\(\int \frac {A+B x}{(c+d x) (c^2-d^2 x^2)^{5/2}} \, dx\) [61]

Optimal result

Integrand size = 29, antiderivative size = 112 \[ \int \frac {A+B x}{(c+d x) \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {(B c+4 A d) x}{15 c^3 d \left (c^2-d^2 x^2\right )^{3/2}}+\frac {B c-A d}{5 c d^2 (c+d x) \left (c^2-d^2 x^2\right )^{3/2}}+\frac {2 (B c+4 A d) x}{15 c^5 d \sqrt {c^2-d^2 x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.16 \[ \int \frac {A+B x}{(c+d x) \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {c^2-d^2 x^2} \left (A d \left (-3 c^4+12 c^3 d x+12 c^2 d^2 x^2-8 c d^3 x^3-8 d^4 x^4\right )+B c \left (3 c^4+3 c^3 d x+3 c^2 d^2 x^2-2 c d^3 x^3-2 d^4 x^4\right )\right )}{15 c^5 d^2 (c-d x)^2 (c+d x)^3} \] Input:

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

Output:

(Sqrt[c^2 - d^2*x^2]*(A*d*(-3*c^4 + 12*c^3*d*x + 12*c^2*d^2*x^2 - 8*c*d^3* 
x^3 - 8*d^4*x^4) + B*c*(3*c^4 + 3*c^3*d*x + 3*c^2*d^2*x^2 - 2*c*d^3*x^3 - 
2*d^4*x^4)))/(15*c^5*d^2*(c - d*x)^2*(c + d*x)^3)
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {671, 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.

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

\(\Big \downarrow \) 671

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

\(\Big \downarrow \) 209

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

\(\Big \downarrow \) 208

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

Input:

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

Output:

(B*c - A*d)/(5*c*d^2*(c + d*x)*(c^2 - d^2*x^2)^(3/2)) + ((B*c + 4*A*d)*(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*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 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 = 0.38 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.12

method	result	size
gosper	\(-\frac {\left (-d x +c \right ) \left (8 A \,d^{5} x^{4}+2 B c \,d^{4} x^{4}+8 A c \,d^{4} x^{3}+2 B \,c^{2} d^{3} x^{3}-12 A \,c^{2} d^{3} x^{2}-3 B \,c^{3} d^{2} x^{2}-12 A \,c^{3} d^{2} x -3 B \,c^{4} d x +3 A \,c^{4} d -3 B \,c^{5}\right )}{15 c^{5} d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}\)	\(126\)
orering	\(-\frac {\left (-d x +c \right ) \left (8 A \,d^{5} x^{4}+2 B c \,d^{4} x^{4}+8 A c \,d^{4} x^{3}+2 B \,c^{2} d^{3} x^{3}-12 A \,c^{2} d^{3} x^{2}-3 B \,c^{3} d^{2} x^{2}-12 A \,c^{3} d^{2} x -3 B \,c^{4} d x +3 A \,c^{4} d -3 B \,c^{5}\right )}{15 c^{5} d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}\)	\(126\)
trager	\(-\frac {\left (8 A \,d^{5} x^{4}+2 B c \,d^{4} x^{4}+8 A c \,d^{4} x^{3}+2 B \,c^{2} d^{3} x^{3}-12 A \,c^{2} d^{3} x^{2}-3 B \,c^{3} d^{2} x^{2}-12 A \,c^{3} d^{2} x -3 B \,c^{4} d x +3 A \,c^{4} d -3 B \,c^{5}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{15 c^{5} \left (d x +c \right )^{3} \left (-d x +c \right )^{2} d^{2}}\)	\(135\)
default	\(\frac {B \left (\frac {x}{3 c^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{4} \sqrt {-d^{2} x^{2}+c^{2}}}\right )}{d}+\frac {\left (A d -B c \right ) \left (-\frac {1}{5 c d \left (x +\frac {c}{d}\right ) \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}+\frac {4 d \left (-\frac {-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d}{6 c^{2} d^{2} \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}-\frac {-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d}{3 d^{2} c^{4} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{5 c}\right )}{d^{2}}\)	\(219\)

Input:

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

Output:

-1/15*(-d*x+c)*(8*A*d^5*x^4+2*B*c*d^4*x^4+8*A*c*d^4*x^3+2*B*c^2*d^3*x^3-12 
*A*c^2*d^3*x^2-3*B*c^3*d^2*x^2-12*A*c^3*d^2*x-3*B*c^4*d*x+3*A*c^4*d-3*B*c^ 
5)/c^5/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. 290 vs. \(2 (100) = 200\).

Time = 0.11 (sec) , antiderivative size = 290, normalized size of antiderivative = 2.59 \[ \int \frac {A+B x}{(c+d x) \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {3 \, B c^{6} - 3 \, A c^{5} d + 3 \, {\left (B c d^{5} - A d^{6}\right )} x^{5} + 3 \, {\left (B c^{2} d^{4} - A c d^{5}\right )} x^{4} - 6 \, {\left (B c^{3} d^{3} - A c^{2} d^{4}\right )} x^{3} - 6 \, {\left (B c^{4} d^{2} - A c^{3} d^{3}\right )} x^{2} + 3 \, {\left (B c^{5} d - A c^{4} d^{2}\right )} x + {\left (3 \, B c^{5} - 3 \, A c^{4} d - 2 \, {\left (B c d^{4} + 4 \, A d^{5}\right )} x^{4} - 2 \, {\left (B c^{2} d^{3} + 4 \, A c d^{4}\right )} x^{3} + 3 \, {\left (B c^{3} d^{2} + 4 \, A c^{2} d^{3}\right )} x^{2} + 3 \, {\left (B c^{4} d + 4 \, A c^{3} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{15 \, {\left (c^{5} d^{7} x^{5} + c^{6} d^{6} x^{4} - 2 \, c^{7} d^{5} x^{3} - 2 \, c^{8} d^{4} x^{2} + c^{9} d^{3} x + c^{10} d^{2}\right )}} \] Input:

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

Output:

1/15*(3*B*c^6 - 3*A*c^5*d + 3*(B*c*d^5 - A*d^6)*x^5 + 3*(B*c^2*d^4 - A*c*d 
^5)*x^4 - 6*(B*c^3*d^3 - A*c^2*d^4)*x^3 - 6*(B*c^4*d^2 - A*c^3*d^3)*x^2 + 
3*(B*c^5*d - A*c^4*d^2)*x + (3*B*c^5 - 3*A*c^4*d - 2*(B*c*d^4 + 4*A*d^5)*x 
^4 - 2*(B*c^2*d^3 + 4*A*c*d^4)*x^3 + 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 + c^6*d^6*x^4 
 - 2*c^7*d^5*x^3 - 2*c^8*d^4*x^2 + c^9*d^3*x + c^10*d^2)
 

Sympy [F]

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

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

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.64 \[ \int \frac {A+B x}{(c+d x) \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {B c}{5 \, {\left ({\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c d^{3} x + {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c^{2} d^{2}\right )}} - \frac {A}{5 \, {\left ({\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c d^{2} x + {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c^{2} d\right )}} + \frac {4 \, A x}{15 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c^{3}} + \frac {B x}{15 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c^{2} d} + \frac {8 \, A x}{15 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{5}} + \frac {2 \, B x}{15 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{4} d} \] Input:

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 10.03 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.20 \[ \int \frac {A+B x}{(c+d x) \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {c^2-d^2\,x^2}\,\left (3\,B\,c^5+3\,B\,c^4\,d\,x-3\,A\,c^4\,d+3\,B\,c^3\,d^2\,x^2+12\,A\,c^3\,d^2\,x-2\,B\,c^2\,d^3\,x^3+12\,A\,c^2\,d^3\,x^2-2\,B\,c\,d^4\,x^4-8\,A\,c\,d^4\,x^3-8\,A\,d^5\,x^4\right )}{15\,c^5\,d^2\,{\left (c+d\,x\right )}^3\,{\left (c-d\,x\right )}^2} \] Input:

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

Output:

((c^2 - d^2*x^2)^(1/2)*(3*B*c^5 - 8*A*d^5*x^4 - 3*A*c^4*d + 3*B*c^4*d*x + 
12*A*c^3*d^2*x - 8*A*c*d^4*x^3 - 2*B*c*d^4*x^4 + 12*A*c^2*d^3*x^2 + 3*B*c^ 
3*d^2*x^2 - 2*B*c^2*d^3*x^3))/(15*c^5*d^2*(c + d*x)^3*(c - d*x)^2)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 325, normalized size of antiderivative = 2.90 \[ \int \frac {A+B x}{(c+d x) \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {12 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{3} d +12 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{2} d^{2} x -12 \sqrt {-d^{2} x^{2}+c^{2}}\, a c \,d^{3} x^{2}-12 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{4} x^{3}+3 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{4}+3 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{3} d x -3 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2} d^{2} x^{2}-3 \sqrt {-d^{2} x^{2}+c^{2}}\, b c \,d^{3} x^{3}-3 a \,c^{4} d +12 a \,c^{3} d^{2} x +12 a \,c^{2} d^{3} x^{2}-8 a c \,d^{4} x^{3}-8 a \,d^{5} x^{4}+3 b \,c^{5}+3 b \,c^{4} d x +3 b \,c^{3} d^{2} x^{2}-2 b \,c^{2} d^{3} x^{3}-2 b c \,d^{4} x^{4}}{15 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{5} d^{2} \left (-d^{3} x^{3}-c \,d^{2} x^{2}+c^{2} d x +c^{3}\right )} \] Input:

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

Output:

(12*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 - 12*sqrt(c**2 - d**2*x**2)* 
a*d**4*x**3 + 3*sqrt(c**2 - d**2*x**2)*b*c**4 + 3*sqrt(c**2 - d**2*x**2)*b 
*c**3*d*x - 3*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 - 3*a*c**4*d + 12*a*c**3*d**2*x + 12*a*c**2*d**3*x**2 
 - 8*a*c*d**4*x**3 - 8*a*d**5*x**4 + 3*b*c**5 + 3*b*c**4*d*x + 3*b*c**3*d* 
*2*x**2 - 2*b*c**2*d**3*x**3 - 2*b*c*d**4*x**4)/(15*sqrt(c**2 - d**2*x**2) 
*c**5*d**2*(c**3 + c**2*d*x - c*d**2*x**2 - d**3*x**3))