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]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0170756, antiderivative size = 25, normalized size of antiderivative = 1., 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 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{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*}

Mathematica [A]  time = 0.0058424, size = 25, normalized size = 1. \[ \frac{1}{4} \log \left (x^2+1\right )-\frac{1}{2} \log (x+1)+\frac{1}{2} \tan ^{-1}(x) \]

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

________________________________________________________________________________________

Maple [A]  time = 0.004, size = 20, normalized size = 0.8 \begin{align*}{\frac{\arctan \left ( x \right ) }{2}}-{\frac{\ln \left ( 1+x \right ) }{2}}+{\frac{\ln \left ({x}^{2}+1 \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.65667, size = 26, normalized size = 1.04 \begin{align*} \frac{1}{2} \, \arctan \left (x\right ) + \frac{1}{4} \, \log \left (x^{2} + 1\right ) - \frac{1}{2} \, \log \left (x + 1\right ) \end{align*}

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)

________________________________________________________________________________________

Fricas [A]  time = 1.27835, size = 69, normalized size = 2.76 \begin{align*} \frac{1}{2} \, \arctan \left (x\right ) + \frac{1}{4} \, \log \left (x^{2} + 1\right ) - \frac{1}{2} \, \log \left (x + 1\right ) \end{align*}

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)

________________________________________________________________________________________

Sympy [A]  time = 0.109413, size = 19, normalized size = 0.76 \begin{align*} - \frac{\log{\left (x + 1 \right )}}{2} + \frac{\log{\left (x^{2} + 1 \right )}}{4} + \frac{\operatorname{atan}{\left (x \right )}}{2} \end{align*}

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

________________________________________________________________________________________

Giac [A]  time = 1.11849, size = 27, normalized size = 1.08 \begin{align*} \frac{1}{2} \, \arctan \left (x\right ) + \frac{1}{4} \, \log \left (x^{2} + 1\right ) - \frac{1}{2} \, \log \left ({\left | x + 1 \right |}\right ) \end{align*}

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))