\(\int \frac {(A+B x^2) (b x^2+c x^4)^{3/2}}{x^{12}} \, dx\) [192]

Optimal result

Integrand size = 26, antiderivative size = 177 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{12}} \, dx=-\frac {c (8 b B-3 A c) \sqrt {b x^2+c x^4}}{64 b x^5}-\frac {c^2 (8 b B-3 A c) \sqrt {b x^2+c x^4}}{128 b^2 x^3}-\frac {(8 b B-3 A c) \left (b x^2+c x^4\right )^{3/2}}{48 b x^9}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{8 b x^{13}}+\frac {c^3 (8 b B-3 A c) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{128 b^{5/2}} \] Output:

-1/64*c*(-3*A*c+8*B*b)*(c*x^4+b*x^2)^(1/2)/b/x^5-1/128*c^2*(-3*A*c+8*B*b)* 
(c*x^4+b*x^2)^(1/2)/b^2/x^3-1/48*(-3*A*c+8*B*b)*(c*x^4+b*x^2)^(3/2)/b/x^9- 
1/8*A*(c*x^4+b*x^2)^(5/2)/b/x^13+1/128*c^3*(-3*A*c+8*B*b)*arctanh(b^(1/2)* 
x/(c*x^4+b*x^2)^(1/2))/b^(5/2)
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.87 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{12}} \, dx=-\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (\sqrt {b} \sqrt {b+c x^2} \left (8 b B x^2 \left (8 b^2+14 b c x^2+3 c^2 x^4\right )+A \left (48 b^3+72 b^2 c x^2+6 b c^2 x^4-9 c^3 x^6\right )\right )+3 c^3 (-8 b B+3 A c) x^8 \text {arctanh}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )\right )}{384 b^{5/2} x^9 \sqrt {b+c x^2}} \] Input:

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^12,x]
 

Output:

-1/384*(Sqrt[x^2*(b + c*x^2)]*(Sqrt[b]*Sqrt[b + c*x^2]*(8*b*B*x^2*(8*b^2 + 
 14*b*c*x^2 + 3*c^2*x^4) + A*(48*b^3 + 72*b^2*c*x^2 + 6*b*c^2*x^4 - 9*c^3* 
x^6)) + 3*c^3*(-8*b*B + 3*A*c)*x^8*ArcTanh[Sqrt[b + c*x^2]/Sqrt[b]]))/(b^( 
5/2)*x^9*Sqrt[b + c*x^2])
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.89, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1944, 1425, 1425, 1430, 1400, 219}

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.

\(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{12}} \, dx\)

\(\Big \downarrow \) 1944

\(\displaystyle \frac {(8 b B-3 A c) \int \frac {\left (c x^4+b x^2\right )^{3/2}}{x^{10}}dx}{8 b}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{8 b x^{13}}\)

\(\Big \downarrow \) 1425

\(\displaystyle \frac {(8 b B-3 A c) \left (\frac {1}{2} c \int \frac {\sqrt {c x^4+b x^2}}{x^6}dx-\frac {\left (b x^2+c x^4\right )^{3/2}}{6 x^9}\right )}{8 b}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{8 b x^{13}}\)

\(\Big \downarrow \) 1425

\(\displaystyle \frac {(8 b B-3 A c) \left (\frac {1}{2} c \left (\frac {1}{4} c \int \frac {1}{x^2 \sqrt {c x^4+b x^2}}dx-\frac {\sqrt {b x^2+c x^4}}{4 x^5}\right )-\frac {\left (b x^2+c x^4\right )^{3/2}}{6 x^9}\right )}{8 b}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{8 b x^{13}}\)

\(\Big \downarrow \) 1430

\(\displaystyle \frac {(8 b B-3 A c) \left (\frac {1}{2} c \left (\frac {1}{4} c \left (-\frac {c \int \frac {1}{\sqrt {c x^4+b x^2}}dx}{2 b}-\frac {\sqrt {b x^2+c x^4}}{2 b x^3}\right )-\frac {\sqrt {b x^2+c x^4}}{4 x^5}\right )-\frac {\left (b x^2+c x^4\right )^{3/2}}{6 x^9}\right )}{8 b}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{8 b x^{13}}\)

\(\Big \downarrow \) 1400

\(\displaystyle \frac {(8 b B-3 A c) \left (\frac {1}{2} c \left (\frac {1}{4} c \left (\frac {c \int \frac {1}{1-\frac {b x^2}{c x^4+b x^2}}d\frac {x}{\sqrt {c x^4+b x^2}}}{2 b}-\frac {\sqrt {b x^2+c x^4}}{2 b x^3}\right )-\frac {\sqrt {b x^2+c x^4}}{4 x^5}\right )-\frac {\left (b x^2+c x^4\right )^{3/2}}{6 x^9}\right )}{8 b}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{8 b x^{13}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {(8 b B-3 A c) \left (\frac {1}{2} c \left (\frac {1}{4} c \left (\frac {c \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{2 b^{3/2}}-\frac {\sqrt {b x^2+c x^4}}{2 b x^3}\right )-\frac {\sqrt {b x^2+c x^4}}{4 x^5}\right )-\frac {\left (b x^2+c x^4\right )^{3/2}}{6 x^9}\right )}{8 b}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{8 b x^{13}}\)

Input:

Int[((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^12,x]
 

Output:

-1/8*(A*(b*x^2 + c*x^4)^(5/2))/(b*x^13) + ((8*b*B - 3*A*c)*(-1/6*(b*x^2 + 
c*x^4)^(3/2)/x^9 + (c*(-1/4*Sqrt[b*x^2 + c*x^4]/x^5 + (c*(-1/2*Sqrt[b*x^2 
+ c*x^4]/(b*x^3) + (c*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4]])/(2*b^(3/2) 
)))/4))/2))/(8*b)
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1400

Int[1/Sqrt[(b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> -Subst[Int[1/(1 - b*x 
^2), x], x, x/Sqrt[b*x^2 + c*x^4]] /; FreeQ[{b, c}, x]
 

rule 1425

Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp 
[(d*x)^(m + 1)*((b*x^2 + c*x^4)^p/(d*(m + 2*p + 1))), x] - Simp[2*c*(p/(d^4 
*(m + 2*p + 1)))   Int[(d*x)^(m + 4)*(b*x^2 + c*x^4)^(p - 1), x], x] /; Fre 
eQ[{b, c, d, m, p}, x] &&  !IntegerQ[p] && GtQ[p, 0] && LtQ[m + 2*p + 1, 0]
 

rule 1430

Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp 
[d*(d*x)^(m - 1)*((b*x^2 + c*x^4)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[c*( 
(m + 4*p + 3)/(b*d^2*(m + 2*p + 1)))   Int[(d*x)^(m + 2)*(b*x^2 + c*x^4)^p, 
 x], x] /; FreeQ[{b, c, d, m, p}, x] &&  !IntegerQ[p] && LtQ[m + 2*p + 1, 0 
]
 

rule 1944

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + 
(d_.)*(x_)^(n_.)), x_Symbol] :> Simp[c*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j 
+ b*x^(j + n))^(p + 1)/(a*(m + j*p + 1))), x] + Simp[(a*d*(m + j*p + 1) - b 
*c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1))   Int[(e*x)^(m + n)*(a*x^ 
j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j 
+ n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m + j*p, -1 
] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, (-n)*p - 1])) && ( 
GtQ[e, 0] || IntegersQ[j, n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1 
, 0]
 

Maple [A] (verified)

Time = 1.45 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.86

method	result	size
risch	\(-\frac {\left (-9 A \,c^{3} x^{6}+24 B b \,c^{2} x^{6}+6 A b \,c^{2} x^{4}+112 x^{4} B \,b^{2} c +72 A \,b^{2} c \,x^{2}+64 x^{2} B \,b^{3}+48 A \,b^{3}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{384 x^{9} b^{2}}-\frac {\left (3 A c -8 B b \right ) c^{3} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{128 b^{\frac {5}{2}} x \sqrt {c \,x^{2}+b}}\)	\(153\)
default	\(-\frac {\left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \left (9 A \,b^{\frac {3}{2}} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) c^{4} x^{8}-3 A \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{4} x^{8}-24 B \,b^{\frac {5}{2}} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) c^{3} x^{8}+8 B \left (c \,x^{2}+b \right )^{\frac {3}{2}} b \,c^{3} x^{8}+3 A \left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{3} x^{6}-9 A \sqrt {c \,x^{2}+b}\, b \,c^{4} x^{8}-8 B \left (c \,x^{2}+b \right )^{\frac {5}{2}} b \,c^{2} x^{6}+24 B \sqrt {c \,x^{2}+b}\, b^{2} c^{3} x^{8}+6 A \left (c \,x^{2}+b \right )^{\frac {5}{2}} b \,c^{2} x^{4}-16 B \left (c \,x^{2}+b \right )^{\frac {5}{2}} b^{2} c \,x^{4}-24 A \left (c \,x^{2}+b \right )^{\frac {5}{2}} b^{2} c \,x^{2}+64 B \left (c \,x^{2}+b \right )^{\frac {5}{2}} b^{3} x^{2}+48 A \left (c \,x^{2}+b \right )^{\frac {5}{2}} b^{3}\right )}{384 x^{11} \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{4}}\)	\(302\)

Input:

int((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^12,x,method=_RETURNVERBOSE)
 

Output:

-1/384*(-9*A*c^3*x^6+24*B*b*c^2*x^6+6*A*b*c^2*x^4+112*B*b^2*c*x^4+72*A*b^2 
*c*x^2+64*B*b^3*x^2+48*A*b^3)/x^9/b^2*(x^2*(c*x^2+b))^(1/2)-1/128*(3*A*c-8 
*B*b)*c^3/b^(5/2)*ln((2*b+2*b^(1/2)*(c*x^2+b)^(1/2))/x)*(x^2*(c*x^2+b))^(1 
/2)/x/(c*x^2+b)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.66 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{12}} \, dx=\left [-\frac {3 \, {\left (8 \, B b c^{3} - 3 \, A c^{4}\right )} \sqrt {b} x^{9} \log \left (-\frac {c x^{3} + 2 \, b x - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) + 2 \, {\left (3 \, {\left (8 \, B b^{2} c^{2} - 3 \, A b c^{3}\right )} x^{6} + 48 \, A b^{4} + 2 \, {\left (56 \, B b^{3} c + 3 \, A b^{2} c^{2}\right )} x^{4} + 8 \, {\left (8 \, B b^{4} + 9 \, A b^{3} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{768 \, b^{3} x^{9}}, -\frac {3 \, {\left (8 \, B b c^{3} - 3 \, A c^{4}\right )} \sqrt {-b} x^{9} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{b x}\right ) + {\left (3 \, {\left (8 \, B b^{2} c^{2} - 3 \, A b c^{3}\right )} x^{6} + 48 \, A b^{4} + 2 \, {\left (56 \, B b^{3} c + 3 \, A b^{2} c^{2}\right )} x^{4} + 8 \, {\left (8 \, B b^{4} + 9 \, A b^{3} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{384 \, b^{3} x^{9}}\right ] \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^12,x, algorithm="fricas")
 

Output:

[-1/768*(3*(8*B*b*c^3 - 3*A*c^4)*sqrt(b)*x^9*log(-(c*x^3 + 2*b*x - 2*sqrt( 
c*x^4 + b*x^2)*sqrt(b))/x^3) + 2*(3*(8*B*b^2*c^2 - 3*A*b*c^3)*x^6 + 48*A*b 
^4 + 2*(56*B*b^3*c + 3*A*b^2*c^2)*x^4 + 8*(8*B*b^4 + 9*A*b^3*c)*x^2)*sqrt( 
c*x^4 + b*x^2))/(b^3*x^9), -1/384*(3*(8*B*b*c^3 - 3*A*c^4)*sqrt(-b)*x^9*ar 
ctan(sqrt(c*x^4 + b*x^2)*sqrt(-b)/(b*x)) + (3*(8*B*b^2*c^2 - 3*A*b*c^3)*x^ 
6 + 48*A*b^4 + 2*(56*B*b^3*c + 3*A*b^2*c^2)*x^4 + 8*(8*B*b^4 + 9*A*b^3*c)* 
x^2)*sqrt(c*x^4 + b*x^2))/(b^3*x^9)]
 

Sympy [F]

\[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{12}} \, dx=\int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}} \left (A + B x^{2}\right )}{x^{12}}\, dx \] Input:

integrate((B*x**2+A)*(c*x**4+b*x**2)**(3/2)/x**12,x)
 

Output:

Integral((x**2*(b + c*x**2))**(3/2)*(A + B*x**2)/x**12, x)
 

Maxima [F]

\[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{12}} \, dx=\int { \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} {\left (B x^{2} + A\right )}}{x^{12}} \,d x } \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^12,x, algorithm="maxima")
 

Output:

integrate((c*x^4 + b*x^2)^(3/2)*(B*x^2 + A)/x^12, x)
 

Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.21 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{12}} \, dx=-\frac {\frac {3 \, {\left (8 \, B b c^{4} \mathrm {sgn}\left (x\right ) - 3 \, A c^{5} \mathrm {sgn}\left (x\right )\right )} \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{2}} + \frac {24 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} B b c^{4} \mathrm {sgn}\left (x\right ) + 40 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} B b^{2} c^{4} \mathrm {sgn}\left (x\right ) - 88 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} B b^{3} c^{4} \mathrm {sgn}\left (x\right ) + 24 \, \sqrt {c x^{2} + b} B b^{4} c^{4} \mathrm {sgn}\left (x\right ) - 9 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} A c^{5} \mathrm {sgn}\left (x\right ) + 33 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} A b c^{5} \mathrm {sgn}\left (x\right ) + 33 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} A b^{2} c^{5} \mathrm {sgn}\left (x\right ) - 9 \, \sqrt {c x^{2} + b} A b^{3} c^{5} \mathrm {sgn}\left (x\right )}{b^{2} c^{4} x^{8}}}{384 \, c} \] Input:

integrate((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^12,x, algorithm="giac")
 

Output:

-1/384*(3*(8*B*b*c^4*sgn(x) - 3*A*c^5*sgn(x))*arctan(sqrt(c*x^2 + b)/sqrt( 
-b))/(sqrt(-b)*b^2) + (24*(c*x^2 + b)^(7/2)*B*b*c^4*sgn(x) + 40*(c*x^2 + b 
)^(5/2)*B*b^2*c^4*sgn(x) - 88*(c*x^2 + b)^(3/2)*B*b^3*c^4*sgn(x) + 24*sqrt 
(c*x^2 + b)*B*b^4*c^4*sgn(x) - 9*(c*x^2 + b)^(7/2)*A*c^5*sgn(x) + 33*(c*x^ 
2 + b)^(5/2)*A*b*c^5*sgn(x) + 33*(c*x^2 + b)^(3/2)*A*b^2*c^5*sgn(x) - 9*sq 
rt(c*x^2 + b)*A*b^3*c^5*sgn(x))/(b^2*c^4*x^8))/c
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{12}} \, dx=\int \frac {\left (B\,x^2+A\right )\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^{12}} \,d x \] Input:

int(((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^12,x)
 

Output:

int(((A + B*x^2)*(b*x^2 + c*x^4)^(3/2))/x^12, x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.51 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{12}} \, dx=\frac {-48 \sqrt {c \,x^{2}+b}\, a \,b^{4}-72 \sqrt {c \,x^{2}+b}\, a \,b^{3} c \,x^{2}-6 \sqrt {c \,x^{2}+b}\, a \,b^{2} c^{2} x^{4}+9 \sqrt {c \,x^{2}+b}\, a b \,c^{3} x^{6}-64 \sqrt {c \,x^{2}+b}\, b^{5} x^{2}-112 \sqrt {c \,x^{2}+b}\, b^{4} c \,x^{4}-24 \sqrt {c \,x^{2}+b}\, b^{3} c^{2} x^{6}+9 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}-\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a \,c^{4} x^{8}-24 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}-\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{2} c^{3} x^{8}-9 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) a \,c^{4} x^{8}+24 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {c \,x^{2}+b}+\sqrt {b}+\sqrt {c}\, x}{\sqrt {b}}\right ) b^{2} c^{3} x^{8}}{384 b^{3} x^{8}} \] Input:

int((B*x^2+A)*(c*x^4+b*x^2)^(3/2)/x^12,x)
 

Output:

( - 48*sqrt(b + c*x**2)*a*b**4 - 72*sqrt(b + c*x**2)*a*b**3*c*x**2 - 6*sqr 
t(b + c*x**2)*a*b**2*c**2*x**4 + 9*sqrt(b + c*x**2)*a*b*c**3*x**6 - 64*sqr 
t(b + c*x**2)*b**5*x**2 - 112*sqrt(b + c*x**2)*b**4*c*x**4 - 24*sqrt(b + c 
*x**2)*b**3*c**2*x**6 + 9*sqrt(b)*log((sqrt(b + c*x**2) - sqrt(b) + sqrt(c 
)*x)/sqrt(b))*a*c**4*x**8 - 24*sqrt(b)*log((sqrt(b + c*x**2) - sqrt(b) + s 
qrt(c)*x)/sqrt(b))*b**2*c**3*x**8 - 9*sqrt(b)*log((sqrt(b + c*x**2) + sqrt 
(b) + sqrt(c)*x)/sqrt(b))*a*c**4*x**8 + 24*sqrt(b)*log((sqrt(b + c*x**2) + 
 sqrt(b) + sqrt(c)*x)/sqrt(b))*b**2*c**3*x**8)/(384*b**3*x**8)