\(\int \frac {A+B x}{(c+d x)^3 (c^2-d^2 x^2)^{3/2}} \, dx\) [54]

Optimal result

Integrand size = 29, antiderivative size = 156 \[ \int \frac {A+B x}{(c+d x)^3 \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {(3 B c+4 A d) x}{35 c^3 d \left (c^2-d^2 x^2\right )^{3/2}}-\frac {2 (3 B c+4 A d)}{35 c d^2 (c+d x) \left (c^2-d^2 x^2\right )^{3/2}}+\frac {2 (3 B c+4 A d) x}{35 c^5 d \sqrt {c^2-d^2 x^2}}+\frac {B c-A d}{7 c d^2 (c+d x)^3 \sqrt {c^2-d^2 x^2}} \] Output:

1/35*(4*A*d+3*B*c)*x/c^3/d/(-d^2*x^2+c^2)^(3/2)-2/35*(4*A*d+3*B*c)/c/d^2/( 
d*x+c)/(-d^2*x^2+c^2)^(3/2)+2/35*(4*A*d+3*B*c)*x/c^5/d/(-d^2*x^2+c^2)^(1/2 
)+1/7*(-A*d+B*c)/c/d^2/(d*x+c)^3/(-d^2*x^2+c^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.80 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.82 \[ \int \frac {A+B x}{(c+d x)^3 \left (c^2-d^2 x^2\right )^{3/2}} \, dx=-\frac {\sqrt {c^2-d^2 x^2} \left (A d \left (13 c^4+4 c^3 d x-20 c^2 d^2 x^2-24 c d^3 x^3-8 d^4 x^4\right )+B c \left (c^4+3 c^3 d x-15 c^2 d^2 x^2-18 c d^3 x^3-6 d^4 x^4\right )\right )}{35 c^5 d^2 (c-d x) (c+d x)^4} \] Input:

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

Output:

-1/35*(Sqrt[c^2 - d^2*x^2]*(A*d*(13*c^4 + 4*c^3*d*x - 20*c^2*d^2*x^2 - 24* 
c*d^3*x^3 - 8*d^4*x^4) + B*c*(c^4 + 3*c^3*d*x - 15*c^2*d^2*x^2 - 18*c*d^3* 
x^3 - 6*d^4*x^4)))/(c^5*d^2*(c - d*x)*(c + d*x)^4)
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {671, 461, 470, 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.

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

\(\Big \downarrow \) 671

\(\displaystyle \frac {(4 A d+3 B c) \int \frac {1}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2}}dx}{7 c d}+\frac {B c-A d}{7 c d^2 (c+d x)^3 \sqrt {c^2-d^2 x^2}}\)

\(\Big \downarrow \) 461

\(\displaystyle \frac {(4 A d+3 B c) \left (\frac {3 \int \frac {1}{(c+d x) \left (c^2-d^2 x^2\right )^{3/2}}dx}{5 c}-\frac {1}{5 c d (c+d x)^2 \sqrt {c^2-d^2 x^2}}\right )}{7 c d}+\frac {B c-A d}{7 c d^2 (c+d x)^3 \sqrt {c^2-d^2 x^2}}\)

\(\Big \downarrow \) 470

\(\displaystyle \frac {(4 A d+3 B c) \left (\frac {3 \left (\frac {2 \int \frac {1}{\left (c^2-d^2 x^2\right )^{3/2}}dx}{3 c}-\frac {1}{3 c d (c+d x) \sqrt {c^2-d^2 x^2}}\right )}{5 c}-\frac {1}{5 c d (c+d x)^2 \sqrt {c^2-d^2 x^2}}\right )}{7 c d}+\frac {B c-A d}{7 c d^2 (c+d x)^3 \sqrt {c^2-d^2 x^2}}\)

\(\Big \downarrow \) 208

\(\displaystyle \frac {B c-A d}{7 c d^2 (c+d x)^3 \sqrt {c^2-d^2 x^2}}+\frac {\left (\frac {3 \left (\frac {2 x}{3 c^3 \sqrt {c^2-d^2 x^2}}-\frac {1}{3 c d (c+d x) \sqrt {c^2-d^2 x^2}}\right )}{5 c}-\frac {1}{5 c d (c+d x)^2 \sqrt {c^2-d^2 x^2}}\right ) (4 A d+3 B c)}{7 c d}\)

Input:

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

Output:

(B*c - A*d)/(7*c*d^2*(c + d*x)^3*Sqrt[c^2 - d^2*x^2]) + ((3*B*c + 4*A*d)*( 
-1/5*1/(c*d*(c + d*x)^2*Sqrt[c^2 - d^2*x^2]) + (3*((2*x)/(3*c^3*Sqrt[c^2 - 
 d^2*x^2]) - 1/(3*c*d*(c + d*x)*Sqrt[c^2 - d^2*x^2])))/(5*c)))/(7*c*d)
 

Defintions of rubi rules used

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 461

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*b*c*(n + p + 1))), x] + Simp[Simpl 
ify[n + 2*p + 2]/(2*c*(n + p + 1))   Int[(c + d*x)^(n + 1)*(a + b*x^2)^p, x 
], x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && ILtQ[Simp 
lify[n + 2*p + 2], 0] && (LtQ[n, -1] || GtQ[n + p, 0])
 

rule 470

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*b*c*(n + p + 1))), x] + Simp[(n + 
2*p + 2)/(2*c*(n + p + 1))   Int[(c + d*x)^(n + 1)*(a + b*x^2)^p, x], x] /; 
 FreeQ[{a, b, c, d, p}, x] && EqQ[b*c^2 + a*d^2, 0] && LtQ[n, 0] && NeQ[n + 
 p + 1, 0] && IntegerQ[2*p]
 

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]
 

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.85

method	result	size
gosper	\(-\frac {\left (-d x +c \right ) \left (-8 A \,d^{5} x^{4}-6 B c \,d^{4} x^{4}-24 A c \,d^{4} x^{3}-18 B \,c^{2} d^{3} x^{3}-20 A \,c^{2} d^{3} x^{2}-15 B \,c^{3} d^{2} x^{2}+4 A \,c^{3} d^{2} x +3 B \,c^{4} d x +13 A \,c^{4} d +B \,c^{5}\right )}{35 \left (d x +c \right )^{2} c^{5} d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}\)	\(132\)
orering	\(-\frac {\left (-d x +c \right ) \left (-8 A \,d^{5} x^{4}-6 B c \,d^{4} x^{4}-24 A c \,d^{4} x^{3}-18 B \,c^{2} d^{3} x^{3}-20 A \,c^{2} d^{3} x^{2}-15 B \,c^{3} d^{2} x^{2}+4 A \,c^{3} d^{2} x +3 B \,c^{4} d x +13 A \,c^{4} d +B \,c^{5}\right )}{35 \left (d x +c \right )^{2} c^{5} d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}\)	\(132\)
trager	\(-\frac {\left (-8 A \,d^{5} x^{4}-6 B c \,d^{4} x^{4}-24 A c \,d^{4} x^{3}-18 B \,c^{2} d^{3} x^{3}-20 A \,c^{2} d^{3} x^{2}-15 B \,c^{3} d^{2} x^{2}+4 A \,c^{3} d^{2} x +3 B \,c^{4} d x +13 A \,c^{4} d +B \,c^{5}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{35 c^{5} \left (d x +c \right )^{4} d^{2} \left (-d x +c \right )}\)	\(134\)
default	\(\frac {B \left (-\frac {1}{5 c d \left (x +\frac {c}{d}\right )^{2} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}+\frac {3 d \left (-\frac {1}{3 c d \left (x +\frac {c}{d}\right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}-\frac {-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d}{3 d \,c^{3} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{5 c}\right )}{d^{3}}+\frac {\left (A d -B c \right ) \left (-\frac {1}{7 c d \left (x +\frac {c}{d}\right )^{3} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}+\frac {4 d \left (-\frac {1}{5 c d \left (x +\frac {c}{d}\right )^{2} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}+\frac {3 d \left (-\frac {1}{3 c d \left (x +\frac {c}{d}\right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}-\frac {-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d}{3 d \,c^{3} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{5 c}\right )}{7 c}\right )}{d^{4}}\)	\(373\)

Input:

int((B*x+A)/(d*x+c)^3/(-d^2*x^2+c^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/35*(-d*x+c)*(-8*A*d^5*x^4-6*B*c*d^4*x^4-24*A*c*d^4*x^3-18*B*c^2*d^3*x^3 
-20*A*c^2*d^3*x^2-15*B*c^3*d^2*x^2+4*A*c^3*d^2*x+3*B*c^4*d*x+13*A*c^4*d+B* 
c^5)/(d*x+c)^2/c^5/d^2/(-d^2*x^2+c^2)^(3/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 294 vs. \(2 (140) = 280\).

Time = 0.19 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.88 \[ \int \frac {A+B x}{(c+d x)^3 \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {B c^{6} + 13 \, A c^{5} d - {\left (B c d^{5} + 13 \, A d^{6}\right )} x^{5} - 3 \, {\left (B c^{2} d^{4} + 13 \, A c d^{5}\right )} x^{4} - 2 \, {\left (B c^{3} d^{3} + 13 \, A c^{2} d^{4}\right )} x^{3} + 2 \, {\left (B c^{4} d^{2} + 13 \, A c^{3} d^{3}\right )} x^{2} + 3 \, {\left (B c^{5} d + 13 \, A c^{4} d^{2}\right )} x + {\left (B c^{5} + 13 \, A c^{4} d - 2 \, {\left (3 \, B c d^{4} + 4 \, A d^{5}\right )} x^{4} - 6 \, {\left (3 \, B c^{2} d^{3} + 4 \, A c d^{4}\right )} x^{3} - 5 \, {\left (3 \, B c^{3} d^{2} + 4 \, A c^{2} d^{3}\right )} x^{2} + {\left (3 \, B c^{4} d + 4 \, A c^{3} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{35 \, {\left (c^{5} d^{7} x^{5} + 3 \, c^{6} d^{6} x^{4} + 2 \, c^{7} d^{5} x^{3} - 2 \, c^{8} d^{4} x^{2} - 3 \, c^{9} d^{3} x - c^{10} d^{2}\right )}} \] Input:

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

Output:

1/35*(B*c^6 + 13*A*c^5*d - (B*c*d^5 + 13*A*d^6)*x^5 - 3*(B*c^2*d^4 + 13*A* 
c*d^5)*x^4 - 2*(B*c^3*d^3 + 13*A*c^2*d^4)*x^3 + 2*(B*c^4*d^2 + 13*A*c^3*d^ 
3)*x^2 + 3*(B*c^5*d + 13*A*c^4*d^2)*x + (B*c^5 + 13*A*c^4*d - 2*(3*B*c*d^4 
 + 4*A*d^5)*x^4 - 6*(3*B*c^2*d^3 + 4*A*c*d^4)*x^3 - 5*(3*B*c^3*d^2 + 4*A*c 
^2*d^3)*x^2 + (3*B*c^4*d + 4*A*c^3*d^2)*x)*sqrt(-d^2*x^2 + c^2))/(c^5*d^7* 
x^5 + 3*c^6*d^6*x^4 + 2*c^7*d^5*x^3 - 2*c^8*d^4*x^2 - 3*c^9*d^3*x - c^10*d 
^2)
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 604 vs. \(2 (140) = 280\).

Time = 0.05 (sec) , antiderivative size = 604, normalized size of antiderivative = 3.87 \[ \int \frac {A+B x}{(c+d x)^3 \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {B c}{7 \, {\left (\sqrt {-d^{2} x^{2} + c^{2}} c d^{5} x^{3} + 3 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{2} d^{4} x^{2} + 3 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{3} d^{3} x + \sqrt {-d^{2} x^{2} + c^{2}} c^{4} d^{2}\right )}} + \frac {4 \, B c}{35 \, {\left (\sqrt {-d^{2} x^{2} + c^{2}} c^{2} d^{4} x^{2} + 2 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{3} d^{3} x + \sqrt {-d^{2} x^{2} + c^{2}} c^{4} d^{2}\right )}} + \frac {4 \, B c}{35 \, {\left (\sqrt {-d^{2} x^{2} + c^{2}} c^{3} d^{3} x + \sqrt {-d^{2} x^{2} + c^{2}} c^{4} d^{2}\right )}} - \frac {A}{7 \, {\left (\sqrt {-d^{2} x^{2} + c^{2}} c d^{4} x^{3} + 3 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{2} d^{3} x^{2} + 3 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{3} d^{2} x + \sqrt {-d^{2} x^{2} + c^{2}} c^{4} d\right )}} - \frac {4 \, A}{35 \, {\left (\sqrt {-d^{2} x^{2} + c^{2}} c^{2} d^{3} x^{2} + 2 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{3} d^{2} x + \sqrt {-d^{2} x^{2} + c^{2}} c^{4} d\right )}} - \frac {4 \, A}{35 \, {\left (\sqrt {-d^{2} x^{2} + c^{2}} c^{3} d^{2} x + \sqrt {-d^{2} x^{2} + c^{2}} c^{4} d\right )}} - \frac {B}{5 \, {\left (\sqrt {-d^{2} x^{2} + c^{2}} c d^{4} x^{2} + 2 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{2} d^{3} x + \sqrt {-d^{2} x^{2} + c^{2}} c^{3} d^{2}\right )}} - \frac {B}{5 \, {\left (\sqrt {-d^{2} x^{2} + c^{2}} c^{2} d^{3} x + \sqrt {-d^{2} x^{2} + c^{2}} c^{3} d^{2}\right )}} + \frac {8 \, A x}{35 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{5}} + \frac {6 \, B x}{35 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{4} d} \] Input:

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

Output:

1/7*B*c/(sqrt(-d^2*x^2 + c^2)*c*d^5*x^3 + 3*sqrt(-d^2*x^2 + c^2)*c^2*d^4*x 
^2 + 3*sqrt(-d^2*x^2 + c^2)*c^3*d^3*x + sqrt(-d^2*x^2 + c^2)*c^4*d^2) + 4/ 
35*B*c/(sqrt(-d^2*x^2 + c^2)*c^2*d^4*x^2 + 2*sqrt(-d^2*x^2 + c^2)*c^3*d^3* 
x + sqrt(-d^2*x^2 + c^2)*c^4*d^2) + 4/35*B*c/(sqrt(-d^2*x^2 + c^2)*c^3*d^3 
*x + sqrt(-d^2*x^2 + c^2)*c^4*d^2) - 1/7*A/(sqrt(-d^2*x^2 + c^2)*c*d^4*x^3 
 + 3*sqrt(-d^2*x^2 + c^2)*c^2*d^3*x^2 + 3*sqrt(-d^2*x^2 + c^2)*c^3*d^2*x + 
 sqrt(-d^2*x^2 + c^2)*c^4*d) - 4/35*A/(sqrt(-d^2*x^2 + c^2)*c^2*d^3*x^2 + 
2*sqrt(-d^2*x^2 + c^2)*c^3*d^2*x + sqrt(-d^2*x^2 + c^2)*c^4*d) - 4/35*A/(s 
qrt(-d^2*x^2 + c^2)*c^3*d^2*x + sqrt(-d^2*x^2 + c^2)*c^4*d) - 1/5*B/(sqrt( 
-d^2*x^2 + c^2)*c*d^4*x^2 + 2*sqrt(-d^2*x^2 + c^2)*c^2*d^3*x + sqrt(-d^2*x 
^2 + c^2)*c^3*d^2) - 1/5*B/(sqrt(-d^2*x^2 + c^2)*c^2*d^3*x + sqrt(-d^2*x^2 
 + c^2)*c^3*d^2) + 8/35*A*x/(sqrt(-d^2*x^2 + c^2)*c^5) + 6/35*B*x/(sqrt(-d 
^2*x^2 + c^2)*c^4*d)
 

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 9.99 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.56 \[ \int \frac {A+B x}{(c+d x)^3 \left (c^2-d^2 x^2\right )^{3/2}} \, dx=-\frac {\sqrt {c^2-d^2\,x^2}\,\left (\frac {4\,B\,c^2+10\,A\,d\,c}{140\,c^4\,d^2}+\frac {3\,A\,d^2-3\,B\,c\,d}{140\,c^3\,d^3}\right )}{{\left (c+d\,x\right )}^3}-\frac {\sqrt {c^2-d^2\,x^2}\,\left (\frac {37\,B\,c^2+61\,A\,d\,c}{840\,c^5\,d^2}+\frac {26\,A\,d^2+2\,B\,c\,d}{840\,c^4\,d^3}\right )}{{\left (c+d\,x\right )}^2}-\frac {\sqrt {c^2-d^2\,x^2}\,\left (\frac {A}{14\,c^2\,d}-\frac {B}{14\,c\,d^2}\right )}{{\left (c+d\,x\right )}^4}-\frac {\sqrt {c^2-d^2\,x^2}\,\left (\frac {29\,A\,d^2+13\,B\,c\,d}{280\,c^4\,d^3}-\frac {x\,\left (48\,B\,c^2+64\,A\,d\,c\right )}{280\,c^6\,d}\right )}{\left (c+d\,x\right )\,\left (c-d\,x\right )} \] Input:

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

Output:

- ((c^2 - d^2*x^2)^(1/2)*((4*B*c^2 + 10*A*c*d)/(140*c^4*d^2) + (3*A*d^2 - 
3*B*c*d)/(140*c^3*d^3)))/(c + d*x)^3 - ((c^2 - d^2*x^2)^(1/2)*((37*B*c^2 + 
 61*A*c*d)/(840*c^5*d^2) + (26*A*d^2 + 2*B*c*d)/(840*c^4*d^3)))/(c + d*x)^ 
2 - ((c^2 - d^2*x^2)^(1/2)*(A/(14*c^2*d) - B/(14*c*d^2)))/(c + d*x)^4 - (( 
c^2 - d^2*x^2)^(1/2)*((29*A*d^2 + 13*B*c*d)/(280*c^4*d^3) - (x*(48*B*c^2 + 
 64*A*c*d))/(280*c^6*d)))/((c + d*x)*(c - d*x))
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 325, normalized size of antiderivative = 2.08 \[ \int \frac {A+B x}{(c+d x)^3 \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {-4 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{3} d -12 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{2} d^{2} x -12 \sqrt {-d^{2} x^{2}+c^{2}}\, a c \,d^{3} x^{2}-4 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{4} x^{3}-3 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{4}-9 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{3} d x -9 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2} d^{2} x^{2}-3 \sqrt {-d^{2} x^{2}+c^{2}}\, b c \,d^{3} x^{3}-39 a \,c^{4} d -12 a \,c^{3} d^{2} x +60 a \,c^{2} d^{3} x^{2}+72 a c \,d^{4} x^{3}+24 a \,d^{5} x^{4}-3 b \,c^{5}-9 b \,c^{4} d x +45 b \,c^{3} d^{2} x^{2}+54 b \,c^{2} d^{3} x^{3}+18 b c \,d^{4} x^{4}}{105 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{5} d^{2} \left (d^{3} x^{3}+3 c \,d^{2} x^{2}+3 c^{2} d x +c^{3}\right )} \] Input:

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

Output:

( - 4*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 - 4*sqrt(c**2 - d**2*x**2) 
*a*d**4*x**3 - 3*sqrt(c**2 - d**2*x**2)*b*c**4 - 9*sqrt(c**2 - d**2*x**2)* 
b*c**3*d*x - 9*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 - 39*a*c**4*d - 12*a*c**3*d**2*x + 60*a*c**2*d**3*x* 
*2 + 72*a*c*d**4*x**3 + 24*a*d**5*x**4 - 3*b*c**5 - 9*b*c**4*d*x + 45*b*c* 
*3*d**2*x**2 + 54*b*c**2*d**3*x**3 + 18*b*c*d**4*x**4)/(105*sqrt(c**2 - d* 
*2*x**2)*c**5*d**2*(c**3 + 3*c**2*d*x + 3*c*d**2*x**2 + d**3*x**3))