\(\int \frac {(A+B x^2) (b x^2+c x^4)^{3/2}}{x^{13}} \, dx\) [180]

Optimal result

Integrand size = 26, antiderivative size = 96 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{13}} \, dx=-\frac {A \left (b x^2+c x^4\right )^{5/2}}{9 b x^{14}}-\frac {(9 b B-4 A c) \left (b x^2+c x^4\right )^{5/2}}{63 b^2 x^{12}}+\frac {2 c (9 b B-4 A c) \left (b x^2+c x^4\right )^{5/2}}{315 b^3 x^{10}} \] Output:

-1/9*A*(c*x^4+b*x^2)^(5/2)/b/x^14-1/63*(-4*A*c+9*B*b)*(c*x^4+b*x^2)^(5/2)/ 
b^2/x^12+2/315*c*(-4*A*c+9*B*b)*(c*x^4+b*x^2)^(5/2)/b^3/x^10
 

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.69 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{13}} \, dx=\frac {\left (x^2 \left (b+c x^2\right )\right )^{5/2} \left (9 b B x^2 \left (-5 b+2 c x^2\right )+A \left (-35 b^2+20 b c x^2-8 c^2 x^4\right )\right )}{315 b^3 x^{14}} \] Input:

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^13,x]
 

Output:

((x^2*(b + c*x^2))^(5/2)*(9*b*B*x^2*(-5*b + 2*c*x^2) + A*(-35*b^2 + 20*b*c 
*x^2 - 8*c^2*x^4)))/(315*b^3*x^14)
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1940, 1220, 1129, 1123}

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.

\(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{13}} \, dx\)

\(\Big \downarrow \) 1940

\(\displaystyle \frac {1}{2} \int \frac {\left (B x^2+A\right ) \left (c x^4+b x^2\right )^{3/2}}{x^{14}}dx^2\)

\(\Big \downarrow \) 1220

\(\displaystyle \frac {1}{2} \left (\frac {(9 b B-4 A c) \int \frac {\left (c x^4+b x^2\right )^{3/2}}{x^{12}}dx^2}{9 b}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{9 b x^{14}}\right )\)

\(\Big \downarrow \) 1129

\(\displaystyle \frac {1}{2} \left (\frac {(9 b B-4 A c) \left (-\frac {2 c \int \frac {\left (c x^4+b x^2\right )^{3/2}}{x^{10}}dx^2}{7 b}-\frac {2 \left (b x^2+c x^4\right )^{5/2}}{7 b x^{12}}\right )}{9 b}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{9 b x^{14}}\right )\)

\(\Big \downarrow \) 1123

\(\displaystyle \frac {1}{2} \left (\frac {\left (\frac {4 c \left (b x^2+c x^4\right )^{5/2}}{35 b^2 x^{10}}-\frac {2 \left (b x^2+c x^4\right )^{5/2}}{7 b x^{12}}\right ) (9 b B-4 A c)}{9 b}-\frac {2 A \left (b x^2+c x^4\right )^{5/2}}{9 b x^{14}}\right )\)

Input:

Int[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^13,x]
 

Output:

((-2*A*(b*x^2 + c*x^4)^(5/2))/(9*b*x^14) + ((9*b*B - 4*A*c)*((-2*(b*x^2 + 
c*x^4)^(5/2))/(7*b*x^12) + (4*c*(b*x^2 + c*x^4)^(5/2))/(35*b^2*x^10)))/(9* 
b))/2
 

Defintions of rubi rules used

rule 1123

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[e*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(2*c*d - b 
*e))), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 
0] && EqQ[m + 2*p + 2, 0]
 

rule 1129

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(-e)*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/((m + p + 1)*(2* 
c*d - b*e))), x] + Simp[c*(Simplify[m + 2*p + 2]/((m + p + 1)*(2*c*d - b*e) 
))   Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d 
, e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ILtQ[Simplify[m + 2*p + 
2], 0]
 

rule 1220

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + b*x + c*x 
^2)^(p + 1)/((2*c*d - b*e)*(m + p + 1))), x] + Simp[(m*(g*(c*d - b*e) + c*e 
*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1))   Int[(d + e*x 
)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, 
 x] && EqQ[c*d^2 - b*d*e + 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 
]
 

rule 1940

Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_) 
^(n_))^(q_.), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1) 
*(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x] /; 
FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] &&  !IntegerQ[p] && NeQ[k, j] && I 
ntegerQ[Simplify[j/n]] && IntegerQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 
 1)/n]] && NeQ[n^2, 1]
 

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.71

method	result	size
pseudoelliptic	\(-\frac {\left (c \,x^{2}+b \right )^{2} \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (\left (\frac {9 B \,x^{2}}{7}+A \right ) b^{2}-\frac {4 c \,x^{2} \left (\frac {9 B \,x^{2}}{10}+A \right ) b}{7}+\frac {8 x^{4} A \,c^{2}}{35}\right )}{9 x^{10} b^{3}}\)	\(68\)
gosper	\(-\frac {\left (c \,x^{2}+b \right ) \left (8 x^{4} A \,c^{2}-18 x^{4} B b c -20 A b c \,x^{2}+45 x^{2} B \,b^{2}+35 b^{2} A \right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{315 x^{12} b^{3}}\)	\(70\)
default	\(-\frac {\left (c \,x^{2}+b \right ) \left (8 x^{4} A \,c^{2}-18 x^{4} B b c -20 A b c \,x^{2}+45 x^{2} B \,b^{2}+35 b^{2} A \right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{315 x^{12} b^{3}}\)	\(70\)
orering	\(-\frac {\left (c \,x^{2}+b \right ) \left (8 x^{4} A \,c^{2}-18 x^{4} B b c -20 A b c \,x^{2}+45 x^{2} B \,b^{2}+35 b^{2} A \right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{315 x^{12} b^{3}}\)	\(70\)
trager	\(-\frac {\left (8 A \,c^{4} x^{8}-18 B b \,c^{3} x^{8}-4 A b \,c^{3} x^{6}+9 B \,b^{2} c^{2} x^{6}+3 A \,b^{2} c^{2} x^{4}+72 B \,b^{3} c \,x^{4}+50 A \,b^{3} c \,x^{2}+45 B \,b^{4} x^{2}+35 A \,b^{4}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{315 b^{3} x^{10}}\)	\(111\)
risch	\(-\frac {\sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (8 A \,c^{4} x^{8}-18 B b \,c^{3} x^{8}-4 A b \,c^{3} x^{6}+9 B \,b^{2} c^{2} x^{6}+3 A \,b^{2} c^{2} x^{4}+72 B \,b^{3} c \,x^{4}+50 A \,b^{3} c \,x^{2}+45 B \,b^{4} x^{2}+35 A \,b^{4}\right )}{315 x^{10} b^{3}}\)	\(111\)

Input:

int((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^13,x,method=_RETURNVERBOSE)
 

Output:

-1/9*(c*x^2+b)^2*(x^2*(c*x^2+b))^(1/2)*((9/7*B*x^2+A)*b^2-4/7*c*x^2*(9/10* 
B*x^2+A)*b+8/35*x^4*A*c^2)/x^10/b^3
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.14 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{13}} \, dx=\frac {{\left (2 \, {\left (9 \, B b c^{3} - 4 \, A c^{4}\right )} x^{8} - {\left (9 \, B b^{2} c^{2} - 4 \, A b c^{3}\right )} x^{6} - 35 \, A b^{4} - 3 \, {\left (24 \, B b^{3} c + A b^{2} c^{2}\right )} x^{4} - 5 \, {\left (9 \, B b^{4} + 10 \, A b^{3} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{315 \, b^{3} x^{10}} \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^13,x, algorithm="fricas")
 

Output:

1/315*(2*(9*B*b*c^3 - 4*A*c^4)*x^8 - (9*B*b^2*c^2 - 4*A*b*c^3)*x^6 - 35*A* 
b^4 - 3*(24*B*b^3*c + A*b^2*c^2)*x^4 - 5*(9*B*b^4 + 10*A*b^3*c)*x^2)*sqrt( 
c*x^4 + b*x^2)/(b^3*x^10)
 

Sympy [F]

\[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{13}} \, dx=\int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}} \left (A + B x^{2}\right )}{x^{13}}\, dx \] Input:

integrate((B*x**2+A)*(c*x**4+b*x**2)**(3/2)/x**13,x)
 

Output:

Integral((x**2*(b + c*x**2))**(3/2)*(A + B*x**2)/x**13, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 241 vs. \(2 (84) = 168\).

Time = 0.05 (sec) , antiderivative size = 241, normalized size of antiderivative = 2.51 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{13}} \, dx=\frac {1}{140} \, B {\left (\frac {8 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{b^{2} x^{2}} - \frac {4 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{b x^{4}} + \frac {3 \, \sqrt {c x^{4} + b x^{2}} c}{x^{6}} + \frac {15 \, \sqrt {c x^{4} + b x^{2}} b}{x^{8}} - \frac {35 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{10}}\right )} - \frac {1}{630} \, A {\left (\frac {16 \, \sqrt {c x^{4} + b x^{2}} c^{4}}{b^{3} x^{2}} - \frac {8 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{b^{2} x^{4}} + \frac {6 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{b x^{6}} - \frac {5 \, \sqrt {c x^{4} + b x^{2}} c}{x^{8}} - \frac {35 \, \sqrt {c x^{4} + b x^{2}} b}{x^{10}} + \frac {105 \, {\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}}{x^{12}}\right )} \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^13,x, algorithm="maxima")
 

Output:

1/140*B*(8*sqrt(c*x^4 + b*x^2)*c^3/(b^2*x^2) - 4*sqrt(c*x^4 + b*x^2)*c^2/( 
b*x^4) + 3*sqrt(c*x^4 + b*x^2)*c/x^6 + 15*sqrt(c*x^4 + b*x^2)*b/x^8 - 35*( 
c*x^4 + b*x^2)^(3/2)/x^10) - 1/630*A*(16*sqrt(c*x^4 + b*x^2)*c^4/(b^3*x^2) 
 - 8*sqrt(c*x^4 + b*x^2)*c^3/(b^2*x^4) + 6*sqrt(c*x^4 + b*x^2)*c^2/(b*x^6) 
 - 5*sqrt(c*x^4 + b*x^2)*c/x^8 - 35*sqrt(c*x^4 + b*x^2)*b/x^10 + 105*(c*x^ 
4 + b*x^2)^(3/2)/x^12)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 430 vs. \(2 (84) = 168\).

Time = 1.46 (sec) , antiderivative size = 430, normalized size of antiderivative = 4.48 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{13}} \, dx=\frac {4 \, {\left (315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{14} B c^{\frac {7}{2}} \mathrm {sgn}\left (x\right ) - 315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{12} B b c^{\frac {7}{2}} \mathrm {sgn}\left (x\right ) + 840 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{12} A c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) + 315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{10} B b^{2} c^{\frac {7}{2}} \mathrm {sgn}\left (x\right ) + 1260 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{10} A b c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) - 819 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} B b^{3} c^{\frac {7}{2}} \mathrm {sgn}\left (x\right ) + 1764 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} A b^{2} c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) + 441 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} B b^{4} c^{\frac {7}{2}} \mathrm {sgn}\left (x\right ) + 504 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} A b^{3} c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) - 9 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} B b^{5} c^{\frac {7}{2}} \mathrm {sgn}\left (x\right ) + 144 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} A b^{4} c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) + 81 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} B b^{6} c^{\frac {7}{2}} \mathrm {sgn}\left (x\right ) - 36 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} A b^{5} c^{\frac {9}{2}} \mathrm {sgn}\left (x\right ) - 9 \, B b^{7} c^{\frac {7}{2}} \mathrm {sgn}\left (x\right ) + 4 \, A b^{6} c^{\frac {9}{2}} \mathrm {sgn}\left (x\right )\right )}}{315 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{9}} \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^13,x, algorithm="giac")
 

Output:

4/315*(315*(sqrt(c)*x - sqrt(c*x^2 + b))^14*B*c^(7/2)*sgn(x) - 315*(sqrt(c 
)*x - sqrt(c*x^2 + b))^12*B*b*c^(7/2)*sgn(x) + 840*(sqrt(c)*x - sqrt(c*x^2 
 + b))^12*A*c^(9/2)*sgn(x) + 315*(sqrt(c)*x - sqrt(c*x^2 + b))^10*B*b^2*c^ 
(7/2)*sgn(x) + 1260*(sqrt(c)*x - sqrt(c*x^2 + b))^10*A*b*c^(9/2)*sgn(x) - 
819*(sqrt(c)*x - sqrt(c*x^2 + b))^8*B*b^3*c^(7/2)*sgn(x) + 1764*(sqrt(c)*x 
 - sqrt(c*x^2 + b))^8*A*b^2*c^(9/2)*sgn(x) + 441*(sqrt(c)*x - sqrt(c*x^2 + 
 b))^6*B*b^4*c^(7/2)*sgn(x) + 504*(sqrt(c)*x - sqrt(c*x^2 + b))^6*A*b^3*c^ 
(9/2)*sgn(x) - 9*(sqrt(c)*x - sqrt(c*x^2 + b))^4*B*b^5*c^(7/2)*sgn(x) + 14 
4*(sqrt(c)*x - sqrt(c*x^2 + b))^4*A*b^4*c^(9/2)*sgn(x) + 81*(sqrt(c)*x - s 
qrt(c*x^2 + b))^2*B*b^6*c^(7/2)*sgn(x) - 36*(sqrt(c)*x - sqrt(c*x^2 + b))^ 
2*A*b^5*c^(9/2)*sgn(x) - 9*B*b^7*c^(7/2)*sgn(x) + 4*A*b^6*c^(9/2)*sgn(x))/ 
((sqrt(c)*x - sqrt(c*x^2 + b))^2 - b)^9
 

Mupad [B] (verification not implemented)

Time = 10.58 (sec) , antiderivative size = 206, normalized size of antiderivative = 2.15 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{13}} \, dx=\frac {4\,A\,c^3\,\sqrt {c\,x^4+b\,x^2}}{315\,b^2\,x^4}-\frac {10\,A\,c\,\sqrt {c\,x^4+b\,x^2}}{63\,x^8}-\frac {B\,b\,\sqrt {c\,x^4+b\,x^2}}{7\,x^8}-\frac {8\,B\,c\,\sqrt {c\,x^4+b\,x^2}}{35\,x^6}-\frac {A\,c^2\,\sqrt {c\,x^4+b\,x^2}}{105\,b\,x^6}-\frac {A\,b\,\sqrt {c\,x^4+b\,x^2}}{9\,x^{10}}-\frac {8\,A\,c^4\,\sqrt {c\,x^4+b\,x^2}}{315\,b^3\,x^2}-\frac {B\,c^2\,\sqrt {c\,x^4+b\,x^2}}{35\,b\,x^4}+\frac {2\,B\,c^3\,\sqrt {c\,x^4+b\,x^2}}{35\,b^2\,x^2} \] Input:

int(((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^13,x)
 

Output:

(4*A*c^3*(b*x^2 + c*x^4)^(1/2))/(315*b^2*x^4) - (10*A*c*(b*x^2 + c*x^4)^(1 
/2))/(63*x^8) - (B*b*(b*x^2 + c*x^4)^(1/2))/(7*x^8) - (8*B*c*(b*x^2 + c*x^ 
4)^(1/2))/(35*x^6) - (A*c^2*(b*x^2 + c*x^4)^(1/2))/(105*b*x^6) - (A*b*(b*x 
^2 + c*x^4)^(1/2))/(9*x^10) - (8*A*c^4*(b*x^2 + c*x^4)^(1/2))/(315*b^3*x^2 
) - (B*c^2*(b*x^2 + c*x^4)^(1/2))/(35*b*x^4) + (2*B*c^3*(b*x^2 + c*x^4)^(1 
/2))/(35*b^2*x^2)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.99 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{13}} \, dx=\frac {-35 \sqrt {c \,x^{2}+b}\, a \,b^{4}-50 \sqrt {c \,x^{2}+b}\, a \,b^{3} c \,x^{2}-3 \sqrt {c \,x^{2}+b}\, a \,b^{2} c^{2} x^{4}+4 \sqrt {c \,x^{2}+b}\, a b \,c^{3} x^{6}-8 \sqrt {c \,x^{2}+b}\, a \,c^{4} x^{8}-45 \sqrt {c \,x^{2}+b}\, b^{5} x^{2}-72 \sqrt {c \,x^{2}+b}\, b^{4} c \,x^{4}-9 \sqrt {c \,x^{2}+b}\, b^{3} c^{2} x^{6}+18 \sqrt {c \,x^{2}+b}\, b^{2} c^{3} x^{8}+8 \sqrt {c}\, a \,c^{4} x^{9}-18 \sqrt {c}\, b^{2} c^{3} x^{9}}{315 b^{3} x^{9}} \] Input:

int((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^13,x)
 

Output:

( - 35*sqrt(b + c*x**2)*a*b**4 - 50*sqrt(b + c*x**2)*a*b**3*c*x**2 - 3*sqr 
t(b + c*x**2)*a*b**2*c**2*x**4 + 4*sqrt(b + c*x**2)*a*b*c**3*x**6 - 8*sqrt 
(b + c*x**2)*a*c**4*x**8 - 45*sqrt(b + c*x**2)*b**5*x**2 - 72*sqrt(b + c*x 
**2)*b**4*c*x**4 - 9*sqrt(b + c*x**2)*b**3*c**2*x**6 + 18*sqrt(b + c*x**2) 
*b**2*c**3*x**8 + 8*sqrt(c)*a*c**4*x**9 - 18*sqrt(c)*b**2*c**3*x**9)/(315* 
b**3*x**9)