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

Optimal. Leaf size=24 \[ -\frac {1}{2 \left (x^2+1\right )}+\frac {1}{2} \log \left (x^2+1\right )+\tan ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {28, 1814, 635, 203, 260} \[ -\frac {1}{2 \left (x^2+1\right )}+\frac {1}{2} \log \left (x^2+1\right )+\tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(2*(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 203

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

Rule 260

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 635

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 1814

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*} \int \frac {1+2 x+x^2+x^3}{1+2 x^2+x^4} \, dx &=\int \frac {1+2 x+x^2+x^3}{\left (1+x^2\right )^2} \, dx\\ &=-\frac {1}{2 \left (1+x^2\right )}-\frac {1}{2} \int \frac {-2-2 x}{1+x^2} \, dx\\ &=-\frac {1}{2 \left (1+x^2\right )}+\int \frac {1}{1+x^2} \, dx+\int \frac {x}{1+x^2} \, dx\\ &=-\frac {1}{2 \left (1+x^2\right )}+\tan ^{-1}(x)+\frac {1}{2} \log \left (1+x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 24, normalized size = 1.00 \[ -\frac {1}{2 \left (x^2+1\right )}+\frac {1}{2} \log \left (x^2+1\right )+\tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.79, size = 32, normalized size = 1.33 \[ \frac {2 \, {\left (x^{2} + 1\right )} \arctan \relax (x) + {\left (x^{2} + 1\right )} \log \left (x^{2} + 1\right ) - 1}{2 \, {\left (x^{2} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+x^2+2*x+1)/(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) - 1)/(x^2 + 1)

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giac [A]  time = 0.40, size = 20, normalized size = 0.83 \[ -\frac {1}{2 \, {\left (x^{2} + 1\right )}} + \arctan \relax (x) + \frac {1}{2} \, \log \left (x^{2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 21, normalized size = 0.88 \[ \arctan \relax (x )+\frac {\ln \left (x^{2}+1\right )}{2}-\frac {1}{2 \left (x^{2}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.18, size = 20, normalized size = 0.83 \[ -\frac {1}{2 \, {\left (x^{2} + 1\right )}} + \arctan \relax (x) + \frac {1}{2} \, \log \left (x^{2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.03, size = 22, normalized size = 0.92 \[ \frac {\ln \left (x^2+1\right )}{2}+\mathrm {atan}\relax (x)-\frac {1}{2\,\left (x^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.12, size = 19, normalized size = 0.79 \[ \frac {\log {\left (x^{2} + 1 \right )}}{2} + \operatorname {atan}{\relax (x )} - \frac {1}{2 x^{2} + 2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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