\(\int \frac {1}{x^2 (a+b x^4+c x^8)} \, dx\) [60]

Optimal result

Integrand size = 18, antiderivative size = 363 \[ \int \frac {1}{x^2 \left (a+b x^4+c x^8\right )} \, dx=-\frac {1}{a x}-\frac {\sqrt [4]{c} \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-b-\sqrt {b^2-4 a c}}}\right )}{2\ 2^{3/4} a \sqrt [4]{-b-\sqrt {b^2-4 a c}}}-\frac {\sqrt [4]{c} \left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-b+\sqrt {b^2-4 a c}}}\right )}{2\ 2^{3/4} a \sqrt [4]{-b+\sqrt {b^2-4 a c}}}+\frac {\sqrt [4]{c} \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-b-\sqrt {b^2-4 a c}}}\right )}{2\ 2^{3/4} a \sqrt [4]{-b-\sqrt {b^2-4 a c}}}+\frac {\sqrt [4]{c} \left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-b+\sqrt {b^2-4 a c}}}\right )}{2\ 2^{3/4} a \sqrt [4]{-b+\sqrt {b^2-4 a c}}} \] Output:

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

Mathematica [C] (verified)

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

Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.20 \[ \int \frac {1}{x^2 \left (a+b x^4+c x^8\right )} \, dx=-\frac {1}{a x}-\frac {\text {RootSum}\left [a+b \text {$\#$1}^4+c \text {$\#$1}^8\&,\frac {b \log (x-\text {$\#$1})+c \log (x-\text {$\#$1}) \text {$\#$1}^4}{b \text {$\#$1}+2 c \text {$\#$1}^5}\&\right ]}{4 a} \] Input:

Integrate[1/(x^2*(a + b*x^4 + c*x^8)),x]
 

Output:

-(1/(a*x)) - RootSum[a + b*#1^4 + c*#1^8 & , (b*Log[x - #1] + c*Log[x - #1 
]*#1^4)/(b*#1 + 2*c*#1^5) & ]/(4*a)
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 330, normalized size of antiderivative = 0.91, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {1704, 25, 1834, 27, 827, 218, 221}

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.

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

\(\Big \downarrow \) 1704

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1834

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 827

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 221

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

Input:

Int[1/(x^2*(a + b*x^4 + c*x^8)),x]
 

Output:

-(1/(a*x)) - (c*(1 - b/Sqrt[b^2 - 4*a*c])*(ArcTan[(2^(1/4)*c^(1/4)*x)/(-b 
- Sqrt[b^2 - 4*a*c])^(1/4)]/(2*2^(3/4)*c^(3/4)*(-b - Sqrt[b^2 - 4*a*c])^(1 
/4)) - ArcTanh[(2^(1/4)*c^(1/4)*x)/(-b - Sqrt[b^2 - 4*a*c])^(1/4)]/(2*2^(3 
/4)*c^(3/4)*(-b - Sqrt[b^2 - 4*a*c])^(1/4))) + c*(1 + b/Sqrt[b^2 - 4*a*c]) 
*(ArcTan[(2^(1/4)*c^(1/4)*x)/(-b + Sqrt[b^2 - 4*a*c])^(1/4)]/(2*2^(3/4)*c^ 
(3/4)*(-b + Sqrt[b^2 - 4*a*c])^(1/4)) - ArcTanh[(2^(1/4)*c^(1/4)*x)/(-b + 
Sqrt[b^2 - 4*a*c])^(1/4)]/(2*2^(3/4)*c^(3/4)*(-b + Sqrt[b^2 - 4*a*c])^(1/4 
))))/a
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 221

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

rule 827

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 
 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b)   Int[1/(r + s*x^2), x], 
x] - Simp[s/(2*b)   Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ 
[a/b, 0]
 

rule 1704

Int[((d_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_ 
Symbol] :> Simp[(d*x)^(m + 1)*((a + b*x^n + c*x^(2*n))^(p + 1)/(a*d*(m + 1) 
)), x] - Simp[1/(a*d^n*(m + 1))   Int[(d*x)^(m + n)*(b*(m + n*(p + 1) + 1) 
+ c*(m + 2*n*(p + 1) + 1)*x^n)*(a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{ 
a, b, c, d, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && 
LtQ[m, -1] && IntegerQ[p]
 

rule 1834

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_)))/((a_) + (b_.)*(x_)^(n_) + 
 (c_.)*(x_)^(n2_)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + 
 (2*c*d - b*e)/(2*q))   Int[(f*x)^m/(b/2 - q/2 + c*x^n), x], x] + Simp[(e/2 
 - (2*c*d - b*e)/(2*q))   Int[(f*x)^m/(b/2 + q/2 + c*x^n), x], x]] /; FreeQ 
[{a, b, c, d, e, f, m}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n 
, 0]
 

Maple [C] (verified)

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

Time = 0.05 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.17

method	result	size
default	\(-\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8} c +\textit {\_Z}^{4} b +a \right )}{\sum }\frac {\left (\textit {\_R}^{6} c +\textit {\_R}^{2} b \right ) \ln \left (x -\textit {\_R} \right )}{2 \textit {\_R}^{7} c +\textit {\_R}^{3} b}}{4 a}-\frac {1}{a x}\)	\(63\)
risch	\(-\frac {1}{a x}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (256 a^{9} c^{4}-256 b^{2} c^{3} a^{8}+96 b^{4} c^{2} a^{7}-16 b^{6} c \,a^{6}+a^{5} b^{8}\right ) \textit {\_Z}^{8}+\left (80 a^{4} b \,c^{4}-120 b^{3} a^{3} c^{3}+61 a^{2} b^{5} c^{2}-13 a \,b^{7} c +b^{9}\right ) \textit {\_Z}^{4}+c^{5}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (1152 a^{9} c^{4}-1184 b^{2} c^{3} a^{8}+456 b^{4} c^{2} a^{7}-78 b^{6} c \,a^{6}+5 a^{5} b^{8}\right ) \textit {\_R}^{8}+\left (332 a^{4} b \,c^{4}-487 b^{3} a^{3} c^{3}+245 a^{2} b^{5} c^{2}-52 a \,b^{7} c +4 b^{9}\right ) \textit {\_R}^{4}+4 c^{5}\right ) x +\left (64 a^{8} c^{4}-112 a^{7} b^{2} c^{3}+60 a^{6} b^{4} c^{2}-13 a^{5} b^{6} c +a^{4} b^{8}\right ) \textit {\_R}^{7}+\left (4 a^{3} b \,c^{4}-a^{2} b^{3} c^{3}\right ) \textit {\_R}^{3}\right )\right )}{4}\)	\(305\)

Input:

int(1/x^2/(c*x^8+b*x^4+a),x,method=_RETURNVERBOSE)
 

Output:

-1/4/a*sum((_R^6*c+_R^2*b)/(2*_R^7*c+_R^3*b)*ln(x-_R),_R=RootOf(_Z^8*c+_Z^ 
4*b+a))-1/a/x
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 5758 vs. \(2 (281) = 562\).

Time = 0.37 (sec) , antiderivative size = 5758, normalized size of antiderivative = 15.86 \[ \int \frac {1}{x^2 \left (a+b x^4+c x^8\right )} \, dx=\text {Too large to display} \] Input:

integrate(1/x^2/(c*x^8+b*x^4+a),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [A] (verification not implemented)

Time = 121.14 (sec) , antiderivative size = 304, normalized size of antiderivative = 0.84 \[ \int \frac {1}{x^2 \left (a+b x^4+c x^8\right )} \, dx=\operatorname {RootSum} {\left (t^{8} \cdot \left (16777216 a^{9} c^{4} - 16777216 a^{8} b^{2} c^{3} + 6291456 a^{7} b^{4} c^{2} - 1048576 a^{6} b^{6} c + 65536 a^{5} b^{8}\right ) + t^{4} \cdot \left (20480 a^{4} b c^{4} - 30720 a^{3} b^{3} c^{3} + 15616 a^{2} b^{5} c^{2} - 3328 a b^{7} c + 256 b^{9}\right ) + c^{5}, \left ( t \mapsto t \log {\left (x + \frac {- 2097152 t^{7} a^{10} c^{5} + 5767168 t^{7} a^{9} b^{2} c^{4} - 4587520 t^{7} a^{8} b^{4} c^{3} + 1605632 t^{7} a^{7} b^{6} c^{2} - 262144 t^{7} a^{6} b^{8} c + 16384 t^{7} a^{5} b^{10} - 2304 t^{3} a^{5} b c^{5} + 8256 t^{3} a^{4} b^{3} c^{4} - 8832 t^{3} a^{3} b^{5} c^{3} + 4032 t^{3} a^{2} b^{7} c^{2} - 832 t^{3} a b^{9} c + 64 t^{3} b^{11}}{a^{2} c^{6} - 3 a b^{2} c^{5} + b^{4} c^{4}} \right )} \right )\right )} - \frac {1}{a x} \] Input:

integrate(1/x**2/(c*x**8+b*x**4+a),x)
 

Output:

RootSum(_t**8*(16777216*a**9*c**4 - 16777216*a**8*b**2*c**3 + 6291456*a**7 
*b**4*c**2 - 1048576*a**6*b**6*c + 65536*a**5*b**8) + _t**4*(20480*a**4*b* 
c**4 - 30720*a**3*b**3*c**3 + 15616*a**2*b**5*c**2 - 3328*a*b**7*c + 256*b 
**9) + c**5, Lambda(_t, _t*log(x + (-2097152*_t**7*a**10*c**5 + 5767168*_t 
**7*a**9*b**2*c**4 - 4587520*_t**7*a**8*b**4*c**3 + 1605632*_t**7*a**7*b** 
6*c**2 - 262144*_t**7*a**6*b**8*c + 16384*_t**7*a**5*b**10 - 2304*_t**3*a* 
*5*b*c**5 + 8256*_t**3*a**4*b**3*c**4 - 8832*_t**3*a**3*b**5*c**3 + 4032*_ 
t**3*a**2*b**7*c**2 - 832*_t**3*a*b**9*c + 64*_t**3*b**11)/(a**2*c**6 - 3* 
a*b**2*c**5 + b**4*c**4)))) - 1/(a*x)
 

Maxima [F]

\[ \int \frac {1}{x^2 \left (a+b x^4+c x^8\right )} \, dx=\int { \frac {1}{{\left (c x^{8} + b x^{4} + a\right )} x^{2}} \,d x } \] Input:

integrate(1/x^2/(c*x^8+b*x^4+a),x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {1}{x^2 \left (a+b x^4+c x^8\right )} \, dx=\int { \frac {1}{{\left (c x^{8} + b x^{4} + a\right )} x^{2}} \,d x } \] Input:

integrate(1/x^2/(c*x^8+b*x^4+a),x, algorithm="giac")
 

Output:

integrate(1/((c*x^8 + b*x^4 + a)*x^2), x)
 

Mupad [B] (verification not implemented)

Time = 20.83 (sec) , antiderivative size = 10509, normalized size of antiderivative = 28.95 \[ \int \frac {1}{x^2 \left (a+b x^4+c x^8\right )} \, dx=\text {Too large to display} \] Input:

int(1/(x^2*(a + b*x^4 + c*x^8)),x)
 

Output:

2*atan((((-(b^9 + b^4*(-(4*a*c - b^2)^5)^(1/2) + 80*a^4*b*c^4 + 61*a^2*b^5 
*c^2 - 120*a^3*b^3*c^3 + a^2*c^2*(-(4*a*c - b^2)^5)^(1/2) - 13*a*b^7*c - 3 
*a*b^2*c*(-(4*a*c - b^2)^5)^(1/2))/(512*(a^5*b^8 + 256*a^9*c^4 - 16*a^6*b^ 
6*c + 96*a^7*b^4*c^2 - 256*a^8*b^2*c^3)))^(3/4)*(4096*a^15*c^8 - x*(-(b^9 
+ b^4*(-(4*a*c - b^2)^5)^(1/2) + 80*a^4*b*c^4 + 61*a^2*b^5*c^2 - 120*a^3*b 
^3*c^3 + a^2*c^2*(-(4*a*c - b^2)^5)^(1/2) - 13*a*b^7*c - 3*a*b^2*c*(-(4*a* 
c - b^2)^5)^(1/2))/(512*(a^5*b^8 + 256*a^9*c^4 - 16*a^6*b^6*c + 96*a^7*b^4 
*c^2 - 256*a^8*b^2*c^3)))^(1/4)*(32768*a^16*c^8 + 1024*a^12*b^8*c^4 - 1228 
8*a^13*b^6*c^5 + 51200*a^14*b^4*c^6 - 81920*a^15*b^2*c^7)*1i + 256*a^11*b^ 
8*c^4 - 2816*a^12*b^6*c^5 + 10496*a^13*b^4*c^6 - 14336*a^14*b^2*c^7)*1i + 
4*a^11*b*c^8*x)*(-(b^9 + b^4*(-(4*a*c - b^2)^5)^(1/2) + 80*a^4*b*c^4 + 61* 
a^2*b^5*c^2 - 120*a^3*b^3*c^3 + a^2*c^2*(-(4*a*c - b^2)^5)^(1/2) - 13*a*b^ 
7*c - 3*a*b^2*c*(-(4*a*c - b^2)^5)^(1/2))/(512*(a^5*b^8 + 256*a^9*c^4 - 16 
*a^6*b^6*c + 96*a^7*b^4*c^2 - 256*a^8*b^2*c^3)))^(1/4) - ((-(b^9 + b^4*(-( 
4*a*c - b^2)^5)^(1/2) + 80*a^4*b*c^4 + 61*a^2*b^5*c^2 - 120*a^3*b^3*c^3 + 
a^2*c^2*(-(4*a*c - b^2)^5)^(1/2) - 13*a*b^7*c - 3*a*b^2*c*(-(4*a*c - b^2)^ 
5)^(1/2))/(512*(a^5*b^8 + 256*a^9*c^4 - 16*a^6*b^6*c + 96*a^7*b^4*c^2 - 25 
6*a^8*b^2*c^3)))^(3/4)*(4096*a^15*c^8 + x*(-(b^9 + b^4*(-(4*a*c - b^2)^5)^ 
(1/2) + 80*a^4*b*c^4 + 61*a^2*b^5*c^2 - 120*a^3*b^3*c^3 + a^2*c^2*(-(4*a*c 
 - b^2)^5)^(1/2) - 13*a*b^7*c - 3*a*b^2*c*(-(4*a*c - b^2)^5)^(1/2))/(51...
 

Reduce [F]

\[ \int \frac {1}{x^2 \left (a+b x^4+c x^8\right )} \, dx=\int \frac {1}{x^{2} \left (c \,x^{8}+b \,x^{4}+a \right )}d x \] Input:

int(1/x^2/(c*x^8+b*x^4+a),x)
 

Output:

int(1/x^2/(c*x^8+b*x^4+a),x)