\(\int \frac {1+x^2}{1+x^4} \, dx\) [74]

Optimal result

Integrand size = 13, antiderivative size = 35 \[ \int \frac {1+x^2}{1+x^4} \, dx=-\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}+\frac {\arctan \left (1+\sqrt {2} x\right )}{\sqrt {2}} \]

[Out]

1/2*arctan(-1+x*2^(1/2))*2^(1/2)+1/2*arctan(1+x*2^(1/2))*2^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1176, 631, 210} \[ \int \frac {1+x^2}{1+x^4} \, dx=\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}} \]

[In]

Int[(1 + x^2)/(1 + x^4),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.86 \[ \int \frac {1+x^2}{1+x^4} \, dx=\frac {-\arctan \left (1-\sqrt {2} x\right )+\arctan \left (1+\sqrt {2} x\right )}{\sqrt {2}} \]

[In]

Integrate[(1 + x^2)/(1 + x^4),x]

[Out]

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

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00

method	result	size
risch	\(\frac {\sqrt {2}\, \arctan \left (\frac {x \sqrt {2}}{2}\right )}{2}+\frac {\sqrt {2}\, \arctan \left (\frac {x^{3} \sqrt {2}}{2}+\frac {x \sqrt {2}}{2}\right )}{2}\)	\(35\)
default	\(\frac {\sqrt {2}\, \left (\ln \left (\frac {1+x^{2}+x \sqrt {2}}{1+x^{2}-x \sqrt {2}}\right )+2 \arctan \left (x \sqrt {2}+1\right )+2 \arctan \left (x \sqrt {2}-1\right )\right )}{8}+\frac {\sqrt {2}\, \left (\ln \left (\frac {1+x^{2}-x \sqrt {2}}{1+x^{2}+x \sqrt {2}}\right )+2 \arctan \left (x \sqrt {2}+1\right )+2 \arctan \left (x \sqrt {2}-1\right )\right )}{8}\)	\(104\)
meijerg	\(\frac {x^{3} \sqrt {2}\, \ln \left (1-\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}+\sqrt {x^{4}}\right )}{8 \left (x^{4}\right )^{\frac {3}{4}}}+\frac {x^{3} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}{2-\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}\right )}{4 \left (x^{4}\right )^{\frac {3}{4}}}-\frac {x^{3} \sqrt {2}\, \ln \left (1+\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}+\sqrt {x^{4}}\right )}{8 \left (x^{4}\right )^{\frac {3}{4}}}+\frac {x^{3} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}{2+\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}\right )}{4 \left (x^{4}\right )^{\frac {3}{4}}}-\frac {x \sqrt {2}\, \ln \left (1-\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}+\sqrt {x^{4}}\right )}{8 \left (x^{4}\right )^{\frac {1}{4}}}+\frac {x \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}{2-\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}\right )}{4 \left (x^{4}\right )^{\frac {1}{4}}}+\frac {x \sqrt {2}\, \ln \left (1+\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}+\sqrt {x^{4}}\right )}{8 \left (x^{4}\right )^{\frac {1}{4}}}+\frac {x \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}{2+\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}\right )}{4 \left (x^{4}\right )^{\frac {1}{4}}}\)	\(268\)

[In]

int((x^2+1)/(x^4+1),x,method=_RETURNVERBOSE)

[Out]

1/2*2^(1/2)*arctan(1/2*x*2^(1/2))+1/2*2^(1/2)*arctan(1/2*x^3*2^(1/2)+1/2*x*2^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int \frac {1+x^2}{1+x^4} \, dx=\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x^{3} + x\right )}\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} x\right ) \]

[In]

integrate((x^2+1)/(x^4+1),x, algorithm="fricas")

[Out]

1/2*sqrt(2)*arctan(1/2*sqrt(2)*(x^3 + x)) + 1/2*sqrt(2)*arctan(1/2*sqrt(2)*x)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.11 \[ \int \frac {1+x^2}{1+x^4} \, dx=\frac {\sqrt {2} \cdot \left (2 \operatorname {atan}{\left (\frac {\sqrt {2} x}{2} \right )} + 2 \operatorname {atan}{\left (\frac {\sqrt {2} x^{3}}{2} + \frac {\sqrt {2} x}{2} \right )}\right )}{4} \]

[In]

integrate((x**2+1)/(x**4+1),x)

[Out]

sqrt(2)*(2*atan(sqrt(2)*x/2) + 2*atan(sqrt(2)*x**3/2 + sqrt(2)*x/2))/4

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.11 \[ \int \frac {1+x^2}{1+x^4} \, dx=\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) \]

[In]

integrate((x^2+1)/(x^4+1),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.11 \[ \int \frac {1+x^2}{1+x^4} \, dx=\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) \]

[In]

integrate((x^2+1)/(x^4+1),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.35 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int \frac {1+x^2}{1+x^4} \, dx=\frac {\sqrt {2}\,\left (\mathrm {atan}\left (\frac {\sqrt {2}\,x^3}{2}+\frac {\sqrt {2}\,x}{2}\right )+\mathrm {atan}\left (\frac {\sqrt {2}\,x}{2}\right )\right )}{2} \]

[In]

int((x^2 + 1)/(x^4 + 1),x)

[Out]

(2^(1/2)*(atan((2^(1/2)*x)/2 + (2^(1/2)*x^3)/2) + atan((2^(1/2)*x)/2)))/2