\(\int \frac {x^{11}}{(a+b x^2+c x^4)^3} \, dx\) [802]

Optimal result

Integrand size = 18, antiderivative size = 209 \[ \int \frac {x^{11}}{\left (a+b x^2+c x^4\right )^3} \, dx=-\frac {b \left (b^2-7 a c\right ) x^2}{2 c^2 \left (b^2-4 a c\right )^2}+\frac {x^8 \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac {x^4 \left (a \left (b^2-16 a c\right )+b \left (b^2-10 a c\right ) x^2\right )}{4 c \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac {b \left (b^4-10 a b^2 c+30 a^2 c^2\right ) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 c^3 \left (b^2-4 a c\right )^{5/2}}+\frac {\log \left (a+b x^2+c x^4\right )}{4 c^3} \] Output:

-1/2*b*(-7*a*c+b^2)*x^2/c^2/(-4*a*c+b^2)^2+1/4*x^8*(b*x^2+2*a)/(-4*a*c+b^2 
)/(c*x^4+b*x^2+a)^2+1/4*x^4*(a*(-16*a*c+b^2)+b*(-10*a*c+b^2)*x^2)/c/(-4*a* 
c+b^2)^2/(c*x^4+b*x^2+a)+1/2*b*(30*a^2*c^2-10*a*b^2*c+b^4)*arctanh((2*c*x^ 
2+b)/(-4*a*c+b^2)^(1/2))/c^3/(-4*a*c+b^2)^(5/2)+1/4*ln(c*x^4+b*x^2+a)/c^3
 

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.17 \[ \int \frac {x^{11}}{\left (a+b x^2+c x^4\right )^3} \, dx=\frac {\frac {-b^6+11 a b^4 c-39 a^2 b^2 c^2+32 a^3 c^3+4 b^5 c x^2-30 a b^3 c^2 x^2+50 a^2 b c^3 x^2}{\left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac {2 a^3 c^2+b^5 x^2+a b^3 \left (b-5 c x^2\right )+a^2 b c \left (-4 b+5 c x^2\right )}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac {2 b c \left (b^4-10 a b^2 c+30 a^2 c^2\right ) \arctan \left (\frac {b+2 c x^2}{\sqrt {-b^2+4 a c}}\right )}{\left (-b^2+4 a c\right )^{5/2}}+c \log \left (a+b x^2+c x^4\right )}{4 c^4} \] Input:

Integrate[x^11/(a + b*x^2 + c*x^4)^3,x]
 

Output:

((-b^6 + 11*a*b^4*c - 39*a^2*b^2*c^2 + 32*a^3*c^3 + 4*b^5*c*x^2 - 30*a*b^3 
*c^2*x^2 + 50*a^2*b*c^3*x^2)/((b^2 - 4*a*c)^2*(a + b*x^2 + c*x^4)) + (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) - (2*b*c*(b^4 - 10*a*b^2*c + 30*a^2*c^2)*Arc 
Tan[(b + 2*c*x^2)/Sqrt[-b^2 + 4*a*c]])/(-b^2 + 4*a*c)^(5/2) + c*Log[a + b* 
x^2 + c*x^4])/(4*c^4)
 

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.14, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1434, 1164, 1233, 27, 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.

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

\(\Big \downarrow \) 1434

\(\displaystyle \frac {1}{2} \int \frac {x^{10}}{\left (c x^4+b x^2+a\right )^3}dx^2\)

\(\Big \downarrow \) 1164

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

\(\Big \downarrow \) 1233

\(\displaystyle \frac {1}{2} \left (\frac {x^8 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac {\frac {\int \frac {2 x^2 \left (b \left (b^2-7 a c\right ) x^2+a \left (b^2-16 a c\right )\right )}{c x^4+b x^2+a}dx^2}{c \left (b^2-4 a c\right )}-\frac {x^4 \left (b x^2 \left (b^2-10 a c\right )+a \left (b^2-16 a c\right )\right )}{c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}}{2 \left (b^2-4 a c\right )}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \left (\frac {x^8 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac {\frac {2 \int \frac {x^2 \left (b \left (b^2-7 a c\right ) x^2+a \left (b^2-16 a c\right )\right )}{c x^4+b x^2+a}dx^2}{c \left (b^2-4 a c\right )}-\frac {x^4 \left (b x^2 \left (b^2-10 a c\right )+a \left (b^2-16 a c\right )\right )}{c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}}{2 \left (b^2-4 a c\right )}\right )\)

\(\Big \downarrow \) 1200

\(\displaystyle \frac {1}{2} \left (\frac {x^8 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac {\frac {2 \int \left (-b \left (7 a-\frac {b^2}{c}\right )-\frac {\left (b^2-4 a c\right )^2 x^2+a b \left (b^2-7 a c\right )}{c \left (c x^4+b x^2+a\right )}\right )dx^2}{c \left (b^2-4 a c\right )}-\frac {x^4 \left (b x^2 \left (b^2-10 a c\right )+a \left (b^2-16 a c\right )\right )}{c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}}{2 \left (b^2-4 a c\right )}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (\frac {x^8 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac {\frac {2 \left (-\frac {b \left (30 a^2 c^2-10 a b^2 c+b^4\right ) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{c^2 \sqrt {b^2-4 a c}}-\frac {\left (b^2-4 a c\right )^2 \log \left (a+b x^2+c x^4\right )}{2 c^2}-b x^2 \left (7 a-\frac {b^2}{c}\right )\right )}{c \left (b^2-4 a c\right )}-\frac {x^4 \left (b x^2 \left (b^2-10 a c\right )+a \left (b^2-16 a c\right )\right )}{c \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}}{2 \left (b^2-4 a c\right )}\right )\)

Input:

Int[x^11/(a + b*x^2 + c*x^4)^3,x]
 

Output:

((x^8*(2*a + b*x^2))/(2*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)^2) - (-((x^4*(a* 
(b^2 - 16*a*c) + b*(b^2 - 10*a*c)*x^2))/(c*(b^2 - 4*a*c)*(a + b*x^2 + c*x^ 
4))) + (2*(-(b*(7*a - b^2/c)*x^2) - (b*(b^4 - 10*a*b^2*c + 30*a^2*c^2)*Arc 
Tanh[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(c^2*Sqrt[b^2 - 4*a*c]) - ((b^2 - 4 
*a*c)^2*Log[a + b*x^2 + c*x^4])/(2*c^2)))/(c*(b^2 - 4*a*c)))/(2*(b^2 - 4*a 
*c)))/2
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 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 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 = 0.32 (sec) , antiderivative size = 367, normalized size of antiderivative = 1.76

method	result	size
default	\(\frac {\frac {b \left (25 a^{2} c^{2}-15 a \,b^{2} c +2 b^{4}\right ) x^{6}}{c^{2} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}+\frac {\left (32 a^{3} c^{3}+11 a^{2} b^{2} c^{2}-19 a \,b^{4} c +3 b^{6}\right ) x^{4}}{2 c^{3} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}+\frac {a b \left (31 a^{2} c^{2}-22 a \,b^{2} c +3 b^{4}\right ) x^{2}}{\left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) c^{3}}+\frac {3 a^{2} \left (8 a^{2} c^{2}-7 a \,b^{2} c +b^{4}\right )}{2 c^{3} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}}{2 \left (c \,x^{4}+b \,x^{2}+a \right )^{2}}+\frac {\frac {\left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \ln \left (c \,x^{4}+b \,x^{2}+a \right )}{2 c}+\frac {2 \left (-7 a^{2} b c +b^{3} a -\frac {\left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) b}{2 c}\right ) \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{2 c^{2} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}\)	\(367\)
risch	\(\text {Expression too large to display}\)	\(1687\)

Input:

int(x^11/(c*x^4+b*x^2+a)^3,x,method=_RETURNVERBOSE)
 

Output:

1/2*(1/c^2*b*(25*a^2*c^2-15*a*b^2*c+2*b^4)/(16*a^2*c^2-8*a*b^2*c+b^4)*x^6+ 
1/2*(32*a^3*c^3+11*a^2*b^2*c^2-19*a*b^4*c+3*b^6)/c^3/(16*a^2*c^2-8*a*b^2*c 
+b^4)*x^4+a*b*(31*a^2*c^2-22*a*b^2*c+3*b^4)/(16*a^2*c^2-8*a*b^2*c+b^4)/c^3 
*x^2+3/2*a^2*(8*a^2*c^2-7*a*b^2*c+b^4)/c^3/(16*a^2*c^2-8*a*b^2*c+b^4))/(c* 
x^4+b*x^2+a)^2+1/2/c^2/(16*a^2*c^2-8*a*b^2*c+b^4)*(1/2*(16*a^2*c^2-8*a*b^2 
*c+b^4)/c*ln(c*x^4+b*x^2+a)+2*(-7*a^2*b*c+b^3*a-1/2*(16*a^2*c^2-8*a*b^2*c+ 
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. 804 vs. \(2 (195) = 390\).

Time = 0.11 (sec) , antiderivative size = 1631, normalized size of antiderivative = 7.80 \[ \int \frac {x^{11}}{\left (a+b x^2+c x^4\right )^3} \, dx=\text {Too large to display} \] Input:

integrate(x^11/(c*x^4+b*x^2+a)^3,x, algorithm="fricas")
 

Output:

[1/4*(3*a^2*b^6 - 33*a^3*b^4*c + 108*a^4*b^2*c^2 - 96*a^5*c^3 + 2*(2*b^7*c 
 - 23*a*b^5*c^2 + 85*a^2*b^3*c^3 - 100*a^3*b*c^4)*x^6 + (3*b^8 - 31*a*b^6* 
c + 87*a^2*b^4*c^2 - 12*a^3*b^2*c^3 - 128*a^4*c^4)*x^4 + 2*(3*a*b^7 - 34*a 
^2*b^5*c + 119*a^3*b^3*c^2 - 124*a^4*b*c^3)*x^2 + ((b^5*c^2 - 10*a*b^3*c^3 
 + 30*a^2*b*c^4)*x^8 + a^2*b^5 - 10*a^3*b^3*c + 30*a^4*b*c^2 + 2*(b^6*c - 
10*a*b^4*c^2 + 30*a^2*b^2*c^3)*x^6 + (b^7 - 8*a*b^5*c + 10*a^2*b^3*c^2 + 6 
0*a^3*b*c^3)*x^4 + 2*(a*b^6 - 10*a^2*b^4*c + 30*a^3*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)) + ((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)*log(c*x^4 + b*x^2 + 
a))/(a^2*b^6*c^3 - 12*a^3*b^4*c^4 + 48*a^4*b^2*c^5 - 64*a^5*c^6 + (b^6*c^5 
 - 12*a*b^4*c^6 + 48*a^2*b^2*c^7 - 64*a^3*c^8)*x^8 + 2*(b^7*c^4 - 12*a*b^5 
*c^5 + 48*a^2*b^3*c^6 - 64*a^3*b*c^7)*x^6 + (b^8*c^3 - 10*a*b^6*c^4 + 24*a 
^2*b^4*c^5 + 32*a^3*b^2*c^6 - 128*a^4*c^7)*x^4 + 2*(a*b^7*c^3 - 12*a^2*b^5 
*c^4 + 48*a^3*b^3*c^5 - 64*a^4*b*c^6)*x^2), 1/4*(3*a^2*b^6 - 33*a^3*b^4*c 
+ 108*a^4*b^2*c^2 - 96*a^5*c^3 + 2*(2*b^7*c - 23*a*b^5*c^2 + 85*a^2*b^3*c^ 
3 - 100*a^3*b*c^4)*x^6 + (3*b^8 - 31*a*b^6*c + 87*a^2*b^4*c^2 - 12*a^3*...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{11}}{\left (a+b x^2+c x^4\right )^3} \, dx=\text {Timed out} \] Input:

integrate(x**11/(c*x**4+b*x**2+a)**3,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^{11}}{\left (a+b x^2+c x^4\right )^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x^11/(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 = 1.07 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.46 \[ \int \frac {x^{11}}{\left (a+b x^2+c x^4\right )^3} \, dx=-\frac {{\left (b^{5} - 10 \, a b^{3} c + 30 \, a^{2} b c^{2}\right )} \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{2 \, {\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {3 \, b^{4} c^{2} x^{8} - 24 \, a b^{2} c^{3} x^{8} + 48 \, a^{2} c^{4} x^{8} - 2 \, b^{5} c x^{6} + 12 \, a b^{3} c^{2} x^{6} - 4 \, a^{2} b c^{3} x^{6} - 3 \, b^{6} x^{4} + 20 \, a b^{4} c x^{4} - 22 \, a^{2} b^{2} c^{2} x^{4} + 32 \, a^{3} c^{3} x^{4} - 6 \, a b^{5} x^{2} + 40 \, a^{2} b^{3} c x^{2} - 28 \, a^{3} b c^{2} x^{2} - 3 \, a^{2} b^{4} + 18 \, a^{3} b^{2} c}{8 \, {\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} {\left (c x^{4} + b x^{2} + a\right )}^{2}} + \frac {\log \left (c x^{4} + b x^{2} + a\right )}{4 \, c^{3}} \] Input:

integrate(x^11/(c*x^4+b*x^2+a)^3,x, algorithm="giac")
 

Output:

-1/2*(b^5 - 10*a*b^3*c + 30*a^2*b*c^2)*arctan((2*c*x^2 + b)/sqrt(-b^2 + 4* 
a*c))/((b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*sqrt(-b^2 + 4*a*c)) - 1/8*(3*b 
^4*c^2*x^8 - 24*a*b^2*c^3*x^8 + 48*a^2*c^4*x^8 - 2*b^5*c*x^6 + 12*a*b^3*c^ 
2*x^6 - 4*a^2*b*c^3*x^6 - 3*b^6*x^4 + 20*a*b^4*c*x^4 - 22*a^2*b^2*c^2*x^4 
+ 32*a^3*c^3*x^4 - 6*a*b^5*x^2 + 40*a^2*b^3*c*x^2 - 28*a^3*b*c^2*x^2 - 3*a 
^2*b^4 + 18*a^3*b^2*c)/((b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*(c*x^4 + b*x^ 
2 + a)^2) + 1/4*log(c*x^4 + b*x^2 + a)/c^3
 

Mupad [B] (verification not implemented)

Time = 20.86 (sec) , antiderivative size = 2588, normalized size of antiderivative = 12.38 \[ \int \frac {x^{11}}{\left (a+b x^2+c x^4\right )^3} \, dx=\text {Too large to display} \] Input:

int(x^11/(a + b*x^2 + c*x^4)^3,x)
 

Output:

((x^4*(3*b^6 + 32*a^3*c^3 + 11*a^2*b^2*c^2 - 19*a*b^4*c))/(4*c^3*(b^4 + 16 
*a^2*c^2 - 8*a*b^2*c)) + (x^2*(3*a*b^5 - 22*a^2*b^3*c + 31*a^3*b*c^2))/(2* 
c^3*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (3*a*(a*b^4 + 8*a^3*c^2 - 7*a^2*b^2* 
c))/(4*c^3*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (b*x^6*(2*b^4 + 25*a^2*c^2 - 
15*a*b^2*c))/(2*c^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) - (log((a/c^4 + ((c^3*(-(b^2*(b^4 + 
 30*a^2*c^2 - 10*a*b^2*c)^2)/(c^6*(4*a*c - b^2)^5))^(1/2) - 1)*((8*a)/c + 
(2*(c^3*(-(b^2*(b^4 + 30*a^2*c^2 - 10*a*b^2*c)^2)/(c^6*(4*a*c - b^2)^5))^( 
1/2) - 1)*(2*a + b*x^2))/c + (2*b*x^2*(3*b^4 + 62*a^2*c^2 - 26*a*b^2*c))/( 
c*(4*a*c - b^2)^2)))/(4*c^3) + (x^2*(b^5 + 23*a^2*b*c^2 - 9*a*b^3*c))/(c^4 
*(4*a*c - b^2)^2))*(a/c^4 - ((c^3*(-(b^2*(b^4 + 30*a^2*c^2 - 10*a*b^2*c)^2 
)/(c^6*(4*a*c - b^2)^5))^(1/2) + 1)*((8*a)/c - (2*(c^3*(-(b^2*(b^4 + 30*a^ 
2*c^2 - 10*a*b^2*c)^2)/(c^6*(4*a*c - b^2)^5))^(1/2) + 1)*(2*a + b*x^2))/c 
+ (2*b*x^2*(3*b^4 + 62*a^2*c^2 - 26*a*b^2*c))/(c*(4*a*c - b^2)^2)))/(4*c^3 
) + (x^2*(b^5 + 23*a^2*b*c^2 - 9*a*b^3*c))/(c^4*(4*a*c - b^2)^2)))*(2*b^10 
 - 2048*a^5*c^5 + 320*a^2*b^6*c^2 - 1280*a^3*b^4*c^3 + 2560*a^4*b^2*c^4 - 
40*a*b^8*c))/(2*(4096*a^5*c^8 - 4*b^10*c^3 + 80*a*b^8*c^4 - 640*a^2*b^6*c^ 
5 + 2560*a^3*b^4*c^6 - 5120*a^4*b^2*c^7)) - (b*atan(((x^2*(((b*((6*b^5*c^3 
 - 52*a*b^3*c^4 + 124*a^2*b*c^5)/(16*a^2*c^6 + b^4*c^4 - 8*a*b^2*c^5) + (( 
8*b^5*c^6 - 64*a*b^3*c^7 + 128*a^2*b*c^8)*(2*b^10 - 2048*a^5*c^5 + 320*...
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 3839, normalized size of antiderivative = 18.37 \[ \int \frac {x^{11}}{\left (a+b x^2+c x^4\right )^3} \, dx =\text {Too large to display} \] Input:

int(x^11/(c*x^4+b*x^2+a)^3,x)
 

Output:

(60*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**4*b*c** 
2 - 20*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* 
*3*c + 120*sqrt(2*sqrt(c)*sqrt(a) + b)*sqrt(2*sqrt(c)*sqrt(a) - b)*atan((s 
qrt(2*sqrt(c)*sqrt(a) - b) - 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*a** 
3*b**2*c**2*x**2 + 120*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**3*x**4 + 2*sqrt(2*sqrt(c)*sqrt(a) + b)*sqrt(2*sqrt(c)*s 
qrt(a) - b)*atan((sqrt(2*sqrt(c)*sqrt(a) - b) - 2*sqrt(c)*x)/sqrt(2*sqrt(c 
)*sqrt(a) + b))*a**2*b**5 - 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*sqrt( 
c)*sqrt(a) + b))*a**2*b**4*c*x**2 + 20*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**3*c**2*x**4 + 120*sqrt(2*sqrt(c)*sqrt(a) 
+ b)*sqrt(2*sqrt(c)*sqrt(a) - b)*atan((sqrt(2*sqrt(c)*sqrt(a) - b) - 2*sqr 
t(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b))*a**2*b**2*c**3*x**6 + 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**4*x**8 + 4*sqrt( 
2*sqrt(c)*sqrt(a) + b)*sqrt(2*sqrt(c)*sqrt(a) - b)*atan((sqrt(2*sqrt(c)...