Integrand size = 16, antiderivative size = 106 \[ \int \frac {1}{x^2 \left (1+2 x^4+x^8\right )} \, dx=-\frac {5}{4 x}+\frac {1}{4 x \left (1+x^4\right )}+\frac {5 \arctan \left (1-\sqrt {2} x\right )}{8 \sqrt {2}}-\frac {5 \arctan \left (1+\sqrt {2} x\right )}{8 \sqrt {2}}-\frac {5 \log \left (1-\sqrt {2} x+x^2\right )}{16 \sqrt {2}}+\frac {5 \log \left (1+\sqrt {2} x+x^2\right )}{16 \sqrt {2}} \]
-5/4/x+1/4/x/(x^4+1)-5/16*arctan(-1+x*2^(1/2))*2^(1/2)-5/16*arctan(1+x*2^( 1/2))*2^(1/2)-5/32*ln(1+x^2-x*2^(1/2))*2^(1/2)+5/32*ln(1+x^2+x*2^(1/2))*2^ (1/2)
Time = 0.05 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.92 \[ \int \frac {1}{x^2 \left (1+2 x^4+x^8\right )} \, dx=\frac {1}{32} \left (-\frac {32}{x}-\frac {8 x^3}{1+x^4}+10 \sqrt {2} \arctan \left (1-\sqrt {2} x\right )-10 \sqrt {2} \arctan \left (1+\sqrt {2} x\right )-5 \sqrt {2} \log \left (1-\sqrt {2} x+x^2\right )+5 \sqrt {2} \log \left (1+\sqrt {2} x+x^2\right )\right ) \]
(-32/x - (8*x^3)/(1 + x^4) + 10*Sqrt[2]*ArcTan[1 - Sqrt[2]*x] - 10*Sqrt[2] *ArcTan[1 + Sqrt[2]*x] - 5*Sqrt[2]*Log[1 - Sqrt[2]*x + x^2] + 5*Sqrt[2]*Lo g[1 + Sqrt[2]*x + x^2])/32
Time = 0.30 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.08, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.688, Rules used = {1380, 819, 847, 826, 1476, 1082, 217, 1479, 25, 27, 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.
\(\displaystyle \int \frac {1}{x^2 \left (x^8+2 x^4+1\right )} \, dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle \int \frac {1}{x^2 \left (x^4+1\right )^2}dx\) |
\(\Big \downarrow \) 819 |
\(\displaystyle \frac {5}{4} \int \frac {1}{x^2 \left (x^4+1\right )}dx+\frac {1}{4 x \left (x^4+1\right )}\) |
\(\Big \downarrow \) 847 |
\(\displaystyle \frac {5}{4} \left (-\int \frac {x^2}{x^4+1}dx-\frac {1}{x}\right )+\frac {1}{4 x \left (x^4+1\right )}\) |
\(\Big \downarrow \) 826 |
\(\displaystyle \frac {5}{4} \left (\frac {1}{2} \int \frac {1-x^2}{x^4+1}dx-\frac {1}{2} \int \frac {x^2+1}{x^4+1}dx-\frac {1}{x}\right )+\frac {1}{4 x \left (x^4+1\right )}\) |
\(\Big \downarrow \) 1476 |
\(\displaystyle \frac {5}{4} \left (\frac {1}{2} \left (-\frac {1}{2} \int \frac {1}{x^2-\sqrt {2} x+1}dx-\frac {1}{2} \int \frac {1}{x^2+\sqrt {2} x+1}dx\right )+\frac {1}{2} \int \frac {1-x^2}{x^4+1}dx-\frac {1}{x}\right )+\frac {1}{4 x \left (x^4+1\right )}\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {5}{4} \left (\frac {1}{2} \int \frac {1-x^2}{x^4+1}dx+\frac {1}{2} \left (\frac {\int \frac {1}{-\left (\sqrt {2} x+1\right )^2-1}d\left (\sqrt {2} x+1\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\left (1-\sqrt {2} x\right )^2-1}d\left (1-\sqrt {2} x\right )}{\sqrt {2}}\right )-\frac {1}{x}\right )+\frac {1}{4 x \left (x^4+1\right )}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {5}{4} \left (\frac {1}{2} \int \frac {1-x^2}{x^4+1}dx+\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}\right )-\frac {1}{x}\right )+\frac {1}{4 x \left (x^4+1\right )}\) |
\(\Big \downarrow \) 1479 |
\(\displaystyle \frac {5}{4} \left (\frac {1}{2} \left (-\frac {\int -\frac {\sqrt {2}-2 x}{x^2-\sqrt {2} x+1}dx}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} x+1\right )}{x^2+\sqrt {2} x+1}dx}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}\right )-\frac {1}{x}\right )+\frac {1}{4 x \left (x^4+1\right )}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {5}{4} \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 x}{x^2-\sqrt {2} x+1}dx}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} x+1\right )}{x^2+\sqrt {2} x+1}dx}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}\right )-\frac {1}{x}\right )+\frac {1}{4 x \left (x^4+1\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {5}{4} \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 x}{x^2-\sqrt {2} x+1}dx}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} x+1}{x^2+\sqrt {2} x+1}dx\right )+\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}\right )-\frac {1}{x}\right )+\frac {1}{4 x \left (x^4+1\right )}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {5}{4} \left (\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (x^2+\sqrt {2} x+1\right )}{2 \sqrt {2}}-\frac {\log \left (x^2-\sqrt {2} x+1\right )}{2 \sqrt {2}}\right )-\frac {1}{x}\right )+\frac {1}{4 x \left (x^4+1\right )}\) |
1/(4*x*(1 + x^4)) + (5*(-x^(-1) + (ArcTan[1 - Sqrt[2]*x]/Sqrt[2] - ArcTan[ 1 + Sqrt[2]*x]/Sqrt[2])/2 + (-1/2*Log[1 - Sqrt[2]*x + x^2]/Sqrt[2] + Log[1 + Sqrt[2]*x + x^2]/(2*Sqrt[2]))/2))/4
3.3.86.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-( c*x)^(m + 1))*((a + b*x^n)^(p + 1)/(a*c*n*(p + 1))), x] + Simp[(m + n*(p + 1) + 1)/(a*n*(p + 1)) Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a , b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p , x]
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s) Int[(r + s*x^2)/(a + b*x^ 4), x], x] - Simp[1/(2*s) 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]]))
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] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[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])] /; Fre eQ[{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[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S imp[1/c^p Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 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]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.08 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.39
method | result | size |
risch | \(\frac {-\frac {5 x^{4}}{4}-1}{\left (x^{4}+1\right ) x}+\frac {5 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left (-\textit {\_R}^{3}+x \right )\right )}{16}\) | \(41\) |
default | \(-\frac {1}{x}-\frac {x^{3}}{4 \left (x^{4}+1\right )}-\frac {5 \sqrt {2}\, \left (\ln \left (\frac {1+x^{2}-x \sqrt {2}}{1+x^{2}+x \sqrt {2}}\right )+2 \arctan \left (x \sqrt {2}+1\right )+2 \arctan \left (x \sqrt {2}-1\right )\right )}{32}\) | \(70\) |
Result contains complex when optimal does not.
Time = 0.25 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.06 \[ \int \frac {1}{x^2 \left (1+2 x^4+x^8\right )} \, dx=-\frac {40 \, x^{4} + 5 \, \sqrt {2} {\left (\left (i - 1\right ) \, x^{5} + \left (i - 1\right ) \, x\right )} \log \left (2 \, x + \left (i + 1\right ) \, \sqrt {2}\right ) + 5 \, \sqrt {2} {\left (-\left (i + 1\right ) \, x^{5} - \left (i + 1\right ) \, x\right )} \log \left (2 \, x - \left (i - 1\right ) \, \sqrt {2}\right ) + 5 \, \sqrt {2} {\left (\left (i + 1\right ) \, x^{5} + \left (i + 1\right ) \, x\right )} \log \left (2 \, x + \left (i - 1\right ) \, \sqrt {2}\right ) + 5 \, \sqrt {2} {\left (-\left (i - 1\right ) \, x^{5} - \left (i - 1\right ) \, x\right )} \log \left (2 \, x - \left (i + 1\right ) \, \sqrt {2}\right ) + 32}{32 \, {\left (x^{5} + x\right )}} \]
-1/32*(40*x^4 + 5*sqrt(2)*((I - 1)*x^5 + (I - 1)*x)*log(2*x + (I + 1)*sqrt (2)) + 5*sqrt(2)*(-(I + 1)*x^5 - (I + 1)*x)*log(2*x - (I - 1)*sqrt(2)) + 5 *sqrt(2)*((I + 1)*x^5 + (I + 1)*x)*log(2*x + (I - 1)*sqrt(2)) + 5*sqrt(2)* (-(I - 1)*x^5 - (I - 1)*x)*log(2*x - (I + 1)*sqrt(2)) + 32)/(x^5 + x)
Time = 0.09 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.92 \[ \int \frac {1}{x^2 \left (1+2 x^4+x^8\right )} \, dx=\frac {- 5 x^{4} - 4}{4 x^{5} + 4 x} - \frac {5 \sqrt {2} \log {\left (x^{2} - \sqrt {2} x + 1 \right )}}{32} + \frac {5 \sqrt {2} \log {\left (x^{2} + \sqrt {2} x + 1 \right )}}{32} - \frac {5 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} x - 1 \right )}}{16} - \frac {5 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} x + 1 \right )}}{16} \]
(-5*x**4 - 4)/(4*x**5 + 4*x) - 5*sqrt(2)*log(x**2 - sqrt(2)*x + 1)/32 + 5* sqrt(2)*log(x**2 + sqrt(2)*x + 1)/32 - 5*sqrt(2)*atan(sqrt(2)*x - 1)/16 - 5*sqrt(2)*atan(sqrt(2)*x + 1)/16
Time = 0.27 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^2 \left (1+2 x^4+x^8\right )} \, dx=-\frac {5}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) - \frac {5}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) + \frac {5}{32} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) - \frac {5}{32} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) - \frac {5 \, x^{4} + 4}{4 \, {\left (x^{5} + x\right )}} \]
-5/16*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) - 5/16*sqrt(2)*arctan(1/ 2*sqrt(2)*(2*x - sqrt(2))) + 5/32*sqrt(2)*log(x^2 + sqrt(2)*x + 1) - 5/32* sqrt(2)*log(x^2 - sqrt(2)*x + 1) - 1/4*(5*x^4 + 4)/(x^5 + x)
Time = 0.31 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^2 \left (1+2 x^4+x^8\right )} \, dx=-\frac {5}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) - \frac {5}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) + \frac {5}{32} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) - \frac {5}{32} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) - \frac {5 \, x^{4} + 4}{4 \, {\left (x^{5} + x\right )}} \]
-5/16*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) - 5/16*sqrt(2)*arctan(1/ 2*sqrt(2)*(2*x - sqrt(2))) + 5/32*sqrt(2)*log(x^2 + sqrt(2)*x + 1) - 5/32* sqrt(2)*log(x^2 - sqrt(2)*x + 1) - 1/4*(5*x^4 + 4)/(x^5 + x)
Time = 8.12 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.46 \[ \int \frac {1}{x^2 \left (1+2 x^4+x^8\right )} \, dx=-\frac {\frac {5\,x^4}{4}+1}{x^5+x}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {5}{16}+\frac {5}{16}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {5}{16}-\frac {5}{16}{}\mathrm {i}\right ) \]