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

Optimal. Leaf size=25 \[ \frac {1}{4} \log \left (x^2+1\right )-\frac {1}{2} \log (x+1)+\frac {1}{2} \tan ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {801, 635, 203, 260} \[ \frac {1}{4} \log \left (x^2+1\right )-\frac {1}{2} \log (x+1)+\frac {1}{2} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[x/((1 + x)*(1 + x^2)),x]

[Out]

ArcTan[x]/2 - Log[1 + x]/2 + Log[1 + x^2]/4

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 25, normalized size = 1.00 \[ \frac {1}{4} \log \left (x^2+1\right )-\frac {1}{2} \log (x+1)+\frac {1}{2} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Integrate[x/((1 + x)*(1 + x^2)),x]

[Out]

ArcTan[x]/2 - Log[1 + x]/2 + Log[1 + x^2]/4

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fricas [A]  time = 0.80, size = 19, normalized size = 0.76 \[ \frac {1}{2} \, \arctan \relax (x) + \frac {1}{4} \, \log \left (x^{2} + 1\right ) - \frac {1}{2} \, \log \left (x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.25, size = 20, normalized size = 0.80 \[ \frac {1}{2} \, \arctan \relax (x) + \frac {1}{4} \, \log \left (x^{2} + 1\right ) - \frac {1}{2} \, \log \left ({\left | x + 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 20, normalized size = 0.80 \[ \frac {\arctan \relax (x )}{2}-\frac {\ln \left (x +1\right )}{2}+\frac {\ln \left (x^{2}+1\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.43, size = 19, normalized size = 0.76 \[ \frac {1}{2} \, \arctan \relax (x) + \frac {1}{4} \, \log \left (x^{2} + 1\right ) - \frac {1}{2} \, \log \left (x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.21, size = 25, normalized size = 1.00 \[ -\frac {\ln \left (x+1\right )}{2}+\ln \left (x-\mathrm {i}\right )\,\left (\frac {1}{4}-\frac {1}{4}{}\mathrm {i}\right )+\ln \left (x+1{}\mathrm {i}\right )\,\left (\frac {1}{4}+\frac {1}{4}{}\mathrm {i}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/((x^2 + 1)*(x + 1)),x)

[Out]

log(x - 1i)*(1/4 - 1i/4) - log(x + 1)/2 + log(x + 1i)*(1/4 + 1i/4)

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sympy [A]  time = 0.12, size = 19, normalized size = 0.76 \[ - \frac {\log {\left (x + 1 \right )}}{2} + \frac {\log {\left (x^{2} + 1 \right )}}{4} + \frac {\operatorname {atan}{\relax (x )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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