\(\int \frac {A+B x}{(c+d x)^2 (c^2-d^2 x^2)^{3/2}} \, dx\) [53]

Optimal result

Integrand size = 29, antiderivative size = 120 \[ \int \frac {A+B x}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {2 (2 B c+3 A d) x}{15 c^4 d \sqrt {c^2-d^2 x^2}}+\frac {B c-A d}{5 c d^2 (c+d x)^2 \sqrt {c^2-d^2 x^2}}-\frac {2 B c+3 A d}{15 c^2 d^2 (c+d x) \sqrt {c^2-d^2 x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.88 \[ \int \frac {A+B x}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {\sqrt {c^2-d^2 x^2} \left (3 A d \left (-2 c^3+c^2 d x+4 c d^2 x^2+2 d^3 x^3\right )+B c \left (c^3+2 c^2 d x+8 c d^2 x^2+4 d^3 x^3\right )\right )}{15 c^4 d^2 (c-d x) (c+d x)^3} \] Input:

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

Output:

(Sqrt[c^2 - d^2*x^2]*(3*A*d*(-2*c^3 + c^2*d*x + 4*c*d^2*x^2 + 2*d^3*x^3) + 
 B*c*(c^3 + 2*c^2*d*x + 8*c*d^2*x^2 + 4*d^3*x^3)))/(15*c^4*d^2*(c - d*x)*( 
c + d*x)^3)
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {671, 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.

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

\(\Big \downarrow \) 671

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

\(\Big \downarrow \) 470

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

\(\Big \downarrow \) 208

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

Input:

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

Output:

(B*c - A*d)/(5*c*d^2*(c + d*x)^2*Sqrt[c^2 - d^2*x^2]) + ((2*B*c + 3*A*d)*( 
(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*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 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 = 109, normalized size of antiderivative = 0.91

method	result	size
gosper	\(-\frac {\left (-d x +c \right ) \left (-6 A \,d^{4} x^{3}-4 B c \,d^{3} x^{3}-12 A c \,d^{3} x^{2}-8 x^{2} c^{2} B \,d^{2}-3 A \,c^{2} d^{2} x -2 B \,c^{3} d x +6 A \,c^{3} d -B \,c^{4}\right )}{15 \left (d x +c \right ) c^{4} d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}\)	\(109\)
orering	\(-\frac {\left (-d x +c \right ) \left (-6 A \,d^{4} x^{3}-4 B c \,d^{3} x^{3}-12 A c \,d^{3} x^{2}-8 x^{2} c^{2} B \,d^{2}-3 A \,c^{2} d^{2} x -2 B \,c^{3} d x +6 A \,c^{3} d -B \,c^{4}\right )}{15 \left (d x +c \right ) c^{4} d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}\)	\(109\)
trager	\(-\frac {\left (-6 A \,d^{4} x^{3}-4 B c \,d^{3} x^{3}-12 A c \,d^{3} x^{2}-8 x^{2} c^{2} B \,d^{2}-3 A \,c^{2} d^{2} x -2 B \,c^{3} d x +6 A \,c^{3} d -B \,c^{4}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{15 c^{4} \left (d x +c \right )^{3} d^{2} \left (-d x +c \right )}\)	\(111\)
default	\(\frac {B \left (-\frac {1}{3 c d \left (x +\frac {c}{d}\right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}-\frac {-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d}{3 d \,c^{3} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{d^{2}}+\frac {\left (A d -B c \right ) \left (-\frac {1}{5 c d \left (x +\frac {c}{d}\right )^{2} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}+\frac {3 d \left (-\frac {1}{3 c d \left (x +\frac {c}{d}\right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}-\frac {-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d}{3 d \,c^{3} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{5 c}\right )}{d^{3}}\)	\(269\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.68 \[ \int \frac {A+B x}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2}} \, dx=-\frac {B c^{5} - 6 \, A c^{4} d - {\left (B c d^{4} - 6 \, A d^{5}\right )} x^{4} - 2 \, {\left (B c^{2} d^{3} - 6 \, A c d^{4}\right )} x^{3} + 2 \, {\left (B c^{4} d - 6 \, A c^{3} d^{2}\right )} x + {\left (B c^{4} - 6 \, A c^{3} d + 2 \, {\left (2 \, B c d^{3} + 3 \, A d^{4}\right )} x^{3} + 4 \, {\left (2 \, B c^{2} d^{2} + 3 \, A c d^{3}\right )} x^{2} + {\left (2 \, B c^{3} d + 3 \, A c^{2} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{15 \, {\left (c^{4} d^{6} x^{4} + 2 \, c^{5} d^{5} x^{3} - 2 \, c^{7} d^{3} x - c^{8} d^{2}\right )}} \] Input:

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

Output:

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

Sympy [F]

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

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 333 vs. \(2 (108) = 216\).

Time = 0.04 (sec) , antiderivative size = 333, normalized size of antiderivative = 2.78 \[ \int \frac {A+B x}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {B c}{5 \, {\left (\sqrt {-d^{2} x^{2} + c^{2}} c d^{4} x^{2} + 2 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{2} d^{3} x + \sqrt {-d^{2} x^{2} + c^{2}} c^{3} d^{2}\right )}} + \frac {B c}{5 \, {\left (\sqrt {-d^{2} x^{2} + c^{2}} c^{2} d^{3} x + \sqrt {-d^{2} x^{2} + c^{2}} c^{3} d^{2}\right )}} - \frac {A}{5 \, {\left (\sqrt {-d^{2} x^{2} + c^{2}} c d^{3} x^{2} + 2 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{2} d^{2} x + \sqrt {-d^{2} x^{2} + c^{2}} c^{3} d\right )}} - \frac {A}{5 \, {\left (\sqrt {-d^{2} x^{2} + c^{2}} c^{2} d^{2} x + \sqrt {-d^{2} x^{2} + c^{2}} c^{3} d\right )}} - \frac {B}{3 \, {\left (\sqrt {-d^{2} x^{2} + c^{2}} c d^{3} x + \sqrt {-d^{2} x^{2} + c^{2}} c^{2} d^{2}\right )}} + \frac {2 \, A x}{5 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{4}} + \frac {4 \, B x}{15 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{3} d} \] Input:

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

Output:

1/5*B*c/(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*c/(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*A/(sqrt(-d^2*x^2 + c^2)*c*d^3*x^2 + 
 2*sqrt(-d^2*x^2 + c^2)*c^2*d^2*x + sqrt(-d^2*x^2 + c^2)*c^3*d) - 1/5*A/(s 
qrt(-d^2*x^2 + c^2)*c^2*d^2*x + sqrt(-d^2*x^2 + c^2)*c^3*d) - 1/3*B/(sqrt( 
-d^2*x^2 + c^2)*c*d^3*x + sqrt(-d^2*x^2 + c^2)*c^2*d^2) + 2/5*A*x/(sqrt(-d 
^2*x^2 + c^2)*c^4) + 4/15*B*x/(sqrt(-d^2*x^2 + c^2)*c^3*d)
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.16 (sec) , antiderivative size = 321, normalized size of antiderivative = 2.68 \[ \int \frac {A+B x}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {\frac {16 \, {\left (2 i \, B c + 3 i \, A d\right )} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right )}{c^{4} d} + \frac {15 \, {\left (B c + A d\right )}}{c^{4} d \sqrt {\frac {2 \, c}{d x + c} - 1} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right )} + \frac {3 \, B c^{17} d^{4} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {5}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right )^{4} \mathrm {sgn}\left (d\right )^{4} - 3 \, A c^{16} d^{5} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {5}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right )^{4} \mathrm {sgn}\left (d\right )^{4} + 5 \, B c^{17} d^{4} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right )^{4} \mathrm {sgn}\left (d\right )^{4} - 15 \, A c^{16} d^{5} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right )^{4} \mathrm {sgn}\left (d\right )^{4} - 15 \, B c^{17} d^{4} \sqrt {\frac {2 \, c}{d x + c} - 1} \mathrm {sgn}\left (\frac {1}{d x + c}\right )^{4} \mathrm {sgn}\left (d\right )^{4} - 45 \, A c^{16} d^{5} \sqrt {\frac {2 \, c}{d x + c} - 1} \mathrm {sgn}\left (\frac {1}{d x + c}\right )^{4} \mathrm {sgn}\left (d\right )^{4}}{c^{20} d^{5} \mathrm {sgn}\left (\frac {1}{d x + c}\right )^{5} \mathrm {sgn}\left (d\right )^{5}}}{120 \, {\left | d \right |}} \] Input:

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

Output:

1/120*(16*(2*I*B*c + 3*I*A*d)*sgn(1/(d*x + c))*sgn(d)/(c^4*d) + 15*(B*c + 
A*d)/(c^4*d*sqrt(2*c/(d*x + c) - 1)*sgn(1/(d*x + c))*sgn(d)) + (3*B*c^17*d 
^4*(2*c/(d*x + c) - 1)^(5/2)*sgn(1/(d*x + c))^4*sgn(d)^4 - 3*A*c^16*d^5*(2 
*c/(d*x + c) - 1)^(5/2)*sgn(1/(d*x + c))^4*sgn(d)^4 + 5*B*c^17*d^4*(2*c/(d 
*x + c) - 1)^(3/2)*sgn(1/(d*x + c))^4*sgn(d)^4 - 15*A*c^16*d^5*(2*c/(d*x + 
 c) - 1)^(3/2)*sgn(1/(d*x + c))^4*sgn(d)^4 - 15*B*c^17*d^4*sqrt(2*c/(d*x + 
 c) - 1)*sgn(1/(d*x + c))^4*sgn(d)^4 - 45*A*c^16*d^5*sqrt(2*c/(d*x + c) - 
1)*sgn(1/(d*x + c))^4*sgn(d)^4)/(c^20*d^5*sgn(1/(d*x + c))^5*sgn(d)^5))/ab 
s(d)
 

Mupad [B] (verification not implemented)

Time = 9.82 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.91 \[ \int \frac {A+B x}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {\sqrt {c^2-d^2\,x^2}\,\left (B\,c^4+2\,B\,c^3\,d\,x-6\,A\,c^3\,d+8\,B\,c^2\,d^2\,x^2+3\,A\,c^2\,d^2\,x+4\,B\,c\,d^3\,x^3+12\,A\,c\,d^3\,x^2+6\,A\,d^4\,x^3\right )}{15\,c^4\,d^2\,{\left (c+d\,x\right )}^3\,\left (c-d\,x\right )} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 240, normalized size of antiderivative = 2.00 \[ \int \frac {A+B x}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {3 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{2} d +6 \sqrt {-d^{2} x^{2}+c^{2}}\, a c \,d^{2} x +3 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{3} x^{2}+2 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{3}+4 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2} d x +2 \sqrt {-d^{2} x^{2}+c^{2}}\, b c \,d^{2} x^{2}-12 a \,c^{3} d +6 a \,c^{2} d^{2} x +24 a c \,d^{3} x^{2}+12 a \,d^{4} x^{3}+2 b \,c^{4}+4 b \,c^{3} d x +16 b \,c^{2} d^{2} x^{2}+8 b c \,d^{3} x^{3}}{30 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{4} d^{2} \left (d^{2} x^{2}+2 c d x +c^{2}\right )} \] Input:

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

Output:

(3*sqrt(c**2 - d**2*x**2)*a*c**2*d + 6*sqrt(c**2 - d**2*x**2)*a*c*d**2*x + 
 3*sqrt(c**2 - d**2*x**2)*a*d**3*x**2 + 2*sqrt(c**2 - d**2*x**2)*b*c**3 + 
4*sqrt(c**2 - d**2*x**2)*b*c**2*d*x + 2*sqrt(c**2 - d**2*x**2)*b*c*d**2*x* 
*2 - 12*a*c**3*d + 6*a*c**2*d**2*x + 24*a*c*d**3*x**2 + 12*a*d**4*x**3 + 2 
*b*c**4 + 4*b*c**3*d*x + 16*b*c**2*d**2*x**2 + 8*b*c*d**3*x**3)/(30*sqrt(c 
**2 - d**2*x**2)*c**4*d**2*(c**2 + 2*c*d*x + d**2*x**2))