\(\int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^5} \, dx\) [9]

Optimal result

Integrand size = 29, antiderivative size = 126 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^5} \, dx=\frac {(B c-A d) \left (c^2-d^2 x^2\right )^{3/2}}{7 c d^2 (c+d x)^5}-\frac {(5 B c+2 A d) \left (c^2-d^2 x^2\right )^{3/2}}{35 c^2 d^2 (c+d x)^4}-\frac {(5 B c+2 A d) \left (c^2-d^2 x^2\right )^{3/2}}{105 c^3 d^2 (c+d x)^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.76 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.65 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^5} \, dx=-\frac {(c-d x) \sqrt {c^2-d^2 x^2} \left (5 B c \left (c^2+5 c d x+d^2 x^2\right )+A d \left (23 c^2+10 c d x+2 d^2 x^2\right )\right )}{105 c^3 d^2 (c+d x)^4} \] Input:

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

Output:

-1/105*((c - d*x)*Sqrt[c^2 - d^2*x^2]*(5*B*c*(c^2 + 5*c*d*x + d^2*x^2) + A 
*d*(23*c^2 + 10*c*d*x + 2*d^2*x^2)))/(c^3*d^2*(c + d*x)^4)
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.02, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {671, 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.

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

\(\Big \downarrow \) 671

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

\(\Big \downarrow \) 461

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

\(\Big \downarrow \) 460

\(\displaystyle \frac {\left (c^2-d^2 x^2\right )^{3/2} (B c-A d)}{7 c d^2 (c+d x)^5}+\frac {\left (-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{15 c^2 d (c+d x)^3}-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{5 c d (c+d x)^4}\right ) (2 A d+5 B c)}{7 c d}\)

Input:

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

Output:

((B*c - A*d)*(c^2 - d^2*x^2)^(3/2))/(7*c*d^2*(c + d*x)^5) + ((5*B*c + 2*A* 
d)*(-1/5*(c^2 - d^2*x^2)^(3/2)/(c*d*(c + d*x)^4) - (c^2 - d^2*x^2)^(3/2)/( 
15*c^2*d*(c + d*x)^3)))/(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 = 0.48 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.67

method	result	size
gosper	\(-\frac {\left (-d x +c \right ) \left (2 A \,d^{3} x^{2}+5 B c \,d^{2} x^{2}+10 A c \,d^{2} x +25 B \,c^{2} d x +23 A \,c^{2} d +5 B \,c^{3}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{105 \left (d x +c \right )^{4} c^{3} d^{2}}\)	\(85\)
orering	\(-\frac {\left (-d x +c \right ) \left (2 A \,d^{3} x^{2}+5 B c \,d^{2} x^{2}+10 A c \,d^{2} x +25 B \,c^{2} d x +23 A \,c^{2} d +5 B \,c^{3}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{105 \left (d x +c \right )^{4} c^{3} d^{2}}\)	\(85\)
trager	\(-\frac {\left (-2 A \,d^{4} x^{3}-5 B c \,d^{3} x^{3}-8 A c \,d^{3} x^{2}-20 x^{2} c^{2} B \,d^{2}-13 A \,c^{2} d^{2} x +20 B \,c^{3} d x +23 A \,c^{3} d +5 B \,c^{4}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{105 c^{3} \left (d x +c \right )^{4} d^{2}}\)	\(103\)
default	\(\frac {B \left (-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{5 c d \left (x +\frac {c}{d}\right )^{4}}-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{15 c^{2} \left (x +\frac {c}{d}\right )^{3}}\right )}{d^{5}}+\frac {\left (A d -B c \right ) \left (-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{7 c d \left (x +\frac {c}{d}\right )^{5}}+\frac {2 d \left (-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{5 c d \left (x +\frac {c}{d}\right )^{4}}-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{15 c^{2} \left (x +\frac {c}{d}\right )^{3}}\right )}{7 c}\right )}{d^{6}}\)	\(247\)

Input:

int((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^5,x,method=_RETURNVERBOSE)
 

Output:

-1/105*(-d*x+c)*(2*A*d^3*x^2+5*B*c*d^2*x^2+10*A*c*d^2*x+25*B*c^2*d*x+23*A* 
c^2*d+5*B*c^3)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^4/c^3/d^2
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 240 vs. \(2 (114) = 228\).

Time = 0.09 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.90 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^5} \, dx=-\frac {5 \, B c^{5} + 23 \, A c^{4} d + {\left (5 \, B c d^{4} + 23 \, A d^{5}\right )} x^{4} + 4 \, {\left (5 \, B c^{2} d^{3} + 23 \, A c d^{4}\right )} x^{3} + 6 \, {\left (5 \, B c^{3} d^{2} + 23 \, A c^{2} d^{3}\right )} x^{2} + 4 \, {\left (5 \, B c^{4} d + 23 \, A c^{3} d^{2}\right )} x + {\left (5 \, B c^{4} + 23 \, A c^{3} d - {\left (5 \, B c d^{3} + 2 \, A d^{4}\right )} x^{3} - 4 \, {\left (5 \, B c^{2} d^{2} + 2 \, A c d^{3}\right )} x^{2} + {\left (20 \, B c^{3} d - 13 \, A c^{2} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{105 \, {\left (c^{3} d^{6} x^{4} + 4 \, c^{4} d^{5} x^{3} + 6 \, c^{5} d^{4} x^{2} + 4 \, c^{6} d^{3} x + c^{7} d^{2}\right )}} \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^5,x, algorithm="fricas")
 

Output:

-1/105*(5*B*c^5 + 23*A*c^4*d + (5*B*c*d^4 + 23*A*d^5)*x^4 + 4*(5*B*c^2*d^3 
 + 23*A*c*d^4)*x^3 + 6*(5*B*c^3*d^2 + 23*A*c^2*d^3)*x^2 + 4*(5*B*c^4*d + 2 
3*A*c^3*d^2)*x + (5*B*c^4 + 23*A*c^3*d - (5*B*c*d^3 + 2*A*d^4)*x^3 - 4*(5* 
B*c^2*d^2 + 2*A*c*d^3)*x^2 + (20*B*c^3*d - 13*A*c^2*d^2)*x)*sqrt(-d^2*x^2 
+ c^2))/(c^3*d^6*x^4 + 4*c^4*d^5*x^3 + 6*c^5*d^4*x^2 + 4*c^6*d^3*x + c^7*d 
^2)
 

Sympy [F]

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

integrate((B*x+A)*(-d**2*x**2+c**2)**(1/2)/(d*x+c)**5,x)
 

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 526 vs. \(2 (114) = 228\).

Time = 0.05 (sec) , antiderivative size = 526, normalized size of antiderivative = 4.17 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^5} \, dx=\frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} B c}{7 \, {\left (d^{6} x^{4} + 4 \, c d^{5} x^{3} + 6 \, c^{2} d^{4} x^{2} + 4 \, c^{3} d^{3} x + c^{4} d^{2}\right )}} - \frac {\sqrt {-d^{2} x^{2} + c^{2}} B c}{35 \, {\left (c d^{5} x^{3} + 3 \, c^{2} d^{4} x^{2} + 3 \, c^{3} d^{3} x + c^{4} d^{2}\right )}} - \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} B c}{105 \, {\left (c^{2} d^{4} x^{2} + 2 \, c^{3} d^{3} x + c^{4} d^{2}\right )}} - \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} B c}{105 \, {\left (c^{3} d^{3} x + c^{4} d^{2}\right )}} - \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} A}{7 \, {\left (d^{5} x^{4} + 4 \, c d^{4} x^{3} + 6 \, c^{2} d^{3} x^{2} + 4 \, c^{3} d^{2} x + c^{4} d\right )}} + \frac {\sqrt {-d^{2} x^{2} + c^{2}} A}{35 \, {\left (c d^{4} x^{3} + 3 \, c^{2} d^{3} x^{2} + 3 \, c^{3} d^{2} x + c^{4} d\right )}} + \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} A}{105 \, {\left (c^{2} d^{3} x^{2} + 2 \, c^{3} d^{2} x + c^{4} d\right )}} + \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} A}{105 \, {\left (c^{3} d^{2} x + c^{4} d\right )}} - \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} B}{5 \, {\left (d^{5} x^{3} + 3 \, c d^{4} x^{2} + 3 \, c^{2} d^{3} x + c^{3} d^{2}\right )}} + \frac {\sqrt {-d^{2} x^{2} + c^{2}} B}{15 \, {\left (c d^{4} x^{2} + 2 \, c^{2} d^{3} x + c^{3} d^{2}\right )}} + \frac {\sqrt {-d^{2} x^{2} + c^{2}} B}{15 \, {\left (c^{2} d^{3} x + c^{3} d^{2}\right )}} \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^5,x, algorithm="maxima")
 

Output:

2/7*sqrt(-d^2*x^2 + c^2)*B*c/(d^6*x^4 + 4*c*d^5*x^3 + 6*c^2*d^4*x^2 + 4*c^ 
3*d^3*x + c^4*d^2) - 1/35*sqrt(-d^2*x^2 + c^2)*B*c/(c*d^5*x^3 + 3*c^2*d^4* 
x^2 + 3*c^3*d^3*x + c^4*d^2) - 2/105*sqrt(-d^2*x^2 + c^2)*B*c/(c^2*d^4*x^2 
 + 2*c^3*d^3*x + c^4*d^2) - 2/105*sqrt(-d^2*x^2 + c^2)*B*c/(c^3*d^3*x + c^ 
4*d^2) - 2/7*sqrt(-d^2*x^2 + c^2)*A/(d^5*x^4 + 4*c*d^4*x^3 + 6*c^2*d^3*x^2 
 + 4*c^3*d^2*x + c^4*d) + 1/35*sqrt(-d^2*x^2 + c^2)*A/(c*d^4*x^3 + 3*c^2*d 
^3*x^2 + 3*c^3*d^2*x + c^4*d) + 2/105*sqrt(-d^2*x^2 + c^2)*A/(c^2*d^3*x^2 
+ 2*c^3*d^2*x + c^4*d) + 2/105*sqrt(-d^2*x^2 + c^2)*A/(c^3*d^2*x + c^4*d) 
- 2/5*sqrt(-d^2*x^2 + c^2)*B/(d^5*x^3 + 3*c*d^4*x^2 + 3*c^2*d^3*x + c^3*d^ 
2) + 1/15*sqrt(-d^2*x^2 + c^2)*B/(c*d^4*x^2 + 2*c^2*d^3*x + c^3*d^2) + 1/1 
5*sqrt(-d^2*x^2 + c^2)*B/(c^2*d^3*x + c^3*d^2)
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.14 (sec) , antiderivative size = 388, normalized size of antiderivative = 3.08 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^5} \, dx=-\frac {1}{420} \, {\left (\frac {\frac {3 \, {\left (5 \, {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {7}{2}} + 21 \, {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {5}{2}} + 35 \, {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {3}{2}} + 35 \, \sqrt {\frac {2 \, c}{d x + c} - 1}\right )} A \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right )}{c^{2}} - \frac {7 \, {\left (3 \, {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {5}{2}} + 10 \, {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {3}{2}} + 15 \, \sqrt {\frac {2 \, c}{d x + c} - 1}\right )} A \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right )}{c^{2}} - \frac {3 \, {\left (5 \, {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {7}{2}} + 21 \, {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {5}{2}} + 35 \, {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {3}{2}} + 35 \, \sqrt {\frac {2 \, c}{d x + c} - 1}\right )} B \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right )}{c d} + \frac {21 \, {\left (3 \, {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {5}{2}} + 10 \, {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {3}{2}} + 15 \, \sqrt {\frac {2 \, c}{d x + c} - 1}\right )} B \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right )}{c d} - \frac {70 \, {\left ({\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {3}{2}} + 3 \, \sqrt {\frac {2 \, c}{d x + c} - 1}\right )} B \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right )}{c d}}{c d^{2}} + \frac {4 \, {\left (5 i \, B c + 2 i \, A d\right )} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right )}{c^{3} d^{3}}\right )} {\left | d \right |} \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^5,x, algorithm="giac")
 

Output:

-1/420*((3*(5*(2*c/(d*x + c) - 1)^(7/2) + 21*(2*c/(d*x + c) - 1)^(5/2) + 3 
5*(2*c/(d*x + c) - 1)^(3/2) + 35*sqrt(2*c/(d*x + c) - 1))*A*sgn(1/(d*x + c 
))*sgn(d)/c^2 - 7*(3*(2*c/(d*x + c) - 1)^(5/2) + 10*(2*c/(d*x + c) - 1)^(3 
/2) + 15*sqrt(2*c/(d*x + c) - 1))*A*sgn(1/(d*x + c))*sgn(d)/c^2 - 3*(5*(2* 
c/(d*x + c) - 1)^(7/2) + 21*(2*c/(d*x + c) - 1)^(5/2) + 35*(2*c/(d*x + c) 
- 1)^(3/2) + 35*sqrt(2*c/(d*x + c) - 1))*B*sgn(1/(d*x + c))*sgn(d)/(c*d) + 
 21*(3*(2*c/(d*x + c) - 1)^(5/2) + 10*(2*c/(d*x + c) - 1)^(3/2) + 15*sqrt( 
2*c/(d*x + c) - 1))*B*sgn(1/(d*x + c))*sgn(d)/(c*d) - 70*((2*c/(d*x + c) - 
 1)^(3/2) + 3*sqrt(2*c/(d*x + c) - 1))*B*sgn(1/(d*x + c))*sgn(d)/(c*d))/(c 
*d^2) + 4*(5*I*B*c + 2*I*A*d)*sgn(1/(d*x + c))*sgn(d)/(c^3*d^3))*abs(d)
 

Mupad [B] (verification not implemented)

Time = 10.57 (sec) , antiderivative size = 589, normalized size of antiderivative = 4.67 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^5} \, dx=\frac {\sqrt {c^2-d^2\,x^2}\,\left (\frac {B}{35\,c^2\,d^2}-\frac {2\,A\,d-22\,B\,c}{105\,c^3\,d^2}\right )}{c+d\,x}-\frac {\sqrt {c^2-d^2\,x^2}\,\left (\frac {B}{21\,c\,d^2}-\frac {3\,A\,d-9\,B\,c}{105\,c^2\,d^2}\right )}{{\left (c+d\,x\right )}^2}-\frac {\sqrt {c^2-d^2\,x^2}\,\left (\frac {A}{7\,d}-\frac {c\,\left (\frac {B}{7\,d}-\frac {A\,d-B\,c}{7\,c\,d}\right )}{d}\right )}{{\left (c+d\,x\right )}^4}-\frac {\sqrt {c^2-d^2\,x^2}\,\left (\frac {8\,B\,c^2-A\,c\,d}{35\,c^2\,d^2}-\frac {c\,\left (\frac {A\,d^2-9\,B\,c\,d}{35\,c^2\,d^2}-\frac {B}{35\,c\,d}\right )}{d}\right )}{{\left (c+d\,x\right )}^3}+\frac {\sqrt {c^2-d^2\,x^2}\,\left (\frac {14\,B\,c^2-A\,c\,d}{105\,c^3\,d^2}-\frac {c\,\left (\frac {A\,d^2-15\,B\,c\,d}{105\,c^3\,d^2}-\frac {B}{105\,c^2\,d}\right )}{d}\right )}{{\left (c+d\,x\right )}^2}-\frac {\sqrt {c^2-d^2\,x^2}\,\left (\frac {18\,B\,c^2-A\,c\,d}{105\,c^4\,d^2}-\frac {c\,\left (\frac {A\,d^2-19\,B\,c\,d}{105\,c^4\,d^2}-\frac {B}{105\,c^3\,d}\right )}{d}\right )}{c+d\,x}+\frac {\sqrt {c^2-d^2\,x^2}\,\left (3\,A\,d+B\,c\right )}{105\,c^3\,d^2\,\left (c+d\,x\right )}+\frac {c\,\sqrt {c^2-d^2\,x^2}\,\left (\frac {B}{35\,c\,d}-\frac {A\,d-2\,B\,c}{35\,c^2\,d}\right )}{d\,{\left (c+d\,x\right )}^3}-\frac {c\,\sqrt {c^2-d^2\,x^2}\,\left (\frac {B}{105\,c^2\,d}-\frac {A\,d-10\,B\,c}{105\,c^3\,d}\right )}{d\,{\left (c+d\,x\right )}^2}+\frac {c\,\sqrt {c^2-d^2\,x^2}\,\left (\frac {B}{105\,c^3\,d}-\frac {A\,d-16\,B\,c}{105\,c^4\,d}\right )}{d\,\left (c+d\,x\right )} \] Input:

int(((c^2 - d^2*x^2)^(1/2)*(A + B*x))/(c + d*x)^5,x)
 

Output:

((c^2 - d^2*x^2)^(1/2)*(B/(35*c^2*d^2) - (2*A*d - 22*B*c)/(105*c^3*d^2)))/ 
(c + d*x) - ((c^2 - d^2*x^2)^(1/2)*(B/(21*c*d^2) - (3*A*d - 9*B*c)/(105*c^ 
2*d^2)))/(c + d*x)^2 - ((c^2 - d^2*x^2)^(1/2)*(A/(7*d) - (c*(B/(7*d) - (A* 
d - B*c)/(7*c*d)))/d))/(c + d*x)^4 - ((c^2 - d^2*x^2)^(1/2)*((8*B*c^2 - A* 
c*d)/(35*c^2*d^2) - (c*((A*d^2 - 9*B*c*d)/(35*c^2*d^2) - B/(35*c*d)))/d))/ 
(c + d*x)^3 + ((c^2 - d^2*x^2)^(1/2)*((14*B*c^2 - A*c*d)/(105*c^3*d^2) - ( 
c*((A*d^2 - 15*B*c*d)/(105*c^3*d^2) - B/(105*c^2*d)))/d))/(c + d*x)^2 - (( 
c^2 - d^2*x^2)^(1/2)*((18*B*c^2 - A*c*d)/(105*c^4*d^2) - (c*((A*d^2 - 19*B 
*c*d)/(105*c^4*d^2) - B/(105*c^3*d)))/d))/(c + d*x) + ((c^2 - d^2*x^2)^(1/ 
2)*(3*A*d + B*c))/(105*c^3*d^2*(c + d*x)) + (c*(c^2 - d^2*x^2)^(1/2)*(B/(3 
5*c*d) - (A*d - 2*B*c)/(35*c^2*d)))/(d*(c + d*x)^3) - (c*(c^2 - d^2*x^2)^( 
1/2)*(B/(105*c^2*d) - (A*d - 10*B*c)/(105*c^3*d)))/(d*(c + d*x)^2) + (c*(c 
^2 - d^2*x^2)^(1/2)*(B/(105*c^3*d) - (A*d - 16*B*c)/(105*c^4*d)))/(d*(c + 
d*x))
 

Reduce [F]

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

int((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^5,x)
 

Output:

int((B*x+A)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^5,x)