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

Optimal. Leaf size=45 \[ -\frac{x}{2 \left (x^2+1\right )}-\frac{1}{2 x^2}+\frac{1}{2} \log \left (x^2+1\right )-\frac{1}{x}-\log (x)-\frac{3}{2} \tan ^{-1}(x) \]

[Out]

-1/(2*x^2) - x^(-1) - x/(2*(1 + x^2)) - (3*ArcTan[x])/2 - Log[x] + Log[1 + x^2]/2

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Rubi [A]  time = 0.0633532, antiderivative size = 45, 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, 1802, 635, 203, 260} \[ -\frac{x}{2 \left (x^2+1\right )}-\frac{1}{2 x^2}+\frac{1}{2} \log \left (x^2+1\right )-\frac{1}{x}-\log (x)-\frac{3}{2} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(2*x^2) - x^(-1) - x/(2*(1 + x^2)) - (3*ArcTan[x])/2 - 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 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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

Mathematica [A]  time = 0.0157861, size = 39, normalized size = 0.87 \[ \frac{1}{2} \left (-\frac{x}{x^2+1}-\frac{1}{x^2}+\log \left (x^2+1\right )-\frac{2}{x}-2 \log (x)-3 \tan ^{-1}(x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-x^(-2) - 2/x - x/(1 + x^2) - 3*ArcTan[x] - 2*Log[x] + Log[1 + x^2])/2

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Maple [A]  time = 0.051, size = 38, normalized size = 0.8 \begin{align*} -{\frac{1}{2\,{x}^{2}}}-{x}^{-1}-{\frac{x}{2\,{x}^{2}+2}}-{\frac{3\,\arctan \left ( x \right ) }{2}}-\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^3/(x^2+1)^2,x)

[Out]

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

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Maxima [A]  time = 1.47377, size = 55, normalized size = 1.22 \begin{align*} -\frac{3 \, x^{3} + x^{2} + 2 \, x + 1}{2 \,{\left (x^{4} + x^{2}\right )}} - \frac{3}{2} \, \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^3/(x^2+1)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.03687, size = 159, normalized size = 3.53 \begin{align*} -\frac{3 \, x^{3} + x^{2} + 3 \,{\left (x^{4} + x^{2}\right )} \arctan \left (x\right ) -{\left (x^{4} + x^{2}\right )} \log \left (x^{2} + 1\right ) + 2 \,{\left (x^{4} + x^{2}\right )} \log \left (x\right ) + 2 \, x + 1}{2 \,{\left (x^{4} + x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.158863, size = 41, normalized size = 0.91 \begin{align*} - \log{\left (x \right )} + \frac{\log{\left (x^{2} + 1 \right )}}{2} - \frac{3 \operatorname{atan}{\left (x \right )}}{2} - \frac{3 x^{3} + x^{2} + 2 x + 1}{2 x^{4} + 2 x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.15525, size = 58, normalized size = 1.29 \begin{align*} -\frac{3 \, x^{3} + x^{2} + 2 \, x + 1}{2 \,{\left (x^{2} + 1\right )} x^{2}} - \frac{3}{2} \, \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^3/(x^2+1)^2,x, algorithm="giac")

[Out]

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