\(\int \frac {1}{1+x^4} \, dx\) [90]

Optimal result

Integrand size = 7, antiderivative size = 65 \[ \int \frac {1}{1+x^4} \, dx=\frac {\arctan \left (\frac {\sqrt {2} x}{1-x^2}\right )}{2 \sqrt {2}}+\frac {\log \left (\frac {1+\sqrt {2} x+x^2}{1-\sqrt {2} x+x^2}\right )}{4 \sqrt {2}} \]

[Out]

1/8*2^(1/2)*ln((1+x*2^(1/2)+x^2)/(1-x*2^(1/2)+x^2))+1/4*2^(1/2)*arctan(x*2^(1/2)/(-x^2+1))

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.31, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {217, 1179, 642, 1176, 631, 210} \[ \int \frac {1}{1+x^4} \, dx=-\frac {\arctan \left (1-\sqrt {2} x\right )}{2 \sqrt {2}}+\frac {\arctan \left (\sqrt {2} x+1\right )}{2 \sqrt {2}}-\frac {\log \left (x^2-\sqrt {2} x+1\right )}{4 \sqrt {2}}+\frac {\log \left (x^2+\sqrt {2} x+1\right )}{4 \sqrt {2}} \]

[In]

Int[(1 + x^4)^(-1),x]

[Out]

-1/2*ArcTan[1 - Sqrt[2]*x]/Sqrt[2] + ArcTan[1 + Sqrt[2]*x]/(2*Sqrt[2]) - Log[1 - Sqrt[2]*x + x^2]/(4*Sqrt[2])
+ Log[1 + Sqrt[2]*x + x^2]/(4*Sqrt[2])

Rule 210

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])

Rule 217

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[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
]]))

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[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])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[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]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[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]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {1-x^2}{1+x^4} \, dx+\frac {1}{2} \int \frac {1+x^2}{1+x^4} \, dx \\ & = \frac {1}{4} \int \frac {1}{1-\sqrt {2} x+x^2} \, dx+\frac {1}{4} \int \frac {1}{1+\sqrt {2} x+x^2} \, dx-\frac {\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx}{4 \sqrt {2}}-\frac {\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx}{4 \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}}+\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} x\right )}{2 \sqrt {2}}-\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} x\right )}{2 \sqrt {2}} \\ & = -\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}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.98 \[ \int \frac {1}{1+x^4} \, dx=\frac {-2 \arctan \left (1-\sqrt {2} x\right )+2 \arctan \left (1+\sqrt {2} x\right )-\log \left (1-\sqrt {2} x+x^2\right )+\log \left (1+\sqrt {2} x+x^2\right )}{4 \sqrt {2}} \]

[In]

Integrate[(1 + x^4)^(-1),x]

[Out]

(-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])

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.34

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	\(\frac {\sqrt {2}\, \left (\ln \left (\frac {1+x \sqrt {2}+x^{2}}{1-x \sqrt {2}+x^{2}}\right )+2 \arctan \left (x \sqrt {2}+1\right )+2 \arctan \left (x \sqrt {2}-1\right )\right )}{8}\)	\(52\)
meijerg	\(-\frac {x \sqrt {2}\, \ln \left (1-\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}+\sqrt {x^{4}}\right )}{8 \left (x^{4}\right )^{\frac {1}{4}}}+\frac {x \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}{2-\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}\right )}{4 \left (x^{4}\right )^{\frac {1}{4}}}+\frac {x \sqrt {2}\, \ln \left (1+\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}+\sqrt {x^{4}}\right )}{8 \left (x^{4}\right )^{\frac {1}{4}}}+\frac {x \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}{2+\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}\right )}{4 \left (x^{4}\right )^{\frac {1}{4}}}\)	\(131\)

[In]

int(1/(x^4+1),x,method=_RETURNVERBOSE)

[Out]

1/4*sum(1/_R^3*ln(x-_R),_R=RootOf(_Z^4+1))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.94 \[ \int \frac {1}{1+x^4} \, dx=\left (\frac {1}{8} i + \frac {1}{8}\right ) \, \sqrt {2} \log \left (2 \, x + \left (i + 1\right ) \, \sqrt {2}\right ) - \left (\frac {1}{8} i - \frac {1}{8}\right ) \, \sqrt {2} \log \left (2 \, x - \left (i - 1\right ) \, \sqrt {2}\right ) + \left (\frac {1}{8} i - \frac {1}{8}\right ) \, \sqrt {2} \log \left (2 \, x + \left (i - 1\right ) \, \sqrt {2}\right ) - \left (\frac {1}{8} i + \frac {1}{8}\right ) \, \sqrt {2} \log \left (2 \, x - \left (i + 1\right ) \, \sqrt {2}\right ) \]

[In]

integrate(1/(x^4+1),x, algorithm="fricas")

[Out]

(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)) + (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))

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.12 \[ \int \frac {1}{1+x^4} \, dx=- \frac {\sqrt {2} \log {\left (x^{2} - \sqrt {2} x + 1 \right )}}{8} + \frac {\sqrt {2} \log {\left (x^{2} + \sqrt {2} x + 1 \right )}}{8} + \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} x - 1 \right )}}{4} + \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} x + 1 \right )}}{4} \]

[In]

integrate(1/(x**4+1),x)

[Out]

-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 + s
qrt(2)*atan(sqrt(2)*x + 1)/4

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.11 \[ \int \frac {1}{1+x^4} \, dx=\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) + \frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) + \frac {1}{8} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) - \frac {1}{8} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) \]

[In]

integrate(1/(x^4+1),x, algorithm="maxima")

[Out]

1/4*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) + 1/4*sqrt(2)*arctan(1/2*sqrt(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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.11 \[ \int \frac {1}{1+x^4} \, dx=\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) + \frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) + \frac {1}{8} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) - \frac {1}{8} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) \]

[In]

integrate(1/(x^4+1),x, algorithm="giac")

[Out]

1/4*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) + 1/4*sqrt(2)*arctan(1/2*sqrt(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)

Mupad [B] (verification not implemented)

Time = 15.20 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.51 \[ \int \frac {1}{1+x^4} \, dx=\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{4}+\frac {1}{4}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{4}-\frac {1}{4}{}\mathrm {i}\right ) \]

[In]

int(1/(x^4 + 1),x)

[Out]

2^(1/2)*atan(2^(1/2)*x*(1/2 - 1i/2))*(1/4 + 1i/4) + 2^(1/2)*atan(2^(1/2)*x*(1/2 + 1i/2))*(1/4 - 1i/4)