\(\int \frac {1}{x^3 (a+b x^4+c x^8)} \, dx\) [52]

Optimal result

Integrand size = 18, antiderivative size = 184 \[ \int \frac {1}{x^3 \left (a+b x^4+c x^8\right )} \, dx=-\frac {1}{2 a x^2}-\frac {\sqrt {c} \left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{2 \sqrt {2} a \sqrt {b-\sqrt {b^2-4 a c}}}-\frac {\sqrt {c} \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x^2}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{2 \sqrt {2} a \sqrt {b+\sqrt {b^2-4 a c}}} \] Output:

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

Mathematica [C] (verified)

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

Time = 0.02 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.41 \[ \int \frac {1}{x^3 \left (a+b x^4+c x^8\right )} \, dx=-\frac {1}{2 a x^2}-\frac {\text {RootSum}\left [a+b \text {$\#$1}^4+c \text {$\#$1}^8\&,\frac {b \log (x-\text {$\#$1})+c \log (x-\text {$\#$1}) \text {$\#$1}^4}{b \text {$\#$1}^2+2 c \text {$\#$1}^6}\&\right ]}{4 a} \] Input:

Integrate[1/(x^3*(a + b*x^4 + c*x^8)),x]
 

Output:

-1/2*1/(a*x^2) - RootSum[a + b*#1^4 + c*#1^8 & , (b*Log[x - #1] + c*Log[x 
- #1]*#1^4)/(b*#1^2 + 2*c*#1^6) & ]/(4*a)
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {1695, 1443, 25, 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.

\(\displaystyle \int \frac {1}{x^3 \left (a+b x^4+c x^8\right )} \, dx\)

\(\Big \downarrow \) 1695

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

\(\Big \downarrow \) 1443

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1480

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

\(\Big \downarrow \) 218

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

Input:

Int[1/(x^3*(a + b*x^4 + c*x^8)),x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 1443

Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] 
:> Simp[(d*x)^(m + 1)*((a + b*x^2 + c*x^4)^(p + 1)/(a*d*(m + 1))), x] - Sim 
p[1/(a*d^2*(m + 1))   Int[(d*x)^(m + 2)*(b*(m + 2*p + 3) + c*(m + 4*p + 5)* 
x^2)*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b^2 
- 4*a*c, 0] && LtQ[m, -1] && 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 1695

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

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.86

method	result	size
default	\(-\frac {1}{2 a \,x^{2}}-\frac {2 c \left (-\frac {\left (b +\sqrt {-4 a c +b^{2}}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {c \,x^{2} \sqrt {2}}{\sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 \sqrt {-4 a c +b^{2}}\, \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}+\frac {\left (-b +\sqrt {-4 a c +b^{2}}\right ) \sqrt {2}\, \arctan \left (\frac {c \,x^{2} \sqrt {2}}{\sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{8 \sqrt {-4 a c +b^{2}}\, \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{a}\)	\(159\)
risch	\(-\frac {1}{2 a \,x^{2}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (16 c^{2} a^{5}-8 a^{4} b^{2} c +a^{3} b^{4}\right ) \textit {\_Z}^{4}+\left (12 a^{2} b \,c^{2}-7 a \,b^{3} c +b^{5}\right ) \textit {\_Z}^{2}+c^{3}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (-72 c^{2} a^{5}+38 a^{4} b^{2} c -5 a^{3} b^{4}\right ) \textit {\_R}^{4}+\left (-49 a^{2} b \,c^{2}+28 a \,b^{3} c -4 b^{5}\right ) \textit {\_R}^{2}-4 c^{3}\right ) x^{2}+\left (-4 a^{4} c^{2}+5 a^{3} b^{2} c -b^{4} a^{2}\right ) \textit {\_R}^{3}\right )\right )}{4}\)	\(173\)

Input:

int(1/x^3/(c*x^8+b*x^4+a),x,method=_RETURNVERBOSE)
 

Output:

-1/2/a/x^2-2/a*c*(-1/8*(b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2)*2^(1/2)/( 
(-b+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctanh(c*x^2*2^(1/2)/((-b+(-4*a*c+b^2)^( 
1/2))*c)^(1/2))+1/8*(-b+(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(1/2)*2^(1/2)/((b 
+(-4*a*c+b^2)^(1/2))*c)^(1/2)*arctan(c*x^2*2^(1/2)/((b+(-4*a*c+b^2)^(1/2)) 
*c)^(1/2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1134 vs. \(2 (141) = 282\).

Time = 0.11 (sec) , antiderivative size = 1134, normalized size of antiderivative = 6.16 \[ \int \frac {1}{x^3 \left (a+b x^4+c x^8\right )} \, dx =\text {Too large to display} \] Input:

integrate(1/x^3/(c*x^8+b*x^4+a),x, algorithm="fricas")
 

Output:

-1/4*(sqrt(1/2)*a*x^2*sqrt(-(b^3 - 3*a*b*c + (a^3*b^2 - 4*a^4*c)*sqrt((b^4 
 - 2*a*b^2*c + a^2*c^2)/(a^6*b^2 - 4*a^7*c)))/(a^3*b^2 - 4*a^4*c))*log(-(b 
^2*c^2 - a*c^3)*x^2 + 1/2*sqrt(1/2)*(b^5 - 5*a*b^3*c + 4*a^2*b*c^2 - (a^3* 
b^4 - 6*a^4*b^2*c + 8*a^5*c^2)*sqrt((b^4 - 2*a*b^2*c + a^2*c^2)/(a^6*b^2 - 
 4*a^7*c)))*sqrt(-(b^3 - 3*a*b*c + (a^3*b^2 - 4*a^4*c)*sqrt((b^4 - 2*a*b^2 
*c + a^2*c^2)/(a^6*b^2 - 4*a^7*c)))/(a^3*b^2 - 4*a^4*c))) - sqrt(1/2)*a*x^ 
2*sqrt(-(b^3 - 3*a*b*c + (a^3*b^2 - 4*a^4*c)*sqrt((b^4 - 2*a*b^2*c + a^2*c 
^2)/(a^6*b^2 - 4*a^7*c)))/(a^3*b^2 - 4*a^4*c))*log(-(b^2*c^2 - a*c^3)*x^2 
- 1/2*sqrt(1/2)*(b^5 - 5*a*b^3*c + 4*a^2*b*c^2 - (a^3*b^4 - 6*a^4*b^2*c + 
8*a^5*c^2)*sqrt((b^4 - 2*a*b^2*c + a^2*c^2)/(a^6*b^2 - 4*a^7*c)))*sqrt(-(b 
^3 - 3*a*b*c + (a^3*b^2 - 4*a^4*c)*sqrt((b^4 - 2*a*b^2*c + a^2*c^2)/(a^6*b 
^2 - 4*a^7*c)))/(a^3*b^2 - 4*a^4*c))) + sqrt(1/2)*a*x^2*sqrt(-(b^3 - 3*a*b 
*c - (a^3*b^2 - 4*a^4*c)*sqrt((b^4 - 2*a*b^2*c + a^2*c^2)/(a^6*b^2 - 4*a^7 
*c)))/(a^3*b^2 - 4*a^4*c))*log(-(b^2*c^2 - a*c^3)*x^2 + 1/2*sqrt(1/2)*(b^5 
 - 5*a*b^3*c + 4*a^2*b*c^2 + (a^3*b^4 - 6*a^4*b^2*c + 8*a^5*c^2)*sqrt((b^4 
 - 2*a*b^2*c + a^2*c^2)/(a^6*b^2 - 4*a^7*c)))*sqrt(-(b^3 - 3*a*b*c - (a^3* 
b^2 - 4*a^4*c)*sqrt((b^4 - 2*a*b^2*c + a^2*c^2)/(a^6*b^2 - 4*a^7*c)))/(a^3 
*b^2 - 4*a^4*c))) - sqrt(1/2)*a*x^2*sqrt(-(b^3 - 3*a*b*c - (a^3*b^2 - 4*a^ 
4*c)*sqrt((b^4 - 2*a*b^2*c + a^2*c^2)/(a^6*b^2 - 4*a^7*c)))/(a^3*b^2 - 4*a 
^4*c))*log(-(b^2*c^2 - a*c^3)*x^2 - 1/2*sqrt(1/2)*(b^5 - 5*a*b^3*c + 4*...
 

Sympy [A] (verification not implemented)

Time = 9.26 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^3 \left (a+b x^4+c x^8\right )} \, dx=\operatorname {RootSum} {\left (t^{4} \cdot \left (4096 a^{5} c^{2} - 2048 a^{4} b^{2} c + 256 a^{3} b^{4}\right ) + t^{2} \cdot \left (192 a^{2} b c^{2} - 112 a b^{3} c + 16 b^{5}\right ) + c^{3}, \left ( t \mapsto t \log {\left (x^{2} + \frac {- 512 t^{3} a^{5} c^{2} + 384 t^{3} a^{4} b^{2} c - 64 t^{3} a^{3} b^{4} - 20 t a^{2} b c^{2} + 20 t a b^{3} c - 4 t b^{5}}{a c^{3} - b^{2} c^{2}} \right )} \right )\right )} - \frac {1}{2 a x^{2}} \] Input:

integrate(1/x**3/(c*x**8+b*x**4+a),x)
 

Output:

RootSum(_t**4*(4096*a**5*c**2 - 2048*a**4*b**2*c + 256*a**3*b**4) + _t**2* 
(192*a**2*b*c**2 - 112*a*b**3*c + 16*b**5) + c**3, Lambda(_t, _t*log(x**2 
+ (-512*_t**3*a**5*c**2 + 384*_t**3*a**4*b**2*c - 64*_t**3*a**3*b**4 - 20* 
_t*a**2*b*c**2 + 20*_t*a*b**3*c - 4*_t*b**5)/(a*c**3 - b**2*c**2)))) - 1/( 
2*a*x**2)
 

Maxima [F]

\[ \int \frac {1}{x^3 \left (a+b x^4+c x^8\right )} \, dx=\int { \frac {1}{{\left (c x^{8} + b x^{4} + a\right )} x^{3}} \,d x } \] Input:

integrate(1/x^3/(c*x^8+b*x^4+a),x, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2055 vs. \(2 (141) = 282\).

Time = 1.25 (sec) , antiderivative size = 2055, normalized size of antiderivative = 11.17 \[ \int \frac {1}{x^3 \left (a+b x^4+c x^8\right )} \, dx=\text {Too large to display} \] Input:

integrate(1/x^3/(c*x^8+b*x^4+a),x, algorithm="giac")
 

Output:

-1/8*(2*a*b^4*c^2 - 8*a^2*b^2*c^3 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + s 
qrt(b^2 - 4*a*c)*c)*a*b^4 + 4*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^ 
2 - 4*a*c)*c)*a^2*b^2*c + 2*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 
- 4*a*c)*c)*a*b^3*c - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a* 
c)*c)*a*b^2*c^2 - 2*(b^2 - 4*a*c)*a*b^2*c^2 + (sqrt(2)*sqrt(b*c + sqrt(b^2 
 - 4*a*c)*c)*b^4*c - 8*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^2*c^2 - 
 2*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^3*c^2 - 2*b^4*c^2 + 16*sqrt(2 
)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*c^3 + 8*sqrt(2)*sqrt(b*c + sqrt(b^2 
- 4*a*c)*c)*a*b*c^3 + sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^2*c^3 + 16 
*a*b^2*c^3 - 4*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*c^4 - 32*a^2*c^4 
+ 2*(b^2 - 4*a*c)*b^2*c^2 - 8*(b^2 - 4*a*c)*a*c^3)*x^4*abs(a) + (2*a*b^3*c 
^3 - 8*a^2*b*c^4 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)* 
c)*a*b^3*c + 4*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a 
^2*b*c^2 + 2*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b 
^2*c^2 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b*c^3 
 - 2*(b^2 - 4*a*c)*a*b*c^3)*x^4 + (sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c) 
*b^5 - 8*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^3*c - 2*sqrt(2)*sqrt( 
b*c + sqrt(b^2 - 4*a*c)*c)*b^4*c - 2*b^5*c + 16*sqrt(2)*sqrt(b*c + sqrt(b^ 
2 - 4*a*c)*c)*a^2*b*c^2 + 8*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^2* 
c^2 + sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^3*c^2 + 16*a*b^3*c^2 - ...
 

Mupad [B] (verification not implemented)

Time = 20.04 (sec) , antiderivative size = 5451, normalized size of antiderivative = 29.62 \[ \int \frac {1}{x^3 \left (a+b x^4+c x^8\right )} \, dx=\text {Too large to display} \] Input:

int(1/(x^3*(a + b*x^4 + c*x^8)),x)
 

Output:

- atan((((64*a^10*c^8 + ((-(b^5 + b^2*(-(4*a*c - b^2)^3)^(1/2) + 12*a^2*b* 
c^2 - 7*a*b^3*c - a*c*(-(4*a*c - b^2)^3)^(1/2))/(32*(a^3*b^4 + 16*a^5*c^2 
- 8*a^4*b^2*c)))^(1/2)*(((-(b^5 + b^2*(-(4*a*c - b^2)^3)^(1/2) + 12*a^2*b* 
c^2 - 7*a*b^3*c - a*c*(-(4*a*c - b^2)^3)^(1/2))/(32*(a^3*b^4 + 16*a^5*c^2 
- 8*a^4*b^2*c)))^(1/2)*(4096*a^12*b^6*c^4 - 32768*a^13*b^4*c^5 + 65536*a^1 
4*b^2*c^6) + x^2*(16384*a^13*b*c^7 - 1024*a^10*b^7*c^4 + 9216*a^11*b^5*c^5 
 - 24576*a^12*b^3*c^6))*(-(b^5 + b^2*(-(4*a*c - b^2)^3)^(1/2) + 12*a^2*b*c 
^2 - 7*a*b^3*c - a*c*(-(4*a*c - b^2)^3)^(1/2))/(32*(a^3*b^4 + 16*a^5*c^2 - 
 8*a^4*b^2*c)))^(1/2) + 4096*a^12*b*c^7 + 512*a^10*b^5*c^5 - 3072*a^11*b^3 
*c^6) + x^2*(512*a^11*c^8 - 64*a^8*b^6*c^5 + 448*a^9*b^4*c^6 - 896*a^10*b^ 
2*c^7))*(-(b^5 + b^2*(-(4*a*c - b^2)^3)^(1/2) + 12*a^2*b*c^2 - 7*a*b^3*c - 
 a*c*(-(4*a*c - b^2)^3)^(1/2))/(32*(a^3*b^4 + 16*a^5*c^2 - 8*a^4*b^2*c)))^ 
(1/2) + 16*a^8*b^4*c^6 - 64*a^9*b^2*c^7)*(b^5 + b^2*(-(4*a*c - b^2)^3)^(1/ 
2) + 12*a^2*b*c^2 - 7*a*b^3*c - a*c*(-(4*a*c - b^2)^3)^(1/2))*1i)/(32*(a^3 
*b^4 + 16*a^5*c^2 - 8*a^4*b^2*c)) - ((64*a^10*c^8 + ((-(b^5 + b^2*(-(4*a*c 
 - b^2)^3)^(1/2) + 12*a^2*b*c^2 - 7*a*b^3*c - a*c*(-(4*a*c - b^2)^3)^(1/2) 
)/(32*(a^3*b^4 + 16*a^5*c^2 - 8*a^4*b^2*c)))^(1/2)*(((-(b^5 + b^2*(-(4*a*c 
 - b^2)^3)^(1/2) + 12*a^2*b*c^2 - 7*a*b^3*c - a*c*(-(4*a*c - b^2)^3)^(1/2) 
)/(32*(a^3*b^4 + 16*a^5*c^2 - 8*a^4*b^2*c)))^(1/2)*(4096*a^12*b^6*c^4 - 32 
768*a^13*b^4*c^5 + 65536*a^14*b^2*c^6) - x^2*(16384*a^13*b*c^7 - 1024*a...
 

Reduce [F]

\[ \int \frac {1}{x^3 \left (a+b x^4+c x^8\right )} \, dx=\int \frac {1}{x^{3} \left (c \,x^{8}+b \,x^{4}+a \right )}d x \] Input:

int(1/x^3/(c*x^8+b*x^4+a),x)
 

Output:

int(1/x^3/(c*x^8+b*x^4+a),x)