\(\int \frac {x^3}{(a+b x^2+c x^4)^3} \, dx\) [806]

Optimal result

Integrand size = 18, antiderivative size = 113 \[ \int \frac {x^3}{\left (a+b x^2+c x^4\right )^3} \, dx=\frac {2 a+b x^2}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac {3 b \left (b+2 c x^2\right )}{4 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac {3 b c \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}} \] Output:

1/4*(b*x^2+2*a)/(-4*a*c+b^2)/(c*x^4+b*x^2+a)^2-3/4*b*(2*c*x^2+b)/(-4*a*c+b 
^2)^2/(c*x^4+b*x^2+a)+3*b*c*arctanh((2*c*x^2+b)/(-4*a*c+b^2)^(1/2))/(-4*a* 
c+b^2)^(5/2)
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.01 \[ \int \frac {x^3}{\left (a+b x^2+c x^4\right )^3} \, dx=\frac {\frac {\left (b^2-4 a c\right ) \left (2 a+b x^2\right )}{\left (a+b x^2+c x^4\right )^2}-\frac {3 b \left (b+2 c x^2\right )}{a+b x^2+c x^4}-\frac {12 b c \arctan \left (\frac {b+2 c x^2}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}}{4 \left (b^2-4 a c\right )^2} \] Input:

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

Output:

(((b^2 - 4*a*c)*(2*a + b*x^2))/(a + b*x^2 + c*x^4)^2 - (3*b*(b + 2*c*x^2)) 
/(a + b*x^2 + c*x^4) - (12*b*c*ArcTan[(b + 2*c*x^2)/Sqrt[-b^2 + 4*a*c]])/S 
qrt[-b^2 + 4*a*c])/(4*(b^2 - 4*a*c)^2)
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.14, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {1434, 1159, 1086, 1083, 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.

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

\(\Big \downarrow \) 1434

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

\(\Big \downarrow \) 1159

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

\(\Big \downarrow \) 1086

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (\frac {3 b \left (\frac {4 c \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}-\frac {b+2 c x^2}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )}{2 \left (b^2-4 a c\right )}+\frac {2 a+b x^2}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}\right )\)

Input:

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

Output:

((2*a + b*x^2)/(2*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)^2) + (3*b*(-((b + 2*c* 
x^2)/((b^2 - 4*a*c)*(a + b*x^2 + c*x^4))) + (4*c*ArcTanh[(b + 2*c*x^2)/Sqr 
t[b^2 - 4*a*c]])/(b^2 - 4*a*c)^(3/2)))/(2*(b^2 - 4*a*c)))/2
 

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 1083

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

rule 1086

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) 
*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] - Simp[2*c*((2*p + 
 3)/((p + 1)*(b^2 - 4*a*c)))   Int[(a + b*x + c*x^2)^(p + 1), x], x] /; Fre 
eQ[{a, b, c}, x] && ILtQ[p, -1]
 

rule 1159

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

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]
 

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.13

method	result	size
default	\(\frac {-b \,x^{2}-2 a}{4 \left (4 a c -b^{2}\right ) \left (c \,x^{4}+b \,x^{2}+a \right )^{2}}-\frac {3 b \left (\frac {2 c \,x^{2}+b}{\left (4 a c -b^{2}\right ) \left (c \,x^{4}+b \,x^{2}+a \right )}+\frac {4 c \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {3}{2}}}\right )}{4 \left (4 a c -b^{2}\right )}\)	\(128\)
risch	\(\frac {-\frac {3 b \,c^{2} x^{6}}{2 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}-\frac {9 b^{2} c \,x^{4}}{4 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}-\frac {\left (5 a c +b^{2}\right ) b \,x^{2}}{2 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}-\frac {a \left (8 a c +b^{2}\right )}{4 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}}{\left (c \,x^{4}+b \,x^{2}+a \right )^{2}}-\frac {3 c b \ln \left (\left (-\left (-4 a c +b^{2}\right )^{\frac {5}{2}}-16 a^{2} b \,c^{2}+8 a \,b^{3} c -b^{5}\right ) x^{2}-32 a^{3} c^{2}+16 a^{2} b^{2} c -2 b^{4} a \right )}{2 \left (-4 a c +b^{2}\right )^{\frac {5}{2}}}+\frac {3 c b \ln \left (\left (-\left (-4 a c +b^{2}\right )^{\frac {5}{2}}+16 a^{2} b \,c^{2}-8 a \,b^{3} c +b^{5}\right ) x^{2}+32 a^{3} c^{2}-16 a^{2} b^{2} c +2 b^{4} a \right )}{2 \left (-4 a c +b^{2}\right )^{\frac {5}{2}}}\)	\(297\)

Input:

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

Output:

1/4*(-b*x^2-2*a)/(4*a*c-b^2)/(c*x^4+b*x^2+a)^2-3/4*b/(4*a*c-b^2)*((2*c*x^2 
+b)/(4*a*c-b^2)/(c*x^4+b*x^2+a)+4*c/(4*a*c-b^2)^(3/2)*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. 392 vs. \(2 (105) = 210\).

Time = 0.09 (sec) , antiderivative size = 808, normalized size of antiderivative = 7.15 \[ \int \frac {x^3}{\left (a+b x^2+c x^4\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/4*(6*(b^3*c^2 - 4*a*b*c^3)*x^6 + a*b^4 + 4*a^2*b^2*c - 32*a^3*c^2 + 9* 
(b^4*c - 4*a*b^2*c^2)*x^4 + 2*(b^5 + a*b^3*c - 20*a^2*b*c^2)*x^2 - 6*(b*c^ 
3*x^8 + 2*b^2*c^2*x^6 + 2*a*b^2*c*x^2 + (b^3*c + 2*a*b*c^2)*x^4 + a^2*b*c) 
*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)))/((b^6*c^2 - 12*a*b^4*c^3 + 48*a 
^2*b^2*c^4 - 64*a^3*c^5)*x^8 + a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 6 
4*a^5*c^3 + 2*(b^7*c - 12*a*b^5*c^2 + 48*a^2*b^3*c^3 - 64*a^3*b*c^4)*x^6 + 
 (b^8 - 10*a*b^6*c + 24*a^2*b^4*c^2 + 32*a^3*b^2*c^3 - 128*a^4*c^4)*x^4 + 
2*(a*b^7 - 12*a^2*b^5*c + 48*a^3*b^3*c^2 - 64*a^4*b*c^3)*x^2), -1/4*(6*(b^ 
3*c^2 - 4*a*b*c^3)*x^6 + a*b^4 + 4*a^2*b^2*c - 32*a^3*c^2 + 9*(b^4*c - 4*a 
*b^2*c^2)*x^4 + 2*(b^5 + a*b^3*c - 20*a^2*b*c^2)*x^2 - 12*(b*c^3*x^8 + 2*b 
^2*c^2*x^6 + 2*a*b^2*c*x^2 + (b^3*c + 2*a*b*c^2)*x^4 + a^2*b*c)*sqrt(-b^2 
+ 4*a*c)*arctan(-(2*c*x^2 + b)*sqrt(-b^2 + 4*a*c)/(b^2 - 4*a*c)))/((b^6*c^ 
2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)*x^8 + a^2*b^6 - 12*a^3*b^4 
*c + 48*a^4*b^2*c^2 - 64*a^5*c^3 + 2*(b^7*c - 12*a*b^5*c^2 + 48*a^2*b^3*c^ 
3 - 64*a^3*b*c^4)*x^6 + (b^8 - 10*a*b^6*c + 24*a^2*b^4*c^2 + 32*a^3*b^2*c^ 
3 - 128*a^4*c^4)*x^4 + 2*(a*b^7 - 12*a^2*b^5*c + 48*a^3*b^3*c^2 - 64*a^4*b 
*c^3)*x^2)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 491 vs. \(2 (105) = 210\).

Time = 1.21 (sec) , antiderivative size = 491, normalized size of antiderivative = 4.35 \[ \int \frac {x^3}{\left (a+b x^2+c x^4\right )^3} \, dx=\frac {3 b c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \log {\left (x^{2} + \frac {- 192 a^{3} b c^{4} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 144 a^{2} b^{3} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} - 36 a b^{5} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 3 b^{7} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 3 b^{2} c}{6 b c^{2}} \right )}}{2} - \frac {3 b c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \log {\left (x^{2} + \frac {192 a^{3} b c^{4} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} - 144 a^{2} b^{3} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 36 a b^{5} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} - 3 b^{7} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 3 b^{2} c}{6 b c^{2}} \right )}}{2} + \frac {- 8 a^{2} c - a b^{2} - 9 b^{2} c x^{4} - 6 b c^{2} x^{6} + x^{2} \left (- 10 a b c - 2 b^{3}\right )}{64 a^{4} c^{2} - 32 a^{3} b^{2} c + 4 a^{2} b^{4} + x^{8} \cdot \left (64 a^{2} c^{4} - 32 a b^{2} c^{3} + 4 b^{4} c^{2}\right ) + x^{6} \cdot \left (128 a^{2} b c^{3} - 64 a b^{3} c^{2} + 8 b^{5} c\right ) + x^{4} \cdot \left (128 a^{3} c^{3} - 24 a b^{4} c + 4 b^{6}\right ) + x^{2} \cdot \left (128 a^{3} b c^{2} - 64 a^{2} b^{3} c + 8 a b^{5}\right )} \] Input:

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

Output:

3*b*c*sqrt(-1/(4*a*c - b**2)**5)*log(x**2 + (-192*a**3*b*c**4*sqrt(-1/(4*a 
*c - b**2)**5) + 144*a**2*b**3*c**3*sqrt(-1/(4*a*c - b**2)**5) - 36*a*b**5 
*c**2*sqrt(-1/(4*a*c - b**2)**5) + 3*b**7*c*sqrt(-1/(4*a*c - b**2)**5) + 3 
*b**2*c)/(6*b*c**2))/2 - 3*b*c*sqrt(-1/(4*a*c - b**2)**5)*log(x**2 + (192* 
a**3*b*c**4*sqrt(-1/(4*a*c - b**2)**5) - 144*a**2*b**3*c**3*sqrt(-1/(4*a*c 
 - b**2)**5) + 36*a*b**5*c**2*sqrt(-1/(4*a*c - b**2)**5) - 3*b**7*c*sqrt(- 
1/(4*a*c - b**2)**5) + 3*b**2*c)/(6*b*c**2))/2 + (-8*a**2*c - a*b**2 - 9*b 
**2*c*x**4 - 6*b*c**2*x**6 + x**2*(-10*a*b*c - 2*b**3))/(64*a**4*c**2 - 32 
*a**3*b**2*c + 4*a**2*b**4 + x**8*(64*a**2*c**4 - 32*a*b**2*c**3 + 4*b**4* 
c**2) + x**6*(128*a**2*b*c**3 - 64*a*b**3*c**2 + 8*b**5*c) + x**4*(128*a** 
3*c**3 - 24*a*b**4*c + 4*b**6) + x**2*(128*a**3*b*c**2 - 64*a**2*b**3*c + 
8*a*b**5))
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^3}{\left (a+b x^2+c x^4\right )^3} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for 
 more deta
 

Giac [A] (verification not implemented)

Time = 1.12 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.27 \[ \int \frac {x^3}{\left (a+b x^2+c x^4\right )^3} \, dx=-\frac {3 \, b c \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {6 \, b c^{2} x^{6} + 9 \, b^{2} c x^{4} + 2 \, b^{3} x^{2} + 10 \, a b c x^{2} + a b^{2} + 8 \, a^{2} c}{4 \, {\left (c x^{4} + b x^{2} + a\right )}^{2} {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )}} \] Input:

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

Output:

-3*b*c*arctan((2*c*x^2 + b)/sqrt(-b^2 + 4*a*c))/((b^4 - 8*a*b^2*c + 16*a^2 
*c^2)*sqrt(-b^2 + 4*a*c)) - 1/4*(6*b*c^2*x^6 + 9*b^2*c*x^4 + 2*b^3*x^2 + 1 
0*a*b*c*x^2 + a*b^2 + 8*a^2*c)/((c*x^4 + b*x^2 + a)^2*(b^4 - 8*a*b^2*c + 1 
6*a^2*c^2))
 

Mupad [B] (verification not implemented)

Time = 18.04 (sec) , antiderivative size = 400, normalized size of antiderivative = 3.54 \[ \int \frac {x^3}{\left (a+b x^2+c x^4\right )^3} \, dx=-\frac {\frac {8\,c\,a^2+a\,b^2}{4\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {x^2\,\left (b^3+5\,a\,c\,b\right )}{2\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {9\,b^2\,c\,x^4}{4\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {3\,b\,c^2\,x^6}{2\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}}{x^4\,\left (b^2+2\,a\,c\right )+a^2+c^2\,x^8+2\,a\,b\,x^2+2\,b\,c\,x^6}-\frac {3\,b\,c\,\mathrm {atan}\left (\frac {\left (x^2\,\left (\frac {9\,b^2\,c^4}{a\,{\left (4\,a\,c-b^2\right )}^{9/2}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {b^3\,c^2\,\left (144\,a^2\,b\,c^4-72\,a\,b^3\,c^3+9\,b^5\,c^2\right )}{a\,{\left (4\,a\,c-b^2\right )}^{15/2}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}\right )+\frac {18\,b^3\,c^4}{{\left (4\,a\,c-b^2\right )}^{15/2}}\right )\,\left (b^4\,{\left (4\,a\,c-b^2\right )}^5+16\,a^2\,c^2\,{\left (4\,a\,c-b^2\right )}^5-8\,a\,b^2\,c\,{\left (4\,a\,c-b^2\right )}^5\right )}{18\,b^2\,c^4}\right )}{{\left (4\,a\,c-b^2\right )}^{5/2}} \] Input:

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

Output:

- ((a*b^2 + 8*a^2*c)/(4*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (x^2*(b^3 + 5*a* 
b*c))/(2*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (9*b^2*c*x^4)/(4*(b^4 + 16*a^2* 
c^2 - 8*a*b^2*c)) + (3*b*c^2*x^6)/(2*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)))/(x^4 
*(2*a*c + b^2) + a^2 + c^2*x^8 + 2*a*b*x^2 + 2*b*c*x^6) - (3*b*c*atan(((x^ 
2*((9*b^2*c^4)/(a*(4*a*c - b^2)^(9/2)*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (b 
^3*c^2*(9*b^5*c^2 - 72*a*b^3*c^3 + 144*a^2*b*c^4))/(a*(4*a*c - b^2)^(15/2) 
*(b^4 + 16*a^2*c^2 - 8*a*b^2*c))) + (18*b^3*c^4)/(4*a*c - b^2)^(15/2))*(b^ 
4*(4*a*c - b^2)^5 + 16*a^2*c^2*(4*a*c - b^2)^5 - 8*a*b^2*c*(4*a*c - b^2)^5 
))/(18*b^2*c^4)))/(4*a*c - b^2)^(5/2)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 1055, normalized size of antiderivative = 9.34 \[ \int \frac {x^3}{\left (a+b x^2+c x^4\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

(12*sqrt(2*sqrt(c)*sqrt(a) + b)*sqrt(2*sqrt(c)*sqrt(a) - b)*atan((sqrt(2*s 
qrt(c)*sqrt(a) - b) - 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*a**2*b*c + 
 24*sqrt(2*sqrt(c)*sqrt(a) + b)*sqrt(2*sqrt(c)*sqrt(a) - b)*atan((sqrt(2*s 
qrt(c)*sqrt(a) - b) - 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*a*b**2*c*x 
**2 + 24*sqrt(2*sqrt(c)*sqrt(a) + b)*sqrt(2*sqrt(c)*sqrt(a) - b)*atan((sqr 
t(2*sqrt(c)*sqrt(a) - b) - 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*a*b*c 
**2*x**4 + 12*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))* 
b**3*c*x**4 + 24*sqrt(2*sqrt(c)*sqrt(a) + b)*sqrt(2*sqrt(c)*sqrt(a) - b)*a 
tan((sqrt(2*sqrt(c)*sqrt(a) - b) - 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b 
))*b**2*c**2*x**6 + 12*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))*b*c**3*x**8 + 12*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)*sq 
rt(a) + b))*a**2*b*c + 24*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)*sq 
rt(a) + b))*a*b**2*c*x**2 + 24*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*c**2*x**4 + 12*sqrt(2*sqrt(c)*sqrt(a) + b)*sqrt(2*sqr 
t(c)*sqrt(a) - b)*atan((sqrt(2*sqrt(c)*sqrt(a) - b) + 2*sqrt(c)*x)/sqrt...