\(\int \frac {A+B x+C x^2+D x^3}{(c+d x) (c^2-d^2 x^2)^{5/2}} \, dx\) [190]

Optimal result

Integrand size = 39, antiderivative size = 183 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x) \left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {c^2 C d-B c d^2+A d^3-c^3 D}{5 c d^4 (c+d x) \left (c^2-d^2 x^2\right )^{3/2}}+\frac {5 c^3 (C d-c D)-d \left (c^2 C d-B c d^2-4 A d^3-6 c^3 D\right ) x}{15 c^3 d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\left (2 c^2 C d-2 B c d^2-8 A d^3+3 c^3 D\right ) x}{15 c^5 d^3 \sqrt {c^2-d^2 x^2}} \] Output:

-1/5*(A*d^3-B*c*d^2+C*c^2*d-D*c^3)/c/d^4/(d*x+c)/(-d^2*x^2+c^2)^(3/2)+1/15 
*(5*c^3*(C*d-D*c)-d*(-4*A*d^3-B*c*d^2+C*c^2*d-6*D*c^3)*x)/c^3/d^4/(-d^2*x^ 
2+c^2)^(3/2)-1/15*(-8*A*d^3-2*B*c*d^2+2*C*c^2*d+3*D*c^3)*x/c^5/d^3/(-d^2*x 
^2+c^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.97 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.99 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x) \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {c^2-d^2 x^2} \left (-2 c^7 D-8 A d^7 x^4-2 c d^6 x^3 (4 A+B x)+2 c^6 d (C-D x)+2 c^2 d^5 x^2 (6 A+x (-B+C x))+c^5 d^2 (3 B+x (2 C+3 D x))-3 c^4 d^3 \left (A-x \left (B-C x+D x^2\right )\right )+c^3 d^4 x \left (12 A+x \left (3 B+2 C x+3 D x^2\right )\right )\right )}{15 c^5 d^4 (c-d x)^2 (c+d x)^3} \] Input:

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

Output:

(Sqrt[c^2 - d^2*x^2]*(-2*c^7*D - 8*A*d^7*x^4 - 2*c*d^6*x^3*(4*A + B*x) + 2 
*c^6*d*(C - D*x) + 2*c^2*d^5*x^2*(6*A + x*(-B + C*x)) + c^5*d^2*(3*B + x*( 
2*C + 3*D*x)) - 3*c^4*d^3*(A - x*(B - C*x + D*x^2)) + c^3*d^4*x*(12*A + x* 
(3*B + 2*C*x + 3*D*x^2))))/(15*c^5*d^4*(c - d*x)^2*(c + d*x)^3)
 

Rubi [A] (verified)

Time = 0.98 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.17, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2170, 2170, 27, 671, 209, 208}

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.

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

\(\Big \downarrow \) 2170

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

\(\Big \downarrow \) 2170

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 671

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

\(\Big \downarrow \) 209

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

\(\Big \downarrow \) 208

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

Input:

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

Output:

(D*(c + d*x))/(2*d^4*(c^2 - d^2*x^2)^(3/2)) + ((d*(2*C*d - 5*c*D))/(3*(c^2 
 - d^2*x^2)^(3/2)) + d^2*((-2*(c^2*C*d - B*c*d^2 + A*d^3 - c^3*D))/(5*c*d* 
(c + d*x)*(c^2 - d^2*x^2)^(3/2)) - ((2*c^2*C*d - 2*B*c*d^2 - 8*A*d^3 + 3*c 
^3*D)*(x/(3*c^2*(c^2 - d^2*x^2)^(3/2)) + (2*x)/(3*c^4*Sqrt[c^2 - d^2*x^2]) 
))/(5*c)))/(2*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 208

Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), 
x] /; FreeQ[{a, b}, x]
 

rule 209

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) 
/(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1))   Int[(a + b*x^2)^(p + 1 
), x], x] /; FreeQ[{a, b}, x] && ILtQ[p + 3/2, 0]
 

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 [A] (verified)

Time = 0.44 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.29

method	result	size
gosper	\(-\frac {\left (-d x +c \right ) \left (8 A \,d^{7} x^{4}+2 B c \,d^{6} x^{4}-2 C \,c^{2} d^{5} x^{4}-3 D c^{3} d^{4} x^{4}+8 A c \,d^{6} x^{3}+2 B \,c^{2} d^{5} x^{3}-2 C \,c^{3} d^{4} x^{3}-3 D c^{4} d^{3} x^{3}-12 A \,c^{2} d^{5} x^{2}-3 B \,c^{3} d^{4} x^{2}+3 C \,c^{4} d^{3} x^{2}-3 D c^{5} d^{2} x^{2}-12 A \,c^{3} d^{4} x -3 B \,c^{4} d^{3} x -2 C \,c^{5} d^{2} x +2 D c^{6} d x +3 A \,c^{4} d^{3}-3 B \,c^{5} d^{2}-2 C \,c^{6} d +2 D c^{7}\right )}{15 c^{5} d^{4} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}\)	\(236\)
orering	\(-\frac {\left (-d x +c \right ) \left (8 A \,d^{7} x^{4}+2 B c \,d^{6} x^{4}-2 C \,c^{2} d^{5} x^{4}-3 D c^{3} d^{4} x^{4}+8 A c \,d^{6} x^{3}+2 B \,c^{2} d^{5} x^{3}-2 C \,c^{3} d^{4} x^{3}-3 D c^{4} d^{3} x^{3}-12 A \,c^{2} d^{5} x^{2}-3 B \,c^{3} d^{4} x^{2}+3 C \,c^{4} d^{3} x^{2}-3 D c^{5} d^{2} x^{2}-12 A \,c^{3} d^{4} x -3 B \,c^{4} d^{3} x -2 C \,c^{5} d^{2} x +2 D c^{6} d x +3 A \,c^{4} d^{3}-3 B \,c^{5} d^{2}-2 C \,c^{6} d +2 D c^{7}\right )}{15 c^{5} d^{4} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}\)	\(236\)
trager	\(-\frac {\left (8 A \,d^{7} x^{4}+2 B c \,d^{6} x^{4}-2 C \,c^{2} d^{5} x^{4}-3 D c^{3} d^{4} x^{4}+8 A c \,d^{6} x^{3}+2 B \,c^{2} d^{5} x^{3}-2 C \,c^{3} d^{4} x^{3}-3 D c^{4} d^{3} x^{3}-12 A \,c^{2} d^{5} x^{2}-3 B \,c^{3} d^{4} x^{2}+3 C \,c^{4} d^{3} x^{2}-3 D c^{5} d^{2} x^{2}-12 A \,c^{3} d^{4} x -3 B \,c^{4} d^{3} x -2 C \,c^{5} d^{2} x +2 D c^{6} d x +3 A \,c^{4} d^{3}-3 B \,c^{5} d^{2}-2 C \,c^{6} d +2 D c^{7}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{15 d^{4} c^{5} \left (d x +c \right )^{3} \left (-d x +c \right )^{2}}\)	\(245\)
default	\(\frac {B \,d^{2} \left (\frac {x}{3 c^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{4} \sqrt {-d^{2} x^{2}+c^{2}}}\right )+D c^{2} \left (\frac {x}{3 c^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{4} \sqrt {-d^{2} x^{2}+c^{2}}}\right )+\frac {C d -D c}{3 d \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+D d^{2} \left (\frac {x}{2 d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}-\frac {c^{2} \left (\frac {x}{3 c^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{4} \sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 d^{2}}\right )-C c d \left (\frac {x}{3 c^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{4} \sqrt {-d^{2} x^{2}+c^{2}}}\right )}{d^{3}}+\frac {\left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right ) \left (-\frac {1}{5 c d \left (x +\frac {c}{d}\right ) \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}+\frac {4 d \left (-\frac {-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d}{6 c^{2} d^{2} \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}-\frac {-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d}{3 d^{2} c^{4} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{5 c}\right )}{d^{4}}\)	\(435\)

Input:

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

Output:

-1/15*(-d*x+c)*(8*A*d^7*x^4+2*B*c*d^6*x^4-2*C*c^2*d^5*x^4-3*D*c^3*d^4*x^4+ 
8*A*c*d^6*x^3+2*B*c^2*d^5*x^3-2*C*c^3*d^4*x^3-3*D*c^4*d^3*x^3-12*A*c^2*d^5 
*x^2-3*B*c^3*d^4*x^2+3*C*c^4*d^3*x^2-3*D*c^5*d^2*x^2-12*A*c^3*d^4*x-3*B*c^ 
4*d^3*x-2*C*c^5*d^2*x+2*D*c^6*d*x+3*A*c^4*d^3-3*B*c^5*d^2-2*C*c^6*d+2*D*c^ 
7)/c^5/d^4/(-d^2*x^2+c^2)^(5/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 491 vs. \(2 (171) = 342\).

Time = 0.12 (sec) , antiderivative size = 491, normalized size of antiderivative = 2.68 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x) \left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {2 \, D c^{8} - 2 \, C c^{7} d - 3 \, B c^{6} d^{2} + 3 \, A c^{5} d^{3} + {\left (2 \, D c^{3} d^{5} - 2 \, C c^{2} d^{6} - 3 \, B c d^{7} + 3 \, A d^{8}\right )} x^{5} + {\left (2 \, D c^{4} d^{4} - 2 \, C c^{3} d^{5} - 3 \, B c^{2} d^{6} + 3 \, A c d^{7}\right )} x^{4} - 2 \, {\left (2 \, D c^{5} d^{3} - 2 \, C c^{4} d^{4} - 3 \, B c^{3} d^{5} + 3 \, A c^{2} d^{6}\right )} x^{3} - 2 \, {\left (2 \, D c^{6} d^{2} - 2 \, C c^{5} d^{3} - 3 \, B c^{4} d^{4} + 3 \, A c^{3} d^{5}\right )} x^{2} + {\left (2 \, D c^{7} d - 2 \, C c^{6} d^{2} - 3 \, B c^{5} d^{3} + 3 \, A c^{4} d^{4}\right )} x + {\left (2 \, D c^{7} - 2 \, C c^{6} d - 3 \, B c^{5} d^{2} + 3 \, A c^{4} d^{3} - {\left (3 \, D c^{3} d^{4} + 2 \, C c^{2} d^{5} - 2 \, B c d^{6} - 8 \, A d^{7}\right )} x^{4} - {\left (3 \, D c^{4} d^{3} + 2 \, C c^{3} d^{4} - 2 \, B c^{2} d^{5} - 8 \, A c d^{6}\right )} x^{3} - 3 \, {\left (D c^{5} d^{2} - C c^{4} d^{3} + B c^{3} d^{4} + 4 \, A c^{2} d^{5}\right )} x^{2} + {\left (2 \, D c^{6} d - 2 \, C c^{5} d^{2} - 3 \, B c^{4} d^{3} - 12 \, A c^{3} d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{15 \, {\left (c^{5} d^{9} x^{5} + c^{6} d^{8} x^{4} - 2 \, c^{7} d^{7} x^{3} - 2 \, c^{8} d^{6} x^{2} + c^{9} d^{5} x + c^{10} d^{4}\right )}} \] Input:

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

Output:

-1/15*(2*D*c^8 - 2*C*c^7*d - 3*B*c^6*d^2 + 3*A*c^5*d^3 + (2*D*c^3*d^5 - 2* 
C*c^2*d^6 - 3*B*c*d^7 + 3*A*d^8)*x^5 + (2*D*c^4*d^4 - 2*C*c^3*d^5 - 3*B*c^ 
2*d^6 + 3*A*c*d^7)*x^4 - 2*(2*D*c^5*d^3 - 2*C*c^4*d^4 - 3*B*c^3*d^5 + 3*A* 
c^2*d^6)*x^3 - 2*(2*D*c^6*d^2 - 2*C*c^5*d^3 - 3*B*c^4*d^4 + 3*A*c^3*d^5)*x 
^2 + (2*D*c^7*d - 2*C*c^6*d^2 - 3*B*c^5*d^3 + 3*A*c^4*d^4)*x + (2*D*c^7 - 
2*C*c^6*d - 3*B*c^5*d^2 + 3*A*c^4*d^3 - (3*D*c^3*d^4 + 2*C*c^2*d^5 - 2*B*c 
*d^6 - 8*A*d^7)*x^4 - (3*D*c^4*d^3 + 2*C*c^3*d^4 - 2*B*c^2*d^5 - 8*A*c*d^6 
)*x^3 - 3*(D*c^5*d^2 - C*c^4*d^3 + B*c^3*d^4 + 4*A*c^2*d^5)*x^2 + (2*D*c^6 
*d - 2*C*c^5*d^2 - 3*B*c^4*d^3 - 12*A*c^3*d^4)*x)*sqrt(-d^2*x^2 + c^2))/(c 
^5*d^9*x^5 + c^6*d^8*x^4 - 2*c^7*d^7*x^3 - 2*c^8*d^6*x^2 + c^9*d^5*x + c^1 
0*d^4)
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 418 vs. \(2 (171) = 342\).

Time = 0.07 (sec) , antiderivative size = 418, normalized size of antiderivative = 2.28 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x) \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {D c^{3}}{5 \, {\left ({\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c d^{5} x + {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c^{2} d^{4}\right )}} - \frac {C c^{2}}{5 \, {\left ({\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c d^{4} x + {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c^{2} d^{3}\right )}} + \frac {B c}{5 \, {\left ({\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c d^{3} x + {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c^{2} d^{2}\right )}} - \frac {A}{5 \, {\left ({\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c d^{2} x + {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c^{2} d\right )}} + \frac {4 \, A x}{15 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c^{3}} + \frac {2 \, D x}{5 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{3}} - \frac {C x}{15 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c d^{2}} + \frac {B x}{15 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c^{2} d} - \frac {D c}{3 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{4}} + \frac {C}{3 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{3}} + \frac {8 \, A x}{15 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{5}} - \frac {D x}{5 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{2} d^{3}} - \frac {2 \, C x}{15 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{3} d^{2}} + \frac {2 \, B x}{15 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{4} d} \] Input:

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

Output:

1/5*D*c^3/((-d^2*x^2 + c^2)^(3/2)*c*d^5*x + (-d^2*x^2 + c^2)^(3/2)*c^2*d^4 
) - 1/5*C*c^2/((-d^2*x^2 + c^2)^(3/2)*c*d^4*x + (-d^2*x^2 + c^2)^(3/2)*c^2 
*d^3) + 1/5*B*c/((-d^2*x^2 + c^2)^(3/2)*c*d^3*x + (-d^2*x^2 + c^2)^(3/2)*c 
^2*d^2) - 1/5*A/((-d^2*x^2 + c^2)^(3/2)*c*d^2*x + (-d^2*x^2 + c^2)^(3/2)*c 
^2*d) + 4/15*A*x/((-d^2*x^2 + c^2)^(3/2)*c^3) + 2/5*D*x/((-d^2*x^2 + c^2)^ 
(3/2)*d^3) - 1/15*C*x/((-d^2*x^2 + c^2)^(3/2)*c*d^2) + 1/15*B*x/((-d^2*x^2 
 + c^2)^(3/2)*c^2*d) - 1/3*D*c/((-d^2*x^2 + c^2)^(3/2)*d^4) + 1/3*C/((-d^2 
*x^2 + c^2)^(3/2)*d^3) + 8/15*A*x/(sqrt(-d^2*x^2 + c^2)*c^5) - 1/5*D*x/(sq 
rt(-d^2*x^2 + c^2)*c^2*d^3) - 2/15*C*x/(sqrt(-d^2*x^2 + c^2)*c^3*d^2) + 2/ 
15*B*x/(sqrt(-d^2*x^2 + c^2)*c^4*d)
                                                                                    
                                                                                    
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 347, normalized size of antiderivative = 1.90 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x) \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {12 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{3} d +12 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{2} d^{2} x -12 \sqrt {-d^{2} x^{2}+c^{2}}\, a c \,d^{3} x^{2}-12 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{4} x^{3}+3 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{4}+3 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{3} d x -3 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2} d^{2} x^{2}-3 \sqrt {-d^{2} x^{2}+c^{2}}\, b c \,d^{3} x^{3}-3 a \,c^{4} d +12 a \,c^{3} d^{2} x +12 a \,c^{2} d^{3} x^{2}-8 a c \,d^{4} x^{3}-8 a \,d^{5} x^{4}+3 b \,c^{5}+3 b \,c^{4} d x +3 b \,c^{3} d^{2} x^{2}-2 b \,c^{2} d^{3} x^{3}-2 b c \,d^{4} x^{4}+5 c^{4} d^{2} x^{3}+5 c^{3} d^{3} x^{4}}{15 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{5} d^{2} \left (-d^{3} x^{3}-c \,d^{2} x^{2}+c^{2} d x +c^{3}\right )} \] Input:

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

Output:

(12*sqrt(c**2 - d**2*x**2)*a*c**3*d + 12*sqrt(c**2 - d**2*x**2)*a*c**2*d** 
2*x - 12*sqrt(c**2 - d**2*x**2)*a*c*d**3*x**2 - 12*sqrt(c**2 - d**2*x**2)* 
a*d**4*x**3 + 3*sqrt(c**2 - d**2*x**2)*b*c**4 + 3*sqrt(c**2 - d**2*x**2)*b 
*c**3*d*x - 3*sqrt(c**2 - d**2*x**2)*b*c**2*d**2*x**2 - 3*sqrt(c**2 - d**2 
*x**2)*b*c*d**3*x**3 - 3*a*c**4*d + 12*a*c**3*d**2*x + 12*a*c**2*d**3*x**2 
 - 8*a*c*d**4*x**3 - 8*a*d**5*x**4 + 3*b*c**5 + 3*b*c**4*d*x + 3*b*c**3*d* 
*2*x**2 - 2*b*c**2*d**3*x**3 - 2*b*c*d**4*x**4 + 5*c**4*d**2*x**3 + 5*c**3 
*d**3*x**4)/(15*sqrt(c**2 - d**2*x**2)*c**5*d**2*(c**3 + c**2*d*x - c*d**2 
*x**2 - d**3*x**3))