\(\int \frac {1+x+x^2+x^3}{1+x^4} \, dx\) [168]

Optimal result

Integrand size = 17, antiderivative size = 53 \[ \int \frac {1+x+x^2+x^3}{1+x^4} \, dx=\frac {\arctan \left (x^2\right )}{2}-\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}+\frac {\arctan \left (1+\sqrt {2} x\right )}{\sqrt {2}}+\frac {1}{4} \log \left (1+x^4\right ) \]

[Out]

1/2*arctan(x^2)+1/4*ln(x^4+1)+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 = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {1890, 1176, 631, 210, 1262, 649, 209, 266} \[ \int \frac {1+x+x^2+x^3}{1+x^4} \, dx=\frac {\arctan \left (x^2\right )}{2}-\frac {\arctan \left (1-\sqrt {2} x\right )}{\sqrt {2}}+\frac {\arctan \left (\sqrt {2} x+1\right )}{\sqrt {2}}+\frac {1}{4} \log \left (x^4+1\right ) \]

[In]

Int[(1 + x + x^2 + x^3)/(1 + x^4),x]

[Out]

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

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

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 649

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[(-a)*c]

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 1262

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(d + e*x)^q
*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x]

Rule 1890

Int[(Pq_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = Sum[x^ii*((Coeff[Pq, x, ii] + Coeff[Pq, x, n/2 + ii
]*x^(n/2))/(a + b*x^n)), {ii, 0, n/2 - 1}]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ
[n/2, 0] && Expon[Pq, x] < n

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.94 \[ \int \frac {1+x+x^2+x^3}{1+x^4} \, dx=\frac {1}{4} \left (-2 \left (1+\sqrt {2}\right ) \arctan \left (1-\sqrt {2} x\right )+2 \left (-1+\sqrt {2}\right ) \arctan \left (1+\sqrt {2} x\right )+\log \left (1+x^4\right )\right ) \]

[In]

Integrate[(1 + x + x^2 + x^3)/(1 + x^4),x]

[Out]

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

Maple [C] (verified)

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

Time = 1.52 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.58

method	result	size
risch	\(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{\sum }\frac {\left (\textit {\_R}^{3}+\textit {\_R}^{2}+\textit {\_R} +1\right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}\right )}{4}\)	\(31\)
default	\(\frac {\sqrt {2}\, \left (\ln \left (\frac {1+x^{2}+\sqrt {2}\, x}{1+x^{2}-\sqrt {2}\, x}\right )+2 \arctan \left (\sqrt {2}\, x +1\right )+2 \arctan \left (\sqrt {2}\, x -1\right )\right )}{8}+\frac {\arctan \left (x^{2}\right )}{2}+\frac {\sqrt {2}\, \left (\ln \left (\frac {1+x^{2}-\sqrt {2}\, x}{1+x^{2}+\sqrt {2}\, x}\right )+2 \arctan \left (\sqrt {2}\, x +1\right )+2 \arctan \left (\sqrt {2}\, x -1\right )\right )}{8}+\frac {\ln \left (x^{4}+1\right )}{4}\)	\(118\)
meijerg	\(\frac {\ln \left (x^{4}+1\right )}{4}+\frac {x^{3} \sqrt {2}\, \ln \left (1-\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}+\sqrt {x^{4}}\right )}{8 \left (x^{4}\right )^{\frac {3}{4}}}+\frac {x^{3} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}{2-\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}\right )}{4 \left (x^{4}\right )^{\frac {3}{4}}}-\frac {x^{3} \sqrt {2}\, \ln \left (1+\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}+\sqrt {x^{4}}\right )}{8 \left (x^{4}\right )^{\frac {3}{4}}}+\frac {x^{3} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}{2+\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}\right )}{4 \left (x^{4}\right )^{\frac {3}{4}}}+\frac {\arctan \left (x^{2}\right )}{2}-\frac {x \sqrt {2}\, \ln \left (1-\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}+\sqrt {x^{4}}\right )}{8 \left (x^{4}\right )^{\frac {1}{4}}}+\frac {x \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}{2-\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}\right )}{4 \left (x^{4}\right )^{\frac {1}{4}}}+\frac {x \sqrt {2}\, \ln \left (1+\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}+\sqrt {x^{4}}\right )}{8 \left (x^{4}\right )^{\frac {1}{4}}}+\frac {x \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}{2+\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}\right )}{4 \left (x^{4}\right )^{\frac {1}{4}}}\)	\(282\)

[In]

int((x^3+x^2+x+1)/(x^4+1),x,method=_RETURNVERBOSE)

[Out]

1/4*sum((_R^3+_R^2+_R+1)/_R^3*ln(x-_R),_R=RootOf(_Z^4+1))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (41) = 82\).

Time = 0.43 (sec) , antiderivative size = 151, normalized size of antiderivative = 2.85 \[ \int \frac {1+x+x^2+x^3}{1+x^4} \, dx=\frac {1}{4} \, {\left (\sqrt {2 \, \sqrt {2} - 3} + 1\right )} \log \left (\sqrt {2 \, \sqrt {2} - 3} {\left (\sqrt {2} + 2\right )} + 2 \, x + \sqrt {2}\right ) - \frac {1}{4} \, {\left (\sqrt {2 \, \sqrt {2} - 3} - 1\right )} \log \left (-\sqrt {2 \, \sqrt {2} - 3} {\left (\sqrt {2} + 2\right )} + 2 \, x + \sqrt {2}\right ) - \frac {1}{4} \, {\left (\sqrt {-2 \, \sqrt {2} - 3} - 1\right )} \log \left ({\left (\sqrt {2} - 2\right )} \sqrt {-2 \, \sqrt {2} - 3} + 2 \, x - \sqrt {2}\right ) + \frac {1}{4} \, {\left (\sqrt {-2 \, \sqrt {2} - 3} + 1\right )} \log \left (-{\left (\sqrt {2} - 2\right )} \sqrt {-2 \, \sqrt {2} - 3} + 2 \, x - \sqrt {2}\right ) \]

[In]

integrate((x^3+x^2+x+1)/(x^4+1),x, algorithm="fricas")

[Out]

1/4*(sqrt(2*sqrt(2) - 3) + 1)*log(sqrt(2*sqrt(2) - 3)*(sqrt(2) + 2) + 2*x + sqrt(2)) - 1/4*(sqrt(2*sqrt(2) - 3
) - 1)*log(-sqrt(2*sqrt(2) - 3)*(sqrt(2) + 2) + 2*x + sqrt(2)) - 1/4*(sqrt(-2*sqrt(2) - 3) - 1)*log((sqrt(2) -
2)*sqrt(-2*sqrt(2) - 3) + 2*x - sqrt(2)) + 1/4*(sqrt(-2*sqrt(2) - 3) + 1)*log(-(sqrt(2) - 2)*sqrt(-2*sqrt(2)
- 3) + 2*x - sqrt(2))

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.38 \[ \int \frac {1+x+x^2+x^3}{1+x^4} \, dx=\frac {\log {\left (x^{2} - \sqrt {2} x + 1 \right )}}{4} + \frac {\log {\left (x^{2} + \sqrt {2} x + 1 \right )}}{4} + 2 \cdot \left (\frac {1}{4} + \frac {\sqrt {2}}{4}\right ) \operatorname {atan}{\left (\sqrt {2} x - 1 \right )} + 2 \left (- \frac {1}{4} + \frac {\sqrt {2}}{4}\right ) \operatorname {atan}{\left (\sqrt {2} x + 1 \right )} \]

[In]

integrate((x**3+x**2+x+1)/(x**4+1),x)

[Out]

log(x**2 - sqrt(2)*x + 1)/4 + log(x**2 + sqrt(2)*x + 1)/4 + 2*(1/4 + sqrt(2)/4)*atan(sqrt(2)*x - 1) + 2*(-1/4
+ sqrt(2)/4)*atan(sqrt(2)*x + 1)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.43 \[ \int \frac {1+x+x^2+x^3}{1+x^4} \, dx=-\frac {1}{4} \, \sqrt {2} {\left (\sqrt {2} - 2\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) + \frac {1}{4} \, \sqrt {2} {\left (\sqrt {2} + 2\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) + \frac {1}{4} \, \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {1}{4} \, \log \left (x^{2} - \sqrt {2} x + 1\right ) \]

[In]

integrate((x^3+x^2+x+1)/(x^4+1),x, algorithm="maxima")

[Out]

-1/4*sqrt(2)*(sqrt(2) - 2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) + 1/4*sqrt(2)*(sqrt(2) + 2)*arctan(1/2*sqrt(2)*
(2*x - sqrt(2))) + 1/4*log(x^2 + sqrt(2)*x + 1) + 1/4*log(x^2 - sqrt(2)*x + 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.32 \[ \int \frac {1+x+x^2+x^3}{1+x^4} \, dx=\frac {1}{2} \, {\left (\sqrt {2} - 1\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) + \frac {1}{2} \, {\left (\sqrt {2} + 1\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) + \frac {1}{4} \, \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {1}{4} \, \log \left (x^{2} - \sqrt {2} x + 1\right ) \]

[In]

integrate((x^3+x^2+x+1)/(x^4+1),x, algorithm="giac")

[Out]

1/2*(sqrt(2) - 1)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) + 1/2*(sqrt(2) + 1)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)))
+ 1/4*log(x^2 + sqrt(2)*x + 1) + 1/4*log(x^2 - sqrt(2)*x + 1)

Mupad [B] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 156, normalized size of antiderivative = 2.94 \[ \int \frac {1+x+x^2+x^3}{1+x^4} \, dx=\ln \left (\left (16\,x-16\right )\,\left (\frac {\sqrt {-2\,\sqrt {2}-3}}{4}+\frac {1}{4}\right )-8\,x\right )\,\left (\frac {\sqrt {-2\,\sqrt {2}-3}}{4}+\frac {1}{4}\right )-\ln \left (8\,x+\left (16\,x-16\right )\,\left (\frac {\sqrt {-2\,\sqrt {2}-3}}{4}-\frac {1}{4}\right )\right )\,\left (\frac {\sqrt {-2\,\sqrt {2}-3}}{4}-\frac {1}{4}\right )-\ln \left (8\,x+\left (16\,x-16\right )\,\left (\frac {\sqrt {2\,\sqrt {2}-3}}{4}-\frac {1}{4}\right )\right )\,\left (\frac {\sqrt {2\,\sqrt {2}-3}}{4}-\frac {1}{4}\right )+\ln \left (8\,x-\left (16\,x-16\right )\,\left (\frac {\sqrt {2\,\sqrt {2}-3}}{4}+\frac {1}{4}\right )\right )\,\left (\frac {\sqrt {2\,\sqrt {2}-3}}{4}+\frac {1}{4}\right ) \]

[In]

int((x + x^2 + x^3 + 1)/(x^4 + 1),x)

[Out]

log((16*x - 16)*((- 2*2^(1/2) - 3)^(1/2)/4 + 1/4) - 8*x)*((- 2*2^(1/2) - 3)^(1/2)/4 + 1/4) - log(8*x + (16*x -
16)*((- 2*2^(1/2) - 3)^(1/2)/4 - 1/4))*((- 2*2^(1/2) - 3)^(1/2)/4 - 1/4) - log(8*x + (16*x - 16)*((2*2^(1/2)
- 3)^(1/2)/4 - 1/4))*((2*2^(1/2) - 3)^(1/2)/4 - 1/4) + log(8*x - (16*x - 16)*((2*2^(1/2) - 3)^(1/2)/4 + 1/4))*
((2*2^(1/2) - 3)^(1/2)/4 + 1/4)