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

Optimal. Leaf size=25 \[ -\frac {1}{10} \log \left (x^2+1\right )+\frac {1}{5} \log (x+2)+\frac {2}{5} \tan ^{-1}(x) \]

[Out]

2/5*arctan(x)+1/5*ln(2+x)-1/10*ln(x^2+1)

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Rubi [A]  time = 0.01, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {706, 31, 635, 203, 260} \[ -\frac {1}{10} \log \left (x^2+1\right )+\frac {1}{5} \log (x+2)+\frac {2}{5} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[1/((2 + x)*(1 + x^2)),x]

[Out]

(2*ArcTan[x])/5 + Log[2 + x]/5 - Log[1 + x^2]/10

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 706

Int[1/(((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 + a*e^2), Int[1/(d + e*x), x],
x] + Dist[1/(c*d^2 + a*e^2), Int[(c*d - c*e*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a
*e^2, 0]

Rubi steps

\begin {align*} \int \frac {1}{(2+x) \left (1+x^2\right )} \, dx &=\frac {1}{5} \int \frac {1}{2+x} \, dx+\frac {1}{5} \int \frac {2-x}{1+x^2} \, dx\\ &=\frac {1}{5} \log (2+x)-\frac {1}{5} \int \frac {x}{1+x^2} \, dx+\frac {2}{5} \int \frac {1}{1+x^2} \, dx\\ &=\frac {2}{5} \tan ^{-1}(x)+\frac {1}{5} \log (2+x)-\frac {1}{10} \log \left (1+x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 25, normalized size = 1.00 \[ -\frac {1}{10} \log \left (x^2+1\right )+\frac {1}{5} \log (x+2)+\frac {2}{5} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Integrate[1/((2 + x)*(1 + x^2)),x]

[Out]

(2*ArcTan[x])/5 + Log[2 + x]/5 - Log[1 + x^2]/10

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fricas [A]  time = 0.69, size = 19, normalized size = 0.76 \[ \frac {2}{5} \, \arctan \relax (x) - \frac {1}{10} \, \log \left (x^{2} + 1\right ) + \frac {1}{5} \, \log \left (x + 2\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*arctan(x) - 1/10*log(x^2 + 1) + 1/5*log(x + 2)

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giac [A]  time = 0.37, size = 20, normalized size = 0.80 \[ \frac {2}{5} \, \arctan \relax (x) - \frac {1}{10} \, \log \left (x^{2} + 1\right ) + \frac {1}{5} \, \log \left ({\left | x + 2 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*arctan(x) - 1/10*log(x^2 + 1) + 1/5*log(abs(x + 2))

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maple [A]  time = 0.00, size = 20, normalized size = 0.80 \[ \frac {2 \arctan \relax (x )}{5}+\frac {\ln \left (x +2\right )}{5}-\frac {\ln \left (x^{2}+1\right )}{10} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*arctan(x)+1/5*ln(x+2)-1/10*ln(x^2+1)

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maxima [A]  time = 0.99, size = 19, normalized size = 0.76 \[ \frac {2}{5} \, \arctan \relax (x) - \frac {1}{10} \, \log \left (x^{2} + 1\right ) + \frac {1}{5} \, \log \left (x + 2\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*arctan(x) - 1/10*log(x^2 + 1) + 1/5*log(x + 2)

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mupad [B]  time = 0.05, size = 25, normalized size = 1.00 \[ \frac {\ln \left (x+2\right )}{5}+\ln \left (x-\mathrm {i}\right )\,\left (-\frac {1}{10}-\frac {1}{5}{}\mathrm {i}\right )+\ln \left (x+1{}\mathrm {i}\right )\,\left (-\frac {1}{10}+\frac {1}{5}{}\mathrm {i}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((x^2 + 1)*(x + 2)),x)

[Out]

log(x + 2)/5 - log(x - 1i)*(1/10 + 1i/5) - log(x + 1i)*(1/10 - 1i/5)

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sympy [A]  time = 0.14, size = 20, normalized size = 0.80 \[ \frac {\log {\left (x + 2 \right )}}{5} - \frac {\log {\left (x^{2} + 1 \right )}}{10} + \frac {2 \operatorname {atan}{\relax (x )}}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x + 2)/5 - log(x**2 + 1)/10 + 2*atan(x)/5

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