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

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

[Out]

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

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Rubi [A]  time = 0.047321, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {1805, 801, 635, 203, 260} \[ -\frac{2 x+1}{2 \left (x^2+1\right )}-\frac{1}{2} \log \left (x^2+1\right )+\log (x)-2 \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0203084, size = 33, normalized size = 1. \[ \frac{-2 x-1}{2 \left (x^2+1\right )}-\frac{1}{2} \log \left (x^2+1\right )+\log (x)-2 \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 28, normalized size = 0.9 \begin{align*} -{\frac{1}{{x}^{2}+1} \left ( x+{\frac{1}{2}} \right ) }-{\frac{\ln \left ({x}^{2}+1 \right ) }{2}}-2\,\arctan \left ( x \right ) +\ln \left ( x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.40388, size = 39, normalized size = 1.18 \begin{align*} -\frac{2 \, x + 1}{2 \,{\left (x^{2} + 1\right )}} - 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^3+2*x^2-3*x+1)/x/(x^2+1)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.90529, size = 130, normalized size = 3.94 \begin{align*} -\frac{4 \,{\left (x^{2} + 1\right )} \arctan \left (x\right ) +{\left (x^{2} + 1\right )} \log \left (x^{2} + 1\right ) - 2 \,{\left (x^{2} + 1\right )} \log \left (x\right ) + 2 \, x + 1}{2 \,{\left (x^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.131393, size = 27, normalized size = 0.82 \begin{align*} - \frac{2 x + 1}{2 x^{2} + 2} + \log{\left (x \right )} - \frac{\log{\left (x^{2} + 1 \right )}}{2} - 2 \operatorname{atan}{\left (x \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.06265, size = 41, normalized size = 1.24 \begin{align*} -\frac{2 \, x + 1}{2 \,{\left (x^{2} + 1\right )}} - 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^3+2*x^2-3*x+1)/x/(x^2+1)^2,x, algorithm="giac")

[Out]

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