\(\int \frac {x^2 (A+B x^2)}{(b x^2+c x^4)^3} \, dx\) [89]

Optimal result

Integrand size = 24, antiderivative size = 117 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {A}{3 b^3 x^3}-\frac {b B-3 A c}{b^4 x}-\frac {c (b B-A c) x}{4 b^3 \left (b+c x^2\right )^2}-\frac {c (7 b B-11 A c) x}{8 b^4 \left (b+c x^2\right )}-\frac {5 \sqrt {c} (3 b B-7 A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{9/2}} \] Output:

-1/3*A/b^3/x^3-(-3*A*c+B*b)/b^4/x-1/4*c*(-A*c+B*b)*x/b^3/(c*x^2+b)^2-1/8*c 
*(-11*A*c+7*B*b)*x/b^4/(c*x^2+b)-5/8*c^(1/2)*(-7*A*c+3*B*b)*arctan(c^(1/2) 
*x/b^(1/2))/b^(9/2)
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.02 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {A}{3 b^3 x^3}+\frac {-b B+3 A c}{b^4 x}-\frac {c (b B-A c) x}{4 b^3 \left (b+c x^2\right )^2}-\frac {\left (7 b B c-11 A c^2\right ) x}{8 b^4 \left (b+c x^2\right )}-\frac {5 \sqrt {c} (3 b B-7 A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{9/2}} \] Input:

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

Output:

-1/3*A/(b^3*x^3) + (-(b*B) + 3*A*c)/(b^4*x) - (c*(b*B - A*c)*x)/(4*b^3*(b 
+ c*x^2)^2) - ((7*b*B*c - 11*A*c^2)*x)/(8*b^4*(b + c*x^2)) - (5*Sqrt[c]*(3 
*b*B - 7*A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(8*b^(9/2))
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {9, 361, 25, 1582, 1584, 2009}

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.

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

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 361

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1582

\(\displaystyle \frac {1}{4} c \left (\frac {\int \frac {-\frac {c^2 (7 b B-11 A c) x^4}{b}+8 c (b B-2 A c) x^2+8 A b c}{x^4 \left (c x^2+b\right )}dx}{2 b^3 c^2}-\frac {x (7 b B-11 A c)}{2 b^4 \left (b+c x^2\right )}\right )-\frac {c x (b B-A c)}{4 b^3 \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 1584

\(\displaystyle \frac {1}{4} c \left (\frac {\int \left (-\frac {5 (3 b B-7 A c) c^2}{b \left (c x^2+b\right )}+\frac {8 (b B-3 A c) c}{b x^2}+\frac {8 A c}{x^4}\right )dx}{2 b^3 c^2}-\frac {x (7 b B-11 A c)}{2 b^4 \left (b+c x^2\right )}\right )-\frac {c x (b B-A c)}{4 b^3 \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{4} c \left (\frac {-\frac {5 c^{3/2} (3 b B-7 A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{b^{3/2}}-\frac {8 c (b B-3 A c)}{b x}-\frac {8 A c}{3 x^3}}{2 b^3 c^2}-\frac {x (7 b B-11 A c)}{2 b^4 \left (b+c x^2\right )}\right )-\frac {c x (b B-A c)}{4 b^3 \left (b+c x^2\right )^2}\)

Input:

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

Output:

-1/4*(c*(b*B - A*c)*x)/(b^3*(b + c*x^2)^2) + (c*(-1/2*((7*b*B - 11*A*c)*x) 
/(b^4*(b + c*x^2)) + ((-8*A*c)/(3*x^3) - (8*c*(b*B - 3*A*c))/(b*x) - (5*c^ 
(3/2)*(3*b*B - 7*A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/b^(3/2))/(2*b^3*c^2)))/ 
4
 

Defintions of rubi rules used

rule 9

Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, 
x, Min]}, Simp[1/e^(p*r)   Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, 
x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && 
  !MonomialQ[Px, x]
 

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 361

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : 
> Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p 
+ 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1))   Int[x^m*(a + b*x^2)^(p + 1)*E 
xpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c 
- a*d)*x^(-m + 2))/(a + b*x^2)] - ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], 
 x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ILtQ[m/ 
2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
 

rule 1582

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^ 
4)^(p_.), x_Symbol] :> Simp[(-d)^(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*((d 
+ e*x^2)^(q + 1)/(2*e^(2*p + m/2)*(q + 1))), x] + Simp[(-d)^(m/2 - 1)/(2*e^ 
(2*p)*(q + 1))   Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e 
*x^2))*(2*(-d)^(-m/2 + 1)*e^(2*p)*(q + 1)*(a + b*x^2 + c*x^4)^p - ((c*d^2 - 
 b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2))], x], x], x] /; Fre 
eQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] 
&& ILtQ[m/2, 0]
 

rule 1584

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( 
c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* 
(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[ 
b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.84

method	result	size
default	\(-\frac {A}{3 b^{3} x^{3}}-\frac {-3 A c +B b}{b^{4} x}+\frac {c \left (\frac {\left (\frac {11}{8} A \,c^{2}-\frac {7}{8} B b c \right ) x^{3}+\frac {b \left (13 A c -9 B b \right ) x}{8}}{\left (c \,x^{2}+b \right )^{2}}+\frac {5 \left (7 A c -3 B b \right ) \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \sqrt {b c}}\right )}{b^{4}}\)	\(98\)
risch	\(\frac {\frac {5 c^{2} \left (7 A c -3 B b \right ) x^{6}}{8 b^{4}}+\frac {25 c \left (7 A c -3 B b \right ) x^{4}}{24 b^{3}}+\frac {\left (7 A c -3 B b \right ) x^{2}}{3 b^{2}}-\frac {A}{3 b}}{x^{3} \left (c \,x^{2}+b \right )^{2}}+\frac {35 \sqrt {-b c}\, \ln \left (-c x -\sqrt {-b c}\right ) A c}{16 b^{5}}-\frac {15 \sqrt {-b c}\, \ln \left (-c x -\sqrt {-b c}\right ) B}{16 b^{4}}-\frac {35 \sqrt {-b c}\, \ln \left (-c x +\sqrt {-b c}\right ) A c}{16 b^{5}}+\frac {15 \sqrt {-b c}\, \ln \left (-c x +\sqrt {-b c}\right ) B}{16 b^{4}}\)	\(179\)

Input:

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

Output:

-1/3*A/b^3/x^3-(-3*A*c+B*b)/b^4/x+1/b^4*c*(((11/8*A*c^2-7/8*B*b*c)*x^3+1/8 
*b*(13*A*c-9*B*b)*x)/(c*x^2+b)^2+5/8*(7*A*c-3*B*b)/(b*c)^(1/2)*arctan(c*x/ 
(b*c)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 368, normalized size of antiderivative = 3.15 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\left [-\frac {30 \, {\left (3 \, B b c^{2} - 7 \, A c^{3}\right )} x^{6} + 50 \, {\left (3 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{4} + 16 \, A b^{3} + 16 \, {\left (3 \, B b^{3} - 7 \, A b^{2} c\right )} x^{2} + 15 \, {\left ({\left (3 \, B b c^{2} - 7 \, A c^{3}\right )} x^{7} + 2 \, {\left (3 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{5} + {\left (3 \, B b^{3} - 7 \, A b^{2} c\right )} x^{3}\right )} \sqrt {-\frac {c}{b}} \log \left (\frac {c x^{2} + 2 \, b x \sqrt {-\frac {c}{b}} - b}{c x^{2} + b}\right )}{48 \, {\left (b^{4} c^{2} x^{7} + 2 \, b^{5} c x^{5} + b^{6} x^{3}\right )}}, -\frac {15 \, {\left (3 \, B b c^{2} - 7 \, A c^{3}\right )} x^{6} + 25 \, {\left (3 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{4} + 8 \, A b^{3} + 8 \, {\left (3 \, B b^{3} - 7 \, A b^{2} c\right )} x^{2} + 15 \, {\left ({\left (3 \, B b c^{2} - 7 \, A c^{3}\right )} x^{7} + 2 \, {\left (3 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{5} + {\left (3 \, B b^{3} - 7 \, A b^{2} c\right )} x^{3}\right )} \sqrt {\frac {c}{b}} \arctan \left (x \sqrt {\frac {c}{b}}\right )}{24 \, {\left (b^{4} c^{2} x^{7} + 2 \, b^{5} c x^{5} + b^{6} x^{3}\right )}}\right ] \] Input:

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

Output:

[-1/48*(30*(3*B*b*c^2 - 7*A*c^3)*x^6 + 50*(3*B*b^2*c - 7*A*b*c^2)*x^4 + 16 
*A*b^3 + 16*(3*B*b^3 - 7*A*b^2*c)*x^2 + 15*((3*B*b*c^2 - 7*A*c^3)*x^7 + 2* 
(3*B*b^2*c - 7*A*b*c^2)*x^5 + (3*B*b^3 - 7*A*b^2*c)*x^3)*sqrt(-c/b)*log((c 
*x^2 + 2*b*x*sqrt(-c/b) - b)/(c*x^2 + b)))/(b^4*c^2*x^7 + 2*b^5*c*x^5 + b^ 
6*x^3), -1/24*(15*(3*B*b*c^2 - 7*A*c^3)*x^6 + 25*(3*B*b^2*c - 7*A*b*c^2)*x 
^4 + 8*A*b^3 + 8*(3*B*b^3 - 7*A*b^2*c)*x^2 + 15*((3*B*b*c^2 - 7*A*c^3)*x^7 
 + 2*(3*B*b^2*c - 7*A*b*c^2)*x^5 + (3*B*b^3 - 7*A*b^2*c)*x^3)*sqrt(c/b)*ar 
ctan(x*sqrt(c/b)))/(b^4*c^2*x^7 + 2*b^5*c*x^5 + b^6*x^3)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 226 vs. \(2 (110) = 220\).

Time = 0.39 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.93 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {5 \sqrt {- \frac {c}{b^{9}}} \left (- 7 A c + 3 B b\right ) \log {\left (- \frac {5 b^{5} \sqrt {- \frac {c}{b^{9}}} \left (- 7 A c + 3 B b\right )}{- 35 A c^{2} + 15 B b c} + x \right )}}{16} - \frac {5 \sqrt {- \frac {c}{b^{9}}} \left (- 7 A c + 3 B b\right ) \log {\left (\frac {5 b^{5} \sqrt {- \frac {c}{b^{9}}} \left (- 7 A c + 3 B b\right )}{- 35 A c^{2} + 15 B b c} + x \right )}}{16} + \frac {- 8 A b^{3} + x^{6} \cdot \left (105 A c^{3} - 45 B b c^{2}\right ) + x^{4} \cdot \left (175 A b c^{2} - 75 B b^{2} c\right ) + x^{2} \cdot \left (56 A b^{2} c - 24 B b^{3}\right )}{24 b^{6} x^{3} + 48 b^{5} c x^{5} + 24 b^{4} c^{2} x^{7}} \] Input:

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

Output:

5*sqrt(-c/b**9)*(-7*A*c + 3*B*b)*log(-5*b**5*sqrt(-c/b**9)*(-7*A*c + 3*B*b 
)/(-35*A*c**2 + 15*B*b*c) + x)/16 - 5*sqrt(-c/b**9)*(-7*A*c + 3*B*b)*log(5 
*b**5*sqrt(-c/b**9)*(-7*A*c + 3*B*b)/(-35*A*c**2 + 15*B*b*c) + x)/16 + (-8 
*A*b**3 + x**6*(105*A*c**3 - 45*B*b*c**2) + x**4*(175*A*b*c**2 - 75*B*b**2 
*c) + x**2*(56*A*b**2*c - 24*B*b**3))/(24*b**6*x**3 + 48*b**5*c*x**5 + 24* 
b**4*c**2*x**7)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.09 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {15 \, {\left (3 \, B b c^{2} - 7 \, A c^{3}\right )} x^{6} + 25 \, {\left (3 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{4} + 8 \, A b^{3} + 8 \, {\left (3 \, B b^{3} - 7 \, A b^{2} c\right )} x^{2}}{24 \, {\left (b^{4} c^{2} x^{7} + 2 \, b^{5} c x^{5} + b^{6} x^{3}\right )}} - \frac {5 \, {\left (3 \, B b c - 7 \, A c^{2}\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} b^{4}} \] Input:

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

Output:

-1/24*(15*(3*B*b*c^2 - 7*A*c^3)*x^6 + 25*(3*B*b^2*c - 7*A*b*c^2)*x^4 + 8*A 
*b^3 + 8*(3*B*b^3 - 7*A*b^2*c)*x^2)/(b^4*c^2*x^7 + 2*b^5*c*x^5 + b^6*x^3) 
- 5/8*(3*B*b*c - 7*A*c^2)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^4)
 

Giac [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.92 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {5 \, {\left (3 \, B b c - 7 \, A c^{2}\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} b^{4}} - \frac {7 \, B b c^{2} x^{3} - 11 \, A c^{3} x^{3} + 9 \, B b^{2} c x - 13 \, A b c^{2} x}{8 \, {\left (c x^{2} + b\right )}^{2} b^{4}} - \frac {3 \, B b x^{2} - 9 \, A c x^{2} + A b}{3 \, b^{4} x^{3}} \] Input:

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

Output:

-5/8*(3*B*b*c - 7*A*c^2)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^4) - 1/8*(7*B* 
b*c^2*x^3 - 11*A*c^3*x^3 + 9*B*b^2*c*x - 13*A*b*c^2*x)/((c*x^2 + b)^2*b^4) 
 - 1/3*(3*B*b*x^2 - 9*A*c*x^2 + A*b)/(b^4*x^3)
 

Mupad [B] (verification not implemented)

Time = 8.95 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.97 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {\frac {x^2\,\left (7\,A\,c-3\,B\,b\right )}{3\,b^2}-\frac {A}{3\,b}+\frac {5\,c^2\,x^6\,\left (7\,A\,c-3\,B\,b\right )}{8\,b^4}+\frac {25\,c\,x^4\,\left (7\,A\,c-3\,B\,b\right )}{24\,b^3}}{b^2\,x^3+2\,b\,c\,x^5+c^2\,x^7}+\frac {5\,\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )\,\left (7\,A\,c-3\,B\,b\right )}{8\,b^{9/2}} \] Input:

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

Output:

((x^2*(7*A*c - 3*B*b))/(3*b^2) - A/(3*b) + (5*c^2*x^6*(7*A*c - 3*B*b))/(8* 
b^4) + (25*c*x^4*(7*A*c - 3*B*b))/(24*b^3))/(b^2*x^3 + c^2*x^7 + 2*b*c*x^5 
) + (5*c^(1/2)*atan((c^(1/2)*x)/b^(1/2))*(7*A*c - 3*B*b))/(8*b^(9/2))
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 248, normalized size of antiderivative = 2.12 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {105 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) a \,b^{2} c \,x^{3}+210 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) a b \,c^{2} x^{5}+105 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) a \,c^{3} x^{7}-45 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{4} x^{3}-90 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{3} c \,x^{5}-45 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{2} c^{2} x^{7}-8 a \,b^{4}+56 a \,b^{3} c \,x^{2}+175 a \,b^{2} c^{2} x^{4}+105 a b \,c^{3} x^{6}-24 b^{5} x^{2}-75 b^{4} c \,x^{4}-45 b^{3} c^{2} x^{6}}{24 b^{5} x^{3} \left (c^{2} x^{4}+2 b c \,x^{2}+b^{2}\right )} \] Input:

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

Output:

(105*sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*a*b**2*c*x**3 + 210*sqr 
t(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*a*b*c**2*x**5 + 105*sqrt(c)*sqr 
t(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*a*c**3*x**7 - 45*sqrt(c)*sqrt(b)*atan(( 
c*x)/(sqrt(c)*sqrt(b)))*b**4*x**3 - 90*sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c) 
*sqrt(b)))*b**3*c*x**5 - 45*sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))* 
b**2*c**2*x**7 - 8*a*b**4 + 56*a*b**3*c*x**2 + 175*a*b**2*c**2*x**4 + 105* 
a*b*c**3*x**6 - 24*b**5*x**2 - 75*b**4*c*x**4 - 45*b**3*c**2*x**6)/(24*b** 
5*x**3*(b**2 + 2*b*c*x**2 + c**2*x**4))