\(\int \frac {(A+B x) \sqrt {c+d x}}{(c^2-d^2 x^2)^{5/2}} \, dx\) [111]

Optimal result

Integrand size = 31, antiderivative size = 154 \[ \int \frac {(A+B x) \sqrt {c+d x}}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {(B c+A d) \sqrt {c+d x}}{3 c d^2 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {(B c-5 A d) (c+3 d x)}{24 c^3 d^2 \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}+\frac {(B c-5 A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{8 \sqrt {2} c^{7/2} d^2} \] Output:

1/3*(A*d+B*c)*(d*x+c)^(1/2)/c/d^2/(-d^2*x^2+c^2)^(3/2)-1/24*(-5*A*d+B*c)*( 
3*d*x+c)/c^3/d^2/(d*x+c)^(1/2)/(-d^2*x^2+c^2)^(1/2)+1/16*(-5*A*d+B*c)*arct 
anh(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 = 1.60 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.97 \[ \int \frac {(A+B x) \sqrt {c+d x}}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {\frac {2 \sqrt {c} \sqrt {c^2-d^2 x^2} \left (A d \left (13 c^2+10 c d x-15 d^2 x^2\right )+B c \left (7 c^2-2 c d x+3 d^2 x^2\right )\right )}{(c-d x)^2 (c+d x)^{3/2}}+3 \sqrt {2} (B c-5 A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{48 c^{7/2} d^2} \] Input:

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

Output:

((2*Sqrt[c]*Sqrt[c^2 - d^2*x^2]*(A*d*(13*c^2 + 10*c*d*x - 15*d^2*x^2) + B* 
c*(7*c^2 - 2*c*d*x + 3*d^2*x^2)))/((c - d*x)^2*(c + d*x)^(3/2)) + 3*Sqrt[2 
]*(B*c - 5*A*d)*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[c + d*x])/Sqrt[c^2 - d^2*x^2 
]])/(48*c^(7/2)*d^2)
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.23, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {669, 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.

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

\(\Big \downarrow \) 669

\(\displaystyle \frac {\sqrt {c+d x} (A d+B c)}{3 c d^2 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {(B c-5 A d) \int \frac {1}{\sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}dx}{6 c d}\)

\(\Big \downarrow \) 470

\(\displaystyle \frac {\sqrt {c+d x} (A d+B c)}{3 c d^2 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {(B c-5 A d) \left (\frac {3 \int \frac {\sqrt {c+d x}}{\left (c^2-d^2 x^2\right )^{3/2}}dx}{4 c}-\frac {1}{2 c d \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}\right )}{6 c d}\)

\(\Big \downarrow \) 467

\(\displaystyle \frac {\sqrt {c+d x} (A d+B c)}{3 c d^2 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {(B c-5 A d) \left (\frac {3 \left (\frac {\int \frac {1}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx}{2 c}+\frac {\sqrt {c+d x}}{c d \sqrt {c^2-d^2 x^2}}\right )}{4 c}-\frac {1}{2 c d \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}\right )}{6 c d}\)

\(\Big \downarrow \) 471

\(\displaystyle \frac {\sqrt {c+d x} (A d+B c)}{3 c d^2 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {(B c-5 A d) \left (\frac {3 \left (\frac {d \int \frac {1}{\frac {d^2 \left (c^2-d^2 x^2\right )}{c+d x}-2 c d^2}d\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}}{c}+\frac {\sqrt {c+d x}}{c d \sqrt {c^2-d^2 x^2}}\right )}{4 c}-\frac {1}{2 c d \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}\right )}{6 c d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\sqrt {c+d x} (A d+B c)}{3 c d^2 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {(B c-5 A d) \left (\frac {3 \left (\frac {\sqrt {c+d x}}{c d \sqrt {c^2-d^2 x^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {c+d x}}\right )}{\sqrt {2} c^{3/2} d}\right )}{4 c}-\frac {1}{2 c d \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}\right )}{6 c d}\)

Input:

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

Output:

((B*c + A*d)*Sqrt[c + d*x])/(3*c*d^2*(c^2 - d^2*x^2)^(3/2)) - ((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)))/(6*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 669

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* 
(p + 1))), x] - Simp[e*((m*(d*g + e*f) + 2*e*f*(p + 1))/(2*c*d*(p + 1))) 
Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, 
g}, x] && EqQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0]
 

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.55

method	result	size
default	\(-\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, \left (-15 A \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) d^{3} x^{2}+3 B \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \,d^{2} x^{2}+15 A \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d +30 A \sqrt {c}\, d^{3} x^{2}-3 B \sqrt {-d x +c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3}-6 B \,c^{\frac {3}{2}} d^{2} x^{2}-20 A \,c^{\frac {3}{2}} d^{2} x +4 B \,c^{\frac {5}{2}} d x -26 A \,c^{\frac {5}{2}} d -14 B \,c^{\frac {7}{2}}\right )}{48 \left (d x +c \right )^{\frac {3}{2}} \left (-d x +c \right )^{2} d^{2} c^{\frac {7}{2}}}\)	\(238\)

Input:

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

Output:

-1/48*(-d^2*x^2+c^2)^(1/2)*(-15*A*(-d*x+c)^(1/2)*2^(1/2)*arctanh(1/2*(-d*x 
+c)^(1/2)*2^(1/2)/c^(1/2))*d^3*x^2+3*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+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))*c^2*d+30*A*c^(1/2)*d^3*x^2-3*B*(-d 
*x+c)^(1/2)*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^3-6*B*c^ 
(3/2)*d^2*x^2-20*A*c^(3/2)*d^2*x+4*B*c^(5/2)*d*x-26*A*c^(5/2)*d-14*B*c^(7/ 
2))/(d*x+c)^(3/2)/(-d*x+c)^2/d^2/c^(7/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 457, normalized size of antiderivative = 2.97 \[ \int \frac {(A+B x) \sqrt {c+d x}}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\left [-\frac {3 \, \sqrt {2} {\left (B c^{5} - 5 \, A c^{4} d + {\left (B c d^{4} - 5 \, A d^{5}\right )} x^{4} - 2 \, {\left (B c^{3} d^{2} - 5 \, A c^{2} d^{3}\right )} x^{2}\right )} \sqrt {c} \log \left (-\frac {d^{2} x^{2} - 2 \, c d x + 2 \, \sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {c} - 3 \, c^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) - 4 \, {\left (7 \, B c^{4} + 13 \, A c^{3} d + 3 \, {\left (B c^{2} d^{2} - 5 \, A c d^{3}\right )} x^{2} - 2 \, {\left (B c^{3} d - 5 \, A c^{2} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{96 \, {\left (c^{4} d^{6} x^{4} - 2 \, c^{6} d^{4} x^{2} + c^{8} d^{2}\right )}}, -\frac {3 \, \sqrt {2} {\left (B c^{5} - 5 \, A c^{4} d + {\left (B c d^{4} - 5 \, A d^{5}\right )} x^{4} - 2 \, {\left (B c^{3} d^{2} - 5 \, A c^{2} d^{3}\right )} x^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {-c}}{2 \, {\left (c d x + c^{2}\right )}}\right ) - 2 \, {\left (7 \, B c^{4} + 13 \, A c^{3} d + 3 \, {\left (B c^{2} d^{2} - 5 \, A c d^{3}\right )} x^{2} - 2 \, {\left (B c^{3} d - 5 \, A c^{2} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{48 \, {\left (c^{4} d^{6} x^{4} - 2 \, c^{6} d^{4} x^{2} + c^{8} d^{2}\right )}}\right ] \] Input:

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

Output:

[-1/96*(3*sqrt(2)*(B*c^5 - 5*A*c^4*d + (B*c*d^4 - 5*A*d^5)*x^4 - 2*(B*c^3* 
d^2 - 5*A*c^2*d^3)*x^2)*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*(7*B*c^4 + 13*A*c^3*d + 3*(B*c^2*d^2 - 5*A*c*d^3)*x^2 - 2*(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^6*d^4 
*x^2 + c^8*d^2), -1/48*(3*sqrt(2)*(B*c^5 - 5*A*c^4*d + (B*c*d^4 - 5*A*d^5) 
*x^4 - 2*(B*c^3*d^2 - 5*A*c^2*d^3)*x^2)*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*(7*B*c^4 + 13*A*c 
^3*d + 3*(B*c^2*d^2 - 5*A*c*d^3)*x^2 - 2*(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^6*d^4*x^2 + c^8*d^2)]
 

Sympy [F]

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

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

Output:

Integral((A + B*x)*sqrt(c + d*x)/(-(-c + d*x)*(c + d*x))**(5/2), x)
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.87 \[ \int \frac {(A+B x) \sqrt {c+d x}}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {\frac {3 \, \sqrt {2} {\left (B c - 5 \, A d\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-d x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c} c^{3} d} - \frac {6 \, {\left (\sqrt {-d x + c} B c - \sqrt {-d x + c} A d\right )}}{{\left (d x + c\right )} c^{3} d} + \frac {8 \, {\left (B c^{2} - 3 \, {\left (d x - c\right )} A d + A c d\right )}}{{\left (d x - c\right )} \sqrt {-d x + c} c^{3} d}}{48 \, d} \] Input:

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

Output:

-1/48*(3*sqrt(2)*(B*c - 5*A*d)*arctan(1/2*sqrt(2)*sqrt(-d*x + c)/sqrt(-c)) 
/(sqrt(-c)*c^3*d) - 6*(sqrt(-d*x + c)*B*c - sqrt(-d*x + c)*A*d)/((d*x + c) 
*c^3*d) + 8*(B*c^2 - 3*(d*x - c)*A*d + A*c*d)/((d*x - c)*sqrt(-d*x + c)*c^ 
3*d))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 353, normalized size of antiderivative = 2.29 \[ \int \frac {(A+B x) \sqrt {c+d x}}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {15 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) a \,c^{2} d -15 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) a \,d^{3} x^{2}-3 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) b \,c^{3}+3 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}-\sqrt {c}\, \sqrt {2}\right ) b c \,d^{2} x^{2}-15 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) a \,c^{2} d +15 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) a \,d^{3} x^{2}+3 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) b \,c^{3}-3 \sqrt {c}\, \sqrt {-d x +c}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-d x +c}+\sqrt {c}\, \sqrt {2}\right ) b c \,d^{2} x^{2}+52 a \,c^{3} d +40 a \,c^{2} d^{2} x -60 a c \,d^{3} x^{2}+28 b \,c^{4}-8 b \,c^{3} d x +12 b \,c^{2} d^{2} x^{2}}{96 \sqrt {-d x +c}\, c^{4} d^{2} \left (-d^{2} x^{2}+c^{2}\right )} \] Input:

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

Output:

(15*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt(2))*a*c 
**2*d - 15*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt(c)*sqrt( 
2))*a*d**3*x**2 - 3*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(sqrt(c - d*x) - sqrt 
(c)*sqrt(2))*b*c**3 + 3*sqrt(c)*sqrt(c - d*x)*sqrt(2)*log(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 + 15*sqrt(c)*sqrt(c - d*x)*sqrt(2)*l 
og(sqrt(c - d*x) + sqrt(c)*sqrt(2))*a*d**3*x**2 + 3*sqrt(c)*sqrt(c - d*x)* 
sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*b*c**3 - 3*sqrt(c)*sqrt(c - d 
*x)*sqrt(2)*log(sqrt(c - d*x) + sqrt(c)*sqrt(2))*b*c*d**2*x**2 + 52*a*c**3 
*d + 40*a*c**2*d**2*x - 60*a*c*d**3*x**2 + 28*b*c**4 - 8*b*c**3*d*x + 12*b 
*c**2*d**2*x**2)/(96*sqrt(c - d*x)*c**4*d**2*(c**2 - d**2*x**2))