\(\int \frac {\sqrt {c^2-d^2 x^2} (A+B x+C x^2+D x^3)}{(c+d x)^4} \, dx\) [135]

Optimal result

Integrand size = 39, antiderivative size = 209 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^4} \, dx=\frac {D \sqrt {c^2-d^2 x^2}}{d^4}-\frac {2 (C d-3 c D) \sqrt {c^2-d^2 x^2}}{d^4 (c+d x)}-\frac {\left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \left (c^2-d^2 x^2\right )^{3/2}}{5 c d^4 (c+d x)^4}+\frac {\left (9 c^2 C d-4 B c d^2-A d^3-14 c^3 D\right ) \left (c^2-d^2 x^2\right )^{3/2}}{15 c^2 d^4 (c+d x)^3}-\frac {(C d-4 c D) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{d^4} \] Output:

D*(-d^2*x^2+c^2)^(1/2)/d^4-2*(C*d-3*D*c)*(-d^2*x^2+c^2)^(1/2)/d^4/(d*x+c)- 
1/5*(A*d^3-B*c*d^2+C*c^2*d-D*c^3)*(-d^2*x^2+c^2)^(3/2)/c/d^4/(d*x+c)^4+1/1 
5*(-A*d^3-4*B*c*d^2+9*C*c^2*d-14*D*c^3)*(-d^2*x^2+c^2)^(3/2)/c^2/d^4/(d*x+ 
c)^3-(C*d-4*D*c)*arctan(d*x/(-d^2*x^2+c^2)^(1/2))/d^4
 

Mathematica [A] (verified)

Time = 1.44 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.80 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^4} \, dx=\frac {\frac {\sqrt {c^2-d^2 x^2} \left (94 c^5 D+A d^5 x^2+c d^4 x (3 A+4 B x)-6 c^4 d (4 C-37 D x)-c^3 d^2 (B+x (57 C-149 D x))-c^2 d^3 \left (4 A+3 x \left (B+13 C x-5 D x^2\right )\right )\right )}{c^2 (c+d x)^3}+30 (C d-4 c D) \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{15 d^4} \] Input:

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

Output:

((Sqrt[c^2 - d^2*x^2]*(94*c^5*D + A*d^5*x^2 + c*d^4*x*(3*A + 4*B*x) - 6*c^ 
4*d*(4*C - 37*D*x) - c^3*d^2*(B + x*(57*C - 149*D*x)) - c^2*d^3*(4*A + 3*x 
*(B + 13*C*x - 5*D*x^2))))/(c^2*(c + d*x)^3) + 30*(C*d - 4*c*D)*ArcTan[(d* 
x)/(Sqrt[c^2] - Sqrt[c^2 - d^2*x^2])])/(15*d^4)
 

Rubi [A] (verified)

Time = 1.09 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.25, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2170, 25, 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.

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

\(\Big \downarrow \) 2170

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2168

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {d \left (c^2-d^2 x^2\right )^{3/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{15 c^2 (c+d x)^3}-\frac {d \left (c^2-d^2 x^2\right )^{3/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{5 c (c+d x)^4}-d \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right ) (C d-4 c D)+\frac {d \left (c^2-d^2 x^2\right )^{3/2} \left (-B d^2-3 c^2 D+2 c C d\right )}{3 c (c+d x)^3}-\frac {2 d \sqrt {c^2-d^2 x^2} (C d-4 c D)}{c+d x}}{d^5}-\frac {D \left (c^2-d^2 x^2\right )^{3/2}}{d^4 (c+d x)^2}\)

Input:

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

Output:

-((D*(c^2 - d^2*x^2)^(3/2))/(d^4*(c + d*x)^2)) + ((-2*d*(C*d - 4*c*D)*Sqrt 
[c^2 - d^2*x^2])/(c + d*x) - (d*(c^2*C*d - B*c*d^2 + A*d^3 - c^3*D)*(c^2 - 
 d^2*x^2)^(3/2))/(5*c*(c + d*x)^4) + (d*(2*c*C*d - B*d^2 - 3*c^2*D)*(c^2 - 
 d^2*x^2)^(3/2))/(3*c*(c + d*x)^3) - (d*(c^2*C*d - B*c*d^2 + A*d^3 - c^3*D 
)*(c^2 - d^2*x^2)^(3/2))/(15*c^2*(c + d*x)^3) - d*(C*d - 4*c*D)*ArcTan[(d* 
x)/Sqrt[c^2 - d^2*x^2]])/d^5
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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]
 

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. \(395\) vs. \(2(195)=390\).

Time = 0.47 (sec) , antiderivative size = 396, normalized size of antiderivative = 1.89

method	result	size
default	\(\frac {D \left (\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}+\frac {c d \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{\sqrt {d^{2}}}\right )}{d^{4}}+\frac {\left (C d -3 D c \right ) \left (-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{c d \left (x +\frac {c}{d}\right )^{2}}-\frac {d \left (\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}+\frac {c d \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{\sqrt {d^{2}}}\right )}{c}\right )}{d^{5}}-\frac {\left (B \,d^{2}-2 C c d +3 D c^{2}\right ) \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{3 d^{7} c \left (x +\frac {c}{d}\right )^{3}}+\frac {\left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right ) \left (-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{5 c d \left (x +\frac {c}{d}\right )^{4}}-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{15 c^{2} \left (x +\frac {c}{d}\right )^{3}}\right )}{d^{7}}\)	\(396\)

Input:

int((-d^2*x^2+c^2)^(1/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^4,x,method=_RETURNVER 
BOSE)
 

Output:

D/d^4*((-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(1/2)+c*d/(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))^(3/2)-d/c*((-d^2*(x+c/d)^2+2*c*d*(x+c 
/d))^(1/2)+c*d/(d^2)^(1/2)*arctan((d^2)^(1/2)*x/(-d^2*(x+c/d)^2+2*c*d*(x+c 
/d))^(1/2))))-1/3*(B*d^2-2*C*c*d+3*D*c^2)/d^7/c/(x+c/d)^3*(-d^2*(x+c/d)^2+ 
2*c*d*(x+c/d))^(3/2)+(A*d^3-B*c*d^2+C*c^2*d-D*c^3)/d^7*(-1/5/c/d/(x+c/d)^4 
*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(3/2)-1/15/c^2/(x+c/d)^3*(-d^2*(x+c/d)^2+2 
*c*d*(x+c/d))^(3/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 427 vs. \(2 (195) = 390\).

Time = 0.12 (sec) , antiderivative size = 427, normalized size of antiderivative = 2.04 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^4} \, dx=\frac {94 \, D c^{6} - 24 \, C c^{5} d - B c^{4} d^{2} - 4 \, A c^{3} d^{3} + {\left (94 \, D c^{3} d^{3} - 24 \, C c^{2} d^{4} - B c d^{5} - 4 \, A d^{6}\right )} x^{3} + 3 \, {\left (94 \, D c^{4} d^{2} - 24 \, C c^{3} d^{3} - B c^{2} d^{4} - 4 \, A c d^{5}\right )} x^{2} + 3 \, {\left (94 \, D c^{5} d - 24 \, C c^{4} d^{2} - B c^{3} d^{3} - 4 \, A c^{2} d^{4}\right )} x - 30 \, {\left (4 \, D c^{6} - C c^{5} d + {\left (4 \, D c^{3} d^{3} - C c^{2} d^{4}\right )} x^{3} + 3 \, {\left (4 \, D c^{4} d^{2} - C c^{3} d^{3}\right )} x^{2} + 3 \, {\left (4 \, D c^{5} d - C c^{4} d^{2}\right )} x\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) + {\left (15 \, D c^{2} d^{3} x^{3} + 94 \, D c^{5} - 24 \, C c^{4} d - B c^{3} d^{2} - 4 \, A c^{2} d^{3} + {\left (149 \, D c^{3} d^{2} - 39 \, C c^{2} d^{3} + 4 \, B c d^{4} + A d^{5}\right )} x^{2} + 3 \, {\left (74 \, D c^{4} d - 19 \, C c^{3} d^{2} - B c^{2} d^{3} + A c d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{15 \, {\left (c^{2} d^{7} x^{3} + 3 \, c^{3} d^{6} x^{2} + 3 \, c^{4} d^{5} x + c^{5} d^{4}\right )}} \] Input:

integrate((-d^2*x^2+c^2)^(1/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^4,x, algorithm= 
"fricas")
 

Output:

1/15*(94*D*c^6 - 24*C*c^5*d - B*c^4*d^2 - 4*A*c^3*d^3 + (94*D*c^3*d^3 - 24 
*C*c^2*d^4 - B*c*d^5 - 4*A*d^6)*x^3 + 3*(94*D*c^4*d^2 - 24*C*c^3*d^3 - B*c 
^2*d^4 - 4*A*c*d^5)*x^2 + 3*(94*D*c^5*d - 24*C*c^4*d^2 - B*c^3*d^3 - 4*A*c 
^2*d^4)*x - 30*(4*D*c^6 - C*c^5*d + (4*D*c^3*d^3 - C*c^2*d^4)*x^3 + 3*(4*D 
*c^4*d^2 - C*c^3*d^3)*x^2 + 3*(4*D*c^5*d - C*c^4*d^2)*x)*arctan(-(c - sqrt 
(-d^2*x^2 + c^2))/(d*x)) + (15*D*c^2*d^3*x^3 + 94*D*c^5 - 24*C*c^4*d - B*c 
^3*d^2 - 4*A*c^2*d^3 + (149*D*c^3*d^2 - 39*C*c^2*d^3 + 4*B*c*d^4 + A*d^5)* 
x^2 + 3*(74*D*c^4*d - 19*C*c^3*d^2 - B*c^2*d^3 + A*c*d^4)*x)*sqrt(-d^2*x^2 
 + c^2))/(c^2*d^7*x^3 + 3*c^3*d^6*x^2 + 3*c^4*d^5*x + c^5*d^4)
 

Sympy [F]

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

integrate((-d**2*x**2+c**2)**(1/2)*(D*x**3+C*x**2+B*x+A)/(d*x+c)**4,x)
 

Output:

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

Maxima [F(-1)]

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

integrate((-d^2*x^2+c^2)^(1/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^4,x, algorithm= 
"maxima")
 

Output:

Timed out
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 613 vs. \(2 (195) = 390\).

Time = 0.14 (sec) , antiderivative size = 613, normalized size of antiderivative = 2.93 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^4} \, dx=\frac {{\left (4 \, D c - C d\right )} \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{d^{3} {\left | d \right |}} + \frac {\sqrt {-d^{2} x^{2} + c^{2}} D}{d^{4}} - \frac {2 \, {\left (79 \, D c^{3} - 24 \, C c^{2} d - B c d^{2} - 4 \, A d^{3} - \frac {5 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} B c}{x} + \frac {335 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} D c^{3}}{d^{2} x} - \frac {105 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} C c^{2}}{d x} - \frac {5 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} A d}{x} + \frac {505 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} D c^{3}}{d^{4} x^{2}} - \frac {165 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} C c^{2}}{d^{3} x^{2}} + \frac {5 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} B c}{d^{2} x^{2}} - \frac {25 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} A}{d x^{2}} + \frac {285 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{3} D c^{3}}{d^{6} x^{3}} - \frac {75 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{3} C c^{2}}{d^{5} x^{3}} - \frac {15 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{3} B c}{d^{4} x^{3}} - \frac {15 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{3} A}{d^{3} x^{3}} + \frac {60 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{4} D c^{3}}{d^{8} x^{4}} - \frac {15 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{4} C c^{2}}{d^{7} x^{4}} - \frac {15 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{4} A}{d^{5} x^{4}}\right )}}{15 \, c^{2} d^{3} {\left (\frac {c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}}{d^{2} x} + 1\right )}^{5} {\left | d \right |}} \] Input:

integrate((-d^2*x^2+c^2)^(1/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^4,x, algorithm= 
"giac")
 

Output:

(4*D*c - C*d)*arcsin(d*x/c)*sgn(c)*sgn(d)/(d^3*abs(d)) + sqrt(-d^2*x^2 + c 
^2)*D/d^4 - 2/15*(79*D*c^3 - 24*C*c^2*d - B*c*d^2 - 4*A*d^3 - 5*(c*d + sqr 
t(-d^2*x^2 + c^2)*abs(d))*B*c/x + 335*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))* 
D*c^3/(d^2*x) - 105*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))*C*c^2/(d*x) - 5*(c 
*d + sqrt(-d^2*x^2 + c^2)*abs(d))*A*d/x + 505*(c*d + sqrt(-d^2*x^2 + c^2)* 
abs(d))^2*D*c^3/(d^4*x^2) - 165*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^2*C*c^ 
2/(d^3*x^2) + 5*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^2*B*c/(d^2*x^2) - 25*( 
c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^2*A/(d*x^2) + 285*(c*d + sqrt(-d^2*x^2 
+ c^2)*abs(d))^3*D*c^3/(d^6*x^3) - 75*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^ 
3*C*c^2/(d^5*x^3) - 15*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^3*B*c/(d^4*x^3) 
 - 15*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^3*A/(d^3*x^3) + 60*(c*d + sqrt(- 
d^2*x^2 + c^2)*abs(d))^4*D*c^3/(d^8*x^4) - 15*(c*d + sqrt(-d^2*x^2 + c^2)* 
abs(d))^4*C*c^2/(d^7*x^4) - 15*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^4*A/(d^ 
5*x^4))/(c^2*d^3*((c*d + sqrt(-d^2*x^2 + c^2)*abs(d))/(d^2*x) + 1)^5*abs(d 
))
 

Mupad [F(-1)]

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

int(((c^2 - d^2*x^2)^(1/2)*(A + B*x + C*x^2 + x^3*D))/(c + d*x)^4,x)
 

Output:

int(((c^2 - d^2*x^2)^(1/2)*(A + B*x + C*x^2 + x^3*D))/(c + d*x)^4, x)
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 525, normalized size of antiderivative = 2.51 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^4} \, dx=\frac {-45 \mathit {atan} \left (\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, c^{2}-2 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{2} x^{2}}{-2 d^{3} x^{3}+2 c^{2} d x}\right ) c^{6}-135 \mathit {atan} \left (\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, c^{2}-2 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{2} x^{2}}{-2 d^{3} x^{3}+2 c^{2} d x}\right ) c^{5} d x -135 \mathit {atan} \left (\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, c^{2}-2 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{2} x^{2}}{-2 d^{3} x^{3}+2 c^{2} d x}\right ) c^{4} d^{2} x^{2}-45 \mathit {atan} \left (\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, c^{2}-2 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{2} x^{2}}{-2 d^{3} x^{3}+2 c^{2} d x}\right ) c^{3} d^{3} x^{3}-8 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{2} d^{2}+6 \sqrt {-d^{2} x^{2}+c^{2}}\, a c \,d^{3} x +2 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{4} x^{2}-2 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{3} d -6 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2} d^{2} x +8 \sqrt {-d^{2} x^{2}+c^{2}}\, b c \,d^{3} x^{2}+140 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{5}+330 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{4} d x +220 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{3} d^{2} x^{2}+30 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{2} d^{3} x^{3}}{30 c^{2} d^{3} \left (d^{3} x^{3}+3 c \,d^{2} x^{2}+3 c^{2} d x +c^{3}\right )} \] Input:

int((-d^2*x^2+c^2)^(1/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^4,x)
 

Output:

( - 45*atan((sqrt(c**2 - d**2*x**2)*c**2 - 2*sqrt(c**2 - d**2*x**2)*d**2*x 
**2)/(2*c**2*d*x - 2*d**3*x**3))*c**6 - 135*atan((sqrt(c**2 - d**2*x**2)*c 
**2 - 2*sqrt(c**2 - d**2*x**2)*d**2*x**2)/(2*c**2*d*x - 2*d**3*x**3))*c**5 
*d*x - 135*atan((sqrt(c**2 - d**2*x**2)*c**2 - 2*sqrt(c**2 - d**2*x**2)*d* 
*2*x**2)/(2*c**2*d*x - 2*d**3*x**3))*c**4*d**2*x**2 - 45*atan((sqrt(c**2 - 
 d**2*x**2)*c**2 - 2*sqrt(c**2 - d**2*x**2)*d**2*x**2)/(2*c**2*d*x - 2*d** 
3*x**3))*c**3*d**3*x**3 - 8*sqrt(c**2 - d**2*x**2)*a*c**2*d**2 + 6*sqrt(c* 
*2 - d**2*x**2)*a*c*d**3*x + 2*sqrt(c**2 - d**2*x**2)*a*d**4*x**2 - 2*sqrt 
(c**2 - d**2*x**2)*b*c**3*d - 6*sqrt(c**2 - d**2*x**2)*b*c**2*d**2*x + 8*s 
qrt(c**2 - d**2*x**2)*b*c*d**3*x**2 + 140*sqrt(c**2 - d**2*x**2)*c**5 + 33 
0*sqrt(c**2 - d**2*x**2)*c**4*d*x + 220*sqrt(c**2 - d**2*x**2)*c**3*d**2*x 
**2 + 30*sqrt(c**2 - d**2*x**2)*c**2*d**3*x**3)/(30*c**2*d**3*(c**3 + 3*c* 
*2*d*x + 3*c*d**2*x**2 + d**3*x**3))