\(\int \frac {(c+d x) (A+B x+C x^2+D x^3)}{(c^2-d^2 x^2)^{5/2}} \, dx\) [188]

Optimal result

Integrand size = 37, antiderivative size = 147 \[ \int \frac {(c+d x) \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {\left (c^2 C d+B c d^2+A d^3+c^3 D\right ) (c+d x)}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {3 c^3 (C d+c D)+d \left (c^2 C d+B c d^2-2 A d^3+4 c^3 D\right ) x}{3 c^3 d^4 \sqrt {c^2-d^2 x^2}}+\frac {D \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{d^4} \] Output:

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

Mathematica [A] (verified)

Time = 0.93 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.06 \[ \int \frac {(c+d x) \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {\frac {\sqrt {c^2-d^2 x^2} \left (-2 c^5 D-2 A d^5 x^2+c d^4 x (2 A+B x)-c^4 d (2 C+D x)+c^2 d^3 (A+x (-B+C x))+c^3 d^2 (B+2 x (C+2 D x))\right )}{c^3 (c-d x)^2 (c+d x)}-6 D \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{3 d^4} \] Input:

Integrate[((c + d*x)*(A + B*x + C*x^2 + D*x^3))/(c^2 - d^2*x^2)^(5/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {2166, 2345, 27, 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.

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

\(\Big \downarrow \) 2166

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

\(\Big \downarrow \) 2345

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 216

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

Input:

Int[((c + d*x)*(A + B*x + C*x^2 + D*x^3))/(c^2 - d^2*x^2)^(5/2),x]
 

Output:

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

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 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 2166

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{Qx = PolynomialQuotient[Pq, a*e + b*d*x, x], R = PolynomialRemainde 
r[Pq, a*e + b*d*x, x]}, Simp[(-d)*R*(d + e*x)^m*((a + b*x^2)^(p + 1)/(2*a*e 
*(p + 1))), x] + Simp[d/(2*a*(p + 1))   Int[(d + e*x)^(m - 1)*(a + b*x^2)^( 
p + 1)*ExpandToSum[2*a*e*(p + 1)*Qx + R*(m + 2*p + 2), x], x], x]] /; FreeQ 
[{a, b, d, e}, x] && PolyQ[Pq, x] && EqQ[b*d^2 + a*e^2, 0] && ILtQ[p + 1/2, 
 0] && GtQ[m, 0]
 

rule 2345

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot 
ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 
 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b 
*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   In 
t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] 
/; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(284\) vs. \(2(137)=274\).

Time = 0.38 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.94

method	result	size
default	\(A c \left (\frac {x}{3 c^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{4} \sqrt {-d^{2} x^{2}+c^{2}}}\right )+\frac {A d +B c}{3 d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+\left (B d +C c \right ) \left (\frac {x}{2 d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}-\frac {c^{2} \left (\frac {x}{3 c^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{4} \sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 d^{2}}\right )+\left (C d +D c \right ) \left (\frac {x^{2}}{d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}-\frac {2 c^{2}}{3 d^{4} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}\right )+D d \left (\frac {x^{3}}{3 d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}-\frac {\frac {x}{\sqrt {-d^{2} x^{2}+c^{2}}\, d^{2}}-\frac {\arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{d^{2} \sqrt {d^{2}}}}{d^{2}}\right )\)	\(285\)

Input:

int((d*x+c)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(5/2),x,method=_RETURNVERBO 
SE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 369 vs. \(2 (137) = 274\).

Time = 0.10 (sec) , antiderivative size = 369, normalized size of antiderivative = 2.51 \[ \int \frac {(c+d x) \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {2 \, D c^{6} + 2 \, C c^{5} d - B c^{4} d^{2} - A c^{3} d^{3} + {\left (2 \, D c^{3} d^{3} + 2 \, C c^{2} d^{4} - B c d^{5} - A d^{6}\right )} x^{3} - {\left (2 \, D c^{4} d^{2} + 2 \, C c^{3} d^{3} - B c^{2} d^{4} - A c d^{5}\right )} x^{2} - {\left (2 \, D c^{5} d + 2 \, C c^{4} d^{2} - B c^{3} d^{3} - A c^{2} d^{4}\right )} x + 6 \, {\left (D c^{3} d^{3} x^{3} - D c^{4} d^{2} x^{2} - D c^{5} d x + D c^{6}\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) + {\left (2 \, D c^{5} + 2 \, C c^{4} d - B c^{3} d^{2} - A c^{2} d^{3} - {\left (4 \, D c^{3} d^{2} + C c^{2} d^{3} + B c d^{4} - 2 \, A d^{5}\right )} x^{2} + {\left (D c^{4} d - 2 \, C c^{3} d^{2} + B c^{2} d^{3} - 2 \, A c d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{3 \, {\left (c^{3} d^{7} x^{3} - c^{4} d^{6} x^{2} - c^{5} d^{5} x + c^{6} d^{4}\right )}} \] Input:

integrate((d*x+c)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(5/2),x, algorithm="f 
ricas")
 

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 9.04 (sec) , antiderivative size = 1550, normalized size of antiderivative = 10.54 \[ \int \frac {(c+d x) \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate((d*x+c)*(D*x**3+C*x**2+B*x+A)/(-d**2*x**2+c**2)**(5/2),x)
 

Output:

A*c*Piecewise((3*I*c**2*x/(-3*c**7*sqrt(-1 + d**2*x**2/c**2) + 3*c**5*d**2 
*x**2*sqrt(-1 + d**2*x**2/c**2)) - 2*I*d**2*x**3/(-3*c**7*sqrt(-1 + d**2*x 
**2/c**2) + 3*c**5*d**2*x**2*sqrt(-1 + d**2*x**2/c**2)), Abs(d**2*x**2/c** 
2) > 1), (-3*c**2*x/(-3*c**7*sqrt(1 - d**2*x**2/c**2) + 3*c**5*d**2*x**2*s 
qrt(1 - d**2*x**2/c**2)) + 2*d**2*x**3/(-3*c**7*sqrt(1 - d**2*x**2/c**2) + 
 3*c**5*d**2*x**2*sqrt(1 - d**2*x**2/c**2)), True)) + A*d*Piecewise((-1/(- 
3*c**2*d**2*sqrt(c**2 - d**2*x**2) + 3*d**4*x**2*sqrt(c**2 - d**2*x**2)), 
Ne(d, 0)), (x**2/(2*(c**2)**(5/2)), True)) + B*c*Piecewise((-1/(-3*c**2*d* 
*2*sqrt(c**2 - d**2*x**2) + 3*d**4*x**2*sqrt(c**2 - d**2*x**2)), Ne(d, 0)) 
, (x**2/(2*(c**2)**(5/2)), True)) + B*d*Piecewise((I*x**3/(-3*c**5*sqrt(-1 
 + d**2*x**2/c**2) + 3*c**3*d**2*x**2*sqrt(-1 + d**2*x**2/c**2)), Abs(d**2 
*x**2/c**2) > 1), (-x**3/(-3*c**5*sqrt(1 - d**2*x**2/c**2) + 3*c**3*d**2*x 
**2*sqrt(1 - d**2*x**2/c**2)), True)) + C*c*Piecewise((I*x**3/(-3*c**5*sqr 
t(-1 + d**2*x**2/c**2) + 3*c**3*d**2*x**2*sqrt(-1 + d**2*x**2/c**2)), Abs( 
d**2*x**2/c**2) > 1), (-x**3/(-3*c**5*sqrt(1 - d**2*x**2/c**2) + 3*c**3*d* 
*2*x**2*sqrt(1 - d**2*x**2/c**2)), True)) + C*d*Piecewise((2*c**2/(-3*c**2 
*d**4*sqrt(c**2 - d**2*x**2) + 3*d**6*x**2*sqrt(c**2 - d**2*x**2)) - 3*d** 
2*x**2/(-3*c**2*d**4*sqrt(c**2 - d**2*x**2) + 3*d**6*x**2*sqrt(c**2 - d**2 
*x**2)), Ne(d, 0)), (x**4/(4*(c**2)**(5/2)), True)) + D*c*Piecewise((2*c** 
2/(-3*c**2*d**4*sqrt(c**2 - d**2*x**2) + 3*d**6*x**2*sqrt(c**2 - d**2*x...
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 281 vs. \(2 (137) = 274\).

Time = 0.12 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.91 \[ \int \frac {(c+d x) \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {1}{3} \, D d x {\left (\frac {3 \, x^{2}}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{2}} - \frac {2 \, c^{2}}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{4}}\right )} + \frac {A x}{3 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c} + \frac {{\left (D c + C d\right )} x^{2}}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{2}} + \frac {B c}{3 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{2}} + \frac {A}{3 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d} + \frac {2 \, A x}{3 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{3}} - \frac {D x}{3 \, \sqrt {-d^{2} x^{2} + c^{2}} d^{3}} + \frac {{\left (C c + B d\right )} x}{3 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{2}} + \frac {D \arcsin \left (\frac {d x}{c}\right )}{d^{4}} - \frac {2 \, {\left (D c + C d\right )} c^{2}}{3 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{4}} - \frac {{\left (C c + B d\right )} x}{3 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{2} d^{2}} \] Input:

integrate((d*x+c)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(5/2),x, algorithm="m 
axima")
 

Output:

1/3*D*d*x*(3*x^2/((-d^2*x^2 + c^2)^(3/2)*d^2) - 2*c^2/((-d^2*x^2 + c^2)^(3 
/2)*d^4)) + 1/3*A*x/((-d^2*x^2 + c^2)^(3/2)*c) + (D*c + C*d)*x^2/((-d^2*x^ 
2 + c^2)^(3/2)*d^2) + 1/3*B*c/((-d^2*x^2 + c^2)^(3/2)*d^2) + 1/3*A/((-d^2* 
x^2 + c^2)^(3/2)*d) + 2/3*A*x/(sqrt(-d^2*x^2 + c^2)*c^3) - 1/3*D*x/(sqrt(- 
d^2*x^2 + c^2)*d^3) + 1/3*(C*c + B*d)*x/((-d^2*x^2 + c^2)^(3/2)*d^2) + D*a 
rcsin(d*x/c)/d^4 - 2/3*(D*c + C*d)*c^2/((-d^2*x^2 + c^2)^(3/2)*d^4) - 1/3* 
(C*c + B*d)*x/(sqrt(-d^2*x^2 + c^2)*c^2*d^2)
 

Giac [F]

\[ \int \frac {(c+d x) \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\int { \frac {{\left (D x^{3} + C x^{2} + B x + A\right )} {\left (d x + c\right )}}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((d*x+c)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(5/2),x, algorithm="g 
iac")
 

Output:

integrate((D*x^3 + C*x^2 + B*x + A)*(d*x + c)/(-d^2*x^2 + c^2)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x) \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\int \frac {\left (c+d\,x\right )\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{{\left (c^2-d^2\,x^2\right )}^{5/2}} \,d x \] Input:

int(((c + d*x)*(A + B*x + C*x^2 + x^3*D))/(c^2 - d^2*x^2)^(5/2),x)
 

Output:

int(((c + d*x)*(A + B*x + C*x^2 + x^3*D))/(c^2 - d^2*x^2)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 577, normalized size of antiderivative = 3.93 \[ \int \frac {(c+d x) \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

int((d*x+c)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(5/2),x)
 

Output:

(3*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*tan(asin((d*x)/c)/2)**4*c**3 - 6*s 
qrt(c**2 - d**2*x**2)*asin((d*x)/c)*tan(asin((d*x)/c)/2)**3*c**3 + 6*sqrt( 
c**2 - d**2*x**2)*asin((d*x)/c)*tan(asin((d*x)/c)/2)*c**3 - 3*sqrt(c**2 - 
d**2*x**2)*asin((d*x)/c)*c**3 - 3*sqrt(c**2 - d**2*x**2)*tan(asin((d*x)/c) 
/2)**4*a*d**2 + 3*sqrt(c**2 - d**2*x**2)*tan(asin((d*x)/c)/2)**4*c**3 + 6* 
sqrt(c**2 - d**2*x**2)*tan(asin((d*x)/c)/2)**2*a*d**2 - 6*sqrt(c**2 - d**2 
*x**2)*tan(asin((d*x)/c)/2)**2*c**3 - 8*sqrt(c**2 - d**2*x**2)*tan(asin((d 
*x)/c)/2)*a*d**2 - 8*sqrt(c**2 - d**2*x**2)*tan(asin((d*x)/c)/2)*b*c*d - 1 
6*sqrt(c**2 - d**2*x**2)*tan(asin((d*x)/c)/2)*c**3 + sqrt(c**2 - d**2*x**2 
)*a*d**2 + 4*sqrt(c**2 - d**2*x**2)*b*c*d + 11*sqrt(c**2 - d**2*x**2)*c**3 
 + 3*tan(asin((d*x)/c)/2)**4*b*c**2*d + 3*tan(asin((d*x)/c)/2)**4*c**4 - 6 
*tan(asin((d*x)/c)/2)**3*b*c**2*d - 6*tan(asin((d*x)/c)/2)**3*c**4 + 6*tan 
(asin((d*x)/c)/2)*b*c**2*d + 6*tan(asin((d*x)/c)/2)*c**4 - 3*b*c**2*d - 3* 
c**4)/(3*sqrt(c**2 - d**2*x**2)*c**3*d**3*(tan(asin((d*x)/c)/2)**4 - 2*tan 
(asin((d*x)/c)/2)**3 + 2*tan(asin((d*x)/c)/2) - 1))