\(\int \frac {A+B x}{(c+d x)^4 \sqrt {c^2-d^2 x^2}} \, dx\) [47]

Optimal result

Integrand size = 29, antiderivative size = 162 \[ \int \frac {A+B x}{(c+d x)^4 \sqrt {c^2-d^2 x^2}} \, dx=\frac {2 (4 B c+3 A d) x}{105 c^4 d \sqrt {c^2-d^2 x^2}}-\frac {2 (4 B c+3 A d)}{35 c d^2 (c+d x)^2 \sqrt {c^2-d^2 x^2}}-\frac {4 B c+3 A d}{105 c^2 d^2 (c+d x) \sqrt {c^2-d^2 x^2}}+\frac {(B c-A d) \sqrt {c^2-d^2 x^2}}{7 c d^2 (c+d x)^4} \] Output:

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

Mathematica [A] (verified)

Time = 0.78 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.62 \[ \int \frac {A+B x}{(c+d x)^4 \sqrt {c^2-d^2 x^2}} \, dx=-\frac {\sqrt {c^2-d^2 x^2} \left (3 A d \left (12 c^3+13 c^2 d x+8 c d^2 x^2+2 d^3 x^3\right )+B c \left (13 c^3+52 c^2 d x+32 c d^2 x^2+8 d^3 x^3\right )\right )}{105 c^4 d^2 (c+d x)^4} \] Input:

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

Output:

-1/105*(Sqrt[c^2 - d^2*x^2]*(3*A*d*(12*c^3 + 13*c^2*d*x + 8*c*d^2*x^2 + 2* 
d^3*x^3) + B*c*(13*c^3 + 52*c^2*d*x + 32*c*d^2*x^2 + 8*d^3*x^3)))/(c^4*d^2 
*(c + d*x)^4)
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {671, 461, 461, 460}

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.

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

\(\Big \downarrow \) 671

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

\(\Big \downarrow \) 461

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

\(\Big \downarrow \) 461

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

\(\Big \downarrow \) 460

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 460

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

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 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 = 0.45 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.64

method	result	size
trager	\(-\frac {\left (6 A \,d^{4} x^{3}+8 B c \,d^{3} x^{3}+24 A c \,d^{3} x^{2}+32 x^{2} c^{2} B \,d^{2}+39 A \,c^{2} d^{2} x +52 B \,c^{3} d x +36 A \,c^{3} d +13 B \,c^{4}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{105 c^{4} \left (d x +c \right )^{4} d^{2}}\)	\(103\)
gosper	\(-\frac {\left (-d x +c \right ) \left (6 A \,d^{4} x^{3}+8 B c \,d^{3} x^{3}+24 A c \,d^{3} x^{2}+32 x^{2} c^{2} B \,d^{2}+39 A \,c^{2} d^{2} x +52 B \,c^{3} d x +36 A \,c^{3} d +13 B \,c^{4}\right )}{105 \left (d x +c \right )^{3} c^{4} d^{2} \sqrt {-d^{2} x^{2}+c^{2}}}\)	\(109\)
orering	\(-\frac {\left (-d x +c \right ) \left (6 A \,d^{4} x^{3}+8 B c \,d^{3} x^{3}+24 A c \,d^{3} x^{2}+32 x^{2} c^{2} B \,d^{2}+39 A \,c^{2} d^{2} x +52 B \,c^{3} d x +36 A \,c^{3} d +13 B \,c^{4}\right )}{105 \left (d x +c \right )^{3} c^{4} d^{2} \sqrt {-d^{2} x^{2}+c^{2}}}\)	\(109\)
default	\(\frac {B \left (-\frac {\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{5 c d \left (x +\frac {c}{d}\right )^{3}}+\frac {2 d \left (-\frac {\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{3 c d \left (x +\frac {c}{d}\right )^{2}}-\frac {\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{3 c^{2} \left (x +\frac {c}{d}\right )}\right )}{5 c}\right )}{d^{4}}+\frac {\left (A d -B c \right ) \left (-\frac {\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{7 c d \left (x +\frac {c}{d}\right )^{4}}+\frac {3 d \left (-\frac {\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{5 c d \left (x +\frac {c}{d}\right )^{3}}+\frac {2 d \left (-\frac {\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{3 c d \left (x +\frac {c}{d}\right )^{2}}-\frac {\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{3 c^{2} \left (x +\frac {c}{d}\right )}\right )}{5 c}\right )}{7 c}\right )}{d^{5}}\)	\(351\)

Input:

int((B*x+A)/(d*x+c)^4/(-d^2*x^2+c^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/105*(6*A*d^4*x^3+8*B*c*d^3*x^3+24*A*c*d^3*x^2+32*B*c^2*d^2*x^2+39*A*c^2 
*d^2*x+52*B*c^3*d*x+36*A*c^3*d+13*B*c^4)/c^4/(d*x+c)^4/d^2*(-d^2*x^2+c^2)^ 
(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.49 \[ \int \frac {A+B x}{(c+d x)^4 \sqrt {c^2-d^2 x^2}} \, dx=-\frac {13 \, B c^{5} + 36 \, A c^{4} d + {\left (13 \, B c d^{4} + 36 \, A d^{5}\right )} x^{4} + 4 \, {\left (13 \, B c^{2} d^{3} + 36 \, A c d^{4}\right )} x^{3} + 6 \, {\left (13 \, B c^{3} d^{2} + 36 \, A c^{2} d^{3}\right )} x^{2} + 4 \, {\left (13 \, B c^{4} d + 36 \, A c^{3} d^{2}\right )} x + {\left (13 \, B c^{4} + 36 \, A c^{3} d + 2 \, {\left (4 \, B c d^{3} + 3 \, A d^{4}\right )} x^{3} + 8 \, {\left (4 \, B c^{2} d^{2} + 3 \, A c d^{3}\right )} x^{2} + 13 \, {\left (4 \, B c^{3} d + 3 \, A c^{2} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{105 \, {\left (c^{4} d^{6} x^{4} + 4 \, c^{5} d^{5} x^{3} + 6 \, c^{6} d^{4} x^{2} + 4 \, c^{7} d^{3} x + c^{8} d^{2}\right )}} \] Input:

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

Output:

-1/105*(13*B*c^5 + 36*A*c^4*d + (13*B*c*d^4 + 36*A*d^5)*x^4 + 4*(13*B*c^2* 
d^3 + 36*A*c*d^4)*x^3 + 6*(13*B*c^3*d^2 + 36*A*c^2*d^3)*x^2 + 4*(13*B*c^4* 
d + 36*A*c^3*d^2)*x + (13*B*c^4 + 36*A*c^3*d + 2*(4*B*c*d^3 + 3*A*d^4)*x^3 
 + 8*(4*B*c^2*d^2 + 3*A*c*d^3)*x^2 + 13*(4*B*c^3*d + 3*A*c^2*d^2)*x)*sqrt( 
-d^2*x^2 + c^2))/(c^4*d^6*x^4 + 4*c^5*d^5*x^3 + 6*c^6*d^4*x^2 + 4*c^7*d^3* 
x + c^8*d^2)
 

Sympy [F]

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

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

Output:

Integral((A + B*x)/(sqrt(-(-c + d*x)*(c + d*x))*(c + d*x)**4), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 541 vs. \(2 (146) = 292\).

Time = 0.12 (sec) , antiderivative size = 541, normalized size of antiderivative = 3.34 \[ \int \frac {A+B x}{(c+d x)^4 \sqrt {c^2-d^2 x^2}} \, dx=\frac {\sqrt {-d^{2} x^{2} + c^{2}} B c}{7 \, {\left (c d^{6} x^{4} + 4 \, c^{2} d^{5} x^{3} + 6 \, c^{3} d^{4} x^{2} + 4 \, c^{4} d^{3} x + c^{5} d^{2}\right )}} + \frac {3 \, \sqrt {-d^{2} x^{2} + c^{2}} B c}{35 \, {\left (c^{2} d^{5} x^{3} + 3 \, c^{3} d^{4} x^{2} + 3 \, c^{4} d^{3} x + c^{5} d^{2}\right )}} + \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} B c}{35 \, {\left (c^{3} d^{4} x^{2} + 2 \, c^{4} d^{3} x + c^{5} d^{2}\right )}} + \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} B c}{35 \, {\left (c^{4} d^{3} x + c^{5} d^{2}\right )}} - \frac {\sqrt {-d^{2} x^{2} + c^{2}} A}{7 \, {\left (c d^{5} x^{4} + 4 \, c^{2} d^{4} x^{3} + 6 \, c^{3} d^{3} x^{2} + 4 \, c^{4} d^{2} x + c^{5} d\right )}} - \frac {3 \, \sqrt {-d^{2} x^{2} + c^{2}} A}{35 \, {\left (c^{2} d^{4} x^{3} + 3 \, c^{3} d^{3} x^{2} + 3 \, c^{4} d^{2} x + c^{5} d\right )}} - \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} A}{35 \, {\left (c^{3} d^{3} x^{2} + 2 \, c^{4} d^{2} x + c^{5} d\right )}} - \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} A}{35 \, {\left (c^{4} d^{2} x + c^{5} d\right )}} - \frac {\sqrt {-d^{2} x^{2} + c^{2}} B}{5 \, {\left (c d^{5} x^{3} + 3 \, c^{2} d^{4} x^{2} + 3 \, c^{3} d^{3} x + c^{4} d^{2}\right )}} - \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} B}{15 \, {\left (c^{2} d^{4} x^{2} + 2 \, c^{3} d^{3} x + c^{4} d^{2}\right )}} - \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} B}{15 \, {\left (c^{3} d^{3} x + c^{4} d^{2}\right )}} \] Input:

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

Output:

1/7*sqrt(-d^2*x^2 + c^2)*B*c/(c*d^6*x^4 + 4*c^2*d^5*x^3 + 6*c^3*d^4*x^2 + 
4*c^4*d^3*x + c^5*d^2) + 3/35*sqrt(-d^2*x^2 + c^2)*B*c/(c^2*d^5*x^3 + 3*c^ 
3*d^4*x^2 + 3*c^4*d^3*x + c^5*d^2) + 2/35*sqrt(-d^2*x^2 + c^2)*B*c/(c^3*d^ 
4*x^2 + 2*c^4*d^3*x + c^5*d^2) + 2/35*sqrt(-d^2*x^2 + c^2)*B*c/(c^4*d^3*x 
+ c^5*d^2) - 1/7*sqrt(-d^2*x^2 + c^2)*A/(c*d^5*x^4 + 4*c^2*d^4*x^3 + 6*c^3 
*d^3*x^2 + 4*c^4*d^2*x + c^5*d) - 3/35*sqrt(-d^2*x^2 + c^2)*A/(c^2*d^4*x^3 
 + 3*c^3*d^3*x^2 + 3*c^4*d^2*x + c^5*d) - 2/35*sqrt(-d^2*x^2 + c^2)*A/(c^3 
*d^3*x^2 + 2*c^4*d^2*x + c^5*d) - 2/35*sqrt(-d^2*x^2 + c^2)*A/(c^4*d^2*x + 
 c^5*d) - 1/5*sqrt(-d^2*x^2 + c^2)*B/(c*d^5*x^3 + 3*c^2*d^4*x^2 + 3*c^3*d^ 
3*x + c^4*d^2) - 2/15*sqrt(-d^2*x^2 + c^2)*B/(c^2*d^4*x^2 + 2*c^3*d^3*x + 
c^4*d^2) - 2/15*sqrt(-d^2*x^2 + c^2)*B/(c^3*d^3*x + c^4*d^2)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 406 vs. \(2 (146) = 292\).

Time = 0.15 (sec) , antiderivative size = 406, normalized size of antiderivative = 2.51 \[ \int \frac {A+B x}{(c+d x)^4 \sqrt {c^2-d^2 x^2}} \, dx=\frac {2 \, {\left (13 \, B c + 36 \, A d + \frac {91 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} B c}{d^{2} x} + \frac {147 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} A}{d x} + \frac {168 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} B c}{d^{4} x^{2}} + \frac {441 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} A}{d^{3} x^{2}} + \frac {280 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{3} B c}{d^{6} x^{3}} + \frac {630 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{3} A}{d^{5} x^{3}} + \frac {175 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{4} B c}{d^{8} x^{4}} + \frac {630 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{4} A}{d^{7} x^{4}} + \frac {105 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{5} B c}{d^{10} x^{5}} + \frac {315 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{5} A}{d^{9} x^{5}} + \frac {105 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{6} A}{d^{11} x^{6}}\right )}}{105 \, c^{4} d {\left (\frac {c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}}{d^{2} x} + 1\right )}^{7} {\left | d \right |}} \] Input:

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

Output:

2/105*(13*B*c + 36*A*d + 91*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))*B*c/(d^2*x 
) + 147*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))*A/(d*x) + 168*(c*d + sqrt(-d^2 
*x^2 + c^2)*abs(d))^2*B*c/(d^4*x^2) + 441*(c*d + sqrt(-d^2*x^2 + c^2)*abs( 
d))^2*A/(d^3*x^2) + 280*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^3*B*c/(d^6*x^3 
) + 630*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^3*A/(d^5*x^3) + 175*(c*d + sqr 
t(-d^2*x^2 + c^2)*abs(d))^4*B*c/(d^8*x^4) + 630*(c*d + sqrt(-d^2*x^2 + c^2 
)*abs(d))^4*A/(d^7*x^4) + 105*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^5*B*c/(d 
^10*x^5) + 315*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^5*A/(d^9*x^5) + 105*(c* 
d + sqrt(-d^2*x^2 + c^2)*abs(d))^6*A/(d^11*x^6))/(c^4*d*((c*d + sqrt(-d^2* 
x^2 + c^2)*abs(d))/(d^2*x) + 1)^7*abs(d))
 

Mupad [B] (verification not implemented)

Time = 9.29 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.94 \[ \int \frac {A+B x}{(c+d x)^4 \sqrt {c^2-d^2 x^2}} \, dx=\frac {\sqrt {c^2-d^2\,x^2}\,\left (\frac {B}{7\,d^2}-\frac {A}{7\,c\,d}\right )}{{\left (c+d\,x\right )}^4}-\frac {\sqrt {c^2-d^2\,x^2}\,\left (3\,A\,d+4\,B\,c\right )}{35\,c^2\,d^2\,{\left (c+d\,x\right )}^3}-\frac {\sqrt {c^2-d^2\,x^2}\,\left (6\,A\,d+8\,B\,c\right )}{105\,c^3\,d^2\,{\left (c+d\,x\right )}^2}-\frac {\sqrt {c^2-d^2\,x^2}\,\left (6\,A\,d+8\,B\,c\right )}{105\,c^4\,d^2\,\left (c+d\,x\right )} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.83 \[ \int \frac {A+B x}{(c+d x)^4 \sqrt {c^2-d^2 x^2}} \, dx=\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, \left (-6 a \,d^{4} x^{3}-8 b c \,d^{3} x^{3}-24 a c \,d^{3} x^{2}-32 b \,c^{2} d^{2} x^{2}-39 a \,c^{2} d^{2} x -52 b \,c^{3} d x -36 a \,c^{3} d -13 b \,c^{4}\right )}{105 c^{4} d^{2} \left (d^{4} x^{4}+4 c \,d^{3} x^{3}+6 c^{2} d^{2} x^{2}+4 c^{3} d x +c^{4}\right )} \] Input:

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

Output:

(sqrt(c**2 - d**2*x**2)*( - 36*a*c**3*d - 39*a*c**2*d**2*x - 24*a*c*d**3*x 
**2 - 6*a*d**4*x**3 - 13*b*c**4 - 52*b*c**3*d*x - 32*b*c**2*d**2*x**2 - 8* 
b*c*d**3*x**3))/(105*c**4*d**2*(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c 
*d**3*x**3 + d**4*x**4))