} + 4 \, {\left (66 \, D c^{4} d - 40 \, C c^{3} d^{2} + 19 \, B c^{2} d^{3} - 4 \, A c d^{4}\right )} x - 6 \, {\left (28 \, D c^{5} - 17 \, C c^{4} d + 8 \, B c^{3} d^{2} - 2 \, A c^{2} d^{3} + {\left (28 \, D c^{3} d^{2} - 17 \, C c^{2} d^{3} + 8 \, B c d^{4} - 2 \, A d^{5}\right )} x^{2} + 2 \, {\left (28 \, D c^{4} d - 17 \, C c^{3} d^{2} + 8 \, B c^{2} d^{3} - 2 \, A c d^{4}\right )} x\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) + {\left (2 \, D d^{4} x^{4} + 132 \, D c^{4} - 80 \, C c^{3} d + 38 \, B c^{2} d^{2} - 8 \, A c d^{3} - {\left (8 \, D c d^{3} - 3 \, C d^{4}\right )} x^{3} + 6 \, {\left (5 \, D c^{2} d^{2} - 3 \, C c d^{3} + B d^{4}\right )} x^{2} + {\left (180 \, D c^{3} d - 109 \, C c^{2} d^{2} + 52 \, B c d^{3} - 16 \, A d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{6 \, {\left (d^{6} x^{2} + 2 \, c d^{5} x + c^{2} d^{4}\right )}} \] Input:

integrate((-d^2*x^2+c^2)^(3/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^4,x, algorithm= 
"fricas")
 

Output:

-1/6*(132*D*c^5 - 80*C*c^4*d + 38*B*c^3*d^2 - 8*A*c^2*d^3 + 2*(66*D*c^3*d^ 
2 - 40*C*c^2*d^3 + 19*B*c*d^4 - 4*A*d^5)*x^2 + 4*(66*D*c^4*d - 40*C*c^3*d^ 
2 + 19*B*c^2*d^3 - 4*A*c*d^4)*x - 6*(28*D*c^5 - 17*C*c^4*d + 8*B*c^3*d^2 - 
 2*A*c^2*d^3 + (28*D*c^3*d^2 - 17*C*c^2*d^3 + 8*B*c*d^4 - 2*A*d^5)*x^2 + 2 
*(28*D*c^4*d - 17*C*c^3*d^2 + 8*B*c^2*d^3 - 2*A*c*d^4)*x)*arctan(-(c - sqr 
t(-d^2*x^2 + c^2))/(d*x)) + (2*D*d^4*x^4 + 132*D*c^4 - 80*C*c^3*d + 38*B*c 
^2*d^2 - 8*A*c*d^3 - (8*D*c*d^3 - 3*C*d^4)*x^3 + 6*(5*D*c^2*d^2 - 3*C*c*d^ 
3 + B*d^4)*x^2 + (180*D*c^3*d - 109*C*c^2*d^2 + 52*B*c*d^3 - 16*A*d^4)*x)* 
sqrt(-d^2*x^2 + c^2))/(d^6*x^2 + 2*c*d^5*x + c^2*d^4)
 

Sympy [F]

\[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^4} \, dx=\int \frac {\left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {3}{2}} \left (A + B x + C x^{2} + D x^{3}\right )}{\left (c + d x\right )^{4}}\, dx \] Input:

integrate((-d**2*x**2+c**2)**(3/2)*(D*x**3+C*x**2+B*x+A)/(d*x+c)**4,x)
 

Output:

Integral((-(-c + d*x)*(c + d*x))**(3/2)*(A + B*x + C*x**2 + D*x**3)/(c + d 
*x)**4, x)
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.16 (sec) , antiderivative size = 894, normalized size of antiderivative = 3.47 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^4} \, dx =\text {Too large to display} \] Input:

integrate((-d^2*x^2+c^2)^(3/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^4,x, algorithm= 
"maxima")
 

Output:

1/3*(-d^2*x^2 + c^2)^(3/2)*D*c^3/(d^7*x^3 + 3*c*d^6*x^2 + 3*c^2*d^5*x + c^ 
3*d^4) + 2/3*sqrt(-d^2*x^2 + c^2)*D*c^4/(d^6*x^2 + 2*c*d^5*x + c^2*d^4) - 
1/3*(-d^2*x^2 + c^2)^(3/2)*C*c^2/(d^6*x^3 + 3*c*d^5*x^2 + 3*c^2*d^4*x + c^ 
3*d^3) + 3*(-d^2*x^2 + c^2)^(3/2)*D*c^2/(d^6*x^2 + 2*c*d^5*x + c^2*d^4) - 
2/3*sqrt(-d^2*x^2 + c^2)*C*c^3/(d^5*x^2 + 2*c*d^4*x + c^2*d^3) - 61/3*sqrt 
(-d^2*x^2 + c^2)*D*c^3/(d^5*x + c*d^4) + 1/3*(-d^2*x^2 + c^2)^(3/2)*B*c/(d 
^5*x^3 + 3*c*d^4*x^2 + 3*c^2*d^3*x + c^3*d^2) - 2*(-d^2*x^2 + c^2)^(3/2)*C 
*c/(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/(d^5*x 
 + c*d^4) + 2/3*sqrt(-d^2*x^2 + c^2)*B*c^2/(d^4*x^2 + 2*c*d^3*x + c^2*d^2) 
 + 43/3*sqrt(-d^2*x^2 + c^2)*C*c^2/(d^4*x + c*d^3) - 1/3*(-d^2*x^2 + c^2)^ 
(3/2)*A/(d^4*x^3 + 3*c*d^3*x^2 + 3*c^2*d^2*x + c^3*d) + (-d^2*x^2 + c^2)^( 
3/2)*B/(d^4*x^2 + 2*c*d^3*x + c^2*d^2) + 1/2*(-d^2*x^2 + c^2)^(3/2)*C/(d^4 
*x + c*d^3) - 2/3*sqrt(-d^2*x^2 + c^2)*A*c/(d^3*x^2 + 2*c*d^2*x + c^2*d) - 
 25/3*sqrt(-d^2*x^2 + c^2)*B*c/(d^3*x + c*d^2) - 1/2*I*D*c^3*arcsin(d*x/c 
+ 2)/d^4 - 29/2*D*c^3*arcsin(d*x/c)/d^4 + 17/2*C*c^2*arcsin(d*x/c)/d^3 - 4 
*B*c*arcsin(d*x/c)/d^2 + A*arcsin(d*x/c)/d + 7/3*sqrt(-d^2*x^2 + c^2)*A/(d 
^2*x + c*d) + 1/2*sqrt(d^2*x^2 + 4*c*d*x + 3*c^2)*D*c*x/d^3 + sqrt(d^2*x^2 
 + 4*c*d*x + 3*c^2)*D*c^2/d^4 - 9/2*sqrt(-d^2*x^2 + c^2)*D*c^2/d^4 + 3/2*s 
qrt(-d^2*x^2 + c^2)*C*c/d^3 + 1/3*(-d^2*x^2 + c^2)^(3/2)*D/d^4
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 415, normalized size of antiderivative = 1.61 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^4} \, dx=-\frac {1}{6} \, \sqrt {-d^{2} x^{2} + c^{2}} {\left (x {\left (\frac {2 \, D x}{d^{2}} - \frac {3 \, {\left (4 \, D c d^{9} - C d^{10}\right )}}{d^{12}}\right )} + \frac {2 \, {\left (26 \, D c^{2} d^{8} - 12 \, C c d^{9} + 3 \, B d^{10}\right )}}{d^{12}}\right )} - \frac {{\left (28 \, D c^{3} - 17 \, C c^{2} d + 8 \, B c d^{2} - 2 \, A d^{3}\right )} \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{2 \, d^{3} {\left | d \right |}} + \frac {8 \, {\left (10 \, D c^{3} - 7 \, C c^{2} d + 4 \, B c d^{2} - A d^{3} + \frac {9 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} B c}{x} + \frac {21 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} D c^{3}}{d^{2} x} - \frac {15 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} C c^{2}}{d x} - \frac {3 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} A d}{x} + \frac {9 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} D c^{3}}{d^{4} x^{2}} - \frac {6 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} C c^{2}}{d^{3} x^{2}} + \frac {3 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} B c}{d^{2} x^{2}}\right )}}{3 \, d^{3} {\left (\frac {c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}}{d^{2} x} + 1\right )}^{3} {\left | d \right |}} \] Input:

integrate((-d^2*x^2+c^2)^(3/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^4,x, algorithm= 
"giac")
 

Output:

-1/6*sqrt(-d^2*x^2 + c^2)*(x*(2*D*x/d^2 - 3*(4*D*c*d^9 - C*d^10)/d^12) + 2 
*(26*D*c^2*d^8 - 12*C*c*d^9 + 3*B*d^10)/d^12) - 1/2*(28*D*c^3 - 17*C*c^2*d 
 + 8*B*c*d^2 - 2*A*d^3)*arcsin(d*x/c)*sgn(c)*sgn(d)/(d^3*abs(d)) + 8/3*(10 
*D*c^3 - 7*C*c^2*d + 4*B*c*d^2 - A*d^3 + 9*(c*d + sqrt(-d^2*x^2 + c^2)*abs 
(d))*B*c/x + 21*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))*D*c^3/(d^2*x) - 15*(c* 
d + sqrt(-d^2*x^2 + c^2)*abs(d))*C*c^2/(d*x) - 3*(c*d + sqrt(-d^2*x^2 + c^ 
2)*abs(d))*A*d/x + 9*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^2*D*c^3/(d^4*x^2) 
 - 6*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^2*C*c^2/(d^3*x^2) + 3*(c*d + sqrt 
(-d^2*x^2 + c^2)*abs(d))^2*B*c/(d^2*x^2))/(d^3*((c*d + sqrt(-d^2*x^2 + c^2 
)*abs(d))/(d^2*x) + 1)^3*abs(d))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^4} \, dx=\int \frac {{\left (c^2-d^2\,x^2\right )}^{3/2}\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{{\left (c+d\,x\right )}^4} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 650, normalized size of antiderivative = 2.52 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^4} \, dx=\frac {6 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) a \,d^{3} x -22 c^{5}+12 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{2} d^{2} x^{2}+22 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{4}+2 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{4} x^{4}+2 d^{5} x^{5}+6 b \,d^{4} x^{3}+6 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) a c \,d^{2}-6 \mathit {asin} \left (\frac {d x}{c}\right ) a \,c^{2} d^{2}-8 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{3} x +16 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2} d +6 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,d^{3} x^{2}+24 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{3} d +66 \mathit {asin} \left (\frac {d x}{c}\right ) c^{4} d x +33 \mathit {asin} \left (\frac {d x}{c}\right ) c^{3} d^{2} x^{2}+30 b \,c^{2} d^{2} x +68 b c \,d^{3} x^{2}-8 a c \,d^{3} x -33 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) c^{4}-24 a \,d^{4} x^{2}-6 \mathit {asin} \left (\frac {d x}{c}\right ) a \,d^{4} x^{2}-24 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{2} d -33 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) c^{3} d x +48 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{2} d^{2} x +24 \mathit {asin} \left (\frac {d x}{c}\right ) b c \,d^{3} x^{2}+30 \sqrt {-d^{2} x^{2}+c^{2}}\, b c \,d^{2} x +41 c^{4} d x +89 c^{3} d^{2} x^{2}+17 c^{2} d^{3} x^{3}-7 c \,d^{4} x^{4}+41 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{3} d x -5 \sqrt {-d^{2} x^{2}+c^{2}}\, c \,d^{3} x^{3}-24 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) b c \,d^{2} x +33 \mathit {asin} \left (\frac {d x}{c}\right ) c^{5}-16 b \,c^{3} d -12 \mathit {asin} \left (\frac {d x}{c}\right ) a c \,d^{3} x}{6 d^{3} \left (\sqrt {-d^{2} x^{2}+c^{2}}\, c +\sqrt {-d^{2} x^{2}+c^{2}}\, d x -c^{2}-2 c d x -d^{2} x^{2}\right )} \] Input:

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

Output:

(6*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*a*c*d**2 + 6*sqrt(c**2 - d**2*x**2 
)*asin((d*x)/c)*a*d**3*x - 24*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*b*c**2* 
d - 24*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*b*c*d**2*x - 33*sqrt(c**2 - d* 
*2*x**2)*asin((d*x)/c)*c**4 - 33*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*c**3 
*d*x - 6*asin((d*x)/c)*a*c**2*d**2 - 12*asin((d*x)/c)*a*c*d**3*x - 6*asin( 
(d*x)/c)*a*d**4*x**2 + 24*asin((d*x)/c)*b*c**3*d + 48*asin((d*x)/c)*b*c**2 
*d**2*x + 24*asin((d*x)/c)*b*c*d**3*x**2 + 33*asin((d*x)/c)*c**5 + 66*asin 
((d*x)/c)*c**4*d*x + 33*asin((d*x)/c)*c**3*d**2*x**2 - 8*sqrt(c**2 - d**2* 
x**2)*a*d**3*x + 16*sqrt(c**2 - d**2*x**2)*b*c**2*d + 30*sqrt(c**2 - d**2* 
x**2)*b*c*d**2*x + 6*sqrt(c**2 - d**2*x**2)*b*d**3*x**2 + 22*sqrt(c**2 - d 
**2*x**2)*c**4 + 41*sqrt(c**2 - d**2*x**2)*c**3*d*x + 12*sqrt(c**2 - d**2* 
x**2)*c**2*d**2*x**2 - 5*sqrt(c**2 - d**2*x**2)*c*d**3*x**3 + 2*sqrt(c**2 
- d**2*x**2)*d**4*x**4 - 8*a*c*d**3*x - 24*a*d**4*x**2 - 16*b*c**3*d + 30* 
b*c**2*d**2*x + 68*b*c*d**3*x**2 + 6*b*d**4*x**3 - 22*c**5 + 41*c**4*d*x + 
 89*c**3*d**2*x**2 + 17*c**2*d**3*x**3 - 7*c*d**4*x**4 + 2*d**5*x**5)/(6*d 
**3*(sqrt(c**2 - d**2*x**2)*c + sqrt(c**2 - d**2*x**2)*d*x - c**2 - 2*c*d* 
x - d**2*x**2))