### 3.74 $$\int \frac{1+x+x^2}{x^2 (1+x^2)^2} \, dx$$

Optimal. Leaf size=33 $\frac{1}{2 \left (x^2+1\right )}-\frac{1}{2} \log \left (x^2+1\right )-\frac{1}{x}+\log (x)-\tan ^{-1}(x)$

[Out]

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

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Rubi [A]  time = 0.0447255, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.294, Rules used = {1805, 801, 635, 203, 260} $\frac{1}{2 \left (x^2+1\right )}-\frac{1}{2} \log \left (x^2+1\right )-\frac{1}{x}+\log (x)-\tan ^{-1}(x)$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*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[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

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 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]

Rubi steps

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

Mathematica [A]  time = 0.0161544, size = 33, normalized size = 1. $\frac{1}{2 \left (x^2+1\right )}-\frac{1}{2} \log \left (x^2+1\right )-\frac{1}{x}+\log (x)-\tan ^{-1}(x)$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.053, size = 30, normalized size = 0.9 \begin{align*} -{x}^{-1}+{\frac{1}{2\,{x}^{2}+2}}-\arctan \left ( x \right ) +\ln \left ( x \right ) -{\frac{\ln \left ({x}^{2}+1 \right ) }{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((x^2+x+1)/x^2/(x^2+1)^2,x)

[Out]

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

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Maxima [A]  time = 1.50181, size = 46, normalized size = 1.39 \begin{align*} -\frac{2 \, x^{2} - x + 2}{2 \,{\left (x^{3} + x\right )}} - \arctan \left (x\right ) - \frac{1}{2} \, \log \left (x^{2} + 1\right ) + \log \left (x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.986168, size = 138, normalized size = 4.18 \begin{align*} -\frac{2 \, x^{2} + 2 \,{\left (x^{3} + x\right )} \arctan \left (x\right ) +{\left (x^{3} + x\right )} \log \left (x^{2} + 1\right ) - 2 \,{\left (x^{3} + x\right )} \log \left (x\right ) - x + 2}{2 \,{\left (x^{3} + x\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.137438, size = 31, normalized size = 0.94 \begin{align*} \log{\left (x \right )} - \frac{\log{\left (x^{2} + 1 \right )}}{2} - \operatorname{atan}{\left (x \right )} - \frac{2 x^{2} - x + 2}{2 x^{3} + 2 x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x) - log(x**2 + 1)/2 - atan(x) - (2*x**2 - x + 2)/(2*x**3 + 2*x)

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Giac [A]  time = 1.18142, size = 47, normalized size = 1.42 \begin{align*} -\frac{2 \, x^{2} - x + 2}{2 \,{\left (x^{3} + x\right )}} - \arctan \left (x\right ) - \frac{1}{2} \, \log \left (x^{2} + 1\right ) + \log \left ({\left | x \right |}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*x^2 - x + 2)/(x^3 + x) - arctan(x) - 1/2*log(x^2 + 1) + log(abs(x))