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

Optimal result

Integrand size = 29, antiderivative size = 147 \[ \int (A+B x) (c+d x)^2 \sqrt {c^2-d^2 x^2} \, dx=\frac {c^2 (2 B c+5 A d) x \sqrt {c^2-d^2 x^2}}{8 d}-\frac {B (c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2}}{5 d^2}-\frac {(2 B c+5 A d) (8 c+3 d x) \left (c^2-d^2 x^2\right )^{3/2}}{60 d^2}+\frac {c^4 (2 B c+5 A d) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{8 d^2} \] Output:

1/8*c^2*(5*A*d+2*B*c)*x*(-d^2*x^2+c^2)^(1/2)/d-1/5*B*(d*x+c)^2*(-d^2*x^2+c 
^2)^(3/2)/d^2-1/60*(5*A*d+2*B*c)*(3*d*x+8*c)*(-d^2*x^2+c^2)^(3/2)/d^2+1/8* 
c^4*(5*A*d+2*B*c)*arctan(d*x/(-d^2*x^2+c^2)^(1/2))/d^2
 

Mathematica [A] (verified)

Time = 0.60 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.07 \[ \int (A+B x) (c+d x)^2 \sqrt {c^2-d^2 x^2} \, dx=\frac {d \sqrt {c^2-d^2 x^2} \left (5 A d \left (-16 c^3+9 c^2 d x+16 c d^2 x^2+6 d^3 x^3\right )+B \left (-56 c^4-30 c^3 d x+32 c^2 d^2 x^2+60 c d^3 x^3+24 d^4 x^4\right )\right )+15 c^4 \sqrt {-d^2} (2 B c+5 A d) \log \left (-\sqrt {-d^2} x+\sqrt {c^2-d^2 x^2}\right )}{120 d^3} \] Input:

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

Output:

(d*Sqrt[c^2 - d^2*x^2]*(5*A*d*(-16*c^3 + 9*c^2*d*x + 16*c*d^2*x^2 + 6*d^3* 
x^3) + B*(-56*c^4 - 30*c^3*d*x + 32*c^2*d^2*x^2 + 60*c*d^3*x^3 + 24*d^4*x^ 
4)) + 15*c^4*Sqrt[-d^2]*(2*B*c + 5*A*d)*Log[-(Sqrt[-d^2]*x) + Sqrt[c^2 - d 
^2*x^2]])/(120*d^3)
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {672, 469, 455, 211, 224, 216}

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.

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

\(\Big \downarrow \) 672

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

\(\Big \downarrow \) 469

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

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 455

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

rule 469

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

rule 672

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_ 
), x_Symbol] :> Simp[g*(d + e*x)^m*((a + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), 
 x] + Simp[(m*(d*g + e*f) + 2*e*f*(p + 1))/(e*(m + 2*p + 2))   Int[(d + e*x 
)^m*(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] && NeQ[m + 2*p + 2, 0]
 

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00

method	result	size
risch	\(-\frac {\left (-24 B \,d^{4} x^{4}-30 A \,d^{4} x^{3}-60 B c \,d^{3} x^{3}-80 A c \,d^{3} x^{2}-32 x^{2} c^{2} B \,d^{2}-45 A \,c^{2} d^{2} x +30 B \,c^{3} d x +80 A \,c^{3} d +56 B \,c^{4}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{120 d^{2}}+\frac {c^{4} \left (5 A d +2 B c \right ) \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{8 d \sqrt {d^{2}}}\)	\(147\)
default	\(A \,c^{2} \left (\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 \sqrt {d^{2}}}\right )+d \left (A d +2 B c \right ) \left (-\frac {x \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{4 d^{2}}+\frac {c^{2} \left (\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 \sqrt {d^{2}}}\right )}{4 d^{2}}\right )-\frac {c \left (2 A d +B c \right ) \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{3 d^{2}}+B \,d^{2} \left (-\frac {x^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{5 d^{2}}-\frac {2 c^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{15 d^{4}}\right )\)	\(224\)

Input:

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

Output:

-1/120/d^2*(-24*B*d^4*x^4-30*A*d^4*x^3-60*B*c*d^3*x^3-80*A*c*d^3*x^2-32*B* 
c^2*d^2*x^2-45*A*c^2*d^2*x+30*B*c^3*d*x+80*A*c^3*d+56*B*c^4)*(-d^2*x^2+c^2 
)^(1/2)+1/8*c^4/d*(5*A*d+2*B*c)/(d^2)^(1/2)*arctan((d^2)^(1/2)*x/(-d^2*x^2 
+c^2)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.01 \[ \int (A+B x) (c+d x)^2 \sqrt {c^2-d^2 x^2} \, dx=-\frac {30 \, {\left (2 \, B c^{5} + 5 \, A c^{4} d\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) - {\left (24 \, B d^{4} x^{4} - 56 \, B c^{4} - 80 \, A c^{3} d + 30 \, {\left (2 \, B c d^{3} + A d^{4}\right )} x^{3} + 16 \, {\left (2 \, B c^{2} d^{2} + 5 \, A c d^{3}\right )} x^{2} - 15 \, {\left (2 \, B c^{3} d - 3 \, A c^{2} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{120 \, d^{2}} \] Input:

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

Output:

-1/120*(30*(2*B*c^5 + 5*A*c^4*d)*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d*x)) 
 - (24*B*d^4*x^4 - 56*B*c^4 - 80*A*c^3*d + 30*(2*B*c*d^3 + A*d^4)*x^3 + 16 
*(2*B*c^2*d^2 + 5*A*c*d^3)*x^2 - 15*(2*B*c^3*d - 3*A*c^2*d^2)*x)*sqrt(-d^2 
*x^2 + c^2))/d^2
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 333 vs. \(2 (131) = 262\).

Time = 0.58 (sec) , antiderivative size = 333, normalized size of antiderivative = 2.27 \[ \int (A+B x) (c+d x)^2 \sqrt {c^2-d^2 x^2} \, dx=\begin {cases} \sqrt {c^{2} - d^{2} x^{2}} \left (\frac {B d^{2} x^{4}}{5} - \frac {x^{3} \left (- A d^{4} - 2 B c d^{3}\right )}{4 d^{2}} - \frac {x^{2} \left (- 2 A c d^{3} - \frac {4 B c^{2} d^{2}}{5}\right )}{3 d^{2}} - \frac {x \left (2 B c^{3} d + \frac {3 c^{2} \left (- A d^{4} - 2 B c d^{3}\right )}{4 d^{2}}\right )}{2 d^{2}} - \frac {2 A c^{3} d + B c^{4} + \frac {2 c^{2} \left (- 2 A c d^{3} - \frac {4 B c^{2} d^{2}}{5}\right )}{3 d^{2}}}{d^{2}}\right ) + \left (A c^{4} + \frac {c^{2} \cdot \left (2 B c^{3} d + \frac {3 c^{2} \left (- A d^{4} - 2 B c d^{3}\right )}{4 d^{2}}\right )}{2 d^{2}}\right ) \left (\begin {cases} \frac {\log {\left (- 2 d^{2} x + 2 \sqrt {- d^{2}} \sqrt {c^{2} - d^{2} x^{2}} \right )}}{\sqrt {- d^{2}}} & \text {for}\: c^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- d^{2} x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: d^{2} \neq 0 \\\left (A c^{2} x + \frac {B d^{2} x^{4}}{4} + \frac {x^{3} \left (A d^{2} + 2 B c d\right )}{3} + \frac {x^{2} \cdot \left (2 A c d + B c^{2}\right )}{2}\right ) \sqrt {c^{2}} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((sqrt(c**2 - d**2*x**2)*(B*d**2*x**4/5 - x**3*(-A*d**4 - 2*B*c*d 
**3)/(4*d**2) - x**2*(-2*A*c*d**3 - 4*B*c**2*d**2/5)/(3*d**2) - x*(2*B*c** 
3*d + 3*c**2*(-A*d**4 - 2*B*c*d**3)/(4*d**2))/(2*d**2) - (2*A*c**3*d + B*c 
**4 + 2*c**2*(-2*A*c*d**3 - 4*B*c**2*d**2/5)/(3*d**2))/d**2) + (A*c**4 + c 
**2*(2*B*c**3*d + 3*c**2*(-A*d**4 - 2*B*c*d**3)/(4*d**2))/(2*d**2))*Piecew 
ise((log(-2*d**2*x + 2*sqrt(-d**2)*sqrt(c**2 - d**2*x**2))/sqrt(-d**2), Ne 
(c**2, 0)), (x*log(x)/sqrt(-d**2*x**2), True)), Ne(d**2, 0)), ((A*c**2*x + 
 B*d**2*x**4/4 + x**3*(A*d**2 + 2*B*c*d)/3 + x**2*(2*A*c*d + B*c**2)/2)*sq 
rt(c**2), True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.31 \[ \int (A+B x) (c+d x)^2 \sqrt {c^2-d^2 x^2} \, dx=\frac {A c^{4} \arcsin \left (\frac {d x}{c}\right )}{2 \, d} + \frac {1}{2} \, \sqrt {-d^{2} x^{2} + c^{2}} A c^{2} x - \frac {1}{5} \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} B x^{2} + \frac {{\left (2 \, B c d + A d^{2}\right )} c^{4} \arcsin \left (\frac {d x}{c}\right )}{8 \, d^{3}} - \frac {7 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} B c^{2}}{15 \, d^{2}} - \frac {2 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} A c}{3 \, d} + \frac {\sqrt {-d^{2} x^{2} + c^{2}} {\left (2 \, B c d + A d^{2}\right )} c^{2} x}{8 \, d^{2}} - \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} {\left (2 \, B c d + A d^{2}\right )} x}{4 \, d^{2}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.07 \[ \int (A+B x) (c+d x)^2 \sqrt {c^2-d^2 x^2} \, dx=\frac {{\left (2 \, B c^{5} + 5 \, A c^{4} d\right )} \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{8 \, d {\left | d \right |}} + \frac {1}{120} \, \sqrt {-d^{2} x^{2} + c^{2}} {\left ({\left (2 \, {\left (3 \, {\left (4 \, B d^{2} x + \frac {5 \, {\left (2 \, B c d^{7} + A d^{8}\right )}}{d^{6}}\right )} x + \frac {8 \, {\left (2 \, B c^{2} d^{6} + 5 \, A c d^{7}\right )}}{d^{6}}\right )} x - \frac {15 \, {\left (2 \, B c^{3} d^{5} - 3 \, A c^{2} d^{6}\right )}}{d^{6}}\right )} x - \frac {8 \, {\left (7 \, B c^{4} d^{4} + 10 \, A c^{3} d^{5}\right )}}{d^{6}}\right )} \] Input:

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

Output:

1/8*(2*B*c^5 + 5*A*c^4*d)*arcsin(d*x/c)*sgn(c)*sgn(d)/(d*abs(d)) + 1/120*s 
qrt(-d^2*x^2 + c^2)*((2*(3*(4*B*d^2*x + 5*(2*B*c*d^7 + A*d^8)/d^6)*x + 8*( 
2*B*c^2*d^6 + 5*A*c*d^7)/d^6)*x - 15*(2*B*c^3*d^5 - 3*A*c^2*d^6)/d^6)*x - 
8*(7*B*c^4*d^4 + 10*A*c^3*d^5)/d^6)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.66 \[ \int (A+B x) (c+d x)^2 \sqrt {c^2-d^2 x^2} \, dx=\frac {75 \mathit {asin} \left (\frac {d x}{c}\right ) a \,c^{4} d +30 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{5}-80 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{3} d +45 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{2} d^{2} x +80 \sqrt {-d^{2} x^{2}+c^{2}}\, a c \,d^{3} x^{2}+30 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{4} x^{3}-56 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{4}-30 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{3} d x +32 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2} d^{2} x^{2}+60 \sqrt {-d^{2} x^{2}+c^{2}}\, b c \,d^{3} x^{3}+24 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,d^{4} x^{4}+80 a \,c^{4} d +56 b \,c^{5}}{120 d^{2}} \] Input:

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

Output:

(75*asin((d*x)/c)*a*c**4*d + 30*asin((d*x)/c)*b*c**5 - 80*sqrt(c**2 - d**2 
*x**2)*a*c**3*d + 45*sqrt(c**2 - d**2*x**2)*a*c**2*d**2*x + 80*sqrt(c**2 - 
 d**2*x**2)*a*c*d**3*x**2 + 30*sqrt(c**2 - d**2*x**2)*a*d**4*x**3 - 56*sqr 
t(c**2 - d**2*x**2)*b*c**4 - 30*sqrt(c**2 - d**2*x**2)*b*c**3*d*x + 32*sqr 
t(c**2 - d**2*x**2)*b*c**2*d**2*x**2 + 60*sqrt(c**2 - d**2*x**2)*b*c*d**3* 
x**3 + 24*sqrt(c**2 - d**2*x**2)*b*d**4*x**4 + 80*a*c**4*d + 56*b*c**5)/(1 
20*d**2)