\(\int \frac {A+B x}{(c+d x)^2 (c^2-d^2 x^2)^{5/2}} \, dx\) [62]

Optimal result

Integrand size = 29, antiderivative size = 156 \[ \int \frac {A+B x}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {4 (2 B c+5 A d) x}{105 c^4 d \left (c^2-d^2 x^2\right )^{3/2}}+\frac {B c-A d}{7 c d^2 (c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {2 B c+5 A d}{35 c^2 d^2 (c+d x) \left (c^2-d^2 x^2\right )^{3/2}}+\frac {8 (2 B c+5 A d) x}{105 c^6 d \sqrt {c^2-d^2 x^2}} \] Output:

4/105*(5*A*d+2*B*c)*x/c^4/d/(-d^2*x^2+c^2)^(3/2)+1/7*(-A*d+B*c)/c/d^2/(d*x 
+c)^2/(-d^2*x^2+c^2)^(3/2)-1/35*(5*A*d+2*B*c)/c^2/d^2/(d*x+c)/(-d^2*x^2+c^ 
2)^(3/2)+8/105*(5*A*d+2*B*c)*x/c^6/d/(-d^2*x^2+c^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.70 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.98 \[ \int \frac {A+B x}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {c^2-d^2 x^2} \left (B c \left (9 c^5+18 c^4 d x+48 c^3 d^2 x^2+8 c^2 d^3 x^3-32 c d^4 x^4-16 d^5 x^5\right )-5 A d \left (6 c^5-9 c^4 d x-24 c^3 d^2 x^2-4 c^2 d^3 x^3+16 c d^4 x^4+8 d^5 x^5\right )\right )}{105 c^6 d^2 (c-d x)^2 (c+d x)^4} \] Input:

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

Output:

(Sqrt[c^2 - d^2*x^2]*(B*c*(9*c^5 + 18*c^4*d*x + 48*c^3*d^2*x^2 + 8*c^2*d^3 
*x^3 - 32*c*d^4*x^4 - 16*d^5*x^5) - 5*A*d*(6*c^5 - 9*c^4*d*x - 24*c^3*d^2* 
x^2 - 4*c^2*d^3*x^3 + 16*c*d^4*x^4 + 8*d^5*x^5)))/(105*c^6*d^2*(c - d*x)^2 
*(c + d*x)^4)
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {671, 470, 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.

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

\(\Big \downarrow \) 671

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

\(\Big \downarrow \) 470

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

\(\Big \downarrow \) 209

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

\(\Big \downarrow \) 208

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

Input:

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

Output:

(B*c - A*d)/(7*c*d^2*(c + d*x)^2*(c^2 - d^2*x^2)^(3/2)) + ((2*B*c + 5*A*d) 
*(-1/5*1/(c*d*(c + d*x)*(c^2 - d^2*x^2)^(3/2)) + (4*(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)))/(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 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 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 = 0.40 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.01

method	result	size
gosper	\(-\frac {\left (-d x +c \right ) \left (40 A \,d^{6} x^{5}+16 B c \,d^{5} x^{5}+80 A c \,d^{5} x^{4}+32 B \,c^{2} d^{4} x^{4}-20 A \,c^{2} d^{4} x^{3}-8 B \,c^{3} d^{3} x^{3}-120 A \,c^{3} d^{3} x^{2}-48 B \,c^{4} d^{2} x^{2}-45 A \,c^{4} d^{2} x -18 B \,c^{5} d x +30 A \,c^{5} d -9 B \,c^{6}\right )}{105 \left (d x +c \right ) c^{6} d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}\)	\(157\)
orering	\(-\frac {\left (-d x +c \right ) \left (40 A \,d^{6} x^{5}+16 B c \,d^{5} x^{5}+80 A c \,d^{5} x^{4}+32 B \,c^{2} d^{4} x^{4}-20 A \,c^{2} d^{4} x^{3}-8 B \,c^{3} d^{3} x^{3}-120 A \,c^{3} d^{3} x^{2}-48 B \,c^{4} d^{2} x^{2}-45 A \,c^{4} d^{2} x -18 B \,c^{5} d x +30 A \,c^{5} d -9 B \,c^{6}\right )}{105 \left (d x +c \right ) c^{6} d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}\)	\(157\)
trager	\(-\frac {\left (40 A \,d^{6} x^{5}+16 B c \,d^{5} x^{5}+80 A c \,d^{5} x^{4}+32 B \,c^{2} d^{4} x^{4}-20 A \,c^{2} d^{4} x^{3}-8 B \,c^{3} d^{3} x^{3}-120 A \,c^{3} d^{3} x^{2}-48 B \,c^{4} d^{2} x^{2}-45 A \,c^{4} d^{2} x -18 B \,c^{5} d x +30 A \,c^{5} d -9 B \,c^{6}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{105 c^{6} \left (d x +c \right )^{4} \left (-d x +c \right )^{2} d^{2}}\)	\(159\)
default	\(\frac {B \left (-\frac {1}{5 c d \left (x +\frac {c}{d}\right ) \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}+\frac {4 d \left (-\frac {-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d}{6 c^{2} d^{2} \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}-\frac {-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d}{3 d^{2} c^{4} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{5 c}\right )}{d^{2}}+\frac {\left (A d -B c \right ) \left (-\frac {1}{7 c d \left (x +\frac {c}{d}\right )^{2} \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}+\frac {5 d \left (-\frac {1}{5 c d \left (x +\frac {c}{d}\right ) \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}+\frac {4 d \left (-\frac {-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d}{6 c^{2} d^{2} \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}-\frac {-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d}{3 d^{2} c^{4} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{5 c}\right )}{7 c}\right )}{d^{3}}\)	\(389\)

Input:

int((B*x+A)/(d*x+c)^2/(-d^2*x^2+c^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/105*(-d*x+c)*(40*A*d^6*x^5+16*B*c*d^5*x^5+80*A*c*d^5*x^4+32*B*c^2*d^4*x 
^4-20*A*c^2*d^4*x^3-8*B*c^3*d^3*x^3-120*A*c^3*d^3*x^2-48*B*c^4*d^2*x^2-45* 
A*c^4*d^2*x-18*B*c^5*d*x+30*A*c^5*d-9*B*c^6)/(d*x+c)/c^6/d^2/(-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. 360 vs. \(2 (140) = 280\).

Time = 0.15 (sec) , antiderivative size = 360, normalized size of antiderivative = 2.31 \[ \int \frac {A+B x}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {9 \, B c^{7} - 30 \, A c^{6} d + 3 \, {\left (3 \, B c d^{6} - 10 \, A d^{7}\right )} x^{6} + 6 \, {\left (3 \, B c^{2} d^{5} - 10 \, A c d^{6}\right )} x^{5} - 3 \, {\left (3 \, B c^{3} d^{4} - 10 \, A c^{2} d^{5}\right )} x^{4} - 12 \, {\left (3 \, B c^{4} d^{3} - 10 \, A c^{3} d^{4}\right )} x^{3} - 3 \, {\left (3 \, B c^{5} d^{2} - 10 \, A c^{4} d^{3}\right )} x^{2} + 6 \, {\left (3 \, B c^{6} d - 10 \, A c^{5} d^{2}\right )} x + {\left (9 \, B c^{6} - 30 \, A c^{5} d - 8 \, {\left (2 \, B c d^{5} + 5 \, A d^{6}\right )} x^{5} - 16 \, {\left (2 \, B c^{2} d^{4} + 5 \, A c d^{5}\right )} x^{4} + 4 \, {\left (2 \, B c^{3} d^{3} + 5 \, A c^{2} d^{4}\right )} x^{3} + 24 \, {\left (2 \, B c^{4} d^{2} + 5 \, A c^{3} d^{3}\right )} x^{2} + 9 \, {\left (2 \, B c^{5} d + 5 \, A c^{4} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{105 \, {\left (c^{6} d^{8} x^{6} + 2 \, c^{7} d^{7} x^{5} - c^{8} d^{6} x^{4} - 4 \, c^{9} d^{5} x^{3} - c^{10} d^{4} x^{2} + 2 \, c^{11} d^{3} x + c^{12} d^{2}\right )}} \] Input:

integrate((B*x+A)/(d*x+c)^2/(-d^2*x^2+c^2)^(5/2),x, algorithm="fricas")
 

Output:

1/105*(9*B*c^7 - 30*A*c^6*d + 3*(3*B*c*d^6 - 10*A*d^7)*x^6 + 6*(3*B*c^2*d^ 
5 - 10*A*c*d^6)*x^5 - 3*(3*B*c^3*d^4 - 10*A*c^2*d^5)*x^4 - 12*(3*B*c^4*d^3 
 - 10*A*c^3*d^4)*x^3 - 3*(3*B*c^5*d^2 - 10*A*c^4*d^3)*x^2 + 6*(3*B*c^6*d - 
 10*A*c^5*d^2)*x + (9*B*c^6 - 30*A*c^5*d - 8*(2*B*c*d^5 + 5*A*d^6)*x^5 - 1 
6*(2*B*c^2*d^4 + 5*A*c*d^5)*x^4 + 4*(2*B*c^3*d^3 + 5*A*c^2*d^4)*x^3 + 24*( 
2*B*c^4*d^2 + 5*A*c^3*d^3)*x^2 + 9*(2*B*c^5*d + 5*A*c^4*d^2)*x)*sqrt(-d^2* 
x^2 + c^2))/(c^6*d^8*x^6 + 2*c^7*d^7*x^5 - c^8*d^6*x^4 - 4*c^9*d^5*x^3 - c 
^10*d^4*x^2 + 2*c^11*d^3*x + c^12*d^2)
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 378 vs. \(2 (140) = 280\).

Time = 0.04 (sec) , antiderivative size = 378, normalized size of antiderivative = 2.42 \[ \int \frac {A+B x}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {B c}{7 \, {\left ({\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c d^{4} x^{2} + 2 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c^{2} d^{3} x + {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c^{3} d^{2}\right )}} + \frac {B c}{7 \, {\left ({\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c^{2} d^{3} x + {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c^{3} d^{2}\right )}} - \frac {A}{7 \, {\left ({\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c d^{3} x^{2} + 2 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c^{2} d^{2} x + {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c^{3} d\right )}} - \frac {A}{7 \, {\left ({\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c^{2} d^{2} x + {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c^{3} d\right )}} - \frac {B}{5 \, {\left ({\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c d^{3} x + {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c^{2} d^{2}\right )}} + \frac {4 \, A x}{21 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c^{4}} + \frac {8 \, B x}{105 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c^{3} d} + \frac {8 \, A x}{21 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{6}} + \frac {16 \, B x}{105 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{5} d} \] Input:

integrate((B*x+A)/(d*x+c)^2/(-d^2*x^2+c^2)^(5/2),x, algorithm="maxima")
 

Output:

1/7*B*c/((-d^2*x^2 + c^2)^(3/2)*c*d^4*x^2 + 2*(-d^2*x^2 + c^2)^(3/2)*c^2*d 
^3*x + (-d^2*x^2 + c^2)^(3/2)*c^3*d^2) + 1/7*B*c/((-d^2*x^2 + c^2)^(3/2)*c 
^2*d^3*x + (-d^2*x^2 + c^2)^(3/2)*c^3*d^2) - 1/7*A/((-d^2*x^2 + c^2)^(3/2) 
*c*d^3*x^2 + 2*(-d^2*x^2 + c^2)^(3/2)*c^2*d^2*x + (-d^2*x^2 + c^2)^(3/2)*c 
^3*d) - 1/7*A/((-d^2*x^2 + c^2)^(3/2)*c^2*d^2*x + (-d^2*x^2 + c^2)^(3/2)*c 
^3*d) - 1/5*B/((-d^2*x^2 + c^2)^(3/2)*c*d^3*x + (-d^2*x^2 + c^2)^(3/2)*c^2 
*d^2) + 4/21*A*x/((-d^2*x^2 + c^2)^(3/2)*c^4) + 8/105*B*x/((-d^2*x^2 + c^2 
)^(3/2)*c^3*d) + 8/21*A*x/(sqrt(-d^2*x^2 + c^2)*c^6) + 16/105*B*x/(sqrt(-d 
^2*x^2 + c^2)*c^5*d)
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.17 (sec) , antiderivative size = 427, normalized size of antiderivative = 2.74 \[ \int \frac {A+B x}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {\frac {256 \, {\left (2 i \, B c + 5 i \, A d\right )} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right )}{c^{6} d} + \frac {35 \, {\left (9 \, B c {\left (\frac {2 \, c}{d x + c} - 1\right )} + 15 \, A d {\left (\frac {2 \, c}{d x + c} - 1\right )} + B c + A d\right )}}{c^{6} d {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right )} + \frac {15 \, B c^{37} d^{6} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {7}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right )^{6} \mathrm {sgn}\left (d\right )^{6} - 15 \, A c^{36} d^{7} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {7}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right )^{6} \mathrm {sgn}\left (d\right )^{6} + 63 \, B c^{37} d^{6} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {5}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right )^{6} \mathrm {sgn}\left (d\right )^{6} - 105 \, A c^{36} d^{7} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {5}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right )^{6} \mathrm {sgn}\left (d\right )^{6} + 70 \, B c^{37} d^{6} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right )^{6} \mathrm {sgn}\left (d\right )^{6} - 350 \, A c^{36} d^{7} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right )^{6} \mathrm {sgn}\left (d\right )^{6} - 210 \, B c^{37} d^{6} \sqrt {\frac {2 \, c}{d x + c} - 1} \mathrm {sgn}\left (\frac {1}{d x + c}\right )^{6} \mathrm {sgn}\left (d\right )^{6} - 1050 \, A c^{36} d^{7} \sqrt {\frac {2 \, c}{d x + c} - 1} \mathrm {sgn}\left (\frac {1}{d x + c}\right )^{6} \mathrm {sgn}\left (d\right )^{6}}{c^{42} d^{7} \mathrm {sgn}\left (\frac {1}{d x + c}\right )^{7} \mathrm {sgn}\left (d\right )^{7}}}{3360 \, {\left | d \right |}} \] Input:

integrate((B*x+A)/(d*x+c)^2/(-d^2*x^2+c^2)^(5/2),x, algorithm="giac")
 

Output:

1/3360*(256*(2*I*B*c + 5*I*A*d)*sgn(1/(d*x + c))*sgn(d)/(c^6*d) + 35*(9*B* 
c*(2*c/(d*x + c) - 1) + 15*A*d*(2*c/(d*x + c) - 1) + B*c + A*d)/(c^6*d*(2* 
c/(d*x + c) - 1)^(3/2)*sgn(1/(d*x + c))*sgn(d)) + (15*B*c^37*d^6*(2*c/(d*x 
 + c) - 1)^(7/2)*sgn(1/(d*x + c))^6*sgn(d)^6 - 15*A*c^36*d^7*(2*c/(d*x + c 
) - 1)^(7/2)*sgn(1/(d*x + c))^6*sgn(d)^6 + 63*B*c^37*d^6*(2*c/(d*x + c) - 
1)^(5/2)*sgn(1/(d*x + c))^6*sgn(d)^6 - 105*A*c^36*d^7*(2*c/(d*x + c) - 1)^ 
(5/2)*sgn(1/(d*x + c))^6*sgn(d)^6 + 70*B*c^37*d^6*(2*c/(d*x + c) - 1)^(3/2 
)*sgn(1/(d*x + c))^6*sgn(d)^6 - 350*A*c^36*d^7*(2*c/(d*x + c) - 1)^(3/2)*s 
gn(1/(d*x + c))^6*sgn(d)^6 - 210*B*c^37*d^6*sqrt(2*c/(d*x + c) - 1)*sgn(1/ 
(d*x + c))^6*sgn(d)^6 - 1050*A*c^36*d^7*sqrt(2*c/(d*x + c) - 1)*sgn(1/(d*x 
 + c))^6*sgn(d)^6)/(c^42*d^7*sgn(1/(d*x + c))^7*sgn(d)^7))/abs(d)
 

Mupad [B] (verification not implemented)

Time = 10.31 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.76 \[ \int \frac {A+B x}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {c^2-d^2\,x^2}\,\left (x\,\left (\frac {29\,B\,c^3+90\,A\,d\,c^2}{420\,c^6\,d}+\frac {20\,A\,d^3-6\,B\,c\,d^2}{420\,c^4\,d^3}\right )-\frac {25\,A\,d-4\,B\,c}{140\,c^3\,d^2}\right )}{{\left (c+d\,x\right )}^2\,{\left (c-d\,x\right )}^2}-\frac {\sqrt {c^2-d^2\,x^2}\,\left (\frac {A}{28\,c^3\,d}-\frac {B}{28\,c^2\,d^2}\right )}{{\left (c+d\,x\right )}^4}-\frac {\sqrt {c^2-d^2\,x^2}\,\left (\frac {4\,B\,c^3+24\,A\,d\,c^2}{560\,c^6\,d^2}+\frac {c\,\left (\frac {13\,\left (A\,d-B\,c\right )}{560\,c^5\,d}+\frac {3\,A\,d^3-3\,B\,c\,d^2}{560\,c^5\,d^3}\right )}{d}\right )}{{\left (c+d\,x\right )}^3}+\frac {x\,\sqrt {c^2-d^2\,x^2}\,\left (40\,A\,d+16\,B\,c\right )}{105\,c^6\,d\,\left (c+d\,x\right )\,\left (c-d\,x\right )} \] Input:

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

Output:

((c^2 - d^2*x^2)^(1/2)*(x*((29*B*c^3 + 90*A*c^2*d)/(420*c^6*d) + (20*A*d^3 
 - 6*B*c*d^2)/(420*c^4*d^3)) - (25*A*d - 4*B*c)/(140*c^3*d^2)))/((c + d*x) 
^2*(c - d*x)^2) - ((c^2 - d^2*x^2)^(1/2)*(A/(28*c^3*d) - B/(28*c^2*d^2)))/ 
(c + d*x)^4 - ((c^2 - d^2*x^2)^(1/2)*((4*B*c^3 + 24*A*c^2*d)/(560*c^6*d^2) 
 + (c*((13*(A*d - B*c))/(560*c^5*d) + (3*A*d^3 - 3*B*c*d^2)/(560*c^5*d^3)) 
)/d))/(c + d*x)^3 + (x*(c^2 - d^2*x^2)^(1/2)*(40*A*d + 16*B*c))/(105*c^6*d 
*(c + d*x)*(c - d*x))
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 350, normalized size of antiderivative = 2.24 \[ \int \frac {A+B x}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {45 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{4} d +90 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{3} d^{2} x -90 \sqrt {-d^{2} x^{2}+c^{2}}\, a c \,d^{4} x^{3}-45 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{5} x^{4}+18 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{5}+36 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{4} d x -36 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2} d^{3} x^{3}-18 \sqrt {-d^{2} x^{2}+c^{2}}\, b c \,d^{4} x^{4}-60 a \,c^{5} d +90 a \,c^{4} d^{2} x +240 a \,c^{3} d^{3} x^{2}+40 a \,c^{2} d^{4} x^{3}-160 a c \,d^{5} x^{4}-80 a \,d^{6} x^{5}+18 b \,c^{6}+36 b \,c^{5} d x +96 b \,c^{4} d^{2} x^{2}+16 b \,c^{3} d^{3} x^{3}-64 b \,c^{2} d^{4} x^{4}-32 b c \,d^{5} x^{5}}{210 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{6} d^{2} \left (-d^{4} x^{4}-2 c \,d^{3} x^{3}+2 c^{3} d x +c^{4}\right )} \] Input:

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

Output:

(45*sqrt(c**2 - d**2*x**2)*a*c**4*d + 90*sqrt(c**2 - d**2*x**2)*a*c**3*d** 
2*x - 90*sqrt(c**2 - d**2*x**2)*a*c*d**4*x**3 - 45*sqrt(c**2 - d**2*x**2)* 
a*d**5*x**4 + 18*sqrt(c**2 - d**2*x**2)*b*c**5 + 36*sqrt(c**2 - d**2*x**2) 
*b*c**4*d*x - 36*sqrt(c**2 - d**2*x**2)*b*c**2*d**3*x**3 - 18*sqrt(c**2 - 
d**2*x**2)*b*c*d**4*x**4 - 60*a*c**5*d + 90*a*c**4*d**2*x + 240*a*c**3*d** 
3*x**2 + 40*a*c**2*d**4*x**3 - 160*a*c*d**5*x**4 - 80*a*d**6*x**5 + 18*b*c 
**6 + 36*b*c**5*d*x + 96*b*c**4*d**2*x**2 + 16*b*c**3*d**3*x**3 - 64*b*c** 
2*d**4*x**4 - 32*b*c*d**5*x**5)/(210*sqrt(c**2 - d**2*x**2)*c**6*d**2*(c** 
4 + 2*c**3*d*x - 2*c*d**3*x**3 - d**4*x**4))