\(\int \frac {x^{10}}{(a x+b x^3+c x^5)^2} \, dx\) [44]

Optimal result

Integrand size = 20, antiderivative size = 303 \[ \int \frac {x^{10}}{\left (a x+b x^3+c x^5\right )^2} \, dx=\frac {x}{c^2}+\frac {x \left (a \left (b^2-2 a c\right )+b \left (b^2-3 a c\right ) x^2\right )}{2 c^2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\left (3 b^3-13 a b c-\frac {3 b^4-19 a b^2 c+20 a^2 c^2}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{2 \sqrt {2} c^{5/2} \left (b^2-4 a c\right ) \sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\left (3 b^3-13 a b c+\frac {3 b^4-19 a b^2 c+20 a^2 c^2}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{2 \sqrt {2} c^{5/2} \left (b^2-4 a c\right ) \sqrt {b+\sqrt {b^2-4 a c}}} \] Output:

x/c^2+1/2*x*(a*(-2*a*c+b^2)+b*(-3*a*c+b^2)*x^2)/c^2/(-4*a*c+b^2)/(c*x^4+b* 
x^2+a)-1/4*(3*b^3-13*a*b*c-(20*a^2*c^2-19*a*b^2*c+3*b^4)/(-4*a*c+b^2)^(1/2 
))*arctan(2^(1/2)*c^(1/2)*x/(b-(-4*a*c+b^2)^(1/2))^(1/2))*2^(1/2)/c^(5/2)/ 
(-4*a*c+b^2)/(b-(-4*a*c+b^2)^(1/2))^(1/2)-1/4*(3*b^3-13*a*b*c+(20*a^2*c^2- 
19*a*b^2*c+3*b^4)/(-4*a*c+b^2)^(1/2))*arctan(2^(1/2)*c^(1/2)*x/(b+(-4*a*c+ 
b^2)^(1/2))^(1/2))*2^(1/2)/c^(5/2)/(-4*a*c+b^2)/(b+(-4*a*c+b^2)^(1/2))^(1/ 
2)
 

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.08 \[ \int \frac {x^{10}}{\left (a x+b x^3+c x^5\right )^2} \, dx=\frac {4 \sqrt {c} x-\frac {2 \sqrt {c} x \left (2 a^2 c-b^3 x^2-a b \left (b-3 c x^2\right )\right )}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\sqrt {2} \left (-3 b^4+19 a b^2 c-20 a^2 c^2+3 b^3 \sqrt {b^2-4 a c}-13 a b c \sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\left (b^2-4 a c\right )^{3/2} \sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\sqrt {2} \left (3 b^4-19 a b^2 c+20 a^2 c^2+3 b^3 \sqrt {b^2-4 a c}-13 a b c \sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\left (b^2-4 a c\right )^{3/2} \sqrt {b+\sqrt {b^2-4 a c}}}}{4 c^{5/2}} \] Input:

Integrate[x^10/(a*x + b*x^3 + c*x^5)^2,x]
 

Output:

(4*Sqrt[c]*x - (2*Sqrt[c]*x*(2*a^2*c - b^3*x^2 - a*b*(b - 3*c*x^2)))/((b^2 
 - 4*a*c)*(a + b*x^2 + c*x^4)) - (Sqrt[2]*(-3*b^4 + 19*a*b^2*c - 20*a^2*c^ 
2 + 3*b^3*Sqrt[b^2 - 4*a*c] - 13*a*b*c*Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]* 
Sqrt[c]*x)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/((b^2 - 4*a*c)^(3/2)*Sqrt[b - Sqr 
t[b^2 - 4*a*c]]) - (Sqrt[2]*(3*b^4 - 19*a*b^2*c + 20*a^2*c^2 + 3*b^3*Sqrt[ 
b^2 - 4*a*c] - 13*a*b*c*Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt 
[b + Sqrt[b^2 - 4*a*c]]])/((b^2 - 4*a*c)^(3/2)*Sqrt[b + Sqrt[b^2 - 4*a*c]] 
))/(4*c^(5/2))
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {9, 1440, 1602, 27, 1602, 1480, 218}

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.

\(\displaystyle \int \frac {x^{10}}{\left (a x+b x^3+c x^5\right )^2} \, dx\)

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 1440

\(\displaystyle \frac {x^5 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\int \frac {x^4 \left (3 b x^2+10 a\right )}{c x^4+b x^2+a}dx}{2 \left (b^2-4 a c\right )}\)

\(\Big \downarrow \) 1602

\(\displaystyle \frac {x^5 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\frac {b x^3}{c}-\frac {\int \frac {3 x^2 \left (\left (3 b^2-10 a c\right ) x^2+3 a b\right )}{c x^4+b x^2+a}dx}{3 c}}{2 \left (b^2-4 a c\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x^5 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\frac {b x^3}{c}-\frac {\int \frac {x^2 \left (\left (3 b^2-10 a c\right ) x^2+3 a b\right )}{c x^4+b x^2+a}dx}{c}}{2 \left (b^2-4 a c\right )}\)

\(\Big \downarrow \) 1602

\(\displaystyle \frac {x^5 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\frac {b x^3}{c}-\frac {\frac {x \left (3 b^2-10 a c\right )}{c}-\frac {\int \frac {b \left (3 b^2-13 a c\right ) x^2+a \left (3 b^2-10 a c\right )}{c x^4+b x^2+a}dx}{c}}{c}}{2 \left (b^2-4 a c\right )}\)

\(\Big \downarrow \) 1480

\(\displaystyle \frac {x^5 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\frac {b x^3}{c}-\frac {\frac {x \left (3 b^2-10 a c\right )}{c}-\frac {\frac {1}{2} \left (-\frac {20 a^2 c^2-19 a b^2 c+3 b^4}{\sqrt {b^2-4 a c}}-13 a b c+3 b^3\right ) \int \frac {1}{c x^2+\frac {1}{2} \left (b-\sqrt {b^2-4 a c}\right )}dx+\frac {1}{2} \left (\frac {20 a^2 c^2-19 a b^2 c+3 b^4}{\sqrt {b^2-4 a c}}-13 a b c+3 b^3\right ) \int \frac {1}{c x^2+\frac {1}{2} \left (b+\sqrt {b^2-4 a c}\right )}dx}{c}}{c}}{2 \left (b^2-4 a c\right )}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {x^5 \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\frac {b x^3}{c}-\frac {\frac {x \left (3 b^2-10 a c\right )}{c}-\frac {\frac {\left (-\frac {20 a^2 c^2-19 a b^2 c+3 b^4}{\sqrt {b^2-4 a c}}-13 a b c+3 b^3\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {c} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {\left (\frac {20 a^2 c^2-19 a b^2 c+3 b^4}{\sqrt {b^2-4 a c}}-13 a b c+3 b^3\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {\sqrt {b^2-4 a c}+b}}\right )}{\sqrt {2} \sqrt {c} \sqrt {\sqrt {b^2-4 a c}+b}}}{c}}{c}}{2 \left (b^2-4 a c\right )}\)

Input:

Int[x^10/(a*x + b*x^3 + c*x^5)^2,x]
 

Output:

(x^5*(2*a + b*x^2))/(2*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) - ((b*x^3)/c - ( 
((3*b^2 - 10*a*c)*x)/c - (((3*b^3 - 13*a*b*c - (3*b^4 - 19*a*b^2*c + 20*a^ 
2*c^2)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b - Sqrt[b^2 - 4 
*a*c]]])/(Sqrt[2]*Sqrt[c]*Sqrt[b - Sqrt[b^2 - 4*a*c]]) + ((3*b^3 - 13*a*b* 
c + (3*b^4 - 19*a*b^2*c + 20*a^2*c^2)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*S 
qrt[c]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*Sqrt[c]*Sqrt[b + Sqrt[b^2 
 - 4*a*c]]))/c)/c)/(2*(b^2 - 4*a*c))
 

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 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 1440

Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] 
:> Simp[(-d^3)*(d*x)^(m - 3)*(2*a + b*x^2)*((a + b*x^2 + c*x^4)^(p + 1)/(2* 
(p + 1)*(b^2 - 4*a*c))), x] + Simp[d^4/(2*(p + 1)*(b^2 - 4*a*c))   Int[(d*x 
)^(m - 4)*(2*a*(m - 3) + b*(m + 4*p + 3)*x^2)*(a + b*x^2 + c*x^4)^(p + 1), 
x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && Gt 
Q[m, 3] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
 

rule 1480

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : 
> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q))   Int[1/( 
b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q))   Int[1/(b/2 
+ q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] 
 && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
 

rule 1602

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*( 
x_)^4)^(p_), x_Symbol] :> Simp[e*f*(f*x)^(m - 1)*((a + b*x^2 + c*x^4)^(p + 
1)/(c*(m + 4*p + 3))), x] - Simp[f^2/(c*(m + 4*p + 3))   Int[(f*x)^(m - 2)* 
(a + b*x^2 + c*x^4)^p*Simp[a*e*(m - 1) + (b*e*(m + 2*p + 1) - c*d*(m + 4*p 
+ 3))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && NeQ[b^2 - 4*a*c 
, 0] && GtQ[m, 1] && NeQ[m + 4*p + 3, 0] && IntegerQ[2*p] && (IntegerQ[p] | 
| IntegerQ[m])
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.14 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.57

method	result	size
risch	\(\frac {x}{c^{2}}+\frac {\frac {b \left (3 a c -b^{2}\right ) x^{3}}{8 a c -2 b^{2}}+\frac {a \left (2 a c -b^{2}\right ) x}{8 a c -2 b^{2}}}{c^{2} \left (c \,x^{4}+b \,x^{2}+a \right )}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{4}+\textit {\_Z}^{2} b +a \right )}{\sum }\frac {\left (-\frac {b \left (13 a c -3 b^{2}\right ) \textit {\_R}^{2}}{4 a c -b^{2}}-\frac {a \left (10 a c -3 b^{2}\right )}{4 a c -b^{2}}\right ) \ln \left (x -\textit {\_R} \right )}{2 \textit {\_R}^{3} c +\textit {\_R} b}}{4 c^{2}}\)	\(174\)
default	\(\frac {x}{c^{2}}-\frac {\frac {-\frac {b \left (3 a c -b^{2}\right ) x^{3}}{2 \left (4 a c -b^{2}\right )}-\frac {a \left (2 a c -b^{2}\right ) x}{2 \left (4 a c -b^{2}\right )}}{c \,x^{4}+b \,x^{2}+a}+\frac {2 c \left (-\frac {\left (13 a b c \sqrt {-4 a c +b^{2}}-3 b^{3} \sqrt {-4 a c +b^{2}}+20 a^{2} c^{2}-19 a \,b^{2} c +3 b^{4}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {c x \sqrt {2}}{\sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 c \sqrt {-4 a c +b^{2}}\, \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}+\frac {\left (13 a b c \sqrt {-4 a c +b^{2}}-3 b^{3} \sqrt {-4 a c +b^{2}}-20 a^{2} c^{2}+19 a \,b^{2} c -3 b^{4}\right ) \sqrt {2}\, \arctan \left (\frac {c x \sqrt {2}}{\sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 c \sqrt {-4 a c +b^{2}}\, \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{4 a c -b^{2}}}{c^{2}}\)	\(319\)

Input:

int(x^10/(c*x^5+b*x^3+a*x)^2,x,method=_RETURNVERBOSE)
 

Output:

x/c^2+(1/2*b*(3*a*c-b^2)/(4*a*c-b^2)*x^3+1/2*a*(2*a*c-b^2)/(4*a*c-b^2)*x)/ 
c^2/(c*x^4+b*x^2+a)+1/4/c^2*sum((-b*(13*a*c-3*b^2)/(4*a*c-b^2)*_R^2-a*(10* 
a*c-3*b^2)/(4*a*c-b^2))/(2*_R^3*c+_R*b)*ln(x-_R),_R=RootOf(_Z^4*c+_Z^2*b+a 
))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2856 vs. \(2 (261) = 522\).

Time = 0.28 (sec) , antiderivative size = 2856, normalized size of antiderivative = 9.43 \[ \int \frac {x^{10}}{\left (a x+b x^3+c x^5\right )^2} \, dx=\text {Too large to display} \] Input:

integrate(x^10/(c*x^5+b*x^3+a*x)^2,x, algorithm="fricas")
 

Output:

1/4*(4*(b^2*c - 4*a*c^2)*x^5 + 2*(3*b^3 - 11*a*b*c)*x^3 + sqrt(1/2)*(a*b^2 
*c^2 - 4*a^2*c^3 + (b^2*c^3 - 4*a*c^4)*x^4 + (b^3*c^2 - 4*a*b*c^3)*x^2)*sq 
rt(-(9*b^7 - 105*a*b^5*c + 385*a^2*b^3*c^2 - 420*a^3*b*c^3 + (b^6*c^5 - 12 
*a*b^4*c^6 + 48*a^2*b^2*c^7 - 64*a^3*c^8)*sqrt((81*b^8 - 918*a*b^6*c + 305 
1*a^2*b^4*c^2 - 2550*a^3*b^2*c^3 + 625*a^4*c^4)/(b^6*c^10 - 12*a*b^4*c^11 
+ 48*a^2*b^2*c^12 - 64*a^3*c^13)))/(b^6*c^5 - 12*a*b^4*c^6 + 48*a^2*b^2*c^ 
7 - 64*a^3*c^8))*log(-(189*a^2*b^6 - 1971*a^3*b^4*c + 5625*a^4*b^2*c^2 - 2 
500*a^5*c^3)*x + 1/2*sqrt(1/2)*(27*b^10 - 459*a*b^8*c + 2961*a^2*b^6*c^2 - 
 8818*a^3*b^4*c^3 + 11360*a^4*b^2*c^4 - 4000*a^5*c^5 - (3*b^9*c^5 - 52*a*b 
^7*c^6 + 336*a^2*b^5*c^7 - 960*a^3*b^3*c^8 + 1024*a^4*b*c^9)*sqrt((81*b^8 
- 918*a*b^6*c + 3051*a^2*b^4*c^2 - 2550*a^3*b^2*c^3 + 625*a^4*c^4)/(b^6*c^ 
10 - 12*a*b^4*c^11 + 48*a^2*b^2*c^12 - 64*a^3*c^13)))*sqrt(-(9*b^7 - 105*a 
*b^5*c + 385*a^2*b^3*c^2 - 420*a^3*b*c^3 + (b^6*c^5 - 12*a*b^4*c^6 + 48*a^ 
2*b^2*c^7 - 64*a^3*c^8)*sqrt((81*b^8 - 918*a*b^6*c + 3051*a^2*b^4*c^2 - 25 
50*a^3*b^2*c^3 + 625*a^4*c^4)/(b^6*c^10 - 12*a*b^4*c^11 + 48*a^2*b^2*c^12 
- 64*a^3*c^13)))/(b^6*c^5 - 12*a*b^4*c^6 + 48*a^2*b^2*c^7 - 64*a^3*c^8))) 
- sqrt(1/2)*(a*b^2*c^2 - 4*a^2*c^3 + (b^2*c^3 - 4*a*c^4)*x^4 + (b^3*c^2 - 
4*a*b*c^3)*x^2)*sqrt(-(9*b^7 - 105*a*b^5*c + 385*a^2*b^3*c^2 - 420*a^3*b*c 
^3 + (b^6*c^5 - 12*a*b^4*c^6 + 48*a^2*b^2*c^7 - 64*a^3*c^8)*sqrt((81*b^8 - 
 918*a*b^6*c + 3051*a^2*b^4*c^2 - 2550*a^3*b^2*c^3 + 625*a^4*c^4)/(b^6*...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{10}}{\left (a x+b x^3+c x^5\right )^2} \, dx=\text {Timed out} \] Input:

integrate(x**10/(c*x**5+b*x**3+a*x)**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {x^{10}}{\left (a x+b x^3+c x^5\right )^2} \, dx=\int { \frac {x^{10}}{{\left (c x^{5} + b x^{3} + a x\right )}^{2}} \,d x } \] Input:

integrate(x^10/(c*x^5+b*x^3+a*x)^2,x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

1/2*((b^3 - 3*a*b*c)*x^3 + (a*b^2 - 2*a^2*c)*x)/(a*b^2*c^2 - 4*a^2*c^3 + ( 
b^2*c^3 - 4*a*c^4)*x^4 + (b^3*c^2 - 4*a*b*c^3)*x^2) + 1/2*integrate(-(3*a* 
b^2 - 10*a^2*c + (3*b^3 - 13*a*b*c)*x^2)/(c*x^4 + b*x^2 + a), x)/(b^2*c^2 
- 4*a*c^3) + x/c^2
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3335 vs. \(2 (261) = 522\).

Time = 0.79 (sec) , antiderivative size = 3335, normalized size of antiderivative = 11.01 \[ \int \frac {x^{10}}{\left (a x+b x^3+c x^5\right )^2} \, dx=\text {Too large to display} \] Input:

integrate(x^10/(c*x^5+b*x^3+a*x)^2,x, algorithm="giac")
 

Output:

1/2*(b^3*x^3 - 3*a*b*c*x^3 + a*b^2*x - 2*a^2*c*x)/((c*x^4 + b*x^2 + a)*(b^ 
2*c^2 - 4*a*c^3)) + x/c^2 + 1/16*(6*b^9*c^6 - 86*a*b^7*c^7 + 440*a^2*b^5*c 
^8 - 928*a^3*b^3*c^9 + 640*a^4*b*c^10 - 3*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b 
*c + sqrt(b^2 - 4*a*c)*c)*b^9*c^4 + 43*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c 
+ sqrt(b^2 - 4*a*c)*c)*a*b^7*c^5 + 6*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + 
sqrt(b^2 - 4*a*c)*c)*b^8*c^5 - 220*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sq 
rt(b^2 - 4*a*c)*c)*a^2*b^5*c^6 - 62*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + s 
qrt(b^2 - 4*a*c)*c)*a*b^6*c^6 - 3*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqr 
t(b^2 - 4*a*c)*c)*b^7*c^6 + 464*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt( 
b^2 - 4*a*c)*c)*a^3*b^3*c^7 + 192*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqr 
t(b^2 - 4*a*c)*c)*a^2*b^4*c^7 + 31*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sq 
rt(b^2 - 4*a*c)*c)*a*b^5*c^7 - 320*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sq 
rt(b^2 - 4*a*c)*c)*a^4*b*c^8 - 160*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sq 
rt(b^2 - 4*a*c)*c)*a^3*b^2*c^8 - 96*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + s 
qrt(b^2 - 4*a*c)*c)*a^2*b^3*c^8 + 80*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + 
sqrt(b^2 - 4*a*c)*c)*a^3*b*c^9 - 6*(b^2 - 4*a*c)*b^7*c^6 + 62*(b^2 - 4*a*c 
)*a*b^5*c^7 - 192*(b^2 - 4*a*c)*a^2*b^3*c^8 + 160*(b^2 - 4*a*c)*a^3*b*c^9 
- (6*b^5*c^2 - 50*a*b^3*c^3 + 104*a^2*b*c^4 - 3*sqrt(2)*sqrt(b^2 - 4*a*c)* 
sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^5 + 25*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b* 
c + sqrt(b^2 - 4*a*c)*c)*a*b^3*c + 6*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c...
 

Mupad [B] (verification not implemented)

Time = 0.96 (sec) , antiderivative size = 7599, normalized size of antiderivative = 25.08 \[ \int \frac {x^{10}}{\left (a x+b x^3+c x^5\right )^2} \, dx=\text {Too large to display} \] Input:

int(x^10/(a*x + b*x^3 + c*x^5)^2,x)
 

Output:

((b*x^3*(3*a*c - b^2))/(2*(4*a*c - b^2)) + (a*x*(2*a*c - b^2))/(2*(4*a*c - 
 b^2)))/(a*c^2 + c^3*x^4 + b*c^2*x^2) - atan(((((10240*a^5*c^7 + 48*a*b^8* 
c^3 - 736*a^2*b^6*c^4 + 4224*a^3*b^4*c^5 - 10752*a^4*b^2*c^6)/(8*(64*a^3*c 
^6 - b^6*c^3 + 12*a*b^4*c^4 - 48*a^2*b^2*c^5)) - (x*(-(9*b^13 + 9*b^4*(-(4 
*a*c - b^2)^9)^(1/2) + 26880*a^6*b*c^6 + 2077*a^2*b^9*c^2 - 10656*a^3*b^7* 
c^3 + 30240*a^4*b^5*c^4 - 44800*a^5*b^3*c^5 + 25*a^2*c^2*(-(4*a*c - b^2)^9 
)^(1/2) - 213*a*b^11*c - 51*a*b^2*c*(-(4*a*c - b^2)^9)^(1/2))/(32*(4096*a^ 
6*c^11 + b^12*c^5 - 24*a*b^10*c^6 + 240*a^2*b^8*c^7 - 1280*a^3*b^6*c^8 + 3 
840*a^4*b^4*c^9 - 6144*a^5*b^2*c^10)))^(1/2)*(16*b^7*c^5 - 192*a*b^5*c^6 - 
 1024*a^3*b*c^8 + 768*a^2*b^3*c^7))/(2*(16*a^2*c^5 + b^4*c^3 - 8*a*b^2*c^4 
)))*(-(9*b^13 + 9*b^4*(-(4*a*c - b^2)^9)^(1/2) + 26880*a^6*b*c^6 + 2077*a^ 
2*b^9*c^2 - 10656*a^3*b^7*c^3 + 30240*a^4*b^5*c^4 - 44800*a^5*b^3*c^5 + 25 
*a^2*c^2*(-(4*a*c - b^2)^9)^(1/2) - 213*a*b^11*c - 51*a*b^2*c*(-(4*a*c - b 
^2)^9)^(1/2))/(32*(4096*a^6*c^11 + b^12*c^5 - 24*a*b^10*c^6 + 240*a^2*b^8* 
c^7 - 1280*a^3*b^6*c^8 + 3840*a^4*b^4*c^9 - 6144*a^5*b^2*c^10)))^(1/2) - ( 
x*(9*b^8 + 200*a^4*c^4 + 481*a^2*b^4*c^2 - 718*a^3*b^2*c^3 - 114*a*b^6*c)) 
/(2*(16*a^2*c^5 + b^4*c^3 - 8*a*b^2*c^4)))*(-(9*b^13 + 9*b^4*(-(4*a*c - b^ 
2)^9)^(1/2) + 26880*a^6*b*c^6 + 2077*a^2*b^9*c^2 - 10656*a^3*b^7*c^3 + 302 
40*a^4*b^5*c^4 - 44800*a^5*b^3*c^5 + 25*a^2*c^2*(-(4*a*c - b^2)^9)^(1/2) - 
 213*a*b^11*c - 51*a*b^2*c*(-(4*a*c - b^2)^9)^(1/2))/(32*(4096*a^6*c^11...
 

Reduce [B] (verification not implemented)

Time = 2.02 (sec) , antiderivative size = 3084, normalized size of antiderivative = 10.18 \[ \int \frac {x^{10}}{\left (a x+b x^3+c x^5\right )^2} \, dx =\text {Too large to display} \] Input:

int(x^10/(c*x^5+b*x^3+a*x)^2,x)
 

Output:

(32*sqrt(a)*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**2 - 6*sqrt(a)*sqrt(2 
*sqrt(c)*sqrt(a) + b)*atan((sqrt(2*sqrt(c)*sqrt(a) - b) - 2*sqrt(c)*x)/sqr 
t(2*sqrt(c)*sqrt(a) + b))*a*b**3*c + 32*sqrt(a)*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**2*c**2*x**2 + 32*sqrt(a)*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* 
*3*x**4 - 6*sqrt(a)*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**4*c*x**2 - 6*sqrt(a 
)*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**2*x**4 + 40*sqrt(c)*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**3*c**2 - 38*sqrt(c)*sqrt(2*sqrt(c)*sqrt(a) + b)*ata 
n((sqrt(2*sqrt(c)*sqrt(a) - b) - 2*sqrt(c)*x)/sqrt(2*sqrt(c)*sqrt(a) + b)) 
*a**2*b**2*c + 40*sqrt(c)*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**2*x**2 
 + 40*sqrt(c)*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*c**3*x**4 + 6*sqrt(c)*s 
qrt(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**4 - 38*sqrt(c)*sqrt(2*sqrt(c)*sqrt(...