\(\int \frac {A+B x}{(c+d x)^{3/2} (c^2-d^2 x^2)^{3/2}} \, dx\) [105]

Optimal result

Integrand size = 31, antiderivative size = 157 \[ \int \frac {A+B x}{(c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {B c-A d}{4 c d^2 (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}+\frac {(3 B c+5 A d) (c+3 d x)}{32 c^3 d^2 \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}-\frac {3 (3 B c+5 A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{32 \sqrt {2} c^{7/2} d^2} \] Output:

1/4*(-A*d+B*c)/c/d^2/(d*x+c)^(3/2)/(-d^2*x^2+c^2)^(1/2)+1/32*(5*A*d+3*B*c) 
*(3*d*x+c)/c^3/d^2/(d*x+c)^(1/2)/(-d^2*x^2+c^2)^(1/2)-3/64*(5*A*d+3*B*c)*a 
rctanh(2^(1/2)*c^(1/2)*(d*x+c)^(1/2)/(-d^2*x^2+c^2)^(1/2))*2^(1/2)/c^(7/2) 
/d^2
 

Mathematica [A] (verified)

Time = 1.48 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.96 \[ \int \frac {A+B x}{(c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {\frac {2 \sqrt {c} \sqrt {c^2-d^2 x^2} \left (B c \left (11 c^2+12 c d x+9 d^2 x^2\right )+A d \left (-3 c^2+20 c d x+15 d^2 x^2\right )\right )}{(c-d x) (c+d x)^{5/2}}-3 \sqrt {2} (3 B c+5 A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{64 c^{7/2} d^2} \] Input:

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

Output:

((2*Sqrt[c]*Sqrt[c^2 - d^2*x^2]*(B*c*(11*c^2 + 12*c*d*x + 9*d^2*x^2) + A*d 
*(-3*c^2 + 20*c*d*x + 15*d^2*x^2)))/((c - d*x)*(c + d*x)^(5/2)) - 3*Sqrt[2 
]*(3*B*c + 5*A*d)*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[c + d*x])/Sqrt[c^2 - d^2*x 
^2]])/(64*c^(7/2)*d^2)
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.22, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {671, 470, 467, 471, 221}

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.

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

\(\Big \downarrow \) 671

\(\displaystyle \frac {(5 A d+3 B c) \int \frac {1}{\sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}dx}{8 c d}+\frac {B c-A d}{4 c d^2 (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}\)

\(\Big \downarrow \) 470

\(\displaystyle \frac {(5 A d+3 B c) \left (\frac {3 \int \frac {\sqrt {c+d x}}{\left (c^2-d^2 x^2\right )^{3/2}}dx}{4 c}-\frac {1}{2 c d \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}\right )}{8 c d}+\frac {B c-A d}{4 c d^2 (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}\)

\(\Big \downarrow \) 467

\(\displaystyle \frac {(5 A d+3 B c) \left (\frac {3 \left (\frac {\int \frac {1}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx}{2 c}+\frac {\sqrt {c+d x}}{c d \sqrt {c^2-d^2 x^2}}\right )}{4 c}-\frac {1}{2 c d \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}\right )}{8 c d}+\frac {B c-A d}{4 c d^2 (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}\)

\(\Big \downarrow \) 471

\(\displaystyle \frac {(5 A d+3 B c) \left (\frac {3 \left (\frac {d \int \frac {1}{\frac {d^2 \left (c^2-d^2 x^2\right )}{c+d x}-2 c d^2}d\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}}{c}+\frac {\sqrt {c+d x}}{c d \sqrt {c^2-d^2 x^2}}\right )}{4 c}-\frac {1}{2 c d \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}\right )}{8 c d}+\frac {B c-A d}{4 c d^2 (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {(5 A d+3 B c) \left (\frac {3 \left (\frac {\sqrt {c+d x}}{c d \sqrt {c^2-d^2 x^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {c+d x}}\right )}{\sqrt {2} c^{3/2} d}\right )}{4 c}-\frac {1}{2 c d \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}\right )}{8 c d}+\frac {B c-A d}{4 c d^2 (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}\)

Input:

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

Output:

(B*c - A*d)/(4*c*d^2*(c + d*x)^(3/2)*Sqrt[c^2 - d^2*x^2]) + ((3*B*c + 5*A* 
d)*(-1/2*1/(c*d*Sqrt[c + d*x]*Sqrt[c^2 - d^2*x^2]) + (3*(Sqrt[c + d*x]/(c* 
d*Sqrt[c^2 - d^2*x^2]) - ArcTanh[Sqrt[c^2 - d^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt 
[c + d*x])]/(Sqrt[2]*c^(3/2)*d)))/(4*c)))/(8*c*d)
 

Defintions of rubi rules used

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 467

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

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 471

Int[1/(Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Sim 
p[2*d   Subst[Int[1/(2*b*c + d^2*x^2), x], x, Sqrt[a + b*x^2]/Sqrt[c + d*x] 
], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(309\) vs. \(2(131)=262\).

Time = 0.36 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.97

method	result	size
default	\(-\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, \left (15 A \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) d^{3} x^{2}+9 B \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \,d^{2} x^{2}+30 A \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \,d^{2} x +18 B \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d x -22 B \,c^{\frac {7}{2}}-24 B \,c^{\frac {5}{2}} d x -18 B \,c^{\frac {3}{2}} d^{2} x^{2}+15 A \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d +6 A \,c^{\frac {5}{2}} d -40 A \,c^{\frac {3}{2}} d^{2} x -30 A \sqrt {c}\, d^{3} x^{2}+9 B \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3}\right )}{64 \left (d x +c \right )^{\frac {5}{2}} \left (-d x +c \right ) d^{2} c^{\frac {7}{2}}}\)	\(310\)

Input:

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

Output:

-1/64/(d*x+c)^(5/2)*(-d^2*x^2+c^2)^(1/2)*(15*A*(-d*x+c)^(1/2)*2^(1/2)*arct 
anh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*d^3*x^2+9*B*(-d*x+c)^(1/2)*2^(1/2) 
*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c*d^2*x^2+30*A*(-d*x+c)^(1/2) 
*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c*d^2*x+18*B*(-d*x+c) 
^(1/2)*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^2*d*x-22*B*c^ 
(7/2)-24*B*c^(5/2)*d*x-18*B*c^(3/2)*d^2*x^2+15*A*(-d*x+c)^(1/2)*2^(1/2)*ar 
ctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^2*d+6*A*c^(5/2)*d-40*A*c^(3/2) 
*d^2*x-30*A*c^(1/2)*d^3*x^2+9*B*(-d*x+c)^(1/2)*2^(1/2)*arctanh(1/2*(-d*x+c 
)^(1/2)*2^(1/2)/c^(1/2))*c^3)/(-d*x+c)/d^2/c^(7/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 525, normalized size of antiderivative = 3.34 \[ \int \frac {A+B x}{(c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\left [-\frac {3 \, \sqrt {2} {\left (3 \, B c^{5} + 5 \, A c^{4} d - {\left (3 \, B c d^{4} + 5 \, A d^{5}\right )} x^{4} - 2 \, {\left (3 \, B c^{2} d^{3} + 5 \, A c d^{4}\right )} x^{3} + 2 \, {\left (3 \, B c^{4} d + 5 \, A c^{3} d^{2}\right )} x\right )} \sqrt {c} \log \left (-\frac {d^{2} x^{2} - 2 \, c d x + 2 \, \sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {c} - 3 \, c^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + 4 \, {\left (11 \, B c^{4} - 3 \, A c^{3} d + 3 \, {\left (3 \, B c^{2} d^{2} + 5 \, A c d^{3}\right )} x^{2} + 4 \, {\left (3 \, B c^{3} d + 5 \, A c^{2} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{128 \, {\left (c^{4} d^{6} x^{4} + 2 \, c^{5} d^{5} x^{3} - 2 \, c^{7} d^{3} x - c^{8} d^{2}\right )}}, -\frac {3 \, \sqrt {2} {\left (3 \, B c^{5} + 5 \, A c^{4} d - {\left (3 \, B c d^{4} + 5 \, A d^{5}\right )} x^{4} - 2 \, {\left (3 \, B c^{2} d^{3} + 5 \, A c d^{4}\right )} x^{3} + 2 \, {\left (3 \, B c^{4} d + 5 \, A c^{3} d^{2}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {-c}}{2 \, {\left (c d x + c^{2}\right )}}\right ) + 2 \, {\left (11 \, B c^{4} - 3 \, A c^{3} d + 3 \, {\left (3 \, B c^{2} d^{2} + 5 \, A c d^{3}\right )} x^{2} + 4 \, {\left (3 \, B c^{3} d + 5 \, A c^{2} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{64 \, {\left (c^{4} d^{6} x^{4} + 2 \, c^{5} d^{5} x^{3} - 2 \, c^{7} d^{3} x - c^{8} d^{2}\right )}}\right ] \] Input:

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

Output:

[-1/128*(3*sqrt(2)*(3*B*c^5 + 5*A*c^4*d - (3*B*c*d^4 + 5*A*d^5)*x^4 - 2*(3 
*B*c^2*d^3 + 5*A*c*d^4)*x^3 + 2*(3*B*c^4*d + 5*A*c^3*d^2)*x)*sqrt(c)*log(- 
(d^2*x^2 - 2*c*d*x + 2*sqrt(2)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)*sqrt(c) 
- 3*c^2)/(d^2*x^2 + 2*c*d*x + c^2)) + 4*(11*B*c^4 - 3*A*c^3*d + 3*(3*B*c^2 
*d^2 + 5*A*c*d^3)*x^2 + 4*(3*B*c^3*d + 5*A*c^2*d^2)*x)*sqrt(-d^2*x^2 + c^2 
)*sqrt(d*x + c))/(c^4*d^6*x^4 + 2*c^5*d^5*x^3 - 2*c^7*d^3*x - c^8*d^2), -1 
/64*(3*sqrt(2)*(3*B*c^5 + 5*A*c^4*d - (3*B*c*d^4 + 5*A*d^5)*x^4 - 2*(3*B*c 
^2*d^3 + 5*A*c*d^4)*x^3 + 2*(3*B*c^4*d + 5*A*c^3*d^2)*x)*sqrt(-c)*arctan(1 
/2*sqrt(2)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)*sqrt(-c)/(c*d*x + c^2)) + 2* 
(11*B*c^4 - 3*A*c^3*d + 3*(3*B*c^2*d^2 + 5*A*c*d^3)*x^2 + 4*(3*B*c^3*d + 5 
*A*c^2*d^2)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(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]

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

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.89 \[ \int \frac {A+B x}{(c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {\frac {3 \, \sqrt {2} {\left (3 \, B c + 5 \, A d\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-d x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c} c^{3} d} + \frac {16 \, {\left (B c + A d\right )}}{\sqrt {-d x + c} c^{3} d} + \frac {2 \, {\left ({\left (-d x + c\right )}^{\frac {3}{2}} B c + 2 \, \sqrt {-d x + c} B c^{2} + 7 \, {\left (-d x + c\right )}^{\frac {3}{2}} A d - 18 \, \sqrt {-d x + c} A c d\right )}}{{\left (d x + c\right )}^{2} c^{3} d}}{64 \, d} \] Input:

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

Output:

1/64*(3*sqrt(2)*(3*B*c + 5*A*d)*arctan(1/2*sqrt(2)*sqrt(-d*x + c)/sqrt(-c) 
)/(sqrt(-c)*c^3*d) + 16*(B*c + A*d)/(sqrt(-d*x + c)*c^3*d) + 2*((-d*x + c) 
^(3/2)*B*c + 2*sqrt(-d*x + c)*B*c^2 + 7*(-d*x + c)^(3/2)*A*d - 18*sqrt(-d* 
x + c)*A*c*d)/((d*x + c)^2*c^3*d))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 491, normalized size of antiderivative = 3.13 \[ \int \frac {A+B x}{(c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {15 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) a \,c^{2} d +30 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) a c \,d^{2} x +15 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) a \,d^{3} x^{2}+9 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) b \,c^{3}+18 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) b \,c^{2} d x +9 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) b c \,d^{2} x^{2}-15 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) a \,c^{2} d -30 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) a c \,d^{2} x -15 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) a \,d^{3} x^{2}-9 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) b \,c^{3}-18 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) b \,c^{2} d x -9 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) b c \,d^{2} x^{2}-12 a \,c^{3} d +80 a \,c^{2} d^{2} x +60 a c \,d^{3} x^{2}+44 b \,c^{4}+48 b \,c^{3} d x +36 b \,c^{2} d^{2} x^{2}}{128 \sqrt {-d x +c}\, c^{4} d^{2} \left (d^{2} x^{2}+2 c d x +c^{2}\right )} \] Input:

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

Output:

(15*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*a*c 
**2*d + 30*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt( 
2))*a*c*d**2*x + 15*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt 
(c)*sqrt(2))*a*d**3*x**2 + 9*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d* 
x) - sqrt(c)*sqrt(2))*b*c**3 + 18*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c 
 - d*x) - sqrt(c)*sqrt(2))*b*c**2*d*x + 9*sqrt(c)*sqrt(c - d*x)*sqrt(2)*lo 
g(sqrt(c - d*x) - sqrt(c)*sqrt(2))*b*c*d**2*x**2 - 15*sqrt(c)*sqrt(c - d*x 
)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*a*c**2*d - 30*sqrt(c)*sqrt( 
c - d*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*a*c*d**2*x - 15*sqrt 
(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*a*d**3*x**2 
 - 9*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*b* 
c**3 - 18*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2 
))*b*c**2*d*x - 9*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c 
)*sqrt(2))*b*c*d**2*x**2 - 12*a*c**3*d + 80*a*c**2*d**2*x + 60*a*c*d**3*x* 
*2 + 44*b*c**4 + 48*b*c**3*d*x + 36*b*c**2*d**2*x**2)/(128*sqrt(c - d*x)*c 
**4*d**2*(c**2 + 2*c*d*x + d**2*x**2))