\(\int \frac {2+x+x^3}{1+2 x^2+x^4} \, dx\) [286]

Optimal result

Integrand size = 19, antiderivative size = 22 \[ \int \frac {2+x+x^3}{1+2 x^2+x^4} \, dx=\frac {x}{1+x^2}+\arctan (x)+\frac {1}{2} \log \left (1+x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {28, 1828, 649, 209, 266} \[ \int \frac {2+x+x^3}{1+2 x^2+x^4} \, dx=\arctan (x)+\frac {x}{x^2+1}+\frac {1}{2} \log \left (x^2+1\right ) \]

[In]

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

[Out]

x/(1 + x^2) + ArcTan[x] + Log[1 + x^2]/2

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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 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 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 1828

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*
g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \int \frac {2+x+x^3}{\left (1+x^2\right )^2} \, dx \\ & = \frac {x}{1+x^2}-\frac {1}{2} \int \frac {-2-2 x}{1+x^2} \, dx \\ & = \frac {x}{1+x^2}+\int \frac {1}{1+x^2} \, dx+\int \frac {x}{1+x^2} \, dx \\ & = \frac {x}{1+x^2}+\tan ^{-1}(x)+\frac {1}{2} \log \left (1+x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {2+x+x^3}{1+2 x^2+x^4} \, dx=\frac {x}{1+x^2}+\arctan (x)+\frac {1}{2} \log \left (1+x^2\right ) \]

[In]

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

[Out]

x/(1 + x^2) + ArcTan[x] + Log[1 + x^2]/2

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95

method	result	size
default	\(\frac {x}{x^{2}+1}+\arctan \left (x \right )+\frac {\ln \left (x^{2}+1\right )}{2}\)	\(21\)
risch	\(\frac {x}{x^{2}+1}+\arctan \left (x \right )+\frac {\ln \left (x^{2}+1\right )}{2}\)	\(21\)
parallelrisch	\(\frac {i \ln \left (x +i\right ) x^{2}-i \ln \left (x -i\right ) x^{2}+\ln \left (x +i\right ) x^{2}+\ln \left (x -i\right ) x^{2}+i \ln \left (x +i\right )-i \ln \left (x -i\right )+\ln \left (x +i\right )+\ln \left (x -i\right )+2 x}{2 x^{2}+2}\)	\(80\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.55 \[ \int \frac {2+x+x^3}{1+2 x^2+x^4} \, dx=\frac {2 \, {\left (x^{2} + 1\right )} \arctan \left (x\right ) + {\left (x^{2} + 1\right )} \log \left (x^{2} + 1\right ) + 2 \, x}{2 \, {\left (x^{2} + 1\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {2+x+x^3}{1+2 x^2+x^4} \, dx=\frac {x}{x^{2} + 1} + \frac {\log {\left (x^{2} + 1 \right )}}{2} + \operatorname {atan}{\left (x \right )} \]

[In]

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

[Out]

x/(x**2 + 1) + log(x**2 + 1)/2 + atan(x)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {2+x+x^3}{1+2 x^2+x^4} \, dx=\frac {x}{x^{2} + 1} + \arctan \left (x\right ) + \frac {1}{2} \, \log \left (x^{2} + 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {2+x+x^3}{1+2 x^2+x^4} \, dx=\frac {x}{x^{2} + 1} + \arctan \left (x\right ) + \frac {1}{2} \, \log \left (x^{2} + 1\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {2+x+x^3}{1+2 x^2+x^4} \, dx=\frac {\ln \left (x^2+1\right )}{2}+\mathrm {atan}\left (x\right )+\frac {x}{x^2+1} \]

[In]

int((x + x^3 + 2)/(2*x^2 + x^4 + 1),x)

[Out]

log(x^2 + 1)/2 + atan(x) + x/(x^2 + 1)