Integrand size = 7, antiderivative size = 85 \[ \int \frac {1}{1+x^4} \, dx=-\frac {\arctan \left (1-\sqrt {2} x\right )}{2 \sqrt {2}}+\frac {\arctan \left (1+\sqrt {2} x\right )}{2 \sqrt {2}}-\frac {\log \left (1-\sqrt {2} x+x^2\right )}{4 \sqrt {2}}+\frac {\log \left (1+\sqrt {2} x+x^2\right )}{4 \sqrt {2}} \]
1/4*arctan(-1+x*2^(1/2))*2^(1/2)+1/4*arctan(1+x*2^(1/2))*2^(1/2)-1/8*ln(1+ x^2-x*2^(1/2))*2^(1/2)+1/8*ln(1+x^2+x*2^(1/2))*2^(1/2)
Time = 0.01 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.75 \[ \int \frac {1}{1+x^4} \, dx=\frac {-2 \arctan \left (1-\sqrt {2} x\right )+2 \arctan \left (1+\sqrt {2} x\right )-\log \left (1-\sqrt {2} x+x^2\right )+\log \left (1+\sqrt {2} x+x^2\right )}{4 \sqrt {2}} \]
(-2*ArcTan[1 - Sqrt[2]*x] + 2*ArcTan[1 + Sqrt[2]*x] - Log[1 - Sqrt[2]*x + x^2] + Log[1 + Sqrt[2]*x + x^2])/(4*Sqrt[2])
Time = 0.24 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.143, Rules used = {755, 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.
\(\displaystyle \int \frac {1}{x^4+1} \, dx\) |
\(\Big \downarrow \) 755 |
\(\displaystyle \frac {1}{2} \int \frac {1-x^2}{x^4+1}dx+\frac {1}{2} \int \frac {x^2+1}{x^4+1}dx\) |
\(\Big \downarrow \) 1476 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x^2-\sqrt {2} x+1}dx+\frac {1}{2} \int \frac {1}{x^2+\sqrt {2} x+1}dx\right )+\frac {1}{2} \int \frac {1-x^2}{x^4+1}dx\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {1}{2} \int \frac {1-x^2}{x^4+1}dx+\frac {1}{2} \left (\frac {\int \frac {1}{-\left (1-\sqrt {2} x\right )^2-1}d\left (1-\sqrt {2} x\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\left (\sqrt {2} x+1\right )^2-1}d\left (\sqrt {2} x+1\right )}{\sqrt {2}}\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \int \frac {1-x^2}{x^4+1}dx+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}\right )\) |
\(\Big \downarrow \) 1479 |
\(\displaystyle \frac {1}{2} \left (-\frac {\int -\frac {\sqrt {2}-2 x}{x^2-\sqrt {2} x+1}dx}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} x+1\right )}{x^2+\sqrt {2} x+1}dx}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 x}{x^2-\sqrt {2} x+1}dx}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} x+1\right )}{x^2+\sqrt {2} x+1}dx}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 x}{x^2-\sqrt {2} x+1}dx}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} x+1}{x^2+\sqrt {2} x+1}dx\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}\right )\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (x^2+\sqrt {2} x+1\right )}{2 \sqrt {2}}-\frac {\log \left (x^2-\sqrt {2} x+1\right )}{2 \sqrt {2}}\right )\) |
(-(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
3.1.18.3.1 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[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] ], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r) Int[(r - s*x^2)/(a + b*x^4) , x], x] + Simp[1/(2*r) 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[((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 = 0.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.26
method | result | size |
risch | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}\right )}{4}\) | \(22\) |
default | \(\frac {\sqrt {2}\, \left (\ln \left (\frac {1+x^{2}+x \sqrt {2}}{1+x^{2}-x \sqrt {2}}\right )+2 \arctan \left (1+x \sqrt {2}\right )+2 \arctan \left (-1+x \sqrt {2}\right )\right )}{8}\) | \(52\) |
meijerg | \(-\frac {x \sqrt {2}\, \ln \left (1-\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}+\sqrt {x^{4}}\right )}{8 \left (x^{4}\right )^{\frac {1}{4}}}+\frac {x \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}{2-\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}\right )}{4 \left (x^{4}\right )^{\frac {1}{4}}}+\frac {x \sqrt {2}\, \ln \left (1+\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}+\sqrt {x^{4}}\right )}{8 \left (x^{4}\right )^{\frac {1}{4}}}+\frac {x \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}{2+\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}\right )}{4 \left (x^{4}\right )^{\frac {1}{4}}}\) | \(131\) |
Result contains complex when optimal does not.
Time = 0.25 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.72 \[ \int \frac {1}{1+x^4} \, dx=\left (\frac {1}{8} i + \frac {1}{8}\right ) \, \sqrt {2} \log \left (2 \, x + \left (i + 1\right ) \, \sqrt {2}\right ) - \left (\frac {1}{8} i - \frac {1}{8}\right ) \, \sqrt {2} \log \left (2 \, x - \left (i - 1\right ) \, \sqrt {2}\right ) + \left (\frac {1}{8} i - \frac {1}{8}\right ) \, \sqrt {2} \log \left (2 \, x + \left (i - 1\right ) \, \sqrt {2}\right ) - \left (\frac {1}{8} i + \frac {1}{8}\right ) \, \sqrt {2} \log \left (2 \, x - \left (i + 1\right ) \, \sqrt {2}\right ) \]
(1/8*I + 1/8)*sqrt(2)*log(2*x + (I + 1)*sqrt(2)) - (1/8*I - 1/8)*sqrt(2)*l og(2*x - (I - 1)*sqrt(2)) + (1/8*I - 1/8)*sqrt(2)*log(2*x + (I - 1)*sqrt(2 )) - (1/8*I + 1/8)*sqrt(2)*log(2*x - (I + 1)*sqrt(2))
Time = 0.09 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.86 \[ \int \frac {1}{1+x^4} \, dx=- \frac {\sqrt {2} \log {\left (x^{2} - \sqrt {2} x + 1 \right )}}{8} + \frac {\sqrt {2} \log {\left (x^{2} + \sqrt {2} x + 1 \right )}}{8} + \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} x - 1 \right )}}{4} + \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} x + 1 \right )}}{4} \]
-sqrt(2)*log(x**2 - sqrt(2)*x + 1)/8 + sqrt(2)*log(x**2 + sqrt(2)*x + 1)/8 + sqrt(2)*atan(sqrt(2)*x - 1)/4 + sqrt(2)*atan(sqrt(2)*x + 1)/4
Time = 0.28 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.85 \[ \int \frac {1}{1+x^4} \, dx=\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) + \frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) + \frac {1}{8} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) - \frac {1}{8} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) \]
1/4*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) + 1/4*sqrt(2)*arctan(1/2*s qrt(2)*(2*x - sqrt(2))) + 1/8*sqrt(2)*log(x^2 + sqrt(2)*x + 1) - 1/8*sqrt( 2)*log(x^2 - sqrt(2)*x + 1)
Time = 0.28 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.85 \[ \int \frac {1}{1+x^4} \, dx=\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) + \frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) + \frac {1}{8} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) - \frac {1}{8} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) \]
1/4*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) + 1/4*sqrt(2)*arctan(1/2*s qrt(2)*(2*x - sqrt(2))) + 1/8*sqrt(2)*log(x^2 + sqrt(2)*x + 1) - 1/8*sqrt( 2)*log(x^2 - sqrt(2)*x + 1)
Time = 0.00 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.39 \[ \int \frac {1}{1+x^4} \, dx=\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{4}+\frac {1}{4}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{4}-\frac {1}{4}{}\mathrm {i}\right ) \]