3.2.13 \(\int \frac {x^5 (A+B x^2)}{(a+b x^2+c x^4)^2} \, dx\) [113]

3.2.13.1 Optimal result
3.2.13.2 Mathematica [A] (verified)
3.2.13.3 Rubi [A] (verified)
3.2.13.4 Maple [A] (verified)
3.2.13.5 Fricas [B] (verification not implemented)
3.2.13.6 Sympy [F(-1)]
3.2.13.7 Maxima [F(-2)]
3.2.13.8 Giac [A] (verification not implemented)
3.2.13.9 Mupad [B] (verification not implemented)

3.2.13.1 Optimal result

Integrand size = 25, antiderivative size = 147 \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=-\frac {x^2 \left (a (b B-2 A c)+\left (b^2 B-A b c-2 a B c\right ) x^2\right )}{2 c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac {\left (b^3 B-6 a b B c+4 a A c^2\right ) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 c^2 \left (b^2-4 a c\right )^{3/2}}+\frac {B \log \left (a+b x^2+c x^4\right )}{4 c^2} \]

output
-1/2*x^2*(a*(-2*A*c+B*b)+(-A*b*c-2*B*a*c+B*b^2)*x^2)/c/(-4*a*c+b^2)/(c*x^4 
+b*x^2+a)+1/2*(4*A*a*c^2-6*B*a*b*c+B*b^3)*arctanh((2*c*x^2+b)/(-4*a*c+b^2) 
^(1/2))/c^2/(-4*a*c+b^2)^(3/2)+1/4*B*ln(c*x^4+b*x^2+a)/c^2
 
3.2.13.2 Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.09 \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\frac {-\frac {2 \left (2 a^2 B c+b^2 (-b B+A c) x^2+a \left (-b^2 B-2 A c^2 x^2+b c \left (A+3 B x^2\right )\right )\right )}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}+\frac {2 \left (b^3 B-6 a b B c+4 a A c^2\right ) \arctan \left (\frac {b+2 c x^2}{\sqrt {-b^2+4 a c}}\right )}{\left (-b^2+4 a c\right )^{3/2}}+B \log \left (a+b x^2+c x^4\right )}{4 c^2} \]

input
Integrate[(x^5*(A + B*x^2))/(a + b*x^2 + c*x^4)^2,x]
 
output
((-2*(2*a^2*B*c + b^2*(-(b*B) + A*c)*x^2 + a*(-(b^2*B) - 2*A*c^2*x^2 + b*c 
*(A + 3*B*x^2))))/((b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) + (2*(b^3*B - 6*a*b* 
B*c + 4*a*A*c^2)*ArcTan[(b + 2*c*x^2)/Sqrt[-b^2 + 4*a*c]])/(-b^2 + 4*a*c)^ 
(3/2) + B*Log[a + b*x^2 + c*x^4])/(4*c^2)
 
3.2.13.3 Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.15, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {1578, 1233, 1142, 1083, 219, 1103}

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.

\(\displaystyle \int \frac {x^5 \left (A+B x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx\)

\(\Big \downarrow \) 1578

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

\(\Big \downarrow \) 1233

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

\(\Big \downarrow \) 1142

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (\frac {\frac {B \left (b^2-4 a c\right ) \int \frac {2 c x^2+b}{c x^4+b x^2+a}dx^2}{2 c}+\frac {\left (4 a A c^2-6 a b B c+b^3 B\right ) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{c \sqrt {b^2-4 a c}}}{c \left (b^2-4 a c\right )}-\frac {x^2 \left (x^2 \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\)

\(\Big \downarrow \) 1103

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

input
Int[(x^5*(A + B*x^2))/(a + b*x^2 + c*x^4)^2,x]
 
output
(-((x^2*(a*(b*B - 2*A*c) + (b^2*B - A*b*c - 2*a*B*c)*x^2))/(c*(b^2 - 4*a*c 
)*(a + b*x^2 + c*x^4))) + (((b^3*B - 6*a*b*B*c + 4*a*A*c^2)*ArcTanh[(b + 2 
*c*x^2)/Sqrt[b^2 - 4*a*c]])/(c*Sqrt[b^2 - 4*a*c]) + (B*(b^2 - 4*a*c)*Log[a 
 + b*x^2 + c*x^4])/(2*c))/(c*(b^2 - 4*a*c)))/2
 

3.2.13.3.1 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 1103
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1142
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

rule 1233
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(d + e*x)^(m - 1))*(a + b*x + c*x^2) 
^(p + 1)*((2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g - c 
*(b*e*f + b*d*g + 2*a*e*g))*x)/(c*(p + 1)*(b^2 - 4*a*c))), x] - Simp[1/(c*( 
p + 1)*(b^2 - 4*a*c))   Int[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Sim 
p[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a*e*(e*f 
*(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*( 
m + p + 1) + 2*c^2*d*f*(m + 2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2* 
p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && LtQ[p, -1] && 
GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, b, c, d, e, f, g]) | 
|  !ILtQ[m + 2*p + 3, 0])
 

rule 1578
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_ 
)^4)^(p_.), x_Symbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a 
+ b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && Int 
egerQ[(m - 1)/2]
 
3.2.13.4 Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.44

method result size
default \(\frac {-\frac {\left (2 A a \,c^{2}-A \,b^{2} c -3 B a b c +B \,b^{3}\right ) x^{2}}{c^{2} \left (4 a c -b^{2}\right )}+\frac {a \left (A b c +2 B a c -B \,b^{2}\right )}{c^{2} \left (4 a c -b^{2}\right )}}{2 c \,x^{4}+2 b \,x^{2}+2 a}+\frac {\frac {\left (4 B a c -B \,b^{2}\right ) \ln \left (c \,x^{4}+b \,x^{2}+a \right )}{2 c}+\frac {2 \left (2 A a c -a b B -\frac {\left (4 B a c -B \,b^{2}\right ) b}{2 c}\right ) \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{2 \left (4 a c -b^{2}\right ) c}\) \(211\)
risch \(\text {Expression too large to display}\) \(1320\)

input
int(x^5*(B*x^2+A)/(c*x^4+b*x^2+a)^2,x,method=_RETURNVERBOSE)
 
output
1/2*(-1/c^2*(2*A*a*c^2-A*b^2*c-3*B*a*b*c+B*b^3)/(4*a*c-b^2)*x^2+a*(A*b*c+2 
*B*a*c-B*b^2)/c^2/(4*a*c-b^2))/(c*x^4+b*x^2+a)+1/2/(4*a*c-b^2)/c*(1/2*(4*B 
*a*c-B*b^2)/c*ln(c*x^4+b*x^2+a)+2*(2*A*a*c-a*b*B-1/2*(4*B*a*c-B*b^2)*b/c)/ 
(4*a*c-b^2)^(1/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2)))
 
3.2.13.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 412 vs. \(2 (137) = 274\).

Time = 0.30 (sec) , antiderivative size = 849, normalized size of antiderivative = 5.78 \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\left [\frac {2 \, B a b^{4} + 8 \, {\left (2 \, B a^{3} + A a^{2} b\right )} c^{2} + 2 \, {\left (B b^{5} - 8 \, A a^{2} c^{3} + 6 \, {\left (2 \, B a^{2} b + A a b^{2}\right )} c^{2} - {\left (7 \, B a b^{3} + A b^{4}\right )} c\right )} x^{2} - {\left (B a b^{3} - 6 \, B a^{2} b c + 4 \, A a^{2} c^{2} + {\left (B b^{3} c - 6 \, B a b c^{2} + 4 \, A a c^{3}\right )} x^{4} + {\left (B b^{4} - 6 \, B a b^{2} c + 4 \, A a b c^{2}\right )} x^{2}\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{4} + 2 \, b c x^{2} + b^{2} - 2 \, a c - {\left (2 \, c x^{2} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{4} + b x^{2} + a}\right ) - 2 \, {\left (6 \, B a^{2} b^{2} + A a b^{3}\right )} c + {\left (B a b^{4} - 8 \, B a^{2} b^{2} c + 16 \, B a^{3} c^{2} + {\left (B b^{4} c - 8 \, B a b^{2} c^{2} + 16 \, B a^{2} c^{3}\right )} x^{4} + {\left (B b^{5} - 8 \, B a b^{3} c + 16 \, B a^{2} b c^{2}\right )} x^{2}\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, {\left (a b^{4} c^{2} - 8 \, a^{2} b^{2} c^{3} + 16 \, a^{3} c^{4} + {\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} x^{4} + {\left (b^{5} c^{2} - 8 \, a b^{3} c^{3} + 16 \, a^{2} b c^{4}\right )} x^{2}\right )}}, \frac {2 \, B a b^{4} + 8 \, {\left (2 \, B a^{3} + A a^{2} b\right )} c^{2} + 2 \, {\left (B b^{5} - 8 \, A a^{2} c^{3} + 6 \, {\left (2 \, B a^{2} b + A a b^{2}\right )} c^{2} - {\left (7 \, B a b^{3} + A b^{4}\right )} c\right )} x^{2} + 2 \, {\left (B a b^{3} - 6 \, B a^{2} b c + 4 \, A a^{2} c^{2} + {\left (B b^{3} c - 6 \, B a b c^{2} + 4 \, A a c^{3}\right )} x^{4} + {\left (B b^{4} - 6 \, B a b^{2} c + 4 \, A a b c^{2}\right )} x^{2}\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {{\left (2 \, c x^{2} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) - 2 \, {\left (6 \, B a^{2} b^{2} + A a b^{3}\right )} c + {\left (B a b^{4} - 8 \, B a^{2} b^{2} c + 16 \, B a^{3} c^{2} + {\left (B b^{4} c - 8 \, B a b^{2} c^{2} + 16 \, B a^{2} c^{3}\right )} x^{4} + {\left (B b^{5} - 8 \, B a b^{3} c + 16 \, B a^{2} b c^{2}\right )} x^{2}\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, {\left (a b^{4} c^{2} - 8 \, a^{2} b^{2} c^{3} + 16 \, a^{3} c^{4} + {\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} x^{4} + {\left (b^{5} c^{2} - 8 \, a b^{3} c^{3} + 16 \, a^{2} b c^{4}\right )} x^{2}\right )}}\right ] \]

input
integrate(x^5*(B*x^2+A)/(c*x^4+b*x^2+a)^2,x, algorithm="fricas")
 
output
[1/4*(2*B*a*b^4 + 8*(2*B*a^3 + A*a^2*b)*c^2 + 2*(B*b^5 - 8*A*a^2*c^3 + 6*( 
2*B*a^2*b + A*a*b^2)*c^2 - (7*B*a*b^3 + A*b^4)*c)*x^2 - (B*a*b^3 - 6*B*a^2 
*b*c + 4*A*a^2*c^2 + (B*b^3*c - 6*B*a*b*c^2 + 4*A*a*c^3)*x^4 + (B*b^4 - 6* 
B*a*b^2*c + 4*A*a*b*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)) - 2 
*(6*B*a^2*b^2 + A*a*b^3)*c + (B*a*b^4 - 8*B*a^2*b^2*c + 16*B*a^3*c^2 + (B* 
b^4*c - 8*B*a*b^2*c^2 + 16*B*a^2*c^3)*x^4 + (B*b^5 - 8*B*a*b^3*c + 16*B*a^ 
2*b*c^2)*x^2)*log(c*x^4 + b*x^2 + a))/(a*b^4*c^2 - 8*a^2*b^2*c^3 + 16*a^3* 
c^4 + (b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*x^4 + (b^5*c^2 - 8*a*b^3*c^3 + 
16*a^2*b*c^4)*x^2), 1/4*(2*B*a*b^4 + 8*(2*B*a^3 + A*a^2*b)*c^2 + 2*(B*b^5 
- 8*A*a^2*c^3 + 6*(2*B*a^2*b + A*a*b^2)*c^2 - (7*B*a*b^3 + A*b^4)*c)*x^2 + 
 2*(B*a*b^3 - 6*B*a^2*b*c + 4*A*a^2*c^2 + (B*b^3*c - 6*B*a*b*c^2 + 4*A*a*c 
^3)*x^4 + (B*b^4 - 6*B*a*b^2*c + 4*A*a*b*c^2)*x^2)*sqrt(-b^2 + 4*a*c)*arct 
an(-(2*c*x^2 + b)*sqrt(-b^2 + 4*a*c)/(b^2 - 4*a*c)) - 2*(6*B*a^2*b^2 + A*a 
*b^3)*c + (B*a*b^4 - 8*B*a^2*b^2*c + 16*B*a^3*c^2 + (B*b^4*c - 8*B*a*b^2*c 
^2 + 16*B*a^2*c^3)*x^4 + (B*b^5 - 8*B*a*b^3*c + 16*B*a^2*b*c^2)*x^2)*log(c 
*x^4 + b*x^2 + a))/(a*b^4*c^2 - 8*a^2*b^2*c^3 + 16*a^3*c^4 + (b^4*c^3 - 8* 
a*b^2*c^4 + 16*a^2*c^5)*x^4 + (b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*x^2)]
 
3.2.13.6 Sympy [F(-1)]

Timed out. \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\text {Timed out} \]

input
integrate(x**5*(B*x**2+A)/(c*x**4+b*x**2+a)**2,x)
 
output
Timed out
 
3.2.13.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\text {Exception raised: ValueError} \]

input
integrate(x^5*(B*x^2+A)/(c*x^4+b*x^2+a)^2,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
 
3.2.13.8 Giac [A] (verification not implemented)

Time = 0.61 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.32 \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=-\frac {{\left (B b^{3} - 6 \, B a b c + 4 \, A a c^{2}\right )} \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{2 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {B \log \left (c x^{4} + b x^{2} + a\right )}{4 \, c^{2}} - \frac {B b^{2} c x^{4} - 4 \, B a c^{2} x^{4} - B b^{3} x^{2} + 2 \, B a b c x^{2} + 2 \, A b^{2} c x^{2} - 4 \, A a c^{2} x^{2} - B a b^{2} + 2 \, A a b c}{4 \, {\left (c x^{4} + b x^{2} + a\right )} {\left (b^{2} c^{2} - 4 \, a c^{3}\right )}} \]

input
integrate(x^5*(B*x^2+A)/(c*x^4+b*x^2+a)^2,x, algorithm="giac")
 
output
-1/2*(B*b^3 - 6*B*a*b*c + 4*A*a*c^2)*arctan((2*c*x^2 + b)/sqrt(-b^2 + 4*a* 
c))/((b^2*c^2 - 4*a*c^3)*sqrt(-b^2 + 4*a*c)) + 1/4*B*log(c*x^4 + b*x^2 + a 
)/c^2 - 1/4*(B*b^2*c*x^4 - 4*B*a*c^2*x^4 - B*b^3*x^2 + 2*B*a*b*c*x^2 + 2*A 
*b^2*c*x^2 - 4*A*a*c^2*x^2 - B*a*b^2 + 2*A*a*b*c)/((c*x^4 + b*x^2 + a)*(b^ 
2*c^2 - 4*a*c^3))
 
3.2.13.9 Mupad [B] (verification not implemented)

Time = 8.16 (sec) , antiderivative size = 1527, normalized size of antiderivative = 10.39 \[ \int \frac {x^5 \left (A+B x^2\right )}{\left (a+b x^2+c x^4\right )^2} \, dx=\text {Too large to display} \]

input
int((x^5*(A + B*x^2))/(a + b*x^2 + c*x^4)^2,x)
 
output
- ((x^2*(B*b^3 + 2*A*a*c^2 - A*b^2*c - 3*B*a*b*c))/(2*c^2*(4*a*c - b^2)) - 
 (a*(A*b*c - B*b^2 + 2*B*a*c))/(2*c^2*(4*a*c - b^2)))/(a + b*x^2 + c*x^4) 
- (log(a + b*x^2 + c*x^4)*(2*B*b^6 - 128*B*a^3*c^3 - 24*B*a*b^4*c + 96*B*a 
^2*b^2*c^2))/(2*(256*a^3*c^5 - 4*b^6*c^2 + 48*a*b^4*c^3 - 192*a^2*b^2*c^4) 
) - (atan(((8*a*c^3*(4*a*c - b^2)^3 - 2*b^2*c^2*(4*a*c - b^2)^3)*((((8*B*a 
 + (8*a*c^2*(2*B*b^6 - 128*B*a^3*c^3 - 24*B*a*b^4*c + 96*B*a^2*b^2*c^2))/( 
256*a^3*c^5 - 4*b^6*c^2 + 48*a*b^4*c^3 - 192*a^2*b^2*c^4))*(B*b^3 + 4*A*a* 
c^2 - 6*B*a*b*c))/(8*c^2*(4*a*c - b^2)^(3/2)) + (a*(B*b^3 + 4*A*a*c^2 - 6* 
B*a*b*c)*(2*B*b^6 - 128*B*a^3*c^3 - 24*B*a*b^4*c + 96*B*a^2*b^2*c^2))/((4* 
a*c - b^2)^(3/2)*(256*a^3*c^5 - 4*b^6*c^2 + 48*a*b^4*c^3 - 192*a^2*b^2*c^4 
)))/(a*(4*a*c - b^2)) - x^2*(((((6*B*b^3*c^2 + 8*A*a*c^4 - 28*B*a*b*c^3)/( 
4*a*c^3 - b^2*c^2) + ((8*b^3*c^4 - 32*a*b*c^5)*(2*B*b^6 - 128*B*a^3*c^3 - 
24*B*a*b^4*c + 96*B*a^2*b^2*c^2))/(2*(4*a*c^3 - b^2*c^2)*(256*a^3*c^5 - 4* 
b^6*c^2 + 48*a*b^4*c^3 - 192*a^2*b^2*c^4)))*(B*b^3 + 4*A*a*c^2 - 6*B*a*b*c 
))/(8*c^2*(4*a*c - b^2)^(3/2)) + ((8*b^3*c^4 - 32*a*b*c^5)*(B*b^3 + 4*A*a* 
c^2 - 6*B*a*b*c)*(2*B*b^6 - 128*B*a^3*c^3 - 24*B*a*b^4*c + 96*B*a^2*b^2*c^ 
2))/(16*c^2*(4*a*c - b^2)^(3/2)*(4*a*c^3 - b^2*c^2)*(256*a^3*c^5 - 4*b^6*c 
^2 + 48*a*b^4*c^3 - 192*a^2*b^2*c^4)))/(a*(4*a*c - b^2)) + (b*((B^2*b^3 + 
2*A*B*a*c^2 - 5*B^2*a*b*c)/(4*a*c^3 - b^2*c^2) + (((6*B*b^3*c^2 + 8*A*a*c^ 
4 - 28*B*a*b*c^3)/(4*a*c^3 - b^2*c^2) + ((8*b^3*c^4 - 32*a*b*c^5)*(2*B*...