\(\int \frac {1}{x^5 (1+x^6)} \, dx\) [1374]

Optimal result

Integrand size = 11, antiderivative size = 56 \[ \int \frac {1}{x^5 \left (1+x^6\right )} \, dx=-\frac {1}{4 x^4}+\frac {\arctan \left (\frac {1-2 x^2}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {1}{6} \log \left (1+x^2\right )+\frac {1}{12} \log \left (1-x^2+x^4\right ) \]

[Out]

-1/4/x^4-1/6*ln(x^2+1)+1/12*ln(x^4-x^2+1)+1/6*arctan(1/3*(-2*x^2+1)*3^(1/2))*3^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.727, Rules used = {281, 331, 206, 31, 648, 632, 210, 642} \[ \int \frac {1}{x^5 \left (1+x^6\right )} \, dx=\frac {\arctan \left (\frac {1-2 x^2}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {1}{4 x^4}-\frac {1}{6} \log \left (x^2+1\right )+\frac {1}{12} \log \left (x^4-x^2+1\right ) \]

[In]

Int[1/(x^5*(1 + x^6)),x]

[Out]

-1/4*1/x^4 + ArcTan[(1 - 2*x^2)/Sqrt[3]]/(2*Sqrt[3]) - Log[1 + x^2]/6 + Log[1 - x^2 + x^4]/12

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x]
/; FreeQ[{a, b}, x]

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 281

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

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 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; 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 648

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1}{x^3 \left (1+x^3\right )} \, dx,x,x^2\right ) \\ & = -\frac {1}{4 x^4}-\frac {1}{2} \text {Subst}\left (\int \frac {1}{1+x^3} \, dx,x,x^2\right ) \\ & = -\frac {1}{4 x^4}-\frac {1}{6} \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,x^2\right )-\frac {1}{6} \text {Subst}\left (\int \frac {2-x}{1-x+x^2} \, dx,x,x^2\right ) \\ & = -\frac {1}{4 x^4}-\frac {1}{6} \log \left (1+x^2\right )+\frac {1}{12} \text {Subst}\left (\int \frac {-1+2 x}{1-x+x^2} \, dx,x,x^2\right )-\frac {1}{4} \text {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,x^2\right ) \\ & = -\frac {1}{4 x^4}-\frac {1}{6} \log \left (1+x^2\right )+\frac {1}{12} \log \left (1-x^2+x^4\right )+\frac {1}{2} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x^2\right ) \\ & = -\frac {1}{4 x^4}+\frac {\tan ^{-1}\left (\frac {1-2 x^2}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {1}{6} \log \left (1+x^2\right )+\frac {1}{12} \log \left (1-x^2+x^4\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.41 \[ \int \frac {1}{x^5 \left (1+x^6\right )} \, dx=\frac {1}{12} \left (-\frac {3}{x^4}+2 \sqrt {3} \arctan \left (\sqrt {3}-2 x\right )+2 \sqrt {3} \arctan \left (\sqrt {3}+2 x\right )-2 \log \left (1+x^2\right )+\log \left (1-\sqrt {3} x+x^2\right )+\log \left (1+\sqrt {3} x+x^2\right )\right ) \]

[In]

Integrate[1/(x^5*(1 + x^6)),x]

[Out]

(-3/x^4 + 2*Sqrt[3]*ArcTan[Sqrt[3] - 2*x] + 2*Sqrt[3]*ArcTan[Sqrt[3] + 2*x] - 2*Log[1 + x^2] + Log[1 - Sqrt[3]
*x + x^2] + Log[1 + Sqrt[3]*x + x^2])/12

Maple [A] (verified)

Time = 4.48 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.79

method	result	size
risch	\(-\frac {1}{4 x^{4}}+\frac {\ln \left (x^{4}-x^{2}+1\right )}{12}-\frac {\sqrt {3}\, \arctan \left (\frac {2 \left (x^{2}-\frac {1}{2}\right ) \sqrt {3}}{3}\right )}{6}-\frac {\ln \left (x^{2}+1\right )}{6}\)	\(44\)
default	\(\frac {\ln \left (x^{4}-x^{2}+1\right )}{12}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x^{2}-1\right ) \sqrt {3}}{3}\right )}{6}-\frac {\ln \left (x^{2}+1\right )}{6}-\frac {1}{4 x^{4}}\)	\(46\)
meijerg	\(-\frac {1}{4 x^{4}}-\frac {x^{2} \left (\frac {\ln \left (1+\left (x^{6}\right )^{\frac {1}{3}}\right )}{\left (x^{6}\right )^{\frac {1}{3}}}-\frac {\ln \left (1-\left (x^{6}\right )^{\frac {1}{3}}+\left (x^{6}\right )^{\frac {2}{3}}\right )}{2 \left (x^{6}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{3}}}{2-\left (x^{6}\right )^{\frac {1}{3}}}\right )}{\left (x^{6}\right )^{\frac {1}{3}}}\right )}{6}\)	\(80\)

[In]

int(1/x^5/(x^6+1),x,method=_RETURNVERBOSE)

[Out]

-1/4/x^4+1/12*ln(x^4-x^2+1)-1/6*3^(1/2)*arctan(2/3*(x^2-1/2)*3^(1/2))-1/6*ln(x^2+1)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.98 \[ \int \frac {1}{x^5 \left (1+x^6\right )} \, dx=-\frac {2 \, \sqrt {3} x^{4} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) - x^{4} \log \left (x^{4} - x^{2} + 1\right ) + 2 \, x^{4} \log \left (x^{2} + 1\right ) + 3}{12 \, x^{4}} \]

[In]

integrate(1/x^5/(x^6+1),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.95 \[ \int \frac {1}{x^5 \left (1+x^6\right )} \, dx=- \frac {\log {\left (x^{2} + 1 \right )}}{6} + \frac {\log {\left (x^{4} - x^{2} + 1 \right )}}{12} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x^{2}}{3} - \frac {\sqrt {3}}{3} \right )}}{6} - \frac {1}{4 x^{4}} \]

[In]

integrate(1/x**5/(x**6+1),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.80 \[ \int \frac {1}{x^5 \left (1+x^6\right )} \, dx=-\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) - \frac {1}{4 \, x^{4}} + \frac {1}{12} \, \log \left (x^{4} - x^{2} + 1\right ) - \frac {1}{6} \, \log \left (x^{2} + 1\right ) \]

[In]

integrate(1/x^5/(x^6+1),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.80 \[ \int \frac {1}{x^5 \left (1+x^6\right )} \, dx=-\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) - \frac {1}{4 \, x^{4}} + \frac {1}{12} \, \log \left (x^{4} - x^{2} + 1\right ) - \frac {1}{6} \, \log \left (x^{2} + 1\right ) \]

[In]

integrate(1/x^5/(x^6+1),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.89 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.02 \[ \int \frac {1}{x^5 \left (1+x^6\right )} \, dx=-\frac {\ln \left (x^2+1\right )}{6}-\frac {1}{4\,x^4}+\ln \left (x^2-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}-\frac {1}{2}\right )\,\left (\frac {1}{12}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )-\ln \left (x^2+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}-\frac {1}{2}\right )\,\left (-\frac {1}{12}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right ) \]

[In]

int(1/(x^5*(x^6 + 1)),x)

[Out]

log(x^2 - (3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/12 + 1/12) - 1/(4*x^4) - log(x^2 + 1)/6 - log((3^(1/2)*1i)/2 + x
^2 - 1/2)*((3^(1/2)*1i)/12 - 1/12)