\(\int \frac {1}{x^4 (1-x^8)} \, dx\) [1487]

Optimal result

Integrand size = 13, antiderivative size = 104 \[ \int \frac {1}{x^4 \left (1-x^8\right )} \, dx=-\frac {1}{3 x^3}+\frac {\arctan (x)}{4}+\frac {\arctan \left (1-\sqrt {2} x\right )}{4 \sqrt {2}}-\frac {\arctan \left (1+\sqrt {2} x\right )}{4 \sqrt {2}}+\frac {\text {arctanh}(x)}{4}+\frac {\log \left (1-\sqrt {2} x+x^2\right )}{8 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} x+x^2\right )}{8 \sqrt {2}} \]

[Out]

-1/3/x^3+1/4*arctan(x)+1/4*arctanh(x)-1/8*arctan(-1+x*2^(1/2))*2^(1/2)-1/8*arctan(1+x*2^(1/2))*2^(1/2)+1/16*ln
(1+x^2-x*2^(1/2))*2^(1/2)-1/16*ln(1+x^2+x*2^(1/2))*2^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.846, Rules used = {331, 307, 217, 1179, 642, 1176, 631, 210, 218, 212, 209} \[ \int \frac {1}{x^4 \left (1-x^8\right )} \, dx=\frac {\arctan (x)}{4}+\frac {\arctan \left (1-\sqrt {2} x\right )}{4 \sqrt {2}}-\frac {\arctan \left (\sqrt {2} x+1\right )}{4 \sqrt {2}}+\frac {\text {arctanh}(x)}{4}-\frac {1}{3 x^3}+\frac {\log \left (x^2-\sqrt {2} x+1\right )}{8 \sqrt {2}}-\frac {\log \left (x^2+\sqrt {2} x+1\right )}{8 \sqrt {2}} \]

[In]

Int[1/(x^4*(1 - x^8)),x]

[Out]

-1/3*1/x^3 + ArcTan[x]/4 + ArcTan[1 - Sqrt[2]*x]/(4*Sqrt[2]) - ArcTan[1 + Sqrt[2]*x]/(4*Sqrt[2]) + ArcTanh[x]/
4 + Log[1 - Sqrt[2]*x + x^2]/(8*Sqrt[2]) - Log[1 + Sqrt[2]*x + x^2]/(8*Sqrt[2])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 210

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 218

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

Rule 307

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b
, 2]]}, Dist[s/(2*b), Int[x^(m - n/2)/(r + s*x^(n/2)), x], x] - Dist[s/(2*b), Int[x^(m - n/2)/(r - s*x^(n/2)),
x], x]] /; FreeQ[{a, b}, x] && IGtQ[n/4, 0] && IGtQ[m, 0] && LeQ[n/2, m] && LtQ[m, n] &&  !GtQ[a/b, 0]

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[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 1176

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

Rule 1179

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

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{3 x^3}+\int \frac {x^4}{1-x^8} \, dx \\ & = -\frac {1}{3 x^3}+\frac {1}{2} \int \frac {1}{1-x^4} \, dx-\frac {1}{2} \int \frac {1}{1+x^4} \, dx \\ & = -\frac {1}{3 x^3}+\frac {1}{4} \int \frac {1}{1-x^2} \, dx+\frac {1}{4} \int \frac {1}{1+x^2} \, dx-\frac {1}{4} \int \frac {1-x^2}{1+x^4} \, dx-\frac {1}{4} \int \frac {1+x^2}{1+x^4} \, dx \\ & = -\frac {1}{3 x^3}+\frac {1}{4} \tan ^{-1}(x)+\frac {1}{4} \tanh ^{-1}(x)-\frac {1}{8} \int \frac {1}{1-\sqrt {2} x+x^2} \, dx-\frac {1}{8} \int \frac {1}{1+\sqrt {2} x+x^2} \, dx+\frac {\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx}{8 \sqrt {2}}+\frac {\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx}{8 \sqrt {2}} \\ & = -\frac {1}{3 x^3}+\frac {1}{4} \tan ^{-1}(x)+\frac {1}{4} \tanh ^{-1}(x)+\frac {\log \left (1-\sqrt {2} x+x^2\right )}{8 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} x+x^2\right )}{8 \sqrt {2}}-\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} x\right )}{4 \sqrt {2}}+\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} x\right )}{4 \sqrt {2}} \\ & = -\frac {1}{3 x^3}+\frac {1}{4} \tan ^{-1}(x)+\frac {\tan ^{-1}\left (1-\sqrt {2} x\right )}{4 \sqrt {2}}-\frac {\tan ^{-1}\left (1+\sqrt {2} x\right )}{4 \sqrt {2}}+\frac {1}{4} \tanh ^{-1}(x)+\frac {\log \left (1-\sqrt {2} x+x^2\right )}{8 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} x+x^2\right )}{8 \sqrt {2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^4 \left (1-x^8\right )} \, dx=\frac {1}{48} \left (-\frac {16}{x^3}+12 \arctan (x)+6 \sqrt {2} \arctan \left (1-\sqrt {2} x\right )-6 \sqrt {2} \arctan \left (1+\sqrt {2} x\right )-6 \log (1-x)+6 \log (1+x)+3 \sqrt {2} \log \left (1-\sqrt {2} x+x^2\right )-3 \sqrt {2} \log \left (1+\sqrt {2} x+x^2\right )\right ) \]

[In]

Integrate[1/(x^4*(1 - x^8)),x]

[Out]

(-16/x^3 + 12*ArcTan[x] + 6*Sqrt[2]*ArcTan[1 - Sqrt[2]*x] - 6*Sqrt[2]*ArcTan[1 + Sqrt[2]*x] - 6*Log[1 - x] + 6
*Log[1 + x] + 3*Sqrt[2]*Log[1 - Sqrt[2]*x + x^2] - 3*Sqrt[2]*Log[1 + Sqrt[2]*x + x^2])/48

Maple [C] (verified)

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

Time = 3.38 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.40

method	result	size
risch	\(-\frac {1}{3 x^{3}}-\frac {\ln \left (-1+x \right )}{8}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left (x -\textit {\_R} \right )\right )}{8}+\frac {\ln \left (1+x \right )}{8}+\frac {\arctan \left (x \right )}{4}\)	\(42\)
default	\(-\frac {\ln \left (-1+x \right )}{8}+\frac {\ln \left (1+x \right )}{8}+\frac {\arctan \left (x \right )}{4}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+x^{2}+\sqrt {2}\, x}{1+x^{2}-\sqrt {2}\, x}\right )+2 \arctan \left (\sqrt {2}\, x +1\right )+2 \arctan \left (\sqrt {2}\, x -1\right )\right )}{16}-\frac {1}{3 x^{3}}\)	\(74\)
meijerg	\(\frac {\left (-1\right )^{\frac {3}{8}} \left (\frac {8 \left (-1\right )^{\frac {5}{8}}}{3 x^{3}}+\frac {x^{5} \left (-1\right )^{\frac {5}{8}} \left (\ln \left (1-\left (x^{8}\right )^{\frac {1}{8}}\right )-\ln \left (1+\left (x^{8}\right )^{\frac {1}{8}}\right )-\frac {\sqrt {2}\, \ln \left (1-\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{8}}+\left (x^{8}\right )^{\frac {1}{4}}\right )}{2}+\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{8}}}{2-\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{8}}}\right )-2 \arctan \left (\left (x^{8}\right )^{\frac {1}{8}}\right )+\frac {\sqrt {2}\, \ln \left (1+\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{8}}+\left (x^{8}\right )^{\frac {1}{4}}\right )}{2}+\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{8}}}{2+\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{8}}}\right )\right )}{\left (x^{8}\right )^{\frac {5}{8}}}\right )}{8}\)	\(159\)

[In]

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

[Out]

-1/3/x^3-1/8*ln(-1+x)+1/8*sum(_R*ln(x-_R),_R=RootOf(_Z^4+1))+1/8*ln(1+x)+1/4*arctan(x)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^4 \left (1-x^8\right )} \, dx=\frac {-\left (3 i + 3\right ) \, \sqrt {2} x^{3} \log \left (2 \, x + \left (i + 1\right ) \, \sqrt {2}\right ) + \left (3 i - 3\right ) \, \sqrt {2} x^{3} \log \left (2 \, x - \left (i - 1\right ) \, \sqrt {2}\right ) - \left (3 i - 3\right ) \, \sqrt {2} x^{3} \log \left (2 \, x + \left (i - 1\right ) \, \sqrt {2}\right ) + \left (3 i + 3\right ) \, \sqrt {2} x^{3} \log \left (2 \, x - \left (i + 1\right ) \, \sqrt {2}\right ) + 12 \, x^{3} \arctan \left (x\right ) + 6 \, x^{3} \log \left (x + 1\right ) - 6 \, x^{3} \log \left (x - 1\right ) - 16}{48 \, x^{3}} \]

[In]

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

[Out]

1/48*(-(3*I + 3)*sqrt(2)*x^3*log(2*x + (I + 1)*sqrt(2)) + (3*I - 3)*sqrt(2)*x^3*log(2*x - (I - 1)*sqrt(2)) - (
3*I - 3)*sqrt(2)*x^3*log(2*x + (I - 1)*sqrt(2)) + (3*I + 3)*sqrt(2)*x^3*log(2*x - (I + 1)*sqrt(2)) + 12*x^3*ar
ctan(x) + 6*x^3*log(x + 1) - 6*x^3*log(x - 1) - 16)/x^3

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 139.52 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.51 \[ \int \frac {1}{x^4 \left (1-x^8\right )} \, dx=- \frac {\log {\left (x - 1 \right )}}{8} + \frac {\log {\left (x + 1 \right )}}{8} - \frac {i \log {\left (x - i \right )}}{8} + \frac {i \log {\left (x + i \right )}}{8} - \operatorname {RootSum} {\left (4096 t^{4} + 1, \left ( t \mapsto t \log {\left (- 32768 t^{5} + x \right )} \right )\right )} - \frac {1}{3 x^{3}} \]

[In]

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

[Out]

-log(x - 1)/8 + log(x + 1)/8 - I*log(x - I)/8 + I*log(x + I)/8 - RootSum(4096*_t**4 + 1, Lambda(_t, _t*log(-32
768*_t**5 + x))) - 1/(3*x**3)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x^4 \left (1-x^8\right )} \, dx=-\frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) - \frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) - \frac {1}{16} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {1}{16} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) - \frac {1}{3 \, x^{3}} + \frac {1}{4} \, \arctan \left (x\right ) + \frac {1}{8} \, \log \left (x + 1\right ) - \frac {1}{8} \, \log \left (x - 1\right ) \]

[In]

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

[Out]

-1/8*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) - 1/8*sqrt(2)*arctan(1/2*sqrt(2)*(2*x - sqrt(2))) - 1/16*sqrt
(2)*log(x^2 + sqrt(2)*x + 1) + 1/16*sqrt(2)*log(x^2 - sqrt(2)*x + 1) - 1/3/x^3 + 1/4*arctan(x) + 1/8*log(x + 1
) - 1/8*log(x - 1)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.91 \[ \int \frac {1}{x^4 \left (1-x^8\right )} \, dx=-\frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) - \frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) - \frac {1}{16} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {1}{16} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) - \frac {1}{3 \, x^{3}} + \frac {1}{4} \, \arctan \left (x\right ) + \frac {1}{8} \, \log \left ({\left | x + 1 \right |}\right ) - \frac {1}{8} \, \log \left ({\left | x - 1 \right |}\right ) \]

[In]

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

[Out]

-1/8*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) - 1/8*sqrt(2)*arctan(1/2*sqrt(2)*(2*x - sqrt(2))) - 1/16*sqrt
(2)*log(x^2 + sqrt(2)*x + 1) + 1/16*sqrt(2)*log(x^2 - sqrt(2)*x + 1) - 1/3/x^3 + 1/4*arctan(x) + 1/8*log(abs(x
+ 1)) - 1/8*log(abs(x - 1))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.48 \[ \int \frac {1}{x^4 \left (1-x^8\right )} \, dx=\frac {\mathrm {atan}\left (x\right )}{4}-\frac {1}{3\,x^3}-\frac {\mathrm {atan}\left (x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{8}-\frac {1}{8}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{8}+\frac {1}{8}{}\mathrm {i}\right ) \]

[In]

int(-1/(x^4*(x^8 - 1)),x)

[Out]

atan(x)/4 - (atan(x*1i)*1i)/4 - 2^(1/2)*atan(2^(1/2)*x*(1/2 - 1i/2))*(1/8 + 1i/8) - 2^(1/2)*atan(2^(1/2)*x*(1/
2 + 1i/2))*(1/8 - 1i/8) - 1/(3*x^3)