Integrand size = 11, antiderivative size = 56 \[ \int \frac {1}{x^3 \left (1+x^6\right )} \, dx=-\frac {1}{2 x^2}+\frac {\arctan \left (\frac {1-2 x^2}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {1}{6} \log \left (1+x^2\right )-\frac {1}{12} \log \left (1-x^2+x^4\right ) \]
Time = 0.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.48 \[ \int \frac {1}{x^3 \left (1+x^6\right )} \, dx=\frac {1}{12} \left (-\frac {6}{x^2}+2 \sqrt {3} \arctan \left (\sqrt {3}-2 x\right )+2 \sqrt {3} \arctan \left (\sqrt {3}+2 x\right )+2 \log \left (1+x^2\right )-\log \left (1-\sqrt {3} x+x^2\right )-\log \left (1+\sqrt {3} x+x^2\right )\right ) \]
(-6/x^2 + 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
Time = 0.22 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.09, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.818, Rules used = {807, 847, 821, 16, 1142, 25, 1083, 217, 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.
\(\displaystyle \int \frac {1}{x^3 \left (x^6+1\right )} \, dx\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {1}{2} \int \frac {1}{x^4 \left (x^6+1\right )}dx^2\) |
\(\Big \downarrow \) 847 |
\(\displaystyle \frac {1}{2} \left (-\int \frac {x^2}{x^6+1}dx^2-\frac {1}{x^2}\right )\) |
\(\Big \downarrow \) 821 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \int \frac {1}{x^2+1}dx^2-\frac {1}{3} \int \frac {x^2+1}{x^4-x^2+1}dx^2-\frac {1}{x^2}\right )\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {1}{2} \left (-\frac {1}{3} \int \frac {x^2+1}{x^4-x^2+1}dx^2-\frac {1}{x^2}+\frac {1}{3} \log \left (x^2+1\right )\right )\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (-\frac {3}{2} \int \frac {1}{x^4-x^2+1}dx^2-\frac {1}{2} \int -\frac {1-2 x^2}{x^4-x^2+1}dx^2\right )-\frac {1}{x^2}+\frac {1}{3} \log \left (x^2+1\right )\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {1}{2} \int \frac {1-2 x^2}{x^4-x^2+1}dx^2-\frac {3}{2} \int \frac {1}{x^4-x^2+1}dx^2\right )-\frac {1}{x^2}+\frac {1}{3} \log \left (x^2+1\right )\right )\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (3 \int \frac {1}{-x^4-3}d\left (2 x^2-1\right )+\frac {1}{2} \int \frac {1-2 x^2}{x^4-x^2+1}dx^2\right )-\frac {1}{x^2}+\frac {1}{3} \log \left (x^2+1\right )\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (\frac {1}{2} \int \frac {1-2 x^2}{x^4-x^2+1}dx^2-\sqrt {3} \arctan \left (\frac {2 x^2-1}{\sqrt {3}}\right )\right )-\frac {1}{x^2}+\frac {1}{3} \log \left (x^2+1\right )\right )\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \left (-\sqrt {3} \arctan \left (\frac {2 x^2-1}{\sqrt {3}}\right )-\frac {1}{2} \log \left (x^4-x^2+1\right )\right )-\frac {1}{x^2}+\frac {1}{3} \log \left (x^2+1\right )\right )\) |
(-x^(-2) + Log[1 + x^2]/3 + (-(Sqrt[3]*ArcTan[(-1 + 2*x^2)/Sqrt[3]]) - Log [1 - x^2 + x^4]/2)/3)/2
3.14.72.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
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])
Int[(x_)^(m_.)*((a_) + (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))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Simp[-(3*Rt[a, 3]*Rt[b, 3])^(- 1) Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]*Rt[b, 3]) Int[(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]
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] - Simp[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))) Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a , b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p , x]
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]
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]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Time = 4.69 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.79
method | result | size |
risch | \(-\frac {1}{2 x^{2}}-\frac {\ln \left (x^{4}-x^{2}+1\right )}{12}-\frac {\sqrt {3}\, \arctan \left (\frac {2 \left (x^{2}-\frac {1}{2}\right ) \sqrt {3}}{3}\right )}{6}+\frac {\ln \left (x^{2}+1\right )}{6}\) | \(44\) |
default | \(-\frac {\ln \left (x^{4}-x^{2}+1\right )}{12}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x^{2}-1\right ) \sqrt {3}}{3}\right )}{6}+\frac {\ln \left (x^{2}+1\right )}{6}-\frac {1}{2 x^{2}}\) | \(46\) |
meijerg | \(-\frac {1}{2 x^{2}}-\frac {x^{4} \left (-\frac {\ln \left (1+\left (x^{6}\right )^{\frac {1}{3}}\right )}{\left (x^{6}\right )^{\frac {2}{3}}}+\frac {\ln \left (1-\left (x^{6}\right )^{\frac {1}{3}}+\left (x^{6}\right )^{\frac {2}{3}}\right )}{2 \left (x^{6}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{3}}}{2-\left (x^{6}\right )^{\frac {1}{3}}}\right )}{\left (x^{6}\right )^{\frac {2}{3}}}\right )}{6}\) | \(81\) |
Time = 0.26 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.96 \[ \int \frac {1}{x^3 \left (1+x^6\right )} \, dx=-\frac {2 \, \sqrt {3} x^{2} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) + x^{2} \log \left (x^{4} - x^{2} + 1\right ) - 2 \, x^{2} \log \left (x^{2} + 1\right ) + 6}{12 \, x^{2}} \]
-1/12*(2*sqrt(3)*x^2*arctan(1/3*sqrt(3)*(2*x^2 - 1)) + x^2*log(x^4 - x^2 + 1) - 2*x^2*log(x^2 + 1) + 6)/x^2
Time = 0.07 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.95 \[ \int \frac {1}{x^3 \left (1+x^6\right )} \, dx=\frac {\log {\left (x^{2} + 1 \right )}}{6} - \frac {\log {\left (x^{4} - x^{2} + 1 \right )}}{12} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x^{2}}{3} - \frac {\sqrt {3}}{3} \right )}}{6} - \frac {1}{2 x^{2}} \]
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/(2*x**2)
Time = 0.29 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.80 \[ \int \frac {1}{x^3 \left (1+x^6\right )} \, dx=-\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) - \frac {1}{2 \, x^{2}} - \frac {1}{12} \, \log \left (x^{4} - x^{2} + 1\right ) + \frac {1}{6} \, \log \left (x^{2} + 1\right ) \]
-1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 - 1)) - 1/2/x^2 - 1/12*log(x^4 - x^ 2 + 1) + 1/6*log(x^2 + 1)
Time = 0.29 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.80 \[ \int \frac {1}{x^3 \left (1+x^6\right )} \, dx=-\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) - \frac {1}{2 \, x^{2}} - \frac {1}{12} \, \log \left (x^{4} - x^{2} + 1\right ) + \frac {1}{6} \, \log \left (x^{2} + 1\right ) \]
-1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^2 - 1)) - 1/2/x^2 - 1/12*log(x^4 - x^ 2 + 1) + 1/6*log(x^2 + 1)
Time = 0.08 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.02 \[ \int \frac {1}{x^3 \left (1+x^6\right )} \, dx=\frac {\ln \left (x^2+1\right )}{6}-\frac {1}{2\,x^2}+\ln \left (x^2-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}-\frac {1}{2}\right )\,\left (-\frac {1}{12}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )-\ln \left (x^2+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}-\frac {1}{2}\right )\,\left (\frac {1}{12}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right ) \]