\(\int \frac {(c^2-d^2 x^2)^{3/2} (A+B x+C x^2+D x^3)}{(c+d x)^7} \, dx\) [150]

Optimal result

Integrand size = 39, antiderivative size = 264 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^7} \, dx=\frac {2 D \sqrt {c^2-d^2 x^2}}{d^4 (c+d x)}-\frac {2 D \left (c^2-d^2 x^2\right )^{3/2}}{3 d^4 (c+d x)^3}-\frac {\left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \left (c^2-d^2 x^2\right )^{5/2}}{9 c d^4 (c+d x)^7}+\frac {\left (16 c^2 C d-7 B c d^2-2 A d^3-25 c^3 D\right ) \left (c^2-d^2 x^2\right )^{5/2}}{63 c^2 d^4 (c+d x)^6}-\frac {\left (47 c^2 C d+7 B c d^2+2 A d^3-164 c^3 D\right ) \left (c^2-d^2 x^2\right )^{5/2}}{315 c^3 d^4 (c+d x)^5}+\frac {D \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{d^4} \] Output:

2*D*(-d^2*x^2+c^2)^(1/2)/d^4/(d*x+c)-2/3*D*(-d^2*x^2+c^2)^(3/2)/d^4/(d*x+c 
)^3-1/9*(A*d^3-B*c*d^2+C*c^2*d-D*c^3)*(-d^2*x^2+c^2)^(5/2)/c/d^4/(d*x+c)^7 
+1/63*(-2*A*d^3-7*B*c*d^2+16*C*c^2*d-25*D*c^3)*(-d^2*x^2+c^2)^(5/2)/c^2/d^ 
4/(d*x+c)^6-1/315*(2*A*d^3+7*B*c*d^2+47*C*c^2*d-164*D*c^3)*(-d^2*x^2+c^2)^ 
(5/2)/c^3/d^4/(d*x+c)^5+D*arctan(d*x/(-d^2*x^2+c^2)^(1/2))/d^4
 

Mathematica [A] (verified)

Time = 2.09 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.84 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^7} \, dx=\frac {\frac {\sqrt {c^2-d^2 x^2} \left (494 c^7 D-2 A d^7 x^4-c d^6 x^3 (10 A+7 B x)+c^6 (-2 C d+2155 d D x)-c^2 d^5 x^2 (21 A+x (35 B+47 C x))+c^5 d^2 (-7 B+2 x (-5 C+1806 D x))-c^4 d^3 \left (47 A+x \left (35 B+21 C x-2815 D x^2\right )\right )+4 c^3 d^4 x \left (20 A+x \left (21 B+20 C x+251 D x^2\right )\right )\right )}{c^3 (c+d x)^5}-630 D \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{315 d^4} \] Input:

Integrate[((c^2 - d^2*x^2)^(3/2)*(A + B*x + C*x^2 + D*x^3))/(c + d*x)^7,x]
 

Output:

((Sqrt[c^2 - d^2*x^2]*(494*c^7*D - 2*A*d^7*x^4 - c*d^6*x^3*(10*A + 7*B*x) 
+ c^6*(-2*C*d + 2155*d*D*x) - c^2*d^5*x^2*(21*A + x*(35*B + 47*C*x)) + c^5 
*d^2*(-7*B + 2*x*(-5*C + 1806*D*x)) - c^4*d^3*(47*A + x*(35*B + 21*C*x - 2 
815*D*x^2)) + 4*c^3*d^4*x*(20*A + x*(21*B + 20*C*x + 251*D*x^2))))/(c^3*(c 
 + d*x)^5) - 630*D*ArcTan[(d*x)/(Sqrt[c^2] - Sqrt[c^2 - d^2*x^2])])/(315*d 
^4)
 

Rubi [A] (verified)

Time = 1.03 (sec) , antiderivative size = 403, normalized size of antiderivative = 1.53, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {2168, 2009}

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.

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

\(\Big \downarrow \) 2168

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 \left (c^2-d^2 x^2\right )^{5/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{315 c^3 d^4 (c+d x)^5}-\frac {2 \left (c^2-d^2 x^2\right )^{5/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{63 c^2 d^4 (c+d x)^6}-\frac {\left (c^2-d^2 x^2\right )^{5/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{9 c d^4 (c+d x)^7}+\frac {D \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{d^4}+\frac {\left (c^2-d^2 x^2\right )^{5/2} \left (-B d^2-3 c^2 D+2 c C d\right )}{35 c^2 d^4 (c+d x)^5}+\frac {\left (c^2-d^2 x^2\right )^{5/2} \left (-B d^2-3 c^2 D+2 c C d\right )}{7 c d^4 (c+d x)^6}-\frac {\left (c^2-d^2 x^2\right )^{5/2} (C d-3 c D)}{5 c d^4 (c+d x)^5}-\frac {2 D \left (c^2-d^2 x^2\right )^{3/2}}{3 d^4 (c+d x)^3}+\frac {2 D \sqrt {c^2-d^2 x^2}}{d^4 (c+d x)}\)

Input:

Int[((c^2 - d^2*x^2)^(3/2)*(A + B*x + C*x^2 + D*x^3))/(c + d*x)^7,x]
 

Output:

(2*D*Sqrt[c^2 - d^2*x^2])/(d^4*(c + d*x)) - (2*D*(c^2 - d^2*x^2)^(3/2))/(3 
*d^4*(c + d*x)^3) - ((c^2*C*d - B*c*d^2 + A*d^3 - c^3*D)*(c^2 - d^2*x^2)^( 
5/2))/(9*c*d^4*(c + d*x)^7) + ((2*c*C*d - B*d^2 - 3*c^2*D)*(c^2 - d^2*x^2) 
^(5/2))/(7*c*d^4*(c + d*x)^6) - (2*(c^2*C*d - B*c*d^2 + A*d^3 - c^3*D)*(c^ 
2 - d^2*x^2)^(5/2))/(63*c^2*d^4*(c + d*x)^6) - ((C*d - 3*c*D)*(c^2 - d^2*x 
^2)^(5/2))/(5*c*d^4*(c + d*x)^5) + ((2*c*C*d - B*d^2 - 3*c^2*D)*(c^2 - d^2 
*x^2)^(5/2))/(35*c^2*d^4*(c + d*x)^5) - (2*(c^2*C*d - B*c*d^2 + A*d^3 - c^ 
3*D)*(c^2 - d^2*x^2)^(5/2))/(315*c^3*d^4*(c + d*x)^5) + (D*ArcTan[(d*x)/Sq 
rt[c^2 - d^2*x^2]])/d^4
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2168

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
 Int[ExpandIntegrand[(a + b*x^2)^p, (d + e*x)^m*Pq, x], x] /; FreeQ[{a, b, 
d, e}, x] && PolyQ[Pq, x] && EqQ[b*d^2 + a*e^2, 0] && EqQ[m + Expon[Pq, x] 
+ 2*p + 1, 0] && ILtQ[m, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(623\) vs. \(2(244)=488\).

Time = 0.66 (sec) , antiderivative size = 624, normalized size of antiderivative = 2.36

method	result	size
default	\(\frac {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}}}{3 c d \left (x +\frac {c}{d}\right )^{4}}-\frac {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 )^{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 )}{3 c}\right )}{d^{7}}-\frac {\left (C d -3 D c \right ) \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {5}{2}}}{5 d^{9} c \left (x +\frac {c}{d}\right )^{5}}+\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}}}{7 c d \left (x +\frac {c}{d}\right )^{6}}-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {5}{2}}}{35 c^{2} \left (x +\frac {c}{d}\right )^{5}}\right )}{d^{9}}+\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}}}{9 c d \left (x +\frac {c}{d}\right )^{7}}+\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}}}{7 c d \left (x +\frac {c}{d}\right )^{6}}-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {5}{2}}}{35 c^{2} \left (x +\frac {c}{d}\right )^{5}}\right )}{9 c}\right )}{d^{10}}\)	\(624\)

Input:

int((-d^2*x^2+c^2)^(3/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^7,x,method=_RETURNVER 
BOSE)
 

Output:

D/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)*arctan((d^2)^(1/2)*x/(-d^2*(x+c/d)^2+2*c*d 
*(x+c/d))^(1/2)))))))-1/5*(C*d-3*D*c)/d^9/c/(x+c/d)^5*(-d^2*(x+c/d)^2+2*c* 
d*(x+c/d))^(5/2)+(B*d^2-2*C*c*d+3*D*c^2)/d^9*(-1/7/c/d/(x+c/d)^6*(-d^2*(x+ 
c/d)^2+2*c*d*(x+c/d))^(5/2)-1/35/c^2/(x+c/d)^5*(-d^2*(x+c/d)^2+2*c*d*(x+c/ 
d))^(5/2))+(A*d^3-B*c*d^2+C*c^2*d-D*c^3)/d^10*(-1/9/c/d/(x+c/d)^7*(-d^2*(x 
+c/d)^2+2*c*d*(x+c/d))^(5/2)+2/9*d/c*(-1/7/c/d/(x+c/d)^6*(-d^2*(x+c/d)^2+2 
*c*d*(x+c/d))^(5/2)-1/35/c^2/(x+c/d)^5*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(5/2 
)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 587 vs. \(2 (244) = 488\).

Time = 0.21 (sec) , antiderivative size = 587, normalized size of antiderivative = 2.22 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^7} \, dx=\frac {494 \, D c^{8} - 2 \, C c^{7} d - 7 \, B c^{6} d^{2} - 47 \, A c^{5} d^{3} + {\left (494 \, D c^{3} d^{5} - 2 \, C c^{2} d^{6} - 7 \, B c d^{7} - 47 \, A d^{8}\right )} x^{5} + 5 \, {\left (494 \, D c^{4} d^{4} - 2 \, C c^{3} d^{5} - 7 \, B c^{2} d^{6} - 47 \, A c d^{7}\right )} x^{4} + 10 \, {\left (494 \, D c^{5} d^{3} - 2 \, C c^{4} d^{4} - 7 \, B c^{3} d^{5} - 47 \, A c^{2} d^{6}\right )} x^{3} + 10 \, {\left (494 \, D c^{6} d^{2} - 2 \, C c^{5} d^{3} - 7 \, B c^{4} d^{4} - 47 \, A c^{3} d^{5}\right )} x^{2} + 5 \, {\left (494 \, D c^{7} d - 2 \, C c^{6} d^{2} - 7 \, B c^{5} d^{3} - 47 \, A c^{4} d^{4}\right )} x - 630 \, {\left (D c^{3} d^{5} x^{5} + 5 \, D c^{4} d^{4} x^{4} + 10 \, D c^{5} d^{3} x^{3} + 10 \, D c^{6} d^{2} x^{2} + 5 \, D c^{7} d x + D c^{8}\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) + {\left (494 \, D c^{7} - 2 \, C c^{6} d - 7 \, B c^{5} d^{2} - 47 \, A c^{4} d^{3} + {\left (1004 \, D c^{3} d^{4} - 47 \, C c^{2} d^{5} - 7 \, B c d^{6} - 2 \, A d^{7}\right )} x^{4} + 5 \, {\left (563 \, D c^{4} d^{3} + 16 \, C c^{3} d^{4} - 7 \, B c^{2} d^{5} - 2 \, A c d^{6}\right )} x^{3} + 21 \, {\left (172 \, D c^{5} d^{2} - C c^{4} d^{3} + 4 \, B c^{3} d^{4} - A c^{2} d^{5}\right )} x^{2} + 5 \, {\left (431 \, D c^{6} d - 2 \, C c^{5} d^{2} - 7 \, B c^{4} d^{3} + 16 \, A c^{3} d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{315 \, {\left (c^{3} d^{9} x^{5} + 5 \, c^{4} d^{8} x^{4} + 10 \, c^{5} d^{7} x^{3} + 10 \, c^{6} d^{6} x^{2} + 5 \, c^{7} d^{5} x + c^{8} d^{4}\right )}} \] Input:

integrate((-d^2*x^2+c^2)^(3/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^7,x, algorithm= 
"fricas")
 

Output:

1/315*(494*D*c^8 - 2*C*c^7*d - 7*B*c^6*d^2 - 47*A*c^5*d^3 + (494*D*c^3*d^5 
 - 2*C*c^2*d^6 - 7*B*c*d^7 - 47*A*d^8)*x^5 + 5*(494*D*c^4*d^4 - 2*C*c^3*d^ 
5 - 7*B*c^2*d^6 - 47*A*c*d^7)*x^4 + 10*(494*D*c^5*d^3 - 2*C*c^4*d^4 - 7*B* 
c^3*d^5 - 47*A*c^2*d^6)*x^3 + 10*(494*D*c^6*d^2 - 2*C*c^5*d^3 - 7*B*c^4*d^ 
4 - 47*A*c^3*d^5)*x^2 + 5*(494*D*c^7*d - 2*C*c^6*d^2 - 7*B*c^5*d^3 - 47*A* 
c^4*d^4)*x - 630*(D*c^3*d^5*x^5 + 5*D*c^4*d^4*x^4 + 10*D*c^5*d^3*x^3 + 10* 
D*c^6*d^2*x^2 + 5*D*c^7*d*x + D*c^8)*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d 
*x)) + (494*D*c^7 - 2*C*c^6*d - 7*B*c^5*d^2 - 47*A*c^4*d^3 + (1004*D*c^3*d 
^4 - 47*C*c^2*d^5 - 7*B*c*d^6 - 2*A*d^7)*x^4 + 5*(563*D*c^4*d^3 + 16*C*c^3 
*d^4 - 7*B*c^2*d^5 - 2*A*c*d^6)*x^3 + 21*(172*D*c^5*d^2 - C*c^4*d^3 + 4*B* 
c^3*d^4 - A*c^2*d^5)*x^2 + 5*(431*D*c^6*d - 2*C*c^5*d^2 - 7*B*c^4*d^3 + 16 
*A*c^3*d^4)*x)*sqrt(-d^2*x^2 + c^2))/(c^3*d^9*x^5 + 5*c^4*d^8*x^4 + 10*c^5 
*d^7*x^3 + 10*c^6*d^6*x^2 + 5*c^7*d^5*x + c^8*d^4)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^7} \, dx=\text {Timed out} \] Input:

integrate((-d**2*x**2+c**2)**(3/2)*(D*x**3+C*x**2+B*x+A)/(d*x+c)**7,x)
                                                                                    
                                                                                    
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2467 vs. \(2 (244) = 488\).

Time = 0.18 (sec) , antiderivative size = 2467, normalized size of antiderivative = 9.34 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^7} \, dx=\text {Too large to display} \] Input:

integrate((-d^2*x^2+c^2)^(3/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^7,x, algorithm= 
"maxima")
 

Output:

1/3*(-d^2*x^2 + c^2)^(3/2)*D*c^3/(d^10*x^6 + 6*c*d^9*x^5 + 15*c^2*d^8*x^4 
+ 20*c^3*d^7*x^3 + 15*c^4*d^6*x^2 + 6*c^5*d^5*x + c^6*d^4) - 2/9*sqrt(-d^2 
*x^2 + c^2)*D*c^4/(d^9*x^5 + 5*c*d^8*x^4 + 10*c^2*d^7*x^3 + 10*c^3*d^6*x^2 
 + 5*c^4*d^5*x + c^5*d^4) - 1/3*(-d^2*x^2 + c^2)^(3/2)*C*c^2/(d^9*x^6 + 6* 
c*d^8*x^5 + 15*c^2*d^7*x^4 + 20*c^3*d^6*x^3 + 15*c^4*d^5*x^2 + 6*c^5*d^4*x 
 + c^6*d^3) - 3/2*(-d^2*x^2 + c^2)^(3/2)*D*c^2/(d^9*x^5 + 5*c*d^8*x^4 + 10 
*c^2*d^7*x^3 + 10*c^3*d^6*x^2 + 5*c^4*d^5*x + c^5*d^4) + 2/9*sqrt(-d^2*x^2 
 + c^2)*C*c^3/(d^8*x^5 + 5*c*d^7*x^4 + 10*c^2*d^6*x^3 + 10*c^3*d^5*x^2 + 5 
*c^4*d^4*x + c^5*d^3) + 82/63*sqrt(-d^2*x^2 + c^2)*D*c^3/(d^8*x^4 + 4*c*d^ 
7*x^3 + 6*c^2*d^6*x^2 + 4*c^3*d^5*x + c^4*d^4) + 1/105*sqrt(-d^2*x^2 + c^2 
)*D*c^3/(c*d^7*x^3 + 3*c^2*d^6*x^2 + 3*c^3*d^5*x + c^4*d^4) + 2/315*sqrt(- 
d^2*x^2 + c^2)*D*c^3/(c^2*d^6*x^2 + 2*c^3*d^5*x + c^4*d^4) + 2/315*sqrt(-d 
^2*x^2 + c^2)*D*c^3/(c^3*d^5*x + c^4*d^4) + 1/3*(-d^2*x^2 + c^2)^(3/2)*B*c 
/(d^8*x^6 + 6*c*d^7*x^5 + 15*c^2*d^6*x^4 + 20*c^3*d^5*x^3 + 15*c^4*d^4*x^2 
 + 6*c^5*d^3*x + c^6*d^2) + (-d^2*x^2 + c^2)^(3/2)*C*c/(d^8*x^5 + 5*c*d^7* 
x^4 + 10*c^2*d^6*x^3 + 10*c^3*d^5*x^2 + 5*c^4*d^4*x + c^5*d^3) + 3*(-d^2*x 
^2 + c^2)^(3/2)*D*c/(d^8*x^4 + 4*c*d^7*x^3 + 6*c^2*d^6*x^2 + 4*c^3*d^5*x + 
 c^4*d^4) - 2/9*sqrt(-d^2*x^2 + c^2)*B*c^2/(d^7*x^5 + 5*c*d^6*x^4 + 10*c^2 
*d^5*x^3 + 10*c^3*d^4*x^2 + 5*c^4*d^3*x + c^5*d^2) - 55/63*sqrt(-d^2*x^2 + 
 c^2)*C*c^2/(d^7*x^4 + 4*c*d^6*x^3 + 6*c^2*d^5*x^2 + 4*c^3*d^4*x + c^4*...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1056 vs. \(2 (244) = 488\).

Time = 0.16 (sec) , antiderivative size = 1056, normalized size of antiderivative = 4.00 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^7} \, dx=\text {Too large to display} \] Input:

integrate((-d^2*x^2+c^2)^(3/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^7,x, algorithm= 
"giac")
 

Output:

D*arcsin(d*x/c)*sgn(c)*sgn(d)/(d^3*abs(d)) - 2/315*(494*D*c^3 - 2*C*c^2*d 
- 7*B*c*d^2 - 47*A*d^3 - 63*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))*B*c/x + 41 
31*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))*D*c^3/(d^2*x) - 18*(c*d + sqrt(-d^2 
*x^2 + c^2)*abs(d))*C*c^2/(d*x) - 108*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))* 
A*d/x + 14949*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^2*D*c^3/(d^4*x^2) - 72*( 
c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^2*C*c^2/(d^3*x^2) + 63*(c*d + sqrt(-d^2 
*x^2 + c^2)*abs(d))^2*B*c/(d^2*x^2) - 1062*(c*d + sqrt(-d^2*x^2 + c^2)*abs 
(d))^2*A/(d*x^2) + 30261*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^3*D*c^3/(d^6* 
x^3) + 252*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^3*C*c^2/(d^5*x^3) - 693*(c* 
d + sqrt(-d^2*x^2 + c^2)*abs(d))^3*B*c/(d^4*x^3) - 1638*(c*d + sqrt(-d^2*x 
^2 + c^2)*abs(d))^3*A/(d^3*x^3) + 37359*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d) 
)^4*D*c^3/(d^8*x^4) - 882*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^4*C*c^2/(d^7 
*x^4) + 63*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^4*B*c/(d^6*x^4) - 3402*(c*d 
 + sqrt(-d^2*x^2 + c^2)*abs(d))^4*A/(d^5*x^4) + 24885*(c*d + sqrt(-d^2*x^2 
 + c^2)*abs(d))^5*D*c^3/(d^10*x^5) + 630*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d 
))^5*C*c^2/(d^9*x^5) - 945*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^5*B*c/(d^8* 
x^5) - 2520*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^5*A/(d^7*x^5) + 11235*(c*d 
 + sqrt(-d^2*x^2 + c^2)*abs(d))^6*D*c^3/(d^12*x^6) - 420*(c*d + sqrt(-d^2* 
x^2 + c^2)*abs(d))^6*C*c^2/(d^11*x^6) + 105*(c*d + sqrt(-d^2*x^2 + c^2)*ab 
s(d))^6*B*c/(d^10*x^6) - 2310*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^6*A/(...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^7} \, dx=\int \frac {{\left (c^2-d^2\,x^2\right )}^{3/2}\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{{\left (c+d\,x\right )}^7} \,d x \] Input:

int(((c^2 - d^2*x^2)^(3/2)*(A + B*x + C*x^2 + x^3*D))/(c + d*x)^7,x)
 

Output:

int(((c^2 - d^2*x^2)^(3/2)*(A + B*x + C*x^2 + x^3*D))/(c + d*x)^7, x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 881, normalized size of antiderivative = 3.34 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^7} \, dx =\text {Too large to display} \] Input:

int((-d^2*x^2+c^2)^(3/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^7,x)
 

Output:

(315*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*c**7 + 1260*sqrt(c**2 - d**2*x** 
2)*asin((d*x)/c)*c**6*d*x + 1890*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*c**5 
*d**2*x**2 + 1260*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*c**4*d**3*x**3 + 31 
5*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*c**3*d**4*x**4 - 315*asin((d*x)/c)* 
c**8 - 1575*asin((d*x)/c)*c**7*d*x - 3150*asin((d*x)/c)*c**6*d**2*x**2 - 3 
150*asin((d*x)/c)*c**5*d**3*x**3 - 1575*asin((d*x)/c)*c**4*d**4*x**4 - 315 
*asin((d*x)/c)*c**3*d**5*x**5 + 70*sqrt(c**2 - d**2*x**2)*a*c**4*d**2 + 12 
*sqrt(c**2 - d**2*x**2)*a*c**3*d**3*x + 159*sqrt(c**2 - d**2*x**2)*a*c**2* 
d**4*x**2 + 102*sqrt(c**2 - d**2*x**2)*a*c*d**5*x**3 + 25*sqrt(c**2 - d**2 
*x**2)*a*d**6*x**4 + 7*sqrt(c**2 - d**2*x**2)*b*c**4*d**2*x - 126*sqrt(c** 
2 - d**2*x**2)*b*c**3*d**3*x**2 + 7*sqrt(c**2 - d**2*x**2)*b*c**2*d**4*x** 
3 - 70*sqrt(c**2 - d**2*x**2)*c**7 - 457*sqrt(c**2 - d**2*x**2)*c**6*d*x - 
 1059*sqrt(c**2 - d**2*x**2)*c**5*d**2*x**2 - 1207*sqrt(c**2 - d**2*x**2)* 
c**4*d**3*x**3 - 535*sqrt(c**2 - d**2*x**2)*c**3*d**4*x**4 - 70*a*c**5*d** 
2 + 12*a*c**4*d**3*x - 331*a*c**3*d**4*x**2 - 219*a*c**2*d**5*x**3 - 107*a 
*c*d**6*x**4 - 21*a*d**7*x**5 + 7*b*c**5*d**2*x + 189*b*c**4*d**3*x**2 - 4 
9*b*c**3*d**4*x**3 + 63*b*c**2*d**5*x**4 + 14*b*c*d**6*x**5 + 70*c**8 - 45 
7*c**7*d*x - 2774*c**6*d**2*x**2 - 4916*c**5*d**3*x**3 - 4048*c**4*d**4*x* 
*4 - 1379*c**3*d**5*x**5)/(315*c**3*d**3*(sqrt(c**2 - d**2*x**2)*c**4 + 4* 
sqrt(c**2 - d**2*x**2)*c**3*d*x + 6*sqrt(c**2 - d**2*x**2)*c**2*d**2*x*...