Integrand size = 11, antiderivative size = 100 \[ \int \frac {1}{x^3 \left (1+x^8\right )} \, dx=-\frac {1}{2 x^2}+\frac {\arctan \left (1-\sqrt {2} x^2\right )}{4 \sqrt {2}}-\frac {\arctan \left (1+\sqrt {2} x^2\right )}{4 \sqrt {2}}-\frac {\log \left (1-\sqrt {2} x^2+x^4\right )}{8 \sqrt {2}}+\frac {\log \left (1+\sqrt {2} x^2+x^4\right )}{8 \sqrt {2}} \]
-1/2/x^2-1/8*arctan(-1+x^2*2^(1/2))*2^(1/2)-1/8*arctan(1+x^2*2^(1/2))*2^(1 /2)-1/16*ln(1+x^4-x^2*2^(1/2))*2^(1/2)+1/16*ln(1+x^4+x^2*2^(1/2))*2^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(208\) vs. \(2(100)=200\).
Time = 0.04 (sec) , antiderivative size = 208, normalized size of antiderivative = 2.08 \[ \int \frac {1}{x^3 \left (1+x^8\right )} \, dx=-\frac {1}{2 x^2}-\frac {\arctan \left (\left (x-\cos \left (\frac {\pi }{8}\right )\right ) \csc \left (\frac {\pi }{8}\right )\right )}{4 \sqrt {2}}+\frac {\arctan \left (\left (x+\cos \left (\frac {\pi }{8}\right )\right ) \csc \left (\frac {\pi }{8}\right )\right )}{4 \sqrt {2}}-\frac {\arctan \left (\sec \left (\frac {\pi }{8}\right ) \left (x-\sin \left (\frac {\pi }{8}\right )\right )\right )}{4 \sqrt {2}}+\frac {\arctan \left (\sec \left (\frac {\pi }{8}\right ) \left (x+\sin \left (\frac {\pi }{8}\right )\right )\right )}{4 \sqrt {2}}-\frac {\log \left (1+x^2-2 x \cos \left (\frac {\pi }{8}\right )\right )}{8 \sqrt {2}}-\frac {\log \left (1+x^2+2 x \cos \left (\frac {\pi }{8}\right )\right )}{8 \sqrt {2}}+\frac {\log \left (1+x^2-2 x \sin \left (\frac {\pi }{8}\right )\right )}{8 \sqrt {2}}+\frac {\log \left (1+x^2+2 x \sin \left (\frac {\pi }{8}\right )\right )}{8 \sqrt {2}} \]
-1/2*1/x^2 - ArcTan[(x - Cos[Pi/8])*Csc[Pi/8]]/(4*Sqrt[2]) + ArcTan[(x + C os[Pi/8])*Csc[Pi/8]]/(4*Sqrt[2]) - ArcTan[Sec[Pi/8]*(x - Sin[Pi/8])]/(4*Sq rt[2]) + ArcTan[Sec[Pi/8]*(x + Sin[Pi/8])]/(4*Sqrt[2]) - Log[1 + x^2 - 2*x *Cos[Pi/8]]/(8*Sqrt[2]) - Log[1 + x^2 + 2*x*Cos[Pi/8]]/(8*Sqrt[2]) + Log[1 + x^2 - 2*x*Sin[Pi/8]]/(8*Sqrt[2]) + Log[1 + x^2 + 2*x*Sin[Pi/8]]/(8*Sqrt [2])
Time = 0.30 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.909, Rules used = {807, 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.
\(\displaystyle \int \frac {1}{x^3 \left (x^8+1\right )} \, dx\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {1}{2} \int \frac {1}{x^4 \left (x^8+1\right )}dx^2\) |
\(\Big \downarrow \) 847 |
\(\displaystyle \frac {1}{2} \left (-\int \frac {x^4}{x^8+1}dx^2-\frac {1}{x^2}\right )\) |
\(\Big \downarrow \) 826 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1-x^4}{x^8+1}dx^2-\frac {1}{2} \int \frac {x^4+1}{x^8+1}dx^2-\frac {1}{x^2}\right )\) |
\(\Big \downarrow \) 1476 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (-\frac {1}{2} \int \frac {1}{x^4-\sqrt {2} x^2+1}dx^2-\frac {1}{2} \int \frac {1}{x^4+\sqrt {2} x^2+1}dx^2\right )+\frac {1}{2} \int \frac {1-x^4}{x^8+1}dx^2-\frac {1}{x^2}\right )\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {\int \frac {1}{-x^4-1}d\left (\sqrt {2} x^2+1\right )}{\sqrt {2}}-\frac {\int \frac {1}{-x^4-1}d\left (1-\sqrt {2} x^2\right )}{\sqrt {2}}\right )+\frac {1}{2} \int \frac {1-x^4}{x^8+1}dx^2-\frac {1}{x^2}\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1-x^4}{x^8+1}dx^2+\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} x^2\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} x^2+1\right )}{\sqrt {2}}\right )-\frac {1}{x^2}\right )\) |
\(\Big \downarrow \) 1479 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (-\frac {\int -\frac {\sqrt {2}-2 x^2}{x^4-\sqrt {2} x^2+1}dx^2}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} x^2+1\right )}{x^4+\sqrt {2} x^2+1}dx^2}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} x^2\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} x^2+1\right )}{\sqrt {2}}\right )-\frac {1}{x^2}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 x^2}{x^4-\sqrt {2} x^2+1}dx^2}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} x^2+1\right )}{x^4+\sqrt {2} x^2+1}dx^2}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} x^2\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} x^2+1\right )}{\sqrt {2}}\right )-\frac {1}{x^2}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 x^2}{x^4-\sqrt {2} x^2+1}dx^2}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} x^2+1}{x^4+\sqrt {2} x^2+1}dx^2\right )+\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} x^2\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} x^2+1\right )}{\sqrt {2}}\right )-\frac {1}{x^2}\right )\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} x^2\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} x^2+1\right )}{\sqrt {2}}\right )-\frac {1}{x^2}+\frac {1}{2} \left (\frac {\log \left (x^4+\sqrt {2} x^2+1\right )}{2 \sqrt {2}}-\frac {\log \left (x^4-\sqrt {2} x^2+1\right )}{2 \sqrt {2}}\right )\right )\) |
(-x^(-2) + (ArcTan[1 - Sqrt[2]*x^2]/Sqrt[2] - ArcTan[1 + Sqrt[2]*x^2]/Sqrt [2])/2 + (-1/2*Log[1 - Sqrt[2]*x^2 + x^4]/Sqrt[2] + Log[1 + Sqrt[2]*x^2 + x^4]/(2*Sqrt[2]))/2)/2
3.15.96.3.1 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[(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_)^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[((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 = 5.32 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.30
method | result | size |
risch | \(-\frac {1}{2 x^{2}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left (-\textit {\_R}^{3}+x^{2}\right )\right )}{8}\) | \(30\) |
default | \(-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+x^{4}-\sqrt {2}\, x^{2}}{1+x^{4}+\sqrt {2}\, x^{2}}\right )+2 \arctan \left (1+\sqrt {2}\, x^{2}\right )+2 \arctan \left (-1+\sqrt {2}\, x^{2}\right )\right )}{16}-\frac {1}{2 x^{2}}\) | \(66\) |
meijerg | \(-\frac {1}{2 x^{2}}-\frac {x^{6} \left (\frac {\sqrt {2}\, \ln \left (1-\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{4}}+\sqrt {x^{8}}\right )}{2 \left (x^{8}\right )^{\frac {3}{4}}}+\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{4}}}{2-\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{4}}}\right )}{\left (x^{8}\right )^{\frac {3}{4}}}-\frac {\sqrt {2}\, \ln \left (1+\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{4}}+\sqrt {x^{8}}\right )}{2 \left (x^{8}\right )^{\frac {3}{4}}}+\frac {\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{4}}}{2+\sqrt {2}\, \left (x^{8}\right )^{\frac {1}{4}}}\right )}{\left (x^{8}\right )^{\frac {3}{4}}}\right )}{8}\) | \(136\) |
Result contains complex when optimal does not.
Time = 0.29 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.87 \[ \int \frac {1}{x^3 \left (1+x^8\right )} \, dx=\frac {-\left (i - 1\right ) \, \sqrt {2} x^{2} \log \left (2 \, x^{2} + \left (i + 1\right ) \, \sqrt {2}\right ) + \left (i + 1\right ) \, \sqrt {2} x^{2} \log \left (2 \, x^{2} - \left (i - 1\right ) \, \sqrt {2}\right ) - \left (i + 1\right ) \, \sqrt {2} x^{2} \log \left (2 \, x^{2} + \left (i - 1\right ) \, \sqrt {2}\right ) + \left (i - 1\right ) \, \sqrt {2} x^{2} \log \left (2 \, x^{2} - \left (i + 1\right ) \, \sqrt {2}\right ) - 8}{16 \, x^{2}} \]
1/16*(-(I - 1)*sqrt(2)*x^2*log(2*x^2 + (I + 1)*sqrt(2)) + (I + 1)*sqrt(2)* x^2*log(2*x^2 - (I - 1)*sqrt(2)) - (I + 1)*sqrt(2)*x^2*log(2*x^2 + (I - 1) *sqrt(2)) + (I - 1)*sqrt(2)*x^2*log(2*x^2 - (I + 1)*sqrt(2)) - 8)/x^2
Time = 0.08 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.87 \[ \int \frac {1}{x^3 \left (1+x^8\right )} \, dx=- \frac {\sqrt {2} \log {\left (x^{4} - \sqrt {2} x^{2} + 1 \right )}}{16} + \frac {\sqrt {2} \log {\left (x^{4} + \sqrt {2} x^{2} + 1 \right )}}{16} - \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} x^{2} - 1 \right )}}{8} - \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} x^{2} + 1 \right )}}{8} - \frac {1}{2 x^{2}} \]
-sqrt(2)*log(x**4 - sqrt(2)*x**2 + 1)/16 + sqrt(2)*log(x**4 + sqrt(2)*x**2 + 1)/16 - sqrt(2)*atan(sqrt(2)*x**2 - 1)/8 - sqrt(2)*atan(sqrt(2)*x**2 + 1)/8 - 1/(2*x**2)
Time = 0.28 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x^3 \left (1+x^8\right )} \, dx=-\frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x^{2} + \sqrt {2}\right )}\right ) - \frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x^{2} - \sqrt {2}\right )}\right ) + \frac {1}{16} \, \sqrt {2} \log \left (x^{4} + \sqrt {2} x^{2} + 1\right ) - \frac {1}{16} \, \sqrt {2} \log \left (x^{4} - \sqrt {2} x^{2} + 1\right ) - \frac {1}{2 \, x^{2}} \]
-1/8*sqrt(2)*arctan(1/2*sqrt(2)*(2*x^2 + sqrt(2))) - 1/8*sqrt(2)*arctan(1/ 2*sqrt(2)*(2*x^2 - sqrt(2))) + 1/16*sqrt(2)*log(x^4 + sqrt(2)*x^2 + 1) - 1 /16*sqrt(2)*log(x^4 - sqrt(2)*x^2 + 1) - 1/2/x^2
Time = 0.29 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.97 \[ \int \frac {1}{x^3 \left (1+x^8\right )} \, dx=-\frac {1}{8} \, \sqrt {2} x^{4} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x^{2} + \sqrt {2}\right )}\right ) - \frac {1}{8} \, \sqrt {2} x^{4} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x^{2} - \sqrt {2}\right )}\right ) - \frac {1}{16} \, \sqrt {2} x^{4} \log \left (x^{4} + \sqrt {2} x^{2} + 1\right ) + \frac {1}{16} \, \sqrt {2} x^{4} \log \left (x^{4} - \sqrt {2} x^{2} + 1\right ) - \frac {1}{2 \, x^{2}} \]
-1/8*sqrt(2)*x^4*arctan(1/2*sqrt(2)*(2*x^2 + sqrt(2))) - 1/8*sqrt(2)*x^4*a rctan(1/2*sqrt(2)*(2*x^2 - sqrt(2))) - 1/16*sqrt(2)*x^4*log(x^4 + sqrt(2)* x^2 + 1) + 1/16*sqrt(2)*x^4*log(x^4 - sqrt(2)*x^2 + 1) - 1/2/x^2
Time = 5.72 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.42 \[ \int \frac {1}{x^3 \left (1+x^8\right )} \, dx=-\frac {1}{2\,x^2}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x^2\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{8}+\frac {1}{8}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x^2\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{8}-\frac {1}{8}{}\mathrm {i}\right ) \]