\(\int \frac {1-x^4}{x^3 (1-x^4+x^8)} \, dx\) [67]

Optimal result

Integrand size = 23, antiderivative size = 66 \[ \int \frac {1-x^4}{x^3 \left (1-x^4+x^8\right )} \, dx=-\frac {1}{2 x^2}+\frac {1}{4} \arctan \left (\sqrt {3}-2 x^2\right )-\frac {1}{4} \arctan \left (\sqrt {3}+2 x^2\right )+\frac {\text {arctanh}\left (\frac {\sqrt {3} x^2}{1+x^4}\right )}{4 \sqrt {3}} \] Output:

-1/2/x^2-1/4*arctan(2*x^2-3^(1/2))-1/4*arctan(3^(1/2)+2*x^2)+1/12*arctanh( 
3^(1/2)*x^2/(x^4+1))*3^(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 = 49, normalized size of antiderivative = 0.74 \[ \int \frac {1-x^4}{x^3 \left (1-x^4+x^8\right )} \, dx=-\frac {1}{2 x^2}-\frac {1}{4} \text {RootSum}\left [1-\text {$\#$1}^4+\text {$\#$1}^8\&,\frac {\log (x-\text {$\#$1}) \text {$\#$1}^2}{-1+2 \text {$\#$1}^4}\&\right ] \] Input:

Integrate[(1 - x^4)/(x^3*(1 - x^4 + x^8)),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.44, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {1814, 1604, 1447, 1475, 1083, 217, 1478, 25, 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.

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

\(\Big \downarrow \) 1814

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

\(\Big \downarrow \) 1604

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

\(\Big \downarrow \) 1447

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

\(\Big \downarrow \) 1475

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (-\frac {1}{2} \int \frac {1}{x^4-\sqrt {3} x^2+1}dx^2-\frac {1}{2} \int \frac {1}{x^4+\sqrt {3} x^2+1}dx^2\right )+\frac {1}{2} \int \frac {1-x^4}{x^8-x^4+1}dx^2-\frac {1}{x^2}\right )\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\int \frac {1}{-x^4-1}d\left (2 x^2-\sqrt {3}\right )+\int \frac {1}{-x^4-1}d\left (2 x^2+\sqrt {3}\right )\right )+\frac {1}{2} \int \frac {1-x^4}{x^8-x^4+1}dx^2-\frac {1}{x^2}\right )\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1-x^4}{x^8-x^4+1}dx^2+\frac {1}{2} \left (\arctan \left (\sqrt {3}-2 x^2\right )-\arctan \left (2 x^2+\sqrt {3}\right )\right )-\frac {1}{x^2}\right )\)

\(\Big \downarrow \) 1478

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (-\frac {\int -\frac {\sqrt {3}-2 x^2}{x^4-\sqrt {3} x^2+1}dx^2}{2 \sqrt {3}}-\frac {\int -\frac {2 x^2+\sqrt {3}}{x^4+\sqrt {3} x^2+1}dx^2}{2 \sqrt {3}}\right )+\frac {1}{2} \left (\arctan \left (\sqrt {3}-2 x^2\right )-\arctan \left (2 x^2+\sqrt {3}\right )\right )-\frac {1}{x^2}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {3}-2 x^2}{x^4-\sqrt {3} x^2+1}dx^2}{2 \sqrt {3}}+\frac {\int \frac {2 x^2+\sqrt {3}}{x^4+\sqrt {3} x^2+1}dx^2}{2 \sqrt {3}}\right )+\frac {1}{2} \left (\arctan \left (\sqrt {3}-2 x^2\right )-\arctan \left (2 x^2+\sqrt {3}\right )\right )-\frac {1}{x^2}\right )\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\arctan \left (\sqrt {3}-2 x^2\right )-\arctan \left (2 x^2+\sqrt {3}\right )\right )-\frac {1}{x^2}+\frac {1}{2} \left (\frac {\log \left (x^4+\sqrt {3} x^2+1\right )}{2 \sqrt {3}}-\frac {\log \left (x^4-\sqrt {3} x^2+1\right )}{2 \sqrt {3}}\right )\right )\)

Input:

Int[(1 - x^4)/(x^3*(1 - x^4 + x^8)),x]
 

Output:

(-x^(-2) + (ArcTan[Sqrt[3] - 2*x^2] - ArcTan[Sqrt[3] + 2*x^2])/2 + (-1/2*L 
og[1 - Sqrt[3]*x^2 + x^4]/Sqrt[3] + Log[1 + Sqrt[3]*x^2 + x^4]/(2*Sqrt[3]) 
)/2)/2
 

Defintions of rubi rules used

rule 25

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

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[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 1447

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

rule 1475

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

rule 1478

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

rule 1604

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

rule 1814

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

Maple [C] (verified)

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

Time = 0.07 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.61

method	result	size
risch	\(-\frac {1}{2 x^{2}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (9 \textit {\_Z}^{4}+3 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (-6 \textit {\_R}^{3}+x^{2}-\textit {\_R} \right )\right )}{4}\)	\(40\)
default	\(-\frac {1}{2 x^{2}}+\frac {\sqrt {3}\, \left (\frac {\ln \left (x^{4}+\sqrt {3}\, x^{2}+1\right )}{2}-\sqrt {3}\, \arctan \left (2 x^{2}+\sqrt {3}\right )\right )}{12}+\frac {\sqrt {3}\, \left (-\frac {\ln \left (x^{4}-\sqrt {3}\, x^{2}+1\right )}{2}-\sqrt {3}\, \arctan \left (2 x^{2}-\sqrt {3}\right )\right )}{12}\)	\(82\)

Input:

int((-x^4+1)/x^3/(x^8-x^4+1),x,method=_RETURNVERBOSE)
 

Output:

-1/2/x^2+1/4*sum(_R*ln(-6*_R^3+x^2-_R),_R=RootOf(9*_Z^4+3*_Z^2+1))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.20 \[ \int \frac {1-x^4}{x^3 \left (1-x^4+x^8\right )} \, dx=\frac {\sqrt {3} x^{2} \log \left (x^{4} + \sqrt {3} x^{2} + 1\right ) - \sqrt {3} x^{2} \log \left (x^{4} - \sqrt {3} x^{2} + 1\right ) - 6 \, x^{2} \arctan \left (2 \, x^{2} + \sqrt {3}\right ) + 6 \, x^{2} \arctan \left (-2 \, x^{2} + \sqrt {3}\right ) - 12}{24 \, x^{2}} \] Input:

integrate((-x^4+1)/x^3/(x^8-x^4+1),x, algorithm="fricas")
 

Output:

1/24*(sqrt(3)*x^2*log(x^4 + sqrt(3)*x^2 + 1) - sqrt(3)*x^2*log(x^4 - sqrt( 
3)*x^2 + 1) - 6*x^2*arctan(2*x^2 + sqrt(3)) + 6*x^2*arctan(-2*x^2 + sqrt(3 
)) - 12)/x^2
 

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.15 \[ \int \frac {1-x^4}{x^3 \left (1-x^4+x^8\right )} \, dx=- \frac {\sqrt {3} \log {\left (x^{4} - \sqrt {3} x^{2} + 1 \right )}}{24} + \frac {\sqrt {3} \log {\left (x^{4} + \sqrt {3} x^{2} + 1 \right )}}{24} - \frac {\operatorname {atan}{\left (2 x^{2} - \sqrt {3} \right )}}{4} - \frac {\operatorname {atan}{\left (2 x^{2} + \sqrt {3} \right )}}{4} - \frac {1}{2 x^{2}} \] Input:

integrate((-x**4+1)/x**3/(x**8-x**4+1),x)
 

Output:

-sqrt(3)*log(x**4 - sqrt(3)*x**2 + 1)/24 + sqrt(3)*log(x**4 + sqrt(3)*x**2 
 + 1)/24 - atan(2*x**2 - sqrt(3))/4 - atan(2*x**2 + sqrt(3))/4 - 1/(2*x**2 
)
                                                                                    
                                                                                    
 

Maxima [F]

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

integrate((-x^4+1)/x^3/(x^8-x^4+1),x, algorithm="maxima")
 

Output:

-1/2/x^2 - integrate(x^5/(x^8 - x^4 + 1), x)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.23 \[ \int \frac {1-x^4}{x^3 \left (1-x^4+x^8\right )} \, dx=-\frac {1}{24} \, \sqrt {3} x^{4} \log \left (x^{4} + \sqrt {3} x^{2} + 1\right ) + \frac {1}{24} \, \sqrt {3} x^{4} \log \left (x^{4} - \sqrt {3} x^{2} + 1\right ) - \frac {1}{4} \, x^{4} \arctan \left (2 \, x^{2} + \sqrt {3}\right ) - \frac {1}{4} \, x^{4} \arctan \left (2 \, x^{2} - \sqrt {3}\right ) - \frac {1}{2 \, x^{2}} \] Input:

integrate((-x^4+1)/x^3/(x^8-x^4+1),x, algorithm="giac")
 

Output:

-1/24*sqrt(3)*x^4*log(x^4 + sqrt(3)*x^2 + 1) + 1/24*sqrt(3)*x^4*log(x^4 - 
sqrt(3)*x^2 + 1) - 1/4*x^4*arctan(2*x^2 + sqrt(3)) - 1/4*x^4*arctan(2*x^2 
- sqrt(3)) - 1/2/x^2
 

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.85 \[ \int \frac {1-x^4}{x^3 \left (1-x^4+x^8\right )} \, dx=\mathrm {atan}\left (\frac {2\,x^2}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )+\mathrm {atan}\left (\frac {2\,x^2}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )-\frac {1}{2\,x^2} \] Input:

int(-(x^4 - 1)/(x^3*(x^8 - x^4 + 1)),x)
 

Output:

atan((2*x^2)/(3^(1/2)*1i - 1))*((3^(1/2)*1i)/12 + 1/4) + atan((2*x^2)/(3^( 
1/2)*1i + 1))*((3^(1/2)*1i)/12 - 1/4) - 1/(2*x^2)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 376, normalized size of antiderivative = 5.70 \[ \int \frac {1-x^4}{x^3 \left (1-x^4+x^8\right )} \, dx=\frac {3 \sqrt {-\sqrt {3}+2}\, \sqrt {6}\, \mathit {atan} \left (\frac {\sqrt {6}+\sqrt {2}-4 x}{2 \sqrt {-\sqrt {3}+2}}\right ) x^{2}+3 \sqrt {-\sqrt {3}+2}\, \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {6}+\sqrt {2}-4 x}{2 \sqrt {-\sqrt {3}+2}}\right ) x^{2}+3 \sqrt {-\sqrt {3}+2}\, \sqrt {6}\, \mathit {atan} \left (\frac {\sqrt {6}+\sqrt {2}+4 x}{2 \sqrt {-\sqrt {3}+2}}\right ) x^{2}+3 \sqrt {-\sqrt {3}+2}\, \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {6}+\sqrt {2}+4 x}{2 \sqrt {-\sqrt {3}+2}}\right ) x^{2}+3 \sqrt {-\sqrt {3}+2}\, \sqrt {6}\, \mathit {atan} \left (\frac {2 \sqrt {-\sqrt {3}+2}-4 x}{\sqrt {6}+\sqrt {2}}\right ) x^{2}+3 \sqrt {-\sqrt {3}+2}\, \sqrt {2}\, \mathit {atan} \left (\frac {2 \sqrt {-\sqrt {3}+2}-4 x}{\sqrt {6}+\sqrt {2}}\right ) x^{2}+3 \sqrt {-\sqrt {3}+2}\, \sqrt {6}\, \mathit {atan} \left (\frac {2 \sqrt {-\sqrt {3}+2}+4 x}{\sqrt {6}+\sqrt {2}}\right ) x^{2}+3 \sqrt {-\sqrt {3}+2}\, \sqrt {2}\, \mathit {atan} \left (\frac {2 \sqrt {-\sqrt {3}+2}+4 x}{\sqrt {6}+\sqrt {2}}\right ) x^{2}+\sqrt {3}\, \mathrm {log}\left (-\sqrt {-\sqrt {3}+2}\, x +x^{2}+1\right ) x^{2}+\sqrt {3}\, \mathrm {log}\left (\sqrt {-\sqrt {3}+2}\, x +x^{2}+1\right ) x^{2}-\sqrt {3}\, \mathrm {log}\left (-\frac {\sqrt {6}\, x}{2}-\frac {\sqrt {2}\, x}{2}+x^{2}+1\right ) x^{2}-\sqrt {3}\, \mathrm {log}\left (\frac {\sqrt {6}\, x}{2}+\frac {\sqrt {2}\, x}{2}+x^{2}+1\right ) x^{2}-12}{24 x^{2}} \] Input:

int((-x^4+1)/x^3/(x^8-x^4+1),x)
 

Output:

(3*sqrt( - sqrt(3) + 2)*sqrt(6)*atan((sqrt(6) + sqrt(2) - 4*x)/(2*sqrt( - 
sqrt(3) + 2)))*x**2 + 3*sqrt( - sqrt(3) + 2)*sqrt(2)*atan((sqrt(6) + sqrt( 
2) - 4*x)/(2*sqrt( - sqrt(3) + 2)))*x**2 + 3*sqrt( - sqrt(3) + 2)*sqrt(6)* 
atan((sqrt(6) + sqrt(2) + 4*x)/(2*sqrt( - sqrt(3) + 2)))*x**2 + 3*sqrt( - 
sqrt(3) + 2)*sqrt(2)*atan((sqrt(6) + sqrt(2) + 4*x)/(2*sqrt( - sqrt(3) + 2 
)))*x**2 + 3*sqrt( - sqrt(3) + 2)*sqrt(6)*atan((2*sqrt( - sqrt(3) + 2) - 4 
*x)/(sqrt(6) + sqrt(2)))*x**2 + 3*sqrt( - sqrt(3) + 2)*sqrt(2)*atan((2*sqr 
t( - sqrt(3) + 2) - 4*x)/(sqrt(6) + sqrt(2)))*x**2 + 3*sqrt( - sqrt(3) + 2 
)*sqrt(6)*atan((2*sqrt( - sqrt(3) + 2) + 4*x)/(sqrt(6) + sqrt(2)))*x**2 + 
3*sqrt( - sqrt(3) + 2)*sqrt(2)*atan((2*sqrt( - sqrt(3) + 2) + 4*x)/(sqrt(6 
) + sqrt(2)))*x**2 + sqrt(3)*log( - sqrt( - sqrt(3) + 2)*x + x**2 + 1)*x** 
2 + sqrt(3)*log(sqrt( - sqrt(3) + 2)*x + x**2 + 1)*x**2 - sqrt(3)*log(( - 
sqrt(6)*x - sqrt(2)*x + 2*x**2 + 2)/2)*x**2 - sqrt(3)*log((sqrt(6)*x + sqr 
t(2)*x + 2*x**2 + 2)/2)*x**2 - 12)/(24*x**2)