Integrand size = 16, antiderivative size = 90 \[ \int \frac {x^9}{1-3 x^4+x^8} \, dx=\frac {x^2}{2}-\frac {1}{2} \sqrt {\frac {1}{5} \left (9+4 \sqrt {5}\right )} \text {arctanh}\left (\sqrt {\frac {2}{3+\sqrt {5}}} x^2\right )+\frac {1}{2} \sqrt {\frac {1}{5} \left (9-4 \sqrt {5}\right )} \text {arctanh}\left (\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} x^2\right ) \]
1/2*x^2+1/2*arctanh(x^2*(1/2+1/2*5^(1/2)))*(1-2/5*5^(1/2))-1/2*arctanh(x^2 *2^(1/2)/(3+5^(1/2))^(1/2))*(1+2/5*5^(1/2))
Time = 0.15 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.14 \[ \int \frac {x^9}{1-3 x^4+x^8} \, dx=\frac {1}{20} \left (10 x^2+\left (-5+2 \sqrt {5}\right ) \log \left (-1+\sqrt {5}-2 x^2\right )+\left (5+2 \sqrt {5}\right ) \log \left (1+\sqrt {5}-2 x^2\right )+\left (5-2 \sqrt {5}\right ) \log \left (-1+\sqrt {5}+2 x^2\right )-\left (5+2 \sqrt {5}\right ) \log \left (1+\sqrt {5}+2 x^2\right )\right ) \]
(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 = 102, normalized size of antiderivative = 1.13, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1695, 1442, 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.
\(\displaystyle \int \frac {x^9}{x^8-3 x^4+1} \, dx\) |
\(\Big \downarrow \) 1695 |
\(\displaystyle \frac {1}{2} \int \frac {x^8}{x^8-3 x^4+1}dx^2\) |
\(\Big \downarrow \) 1442 |
\(\displaystyle \frac {1}{2} \left (x^2-\int \frac {1-3 x^4}{x^8-3 x^4+1}dx^2\right )\) |
\(\Big \downarrow \) 1480 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{10} \left (15+7 \sqrt {5}\right ) \int \frac {1}{x^4+\frac {1}{2} \left (-3-\sqrt {5}\right )}dx^2+\frac {1}{10} \left (15-7 \sqrt {5}\right ) \int \frac {1}{x^4+\frac {1}{2} \left (-3+\sqrt {5}\right )}dx^2+x^2\right )\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \frac {1}{2} \left (-\frac {\left (15+7 \sqrt {5}\right ) \text {arctanh}\left (\sqrt {\frac {2}{3+\sqrt {5}}} x^2\right )}{5 \sqrt {2 \left (3+\sqrt {5}\right )}}-\frac {1}{10} \left (15-7 \sqrt {5}\right ) \sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} \text {arctanh}\left (\sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} x^2\right )+x^2\right )\) |
(x^2 - ((15 + 7*Sqrt[5])*ArcTanh[Sqrt[2/(3 + Sqrt[5])]*x^2])/(5*Sqrt[2*(3 + Sqrt[5])]) - ((15 - 7*Sqrt[5])*Sqrt[(3 + Sqrt[5])/2]*ArcTanh[Sqrt[(3 + S qrt[5])/2]*x^2])/10)/2
3.4.87.3.1 Defintions of rubi rules used
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^3*(d*x)^(m - 3)*((a + b*x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 1))), x] - Simp[d^4/(c*(m + 4*p + 1)) Int[(d*x)^(m - 4)*Simp[a*(m - 3) + b*(m + 2*p - 1)*x^2, x]*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, p}, x ] && NeQ[b^2 - 4*a*c, 0] && GtQ[m, 3] && NeQ[m + 4*p + 1, 0] && 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 = 0.07 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.74
method | result | size |
default | \(\frac {x^{2}}{2}-\frac {\ln \left (x^{4}+x^{2}-1\right )}{4}-\frac {\operatorname {arctanh}\left (\frac {\left (2 x^{2}+1\right ) \sqrt {5}}{5}\right ) \sqrt {5}}{5}+\frac {\ln \left (x^{4}-x^{2}-1\right )}{4}-\frac {\sqrt {5}\, \operatorname {arctanh}\left (\frac {\left (2 x^{2}-1\right ) \sqrt {5}}{5}\right )}{5}\) | \(67\) |
risch | \(\frac {x^{2}}{2}-\frac {\ln \left (2 x^{2}-\sqrt {5}+1\right )}{4}+\frac {\ln \left (2 x^{2}-\sqrt {5}+1\right ) \sqrt {5}}{10}-\frac {\ln \left (2 x^{2}+\sqrt {5}+1\right )}{4}-\frac {\ln \left (2 x^{2}+\sqrt {5}+1\right ) \sqrt {5}}{10}+\frac {\ln \left (2 x^{2}-\sqrt {5}-1\right )}{4}+\frac {\ln \left (2 x^{2}-\sqrt {5}-1\right ) \sqrt {5}}{10}+\frac {\ln \left (2 x^{2}+\sqrt {5}-1\right )}{4}-\frac {\ln \left (2 x^{2}+\sqrt {5}-1\right ) \sqrt {5}}{10}\) | \(131\) |
1/2*x^2-1/4*ln(x^4+x^2-1)-1/5*arctanh(1/5*(2*x^2+1)*5^(1/2))*5^(1/2)+1/4*l n(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. 114 vs. \(2 (50) = 100\).
Time = 0.25 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.27 \[ \int \frac {x^9}{1-3 x^4+x^8} \, dx=\frac {1}{2} \, x^{2} + \frac {1}{10} \, \sqrt {5} \log \left (\frac {2 \, x^{4} + 2 \, x^{2} - \sqrt {5} {\left (2 \, x^{2} + 1\right )} + 3}{x^{4} + x^{2} - 1}\right ) + \frac {1}{10} \, \sqrt {5} \log \left (\frac {2 \, x^{4} - 2 \, x^{2} - \sqrt {5} {\left (2 \, x^{2} - 1\right )} + 3}{x^{4} - x^{2} - 1}\right ) - \frac {1}{4} \, \log \left (x^{4} + x^{2} - 1\right ) + \frac {1}{4} \, \log \left (x^{4} - x^{2} - 1\right ) \]
1/2*x^2 + 1/10*sqrt(5)*log((2*x^4 + 2*x^2 - sqrt(5)*(2*x^2 + 1) + 3)/(x^4 + x^2 - 1)) + 1/10*sqrt(5)*log((2*x^4 - 2*x^2 - sqrt(5)*(2*x^2 - 1) + 3)/( x^4 - x^2 - 1)) - 1/4*log(x^4 + x^2 - 1) + 1/4*log(x^4 - x^2 - 1)
Leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (63) = 126\).
Time = 0.20 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.89 \[ \int \frac {x^9}{1-3 x^4+x^8} \, dx=\frac {x^{2}}{2} + \left (- \frac {1}{4} - \frac {\sqrt {5}}{10}\right ) \log {\left (x^{2} - \frac {47}{8} - \frac {47 \sqrt {5}}{20} - 120 \left (- \frac {1}{4} - \frac {\sqrt {5}}{10}\right )^{3} \right )} + \left (- \frac {1}{4} + \frac {\sqrt {5}}{10}\right ) \log {\left (x^{2} - \frac {47}{8} - 120 \left (- \frac {1}{4} + \frac {\sqrt {5}}{10}\right )^{3} + \frac {47 \sqrt {5}}{20} \right )} + \left (\frac {1}{4} - \frac {\sqrt {5}}{10}\right ) \log {\left (x^{2} - \frac {47 \sqrt {5}}{20} - 120 \left (\frac {1}{4} - \frac {\sqrt {5}}{10}\right )^{3} + \frac {47}{8} \right )} + \left (\frac {\sqrt {5}}{10} + \frac {1}{4}\right ) \log {\left (x^{2} - 120 \left (\frac {\sqrt {5}}{10} + \frac {1}{4}\right )^{3} + \frac {47 \sqrt {5}}{20} + \frac {47}{8} \right )} \]
x**2/2 + (-1/4 - sqrt(5)/10)*log(x**2 - 47/8 - 47*sqrt(5)/20 - 120*(-1/4 - sqrt(5)/10)**3) + (-1/4 + sqrt(5)/10)*log(x**2 - 47/8 - 120*(-1/4 + sqrt( 5)/10)**3 + 47*sqrt(5)/20) + (1/4 - sqrt(5)/10)*log(x**2 - 47*sqrt(5)/20 - 120*(1/4 - sqrt(5)/10)**3 + 47/8) + (sqrt(5)/10 + 1/4)*log(x**2 - 120*(sq rt(5)/10 + 1/4)**3 + 47*sqrt(5)/20 + 47/8)
Time = 0.27 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.02 \[ \int \frac {x^9}{1-3 x^4+x^8} \, dx=\frac {1}{2} \, x^{2} + \frac {1}{10} \, \sqrt {5} \log \left (\frac {2 \, x^{2} - \sqrt {5} + 1}{2 \, x^{2} + \sqrt {5} + 1}\right ) + \frac {1}{10} \, \sqrt {5} \log \left (\frac {2 \, x^{2} - \sqrt {5} - 1}{2 \, x^{2} + \sqrt {5} - 1}\right ) - \frac {1}{4} \, \log \left (x^{4} + x^{2} - 1\right ) + \frac {1}{4} \, \log \left (x^{4} - x^{2} - 1\right ) \]
1/2*x^2 + 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)) - 1/4*log(x^ 4 + x^2 - 1) + 1/4*log(x^4 - x^2 - 1)
Time = 0.34 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.08 \[ \int \frac {x^9}{1-3 x^4+x^8} \, dx=\frac {1}{2} \, x^{2} + \frac {1}{10} \, \sqrt {5} \log \left (\frac {{\left | 2 \, x^{2} - \sqrt {5} + 1 \right |}}{2 \, x^{2} + \sqrt {5} + 1}\right ) + \frac {1}{10} \, \sqrt {5} \log \left (\frac {{\left | 2 \, x^{2} - \sqrt {5} - 1 \right |}}{{\left | 2 \, x^{2} + \sqrt {5} - 1 \right |}}\right ) - \frac {1}{4} \, \log \left ({\left | x^{4} + x^{2} - 1 \right |}\right ) + \frac {1}{4} \, \log \left ({\left | x^{4} - x^{2} - 1 \right |}\right ) \]
1/2*x^2 + 1/10*sqrt(5)*log(abs(2*x^2 - sqrt(5) + 1)/(2*x^2 + sqrt(5) + 1)) + 1/10*sqrt(5)*log(abs(2*x^2 - sqrt(5) - 1)/abs(2*x^2 + sqrt(5) - 1)) - 1 /4*log(abs(x^4 + x^2 - 1)) + 1/4*log(abs(x^4 - x^2 - 1))
Time = 0.06 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00 \[ \int \frac {x^9}{1-3 x^4+x^8} \, dx=\frac {x^2}{2}-\mathrm {atanh}\left (\frac {64\,x^2}{64\,\sqrt {5}+192}+\frac {64\,\sqrt {5}\,x^2}{64\,\sqrt {5}+192}\right )\,\left (\frac {\sqrt {5}}{5}+\frac {1}{2}\right )-\mathrm {atanh}\left (\frac {64\,x^2}{64\,\sqrt {5}-192}-\frac {64\,\sqrt {5}\,x^2}{64\,\sqrt {5}-192}\right )\,\left (\frac {\sqrt {5}}{5}-\frac {1}{2}\right ) \]