Integrand size = 16, antiderivative size = 88 \[ \int \frac {1}{x^3 \left (1-3 x^4+x^8\right )} \, dx=-\frac {1}{2 x^2}-\frac {\text {arctanh}\left (\sqrt {\frac {1}{2} \left (3-\sqrt {5}\right )} x^2\right )}{2 \sqrt {5 \left (9+4 \sqrt {5}\right )}}+\frac {\text {arctanh}\left (\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} x^2\right )}{2 \sqrt {5 \left (9-4 \sqrt {5}\right )}} \] Output:
-1/2/x^2-1/2*arctanh((1/2*5^(1/2)-1/2)*x^2)/(5+2*5^(1/2))+1/2*arctanh((1/2 +1/2*5^(1/2))*x^2)/(5-2*5^(1/2))
Time = 0.04 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.17 \[ \int \frac {1}{x^3 \left (1-3 x^4+x^8\right )} \, dx=\frac {1}{20} \left (-\frac {10}{x^2}-\left (5+2 \sqrt {5}\right ) \log \left (-1+\sqrt {5}-2 x^2\right )+\left (5-2 \sqrt {5}\right ) \log \left (1+\sqrt {5}-2 x^2\right )+\left (5+2 \sqrt {5}\right ) \log \left (-1+\sqrt {5}+2 x^2\right )+\left (-5+2 \sqrt {5}\right ) \log \left (1+\sqrt {5}+2 x^2\right )\right ) \] Input:
Integrate[1/(x^3*(1 - 3*x^4 + x^8)),x]
Output:
(-10/x^2 - (5 + 2*Sqrt[5])*Log[-1 + Sqrt[5] - 2*x^2] + (5 - 2*Sqrt[5])*Log [1 + Sqrt[5] - 2*x^2] + (5 + 2*Sqrt[5])*Log[-1 + Sqrt[5] + 2*x^2] + (-5 + 2*Sqrt[5])*Log[1 + Sqrt[5] + 2*x^2])/20
Time = 0.23 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.18, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1695, 1443, 1480, 220}
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.
\(\displaystyle \int \frac {1}{x^3 \left (x^8-3 x^4+1\right )} \, dx\) |
\(\Big \downarrow \) 1695 |
\(\displaystyle \frac {1}{2} \int \frac {1}{x^4 \left (x^8-3 x^4+1\right )}dx^2\) |
\(\Big \downarrow \) 1443 |
\(\displaystyle \frac {1}{2} \left (\int \frac {3-x^4}{x^8-3 x^4+1}dx^2-\frac {1}{x^2}\right )\) |
\(\Big \downarrow \) 1480 |
\(\displaystyle \frac {1}{2} \left (-\frac {1}{10} \left (5-3 \sqrt {5}\right ) \int \frac {1}{x^4+\frac {1}{2} \left (-3-\sqrt {5}\right )}dx^2-\frac {1}{10} \left (5+3 \sqrt {5}\right ) \int \frac {1}{x^4+\frac {1}{2} \left (-3+\sqrt {5}\right )}dx^2-\frac {1}{x^2}\right )\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \frac {1}{2} \left (\frac {\left (5-3 \sqrt {5}\right ) \text {arctanh}\left (\sqrt {\frac {2}{3+\sqrt {5}}} x^2\right )}{5 \sqrt {2 \left (3+\sqrt {5}\right )}}+\frac {1}{10} \sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} \left (5+3 \sqrt {5}\right ) \text {arctanh}\left (\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} x^2\right )-\frac {1}{x^2}\right )\) |
Input:
Int[1/(x^3*(1 - 3*x^4 + x^8)),x]
Output:
(-x^(-2) + ((5 - 3*Sqrt[5])*ArcTanh[Sqrt[2/(3 + Sqrt[5])]*x^2])/(5*Sqrt[2* (3 + Sqrt[5])]) + (Sqrt[(3 + Sqrt[5])/2]*(5 + 3*Sqrt[5])*ArcTanh[Sqrt[(3 + Sqrt[5])/2]*x^2])/10)/2
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
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])
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]
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]
Time = 0.05 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.76
method | result | size |
default | \(-\frac {1}{2 x^{2}}+\frac {\ln \left (x^{4}-x^{2}-1\right )}{4}+\frac {\sqrt {5}\, \operatorname {arctanh}\left (\frac {\left (2 x^{2}-1\right ) \sqrt {5}}{5}\right )}{5}-\frac {\ln \left (x^{4}+x^{2}-1\right )}{4}+\frac {\sqrt {5}\, \operatorname {arctanh}\left (\frac {\left (2 x^{2}+1\right ) \sqrt {5}}{5}\right )}{5}\) | \(67\) |
risch | \(-\frac {1}{2 x^{2}}-\frac {\ln \left (4 x^{2}+2+2 \sqrt {5}\right )}{4}+\frac {\ln \left (4 x^{2}+2+2 \sqrt {5}\right ) \sqrt {5}}{10}-\frac {\ln \left (4 x^{2}+2-2 \sqrt {5}\right )}{4}-\frac {\ln \left (4 x^{2}+2-2 \sqrt {5}\right ) \sqrt {5}}{10}+\frac {\ln \left (4 x^{2}-2+2 \sqrt {5}\right )}{4}+\frac {\ln \left (4 x^{2}-2+2 \sqrt {5}\right ) \sqrt {5}}{10}+\frac {\ln \left (4 x^{2}-2-2 \sqrt {5}\right )}{4}-\frac {\ln \left (4 x^{2}-2-2 \sqrt {5}\right ) \sqrt {5}}{10}\) | \(139\) |
Input:
int(1/x^3/(x^8-3*x^4+1),x,method=_RETURNVERBOSE)
Output:
-1/2/x^2+1/4*ln(x^4-x^2-1)+1/5*5^(1/2)*arctanh(1/5*(2*x^2-1)*5^(1/2))-1/4* ln(x^4+x^2-1)+1/5*5^(1/2)*arctanh(1/5*(2*x^2+1)*5^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 125 vs. \(2 (50) = 100\).
Time = 0.07 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.42 \[ \int \frac {1}{x^3 \left (1-3 x^4+x^8\right )} \, dx=\frac {2 \, \sqrt {5} x^{2} \log \left (\frac {2 \, x^{4} + 2 \, x^{2} + \sqrt {5} {\left (2 \, x^{2} + 1\right )} + 3}{x^{4} + x^{2} - 1}\right ) + 2 \, \sqrt {5} x^{2} \log \left (\frac {2 \, x^{4} - 2 \, x^{2} + \sqrt {5} {\left (2 \, x^{2} - 1\right )} + 3}{x^{4} - x^{2} - 1}\right ) - 5 \, x^{2} \log \left (x^{4} + x^{2} - 1\right ) + 5 \, x^{2} \log \left (x^{4} - x^{2} - 1\right ) - 10}{20 \, x^{2}} \] Input:
integrate(1/x^3/(x^8-3*x^4+1),x, algorithm="fricas")
Output:
1/20*(2*sqrt(5)*x^2*log((2*x^4 + 2*x^2 + sqrt(5)*(2*x^2 + 1) + 3)/(x^4 + x ^2 - 1)) + 2*sqrt(5)*x^2*log((2*x^4 - 2*x^2 + sqrt(5)*(2*x^2 - 1) + 3)/(x^ 4 - x^2 - 1)) - 5*x^2*log(x^4 + x^2 - 1) + 5*x^2*log(x^4 - x^2 - 1) - 10)/ x^2
Leaf count of result is larger than twice the leaf count of optimal. 172 vs. \(2 (53) = 106\).
Time = 0.24 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.95 \[ \int \frac {1}{x^3 \left (1-3 x^4+x^8\right )} \, dx=\left (\frac {\sqrt {5}}{10} + \frac {1}{4}\right ) \log {\left (x^{2} - \frac {123}{8} - \frac {123 \sqrt {5}}{20} + 280 \left (\frac {\sqrt {5}}{10} + \frac {1}{4}\right )^{3} \right )} + \left (\frac {1}{4} - \frac {\sqrt {5}}{10}\right ) \log {\left (x^{2} - \frac {123}{8} + 280 \left (\frac {1}{4} - \frac {\sqrt {5}}{10}\right )^{3} + \frac {123 \sqrt {5}}{20} \right )} + \left (- \frac {1}{4} + \frac {\sqrt {5}}{10}\right ) \log {\left (x^{2} - \frac {123 \sqrt {5}}{20} + 280 \left (- \frac {1}{4} + \frac {\sqrt {5}}{10}\right )^{3} + \frac {123}{8} \right )} + \left (- \frac {1}{4} - \frac {\sqrt {5}}{10}\right ) \log {\left (x^{2} + 280 \left (- \frac {1}{4} - \frac {\sqrt {5}}{10}\right )^{3} + \frac {123 \sqrt {5}}{20} + \frac {123}{8} \right )} - \frac {1}{2 x^{2}} \] Input:
integrate(1/x**3/(x**8-3*x**4+1),x)
Output:
(sqrt(5)/10 + 1/4)*log(x**2 - 123/8 - 123*sqrt(5)/20 + 280*(sqrt(5)/10 + 1 /4)**3) + (1/4 - sqrt(5)/10)*log(x**2 - 123/8 + 280*(1/4 - sqrt(5)/10)**3 + 123*sqrt(5)/20) + (-1/4 + sqrt(5)/10)*log(x**2 - 123*sqrt(5)/20 + 280*(- 1/4 + sqrt(5)/10)**3 + 123/8) + (-1/4 - sqrt(5)/10)*log(x**2 + 280*(-1/4 - sqrt(5)/10)**3 + 123*sqrt(5)/20 + 123/8) - 1/(2*x**2)
Time = 0.10 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.05 \[ \int \frac {1}{x^3 \left (1-3 x^4+x^8\right )} \, dx=-\frac {1}{10} \, \sqrt {5} \log \left (\frac {2 \, x^{2} - \sqrt {5} + 1}{2 \, x^{2} + \sqrt {5} + 1}\right ) - \frac {1}{10} \, \sqrt {5} \log \left (\frac {2 \, x^{2} - \sqrt {5} - 1}{2 \, x^{2} + \sqrt {5} - 1}\right ) - \frac {1}{2 \, x^{2}} - \frac {1}{4} \, \log \left (x^{4} + x^{2} - 1\right ) + \frac {1}{4} \, \log \left (x^{4} - x^{2} - 1\right ) \] Input:
integrate(1/x^3/(x^8-3*x^4+1),x, algorithm="maxima")
Output:
-1/10*sqrt(5)*log((2*x^2 - sqrt(5) + 1)/(2*x^2 + sqrt(5) + 1)) - 1/10*sqrt (5)*log((2*x^2 - sqrt(5) - 1)/(2*x^2 + sqrt(5) - 1)) - 1/2/x^2 - 1/4*log(x ^4 + x^2 - 1) + 1/4*log(x^4 - x^2 - 1)
Time = 0.13 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.10 \[ \int \frac {1}{x^3 \left (1-3 x^4+x^8\right )} \, dx=-\frac {1}{10} \, \sqrt {5} \log \left (\frac {{\left | 2 \, x^{2} - \sqrt {5} + 1 \right |}}{2 \, x^{2} + \sqrt {5} + 1}\right ) - \frac {1}{10} \, \sqrt {5} \log \left (\frac {{\left | 2 \, x^{2} - \sqrt {5} - 1 \right |}}{{\left | 2 \, x^{2} + \sqrt {5} - 1 \right |}}\right ) - \frac {1}{2 \, x^{2}} - \frac {1}{4} \, \log \left ({\left | x^{4} + x^{2} - 1 \right |}\right ) + \frac {1}{4} \, \log \left ({\left | x^{4} - x^{2} - 1 \right |}\right ) \] Input:
integrate(1/x^3/(x^8-3*x^4+1),x, algorithm="giac")
Output:
-1/10*sqrt(5)*log(abs(2*x^2 - sqrt(5) + 1)/(2*x^2 + sqrt(5) + 1)) - 1/10*s qrt(5)*log(abs(2*x^2 - sqrt(5) - 1)/abs(2*x^2 + sqrt(5) - 1)) - 1/2/x^2 - 1/4*log(abs(x^4 + x^2 - 1)) + 1/4*log(abs(x^4 - x^2 - 1))
Time = 19.40 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^3 \left (1-3 x^4+x^8\right )} \, dx=\mathrm {atanh}\left (\frac {12736\,x^2}{3520\,\sqrt {5}-7872}-\frac {5696\,\sqrt {5}\,x^2}{3520\,\sqrt {5}-7872}\right )\,\left (\frac {\sqrt {5}}{5}-\frac {1}{2}\right )+\mathrm {atanh}\left (\frac {12736\,x^2}{3520\,\sqrt {5}+7872}+\frac {5696\,\sqrt {5}\,x^2}{3520\,\sqrt {5}+7872}\right )\,\left (\frac {\sqrt {5}}{5}+\frac {1}{2}\right )-\frac {1}{2\,x^2} \] Input:
int(1/(x^3*(x^8 - 3*x^4 + 1)),x)
Output:
atanh((12736*x^2)/(3520*5^(1/2) - 7872) - (5696*5^(1/2)*x^2)/(3520*5^(1/2) - 7872))*(5^(1/2)/5 - 1/2) + atanh((12736*x^2)/(3520*5^(1/2) + 7872) + (5 696*5^(1/2)*x^2)/(3520*5^(1/2) + 7872))*(5^(1/2)/5 + 1/2) - 1/(2*x^2)
\[ \int \frac {1}{x^3 \left (1-3 x^4+x^8\right )} \, dx=\int \frac {1}{x^{11}-3 x^{7}+x^{3}}d x \] Input:
int(1/x^3/(x^8-3*x^4+1),x)
Output:
int(1/(x**11 - 3*x**7 + x**3),x)