\(\int \frac {A+B x^2}{x^9 \sqrt {b x^2+c x^4}} \, dx\) [203]

Optimal result

Integrand size = 26, antiderivative size = 170 \[ \int \frac {A+B x^2}{x^9 \sqrt {b x^2+c x^4}} \, dx=-\frac {A \sqrt {b x^2+c x^4}}{9 b x^{10}}-\frac {(9 b B-8 A c) \sqrt {b x^2+c x^4}}{63 b^2 x^8}+\frac {2 c (9 b B-8 A c) \sqrt {b x^2+c x^4}}{105 b^3 x^6}-\frac {8 c^2 (9 b B-8 A c) \sqrt {b x^2+c x^4}}{315 b^4 x^4}+\frac {16 c^3 (9 b B-8 A c) \sqrt {b x^2+c x^4}}{315 b^5 x^2} \] Output:

-1/9*A*(c*x^4+b*x^2)^(1/2)/b/x^10-1/63*(-8*A*c+9*B*b)*(c*x^4+b*x^2)^(1/2)/ 
b^2/x^8+2/105*c*(-8*A*c+9*B*b)*(c*x^4+b*x^2)^(1/2)/b^3/x^6-8/315*c^2*(-8*A 
*c+9*B*b)*(c*x^4+b*x^2)^(1/2)/b^4/x^4+16/315*c^3*(-8*A*c+9*B*b)*(c*x^4+b*x 
^2)^(1/2)/b^5/x^2
 

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.65 \[ \int \frac {A+B x^2}{x^9 \sqrt {b x^2+c x^4}} \, dx=\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (9 b B x^2 \left (-5 b^3+6 b^2 c x^2-8 b c^2 x^4+16 c^3 x^6\right )+A \left (-35 b^4+40 b^3 c x^2-48 b^2 c^2 x^4+64 b c^3 x^6-128 c^4 x^8\right )\right )}{315 b^5 x^{10}} \] Input:

Integrate[(A + B*x^2)/(x^9*Sqrt[b*x^2 + c*x^4]),x]
 

Output:

(Sqrt[x^2*(b + c*x^2)]*(9*b*B*x^2*(-5*b^3 + 6*b^2*c*x^2 - 8*b*c^2*x^4 + 16 
*c^3*x^6) + A*(-35*b^4 + 40*b^3*c*x^2 - 48*b^2*c^2*x^4 + 64*b*c^3*x^6 - 12 
8*c^4*x^8)))/(315*b^5*x^10)
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1940, 1220, 1129, 1129, 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.

\(\displaystyle \int \frac {A+B x^2}{x^9 \sqrt {b x^2+c x^4}} \, dx\)

\(\Big \downarrow \) 1940

\(\displaystyle \frac {1}{2} \int \frac {B x^2+A}{x^{10} \sqrt {c x^4+b x^2}}dx^2\)

\(\Big \downarrow \) 1220

\(\displaystyle \frac {1}{2} \left (\frac {(9 b B-8 A c) \int \frac {1}{x^8 \sqrt {c x^4+b x^2}}dx^2}{9 b}-\frac {2 A \sqrt {b x^2+c x^4}}{9 b x^{10}}\right )\)

\(\Big \downarrow \) 1129

\(\displaystyle \frac {1}{2} \left (\frac {(9 b B-8 A c) \left (-\frac {6 c \int \frac {1}{x^6 \sqrt {c x^4+b x^2}}dx^2}{7 b}-\frac {2 \sqrt {b x^2+c x^4}}{7 b x^8}\right )}{9 b}-\frac {2 A \sqrt {b x^2+c x^4}}{9 b x^{10}}\right )\)

\(\Big \downarrow \) 1129

\(\displaystyle \frac {1}{2} \left (\frac {(9 b B-8 A c) \left (-\frac {6 c \left (-\frac {4 c \int \frac {1}{x^4 \sqrt {c x^4+b x^2}}dx^2}{5 b}-\frac {2 \sqrt {b x^2+c x^4}}{5 b x^6}\right )}{7 b}-\frac {2 \sqrt {b x^2+c x^4}}{7 b x^8}\right )}{9 b}-\frac {2 A \sqrt {b x^2+c x^4}}{9 b x^{10}}\right )\)

\(\Big \downarrow \) 1129

\(\displaystyle \frac {1}{2} \left (\frac {(9 b B-8 A c) \left (-\frac {6 c \left (-\frac {4 c \left (-\frac {2 c \int \frac {1}{x^2 \sqrt {c x^4+b x^2}}dx^2}{3 b}-\frac {2 \sqrt {b x^2+c x^4}}{3 b x^4}\right )}{5 b}-\frac {2 \sqrt {b x^2+c x^4}}{5 b x^6}\right )}{7 b}-\frac {2 \sqrt {b x^2+c x^4}}{7 b x^8}\right )}{9 b}-\frac {2 A \sqrt {b x^2+c x^4}}{9 b x^{10}}\right )\)

\(\Big \downarrow \) 1123

\(\displaystyle \frac {1}{2} \left (\frac {\left (-\frac {6 c \left (-\frac {4 c \left (\frac {4 c \sqrt {b x^2+c x^4}}{3 b^2 x^2}-\frac {2 \sqrt {b x^2+c x^4}}{3 b x^4}\right )}{5 b}-\frac {2 \sqrt {b x^2+c x^4}}{5 b x^6}\right )}{7 b}-\frac {2 \sqrt {b x^2+c x^4}}{7 b x^8}\right ) (9 b B-8 A c)}{9 b}-\frac {2 A \sqrt {b x^2+c x^4}}{9 b x^{10}}\right )\)

Input:

Int[(A + B*x^2)/(x^9*Sqrt[b*x^2 + c*x^4]),x]
 

Output:

((-2*A*Sqrt[b*x^2 + c*x^4])/(9*b*x^10) + ((9*b*B - 8*A*c)*((-2*Sqrt[b*x^2 
+ c*x^4])/(7*b*x^8) - (6*c*((-2*Sqrt[b*x^2 + c*x^4])/(5*b*x^6) - (4*c*((-2 
*Sqrt[b*x^2 + c*x^4])/(3*b*x^4) + (4*c*Sqrt[b*x^2 + c*x^4])/(3*b^2*x^2)))/ 
(5*b)))/(7*b)))/(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 = 0.44 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.61

method	result	size
pseudoelliptic	\(-\frac {\left (c \,x^{2}+b \right ) \left (\left (\frac {9 B \,x^{2}}{7}+A \right ) b^{4}-\frac {8 c \,x^{2} \left (\frac {27 B \,x^{2}}{20}+A \right ) b^{3}}{7}+\frac {48 c^{2} x^{4} \left (\frac {3 B \,x^{2}}{2}+A \right ) b^{2}}{35}-\frac {64 c^{3} x^{6} \left (\frac {9 B \,x^{2}}{4}+A \right ) b}{35}+\frac {128 A \,c^{4} x^{8}}{35}\right )}{9 \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, x^{8} b^{5}}\)	\(104\)
trager	\(-\frac {\left (128 A \,c^{4} x^{8}-144 B b \,c^{3} x^{8}-64 A b \,c^{3} x^{6}+72 B \,b^{2} c^{2} x^{6}+48 A \,b^{2} c^{2} x^{4}-54 B \,b^{3} c \,x^{4}-40 A \,b^{3} c \,x^{2}+45 B \,b^{4} x^{2}+35 A \,b^{4}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{315 b^{5} x^{10}}\)	\(111\)
gosper	\(-\frac {\left (c \,x^{2}+b \right ) \left (128 A \,c^{4} x^{8}-144 B b \,c^{3} x^{8}-64 A b \,c^{3} x^{6}+72 B \,b^{2} c^{2} x^{6}+48 A \,b^{2} c^{2} x^{4}-54 B \,b^{3} c \,x^{4}-40 A \,b^{3} c \,x^{2}+45 B \,b^{4} x^{2}+35 A \,b^{4}\right )}{315 x^{8} b^{5} \sqrt {c \,x^{4}+b \,x^{2}}}\)	\(118\)
default	\(-\frac {\left (c \,x^{2}+b \right ) \left (128 A \,c^{4} x^{8}-144 B b \,c^{3} x^{8}-64 A b \,c^{3} x^{6}+72 B \,b^{2} c^{2} x^{6}+48 A \,b^{2} c^{2} x^{4}-54 B \,b^{3} c \,x^{4}-40 A \,b^{3} c \,x^{2}+45 B \,b^{4} x^{2}+35 A \,b^{4}\right )}{315 x^{8} b^{5} \sqrt {c \,x^{4}+b \,x^{2}}}\)	\(118\)
risch	\(-\frac {\left (c \,x^{2}+b \right ) \left (128 A \,c^{4} x^{8}-144 B b \,c^{3} x^{8}-64 A b \,c^{3} x^{6}+72 B \,b^{2} c^{2} x^{6}+48 A \,b^{2} c^{2} x^{4}-54 B \,b^{3} c \,x^{4}-40 A \,b^{3} c \,x^{2}+45 B \,b^{4} x^{2}+35 A \,b^{4}\right )}{315 x^{8} \sqrt {x^{2} \left (c \,x^{2}+b \right )}\, b^{5}}\)	\(118\)
orering	\(-\frac {\left (c \,x^{2}+b \right ) \left (128 A \,c^{4} x^{8}-144 B b \,c^{3} x^{8}-64 A b \,c^{3} x^{6}+72 B \,b^{2} c^{2} x^{6}+48 A \,b^{2} c^{2} x^{4}-54 B \,b^{3} c \,x^{4}-40 A \,b^{3} c \,x^{2}+45 B \,b^{4} x^{2}+35 A \,b^{4}\right )}{315 x^{8} b^{5} \sqrt {c \,x^{4}+b \,x^{2}}}\)	\(118\)

Input:

int((B*x^2+A)/x^9/(c*x^4+b*x^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/9*(c*x^2+b)/(x^2*(c*x^2+b))^(1/2)*((9/7*B*x^2+A)*b^4-8/7*c*x^2*(27/20*B 
*x^2+A)*b^3+48/35*c^2*x^4*(3/2*B*x^2+A)*b^2-64/35*c^3*x^6*(9/4*B*x^2+A)*b+ 
128/35*A*c^4*x^8)/x^8/b^5
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.65 \[ \int \frac {A+B x^2}{x^9 \sqrt {b x^2+c x^4}} \, dx=\frac {{\left (16 \, {\left (9 \, B b c^{3} - 8 \, A c^{4}\right )} x^{8} - 8 \, {\left (9 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} x^{6} - 35 \, A b^{4} + 6 \, {\left (9 \, B b^{3} c - 8 \, A b^{2} c^{2}\right )} x^{4} - 5 \, {\left (9 \, B b^{4} - 8 \, A b^{3} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{315 \, b^{5} x^{10}} \] Input:

integrate((B*x^2+A)/x^9/(c*x^4+b*x^2)^(1/2),x, algorithm="fricas")
 

Output:

1/315*(16*(9*B*b*c^3 - 8*A*c^4)*x^8 - 8*(9*B*b^2*c^2 - 8*A*b*c^3)*x^6 - 35 
*A*b^4 + 6*(9*B*b^3*c - 8*A*b^2*c^2)*x^4 - 5*(9*B*b^4 - 8*A*b^3*c)*x^2)*sq 
rt(c*x^4 + b*x^2)/(b^5*x^10)
 

Sympy [F]

\[ \int \frac {A+B x^2}{x^9 \sqrt {b x^2+c x^4}} \, dx=\int \frac {A + B x^{2}}{x^{9} \sqrt {x^{2} \left (b + c x^{2}\right )}}\, dx \] Input:

integrate((B*x**2+A)/x**9/(c*x**4+b*x**2)**(1/2),x)
 

Output:

Integral((A + B*x**2)/(x**9*sqrt(x**2*(b + c*x**2))), x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.26 \[ \int \frac {A+B x^2}{x^9 \sqrt {b x^2+c x^4}} \, dx=\frac {1}{35} \, B {\left (\frac {16 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{b^{4} x^{2}} - \frac {8 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{b^{3} x^{4}} + \frac {6 \, \sqrt {c x^{4} + b x^{2}} c}{b^{2} x^{6}} - \frac {5 \, \sqrt {c x^{4} + b x^{2}}}{b x^{8}}\right )} - \frac {1}{315} \, A {\left (\frac {128 \, \sqrt {c x^{4} + b x^{2}} c^{4}}{b^{5} x^{2}} - \frac {64 \, \sqrt {c x^{4} + b x^{2}} c^{3}}{b^{4} x^{4}} + \frac {48 \, \sqrt {c x^{4} + b x^{2}} c^{2}}{b^{3} x^{6}} - \frac {40 \, \sqrt {c x^{4} + b x^{2}} c}{b^{2} x^{8}} + \frac {35 \, \sqrt {c x^{4} + b x^{2}}}{b x^{10}}\right )} \] Input:

integrate((B*x^2+A)/x^9/(c*x^4+b*x^2)^(1/2),x, algorithm="maxima")
 

Output:

1/35*B*(16*sqrt(c*x^4 + b*x^2)*c^3/(b^4*x^2) - 8*sqrt(c*x^4 + b*x^2)*c^2/( 
b^3*x^4) + 6*sqrt(c*x^4 + b*x^2)*c/(b^2*x^6) - 5*sqrt(c*x^4 + b*x^2)/(b*x^ 
8)) - 1/315*A*(128*sqrt(c*x^4 + b*x^2)*c^4/(b^5*x^2) - 64*sqrt(c*x^4 + b*x 
^2)*c^3/(b^4*x^4) + 48*sqrt(c*x^4 + b*x^2)*c^2/(b^3*x^6) - 40*sqrt(c*x^4 + 
 b*x^2)*c/(b^2*x^8) + 35*sqrt(c*x^4 + b*x^2)/(b*x^10))
 

Giac [A] (verification not implemented)

Time = 0.92 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.72 \[ \int \frac {A+B x^2}{x^9 \sqrt {b x^2+c x^4}} \, dx=\frac {32 \, {\left (315 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{10} B c^{\frac {7}{2}} - 819 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} B b c^{\frac {7}{2}} + 1008 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{8} A c^{\frac {9}{2}} + 756 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} B b^{2} c^{\frac {7}{2}} - 672 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{6} A b c^{\frac {9}{2}} - 324 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} B b^{3} c^{\frac {7}{2}} + 288 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{4} A b^{2} c^{\frac {9}{2}} + 81 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} B b^{4} c^{\frac {7}{2}} - 72 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} A b^{3} c^{\frac {9}{2}} - 9 \, B b^{5} c^{\frac {7}{2}} + 8 \, A b^{4} c^{\frac {9}{2}}\right )}}{315 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b}\right )}^{2} - b\right )}^{9} \mathrm {sgn}\left (x\right )} \] Input:

integrate((B*x^2+A)/x^9/(c*x^4+b*x^2)^(1/2),x, algorithm="giac")
 

Output:

32/315*(315*(sqrt(c)*x - sqrt(c*x^2 + b))^10*B*c^(7/2) - 819*(sqrt(c)*x - 
sqrt(c*x^2 + b))^8*B*b*c^(7/2) + 1008*(sqrt(c)*x - sqrt(c*x^2 + b))^8*A*c^ 
(9/2) + 756*(sqrt(c)*x - sqrt(c*x^2 + b))^6*B*b^2*c^(7/2) - 672*(sqrt(c)*x 
 - sqrt(c*x^2 + b))^6*A*b*c^(9/2) - 324*(sqrt(c)*x - sqrt(c*x^2 + b))^4*B* 
b^3*c^(7/2) + 288*(sqrt(c)*x - sqrt(c*x^2 + b))^4*A*b^2*c^(9/2) + 81*(sqrt 
(c)*x - sqrt(c*x^2 + b))^2*B*b^4*c^(7/2) - 72*(sqrt(c)*x - sqrt(c*x^2 + b) 
)^2*A*b^3*c^(9/2) - 9*B*b^5*c^(7/2) + 8*A*b^4*c^(9/2))/(((sqrt(c)*x - sqrt 
(c*x^2 + b))^2 - b)^9*sgn(x))
 

Mupad [B] (verification not implemented)

Time = 9.43 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.92 \[ \int \frac {A+B x^2}{x^9 \sqrt {b x^2+c x^4}} \, dx=\frac {\left (8\,A\,c-9\,B\,b\right )\,\sqrt {c\,x^4+b\,x^2}}{63\,b^2\,x^8}-\frac {A\,\sqrt {c\,x^4+b\,x^2}}{9\,b\,x^{10}}-\frac {\left (16\,A\,c^2-18\,B\,b\,c\right )\,\sqrt {c\,x^4+b\,x^2}}{105\,b^3\,x^6}+\frac {\left (64\,A\,c^3-72\,B\,b\,c^2\right )\,\sqrt {c\,x^4+b\,x^2}}{315\,b^4\,x^4}-\frac {\left (128\,A\,c^4-144\,B\,b\,c^3\right )\,\sqrt {c\,x^4+b\,x^2}}{315\,b^5\,x^2} \] Input:

int((A + B*x^2)/(x^9*(b*x^2 + c*x^4)^(1/2)),x)
 

Output:

((8*A*c - 9*B*b)*(b*x^2 + c*x^4)^(1/2))/(63*b^2*x^8) - (A*(b*x^2 + c*x^4)^ 
(1/2))/(9*b*x^10) - ((16*A*c^2 - 18*B*b*c)*(b*x^2 + c*x^4)^(1/2))/(105*b^3 
*x^6) + ((64*A*c^3 - 72*B*b*c^2)*(b*x^2 + c*x^4)^(1/2))/(315*b^4*x^4) - (( 
128*A*c^4 - 144*B*b*c^3)*(b*x^2 + c*x^4)^(1/2))/(315*b^5*x^2)
 

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.12 \[ \int \frac {A+B x^2}{x^9 \sqrt {b x^2+c x^4}} \, dx=\frac {-35 \sqrt {c \,x^{2}+b}\, a \,b^{4}+40 \sqrt {c \,x^{2}+b}\, a \,b^{3} c \,x^{2}-48 \sqrt {c \,x^{2}+b}\, a \,b^{2} c^{2} x^{4}+64 \sqrt {c \,x^{2}+b}\, a b \,c^{3} x^{6}-128 \sqrt {c \,x^{2}+b}\, a \,c^{4} x^{8}-45 \sqrt {c \,x^{2}+b}\, b^{5} x^{2}+54 \sqrt {c \,x^{2}+b}\, b^{4} c \,x^{4}-72 \sqrt {c \,x^{2}+b}\, b^{3} c^{2} x^{6}+144 \sqrt {c \,x^{2}+b}\, b^{2} c^{3} x^{8}+128 \sqrt {c}\, a \,c^{4} x^{9}-144 \sqrt {c}\, b^{2} c^{3} x^{9}}{315 b^{5} x^{9}} \] Input:

int((B*x^2+A)/x^9/(c*x^4+b*x^2)^(1/2),x)
 

Output:

( - 35*sqrt(b + c*x**2)*a*b**4 + 40*sqrt(b + c*x**2)*a*b**3*c*x**2 - 48*sq 
rt(b + c*x**2)*a*b**2*c**2*x**4 + 64*sqrt(b + c*x**2)*a*b*c**3*x**6 - 128* 
sqrt(b + c*x**2)*a*c**4*x**8 - 45*sqrt(b + c*x**2)*b**5*x**2 + 54*sqrt(b + 
 c*x**2)*b**4*c*x**4 - 72*sqrt(b + c*x**2)*b**3*c**2*x**6 + 144*sqrt(b + c 
*x**2)*b**2*c**3*x**8 + 128*sqrt(c)*a*c**4*x**9 - 144*sqrt(c)*b**2*c**3*x* 
*9)/(315*b**5*x**9)