\(\int \frac {x^{13}}{(a x+b x^3+c x^5)^2} \, dx\) [35]

Optimal result

Integrand size = 20, antiderivative size = 216 \[ \int \frac {x^{13}}{\left (a x+b x^3+c x^5\right )^2} \, dx=-\frac {b \left (3 b^2-11 a c\right ) x^2}{2 c^3 \left (b^2-4 a c\right )}+\frac {\left (3 b^2-8 a c\right ) x^4}{4 c^2 \left (b^2-4 a c\right )}-\frac {b x^6}{2 c \left (b^2-4 a c\right )}+\frac {x^8 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac {b \left (3 b^4-20 a b^2 c+30 a^2 c^2\right ) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 c^4 \left (b^2-4 a c\right )^{3/2}}+\frac {\left (3 b^2-2 a c\right ) \log \left (a+b x^2+c x^4\right )}{4 c^4} \] Output:

-1/2*b*(-11*a*c+3*b^2)*x^2/c^3/(-4*a*c+b^2)+1/4*(-8*a*c+3*b^2)*x^4/c^2/(-4 
*a*c+b^2)-1/2*b*x^6/c/(-4*a*c+b^2)+1/2*x^8*(b*x^2+2*a)/(-4*a*c+b^2)/(c*x^4 
+b*x^2+a)+1/2*b*(30*a^2*c^2-20*a*b^2*c+3*b^4)*arctanh((2*c*x^2+b)/(-4*a*c+ 
b^2)^(1/2))/c^4/(-4*a*c+b^2)^(3/2)+1/4*(-2*a*c+3*b^2)*ln(c*x^4+b*x^2+a)/c^ 
4
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.83 \[ \int \frac {x^{13}}{\left (a x+b x^3+c x^5\right )^2} \, dx=\frac {-4 b c x^2+c^2 x^4+\frac {2 \left (2 a^3 c^2+b^5 x^2+a b^3 \left (b-5 c x^2\right )+a^2 b c \left (-4 b+5 c x^2\right )\right )}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac {2 b \left (3 b^4-20 a b^2 c+30 a^2 c^2\right ) \arctan \left (\frac {b+2 c x^2}{\sqrt {-b^2+4 a c}}\right )}{\left (-b^2+4 a c\right )^{3/2}}+\left (3 b^2-2 a c\right ) \log \left (a+b x^2+c x^4\right )}{4 c^4} \] Input:

Integrate[x^13/(a*x + b*x^3 + c*x^5)^2,x]
 

Output:

(-4*b*c*x^2 + c^2*x^4 + (2*(2*a^3*c^2 + b^5*x^2 + a*b^3*(b - 5*c*x^2) + a^ 
2*b*c*(-4*b + 5*c*x^2)))/((b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) + (2*b*(3*b^4 
 - 20*a*b^2*c + 30*a^2*c^2)*ArcTan[(b + 2*c*x^2)/Sqrt[-b^2 + 4*a*c]])/(-b^ 
2 + 4*a*c)^(3/2) + (3*b^2 - 2*a*c)*Log[a + b*x^2 + c*x^4])/(4*c^4)
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {9, 1434, 1164, 1200, 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.

\(\displaystyle \int \frac {x^{13}}{\left (a x+b x^3+c x^5\right )^2} \, dx\)

\(\Big \downarrow \) 9

\(\displaystyle \int \frac {x^{11}}{\left (a+b x^2+c x^4\right )^2}dx\)

\(\Big \downarrow \) 1434

\(\displaystyle \frac {1}{2} \int \frac {x^{10}}{\left (c x^4+b x^2+a\right )^2}dx^2\)

\(\Big \downarrow \) 1164

\(\displaystyle \frac {1}{2} \left (\frac {x^8 \left (2 a+b x^2\right )}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\int \frac {x^6 \left (3 b x^2+8 a\right )}{c x^4+b x^2+a}dx^2}{b^2-4 a c}\right )\)

\(\Big \downarrow \) 1200

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (\frac {x^8 \left (2 a+b x^2\right )}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {-\frac {b \left (30 a^2 c^2-20 a b^2 c+3 b^4\right ) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{c^4 \sqrt {b^2-4 a c}}-\frac {\left (b^2-4 a c\right ) \left (3 b^2-2 a c\right ) \log \left (a+b x^2+c x^4\right )}{2 c^4}+\frac {b x^2 \left (3 b^2-11 a c\right )}{c^3}-\frac {x^4 \left (3 b^2-8 a c\right )}{2 c^2}+\frac {b x^6}{c}}{b^2-4 a c}\right )\)

Input:

Int[x^13/(a*x + b*x^3 + c*x^5)^2,x]
 

Output:

((x^8*(2*a + b*x^2))/((b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) - ((b*(3*b^2 - 11 
*a*c)*x^2)/c^3 - ((3*b^2 - 8*a*c)*x^4)/(2*c^2) + (b*x^6)/c - (b*(3*b^4 - 2 
0*a*b^2*c + 30*a^2*c^2)*ArcTanh[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(c^4*Sqr 
t[b^2 - 4*a*c]) - ((b^2 - 4*a*c)*(3*b^2 - 2*a*c)*Log[a + b*x^2 + c*x^4])/( 
2*c^4))/(b^2 - 4*a*c))/2
 

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 1164

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(d + e*x)^(m - 1)*(d*b - 2*a*e + (2*c*d - b*e)*x)*((a + b*x 
+ c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + Simp[1/((p + 1)*(b^2 - 4*a* 
c))   Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2* 
c*d^2*(2*p + 3) + e*(b*e - 2*d*c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p 
+ 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && LtQ[p, -1] && GtQ[m, 1] && Int 
QuadraticQ[a, b, c, d, e, m, p, x]
 

rule 1200

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* 
(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g* 
x)^n/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && In 
tegersQ[n]
 

rule 1434

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp 
[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x + c*x^2)^p, x], x, x^2], x] /; Free 
Q[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]
 

rule 2009

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

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.15

method	result	size
default	\(\frac {\left (-c \,x^{2}+2 b \right )^{2}}{4 c^{4}}+\frac {\frac {-\frac {b \left (5 a^{2} c^{2}-5 a \,b^{2} c +b^{4}\right ) x^{2}}{c \left (4 a c -b^{2}\right )}-\frac {a \left (2 a^{2} c^{2}-4 a \,b^{2} c +b^{4}\right )}{c \left (4 a c -b^{2}\right )}}{c \,x^{4}+b \,x^{2}+a}+\frac {\frac {\left (-8 a^{2} c^{2}+14 a \,b^{2} c -3 b^{4}\right ) \ln \left (c \,x^{4}+b \,x^{2}+a \right )}{2 c}+\frac {2 \left (11 a^{2} b c -3 b^{3} a -\frac {\left (-8 a^{2} c^{2}+14 a \,b^{2} c -3 b^{4}\right ) b}{2 c}\right ) \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{4 a c -b^{2}}}{2 c^{3}}\)	\(248\)
risch	\(\text {Expression too large to display}\)	\(1649\)

Input:

int(x^13/(c*x^5+b*x^3+a*x)^2,x,method=_RETURNVERBOSE)
 

Output:

1/4*(-c*x^2+2*b)^2/c^4+1/2/c^3*((-b*(5*a^2*c^2-5*a*b^2*c+b^4)/c/(4*a*c-b^2 
)*x^2-a/c*(2*a^2*c^2-4*a*b^2*c+b^4)/(4*a*c-b^2))/(c*x^4+b*x^2+a)+1/(4*a*c- 
b^2)*(1/2*(-8*a^2*c^2+14*a*b^2*c-3*b^4)/c*ln(c*x^4+b*x^2+a)+2*(11*a^2*b*c- 
3*b^3*a-1/2*(-8*a^2*c^2+14*a*b^2*c-3*b^4)*b/c)/(4*a*c-b^2)^(1/2)*arctan((2 
*c*x^2+b)/(4*a*c-b^2)^(1/2))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 516 vs. \(2 (200) = 400\).

Time = 0.16 (sec) , antiderivative size = 1057, normalized size of antiderivative = 4.89 \[ \int \frac {x^{13}}{\left (a x+b x^3+c x^5\right )^2} \, dx =\text {Too large to display} \] Input:

integrate(x^13/(c*x^5+b*x^3+a*x)^2,x, algorithm="fricas")
 

Output:

[1/4*((b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*x^8 + 2*a*b^6 - 16*a^2*b^4*c + 
36*a^3*b^2*c^2 - 16*a^4*c^3 - 3*(b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*x^6 
 - (4*b^6*c - 33*a*b^4*c^2 + 72*a^2*b^2*c^3 - 16*a^3*c^4)*x^4 + 2*(b^7 - 1 
1*a*b^5*c + 41*a^2*b^3*c^2 - 52*a^3*b*c^3)*x^2 - (3*a*b^5 - 20*a^2*b^3*c + 
 30*a^3*b*c^2 + (3*b^5*c - 20*a*b^3*c^2 + 30*a^2*b*c^3)*x^4 + (3*b^6 - 20* 
a*b^4*c + 30*a^2*b^2*c^2)*x^2)*sqrt(b^2 - 4*a*c)*log((2*c^2*x^4 + 2*b*c*x^ 
2 + b^2 - 2*a*c - (2*c*x^2 + b)*sqrt(b^2 - 4*a*c))/(c*x^4 + b*x^2 + a)) + 
(3*a*b^6 - 26*a^2*b^4*c + 64*a^3*b^2*c^2 - 32*a^4*c^3 + (3*b^6*c - 26*a*b^ 
4*c^2 + 64*a^2*b^2*c^3 - 32*a^3*c^4)*x^4 + (3*b^7 - 26*a*b^5*c + 64*a^2*b^ 
3*c^2 - 32*a^3*b*c^3)*x^2)*log(c*x^4 + b*x^2 + a))/(a*b^4*c^4 - 8*a^2*b^2* 
c^5 + 16*a^3*c^6 + (b^4*c^5 - 8*a*b^2*c^6 + 16*a^2*c^7)*x^4 + (b^5*c^4 - 8 
*a*b^3*c^5 + 16*a^2*b*c^6)*x^2), 1/4*((b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5) 
*x^8 + 2*a*b^6 - 16*a^2*b^4*c + 36*a^3*b^2*c^2 - 16*a^4*c^3 - 3*(b^5*c^2 - 
 8*a*b^3*c^3 + 16*a^2*b*c^4)*x^6 - (4*b^6*c - 33*a*b^4*c^2 + 72*a^2*b^2*c^ 
3 - 16*a^3*c^4)*x^4 + 2*(b^7 - 11*a*b^5*c + 41*a^2*b^3*c^2 - 52*a^3*b*c^3) 
*x^2 + 2*(3*a*b^5 - 20*a^2*b^3*c + 30*a^3*b*c^2 + (3*b^5*c - 20*a*b^3*c^2 
+ 30*a^2*b*c^3)*x^4 + (3*b^6 - 20*a*b^4*c + 30*a^2*b^2*c^2)*x^2)*sqrt(-b^2 
 + 4*a*c)*arctan(-(2*c*x^2 + b)*sqrt(-b^2 + 4*a*c)/(b^2 - 4*a*c)) + (3*a*b 
^6 - 26*a^2*b^4*c + 64*a^3*b^2*c^2 - 32*a^4*c^3 + (3*b^6*c - 26*a*b^4*c^2 
+ 64*a^2*b^2*c^3 - 32*a^3*c^4)*x^4 + (3*b^7 - 26*a*b^5*c + 64*a^2*b^3*c...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{13}}{\left (a x+b x^3+c x^5\right )^2} \, dx=\text {Timed out} \] Input:

integrate(x**13/(c*x**5+b*x**3+a*x)**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {x^{13}}{\left (a x+b x^3+c x^5\right )^2} \, dx=\int { \frac {x^{13}}{{\left (c x^{5} + b x^{3} + a x\right )}^{2}} \,d x } \] Input:

integrate(x^13/(c*x^5+b*x^3+a*x)^2,x, algorithm="maxima")
 

Output:

1/2*(a*b^4 - 4*a^2*b^2*c + 2*a^3*c^2 + (b^5 - 5*a*b^3*c + 5*a^2*b*c^2)*x^2 
)/(a*b^2*c^4 - 4*a^2*c^5 + (b^2*c^5 - 4*a*c^6)*x^4 + (b^3*c^4 - 4*a*b*c^5) 
*x^2) + integrate(((3*b^4 - 14*a*b^2*c + 8*a^2*c^2)*x^3 + (3*a*b^3 - 11*a^ 
2*b*c)*x)/(c*x^4 + b*x^2 + a), x)/(b^2*c^3 - 4*a*c^4) + 1/4*(c*x^4 - 4*b*x 
^2)/c^3
 

Giac [A] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.11 \[ \int \frac {x^{13}}{\left (a x+b x^3+c x^5\right )^2} \, dx=-\frac {{\left (3 \, b^{5} - 20 \, a b^{3} c + 30 \, a^{2} b c^{2}\right )} \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{2 \, {\left (b^{2} c^{4} - 4 \, a c^{5}\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {3 \, b^{4} c x^{4} - 14 \, a b^{2} c^{2} x^{4} + 8 \, a^{2} c^{3} x^{4} + b^{5} x^{2} - 4 \, a b^{3} c x^{2} - 2 \, a^{2} b c^{2} x^{2} + a b^{4} - 6 \, a^{2} b^{2} c + 4 \, a^{3} c^{2}}{4 \, {\left (b^{2} c^{4} - 4 \, a c^{5}\right )} {\left (c x^{4} + b x^{2} + a\right )}} + \frac {{\left (3 \, b^{2} - 2 \, a c\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, c^{4}} + \frac {c^{2} x^{4} - 4 \, b c x^{2}}{4 \, c^{4}} \] Input:

integrate(x^13/(c*x^5+b*x^3+a*x)^2,x, algorithm="giac")
 

Output:

-1/2*(3*b^5 - 20*a*b^3*c + 30*a^2*b*c^2)*arctan((2*c*x^2 + b)/sqrt(-b^2 + 
4*a*c))/((b^2*c^4 - 4*a*c^5)*sqrt(-b^2 + 4*a*c)) - 1/4*(3*b^4*c*x^4 - 14*a 
*b^2*c^2*x^4 + 8*a^2*c^3*x^4 + b^5*x^2 - 4*a*b^3*c*x^2 - 2*a^2*b*c^2*x^2 + 
 a*b^4 - 6*a^2*b^2*c + 4*a^3*c^2)/((b^2*c^4 - 4*a*c^5)*(c*x^4 + b*x^2 + a) 
) + 1/4*(3*b^2 - 2*a*c)*log(c*x^4 + b*x^2 + a)/c^4 + 1/4*(c^2*x^4 - 4*b*c* 
x^2)/c^4
 

Mupad [B] (verification not implemented)

Time = 12.32 (sec) , antiderivative size = 1691, normalized size of antiderivative = 7.83 \[ \int \frac {x^{13}}{\left (a x+b x^3+c x^5\right )^2} \, dx=\text {Too large to display} \] Input:

int(x^13/(a*x + b*x^3 + c*x^5)^2,x)
 

Output:

x^4/(4*c^2) - ((x^2*(b^5 + 5*a^2*b*c^2 - 5*a*b^3*c))/(2*c*(4*a*c - b^2)) + 
 (a*(b^4 + 2*a^2*c^2 - 4*a*b^2*c))/(2*c*(4*a*c - b^2)))/(a*c^3 + c^4*x^4 + 
 b*c^3*x^2) - (b*x^2)/c^3 - (log(a + b*x^2 + c*x^4)*(6*b^8 + 256*a^4*c^4 + 
 336*a^2*b^4*c^2 - 576*a^3*b^2*c^3 - 76*a*b^6*c))/(2*(256*a^3*c^7 - 4*b^6* 
c^4 + 48*a*b^4*c^5 - 192*a^2*b^2*c^6)) + (b*atan(((8*a*c^7*(4*a*c - b^2)^3 
 - 2*b^2*c^6*(4*a*c - b^2)^3)*(x^2*(((b*((18*b^5*c^4 - 96*a*b^3*c^5 + 92*a 
^2*b*c^6)/(4*a*c^7 - b^2*c^6) + ((8*b^3*c^8 - 32*a*b*c^9)*(6*b^8 + 256*a^4 
*c^4 + 336*a^2*b^4*c^2 - 576*a^3*b^2*c^3 - 76*a*b^6*c))/(2*(4*a*c^7 - b^2* 
c^6)*(256*a^3*c^7 - 4*b^6*c^4 + 48*a*b^4*c^5 - 192*a^2*b^2*c^6)))*(3*b^4 + 
 30*a^2*c^2 - 20*a*b^2*c))/(8*c^4*(4*a*c - b^2)^(3/2)) + (b*(8*b^3*c^8 - 3 
2*a*b*c^9)*(3*b^4 + 30*a^2*c^2 - 20*a*b^2*c)*(6*b^8 + 256*a^4*c^4 + 336*a^ 
2*b^4*c^2 - 576*a^3*b^2*c^3 - 76*a*b^6*c))/(16*c^4*(4*a*c - b^2)^(3/2)*(4* 
a*c^7 - b^2*c^6)*(256*a^3*c^7 - 4*b^6*c^4 + 48*a*b^4*c^5 - 192*a^2*b^2*c^6 
)))/(a*(4*a*c - b^2)) + (b*((9*b^7 - 38*a^3*b*c^3 + 91*a^2*b^3*c^2 - 57*a* 
b^5*c)/(4*a*c^7 - b^2*c^6) + (((18*b^5*c^4 - 96*a*b^3*c^5 + 92*a^2*b*c^6)/ 
(4*a*c^7 - b^2*c^6) + ((8*b^3*c^8 - 32*a*b*c^9)*(6*b^8 + 256*a^4*c^4 + 336 
*a^2*b^4*c^2 - 576*a^3*b^2*c^3 - 76*a*b^6*c))/(2*(4*a*c^7 - b^2*c^6)*(256* 
a^3*c^7 - 4*b^6*c^4 + 48*a*b^4*c^5 - 192*a^2*b^2*c^6)))*(6*b^8 + 256*a^4*c 
^4 + 336*a^2*b^4*c^2 - 576*a^3*b^2*c^3 - 76*a*b^6*c))/(2*(256*a^3*c^7 - 4* 
b^6*c^4 + 48*a*b^4*c^5 - 192*a^2*b^2*c^6)) - (b^2*((b^3*c^8)/2 - 2*a*b*...
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 2123, normalized size of antiderivative = 9.83 \[ \int \frac {x^{13}}{\left (a x+b x^3+c x^5\right )^2} \, dx =\text {Too large to display} \] Input:

int(x^13/(c*x^5+b*x^3+a*x)^2,x)
 

Output:

( - 60*sqrt(2*sqrt(c)*sqrt(a) + b)*sqrt(2*sqrt(c)*sqrt(a) - b)*atan((sqrt( 
2*sqrt(c)*sqrt(a) - b) - 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*a**3*b* 
c**2 + 40*sqrt(2*sqrt(c)*sqrt(a) + b)*sqrt(2*sqrt(c)*sqrt(a) - b)*atan((sq 
rt(2*sqrt(c)*sqrt(a) - b) - 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*a**2 
*b**3*c - 60*sqrt(2*sqrt(c)*sqrt(a) + b)*sqrt(2*sqrt(c)*sqrt(a) - b)*atan( 
(sqrt(2*sqrt(c)*sqrt(a) - b) - 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*a 
**2*b**2*c**2*x**2 - 60*sqrt(2*sqrt(c)*sqrt(a) + b)*sqrt(2*sqrt(c)*sqrt(a) 
 - b)*atan((sqrt(2*sqrt(c)*sqrt(a) - b) - 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt 
(a) + b))*a**2*b*c**3*x**4 - 6*sqrt(2*sqrt(c)*sqrt(a) + b)*sqrt(2*sqrt(c)* 
sqrt(a) - b)*atan((sqrt(2*sqrt(c)*sqrt(a) - b) - 2*sqrt(c)*x)/sqrt(2*sqrt( 
c)*sqrt(a) + b))*a*b**5 + 40*sqrt(2*sqrt(c)*sqrt(a) + b)*sqrt(2*sqrt(c)*sq 
rt(a) - b)*atan((sqrt(2*sqrt(c)*sqrt(a) - b) - 2*sqrt(c)*x)/sqrt(2*sqrt(c) 
*sqrt(a) + b))*a*b**4*c*x**2 + 40*sqrt(2*sqrt(c)*sqrt(a) + b)*sqrt(2*sqrt( 
c)*sqrt(a) - b)*atan((sqrt(2*sqrt(c)*sqrt(a) - b) - 2*sqrt(c)*x)/sqrt(2*sq 
rt(c)*sqrt(a) + b))*a*b**3*c**2*x**4 - 6*sqrt(2*sqrt(c)*sqrt(a) + b)*sqrt( 
2*sqrt(c)*sqrt(a) - b)*atan((sqrt(2*sqrt(c)*sqrt(a) - b) - 2*sqrt(c)*x)/sq 
rt(2*sqrt(c)*sqrt(a) + b))*b**6*x**2 - 6*sqrt(2*sqrt(c)*sqrt(a) + b)*sqrt( 
2*sqrt(c)*sqrt(a) - b)*atan((sqrt(2*sqrt(c)*sqrt(a) - b) - 2*sqrt(c)*x)/sq 
rt(2*sqrt(c)*sqrt(a) + b))*b**5*c*x**4 - 60*sqrt(2*sqrt(c)*sqrt(a) + b)*sq 
rt(2*sqrt(c)*sqrt(a) - b)*atan((sqrt(2*sqrt(c)*sqrt(a) - b) + 2*sqrt(c)...