\(\int \frac {1}{x (1+x) (1+x^2)} \, dx\) [108]

Optimal result

Integrand size = 16, antiderivative size = 27 \[ \int \frac {1}{x (1+x) \left (1+x^2\right )} \, dx=-\frac {\arctan (x)}{2}+\log (x)-\frac {1}{2} \log (1+x)-\frac {1}{4} \log \left (1+x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {908, 649, 209, 266} \[ \int \frac {1}{x (1+x) \left (1+x^2\right )} \, dx=-\frac {\arctan (x)}{2}-\frac {1}{4} \log \left (x^2+1\right )+\log (x)-\frac {1}{2} \log (x+1) \]

[In]

Int[1/(x*(1 + x)*(1 + x^2)),x]

[Out]

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

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 266

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 649

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 908

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x (1+x) \left (1+x^2\right )} \, dx=-\frac {\arctan (x)}{2}+\log (x)-\frac {1}{2} \log (1+x)-\frac {1}{4} \log \left (1+x^2\right ) \]

[In]

Integrate[1/(x*(1 + x)*(1 + x^2)),x]

[Out]

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

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81

method	result	size
default	\(-\frac {\arctan \left (x \right )}{2}+\ln \left (x \right )-\frac {\ln \left (1+x \right )}{2}-\frac {\ln \left (x^{2}+1\right )}{4}\)	\(22\)
risch	\(-\frac {\arctan \left (x \right )}{2}+\ln \left (x \right )-\frac {\ln \left (1+x \right )}{2}-\frac {\ln \left (x^{2}+1\right )}{4}\)	\(22\)
parallelrisch	\(\ln \left (x \right )-\frac {\ln \left (1+x \right )}{2}-\frac {\ln \left (x -i\right )}{4}+\frac {i \ln \left (x -i\right )}{4}-\frac {\ln \left (x +i\right )}{4}-\frac {i \ln \left (x +i\right )}{4}\)	\(40\)

[In]

int(1/x/(1+x)/(x^2+1),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x (1+x) \left (1+x^2\right )} \, dx=-\frac {1}{2} \, \arctan \left (x\right ) - \frac {1}{4} \, \log \left (x^{2} + 1\right ) - \frac {1}{2} \, \log \left (x + 1\right ) + \log \left (x\right ) \]

[In]

integrate(1/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) + log(x)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {1}{x (1+x) \left (1+x^2\right )} \, dx=\log {\left (x \right )} - \frac {\log {\left (x + 1 \right )}}{2} - \frac {\log {\left (x^{2} + 1 \right )}}{4} - \frac {\operatorname {atan}{\left (x \right )}}{2} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x (1+x) \left (1+x^2\right )} \, dx=-\frac {1}{2} \, \arctan \left (x\right ) - \frac {1}{4} \, \log \left (x^{2} + 1\right ) - \frac {1}{2} \, \log \left (x + 1\right ) + \log \left (x\right ) \]

[In]

integrate(1/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) + log(x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x (1+x) \left (1+x^2\right )} \, dx=-\frac {1}{2} \, \arctan \left (x\right ) - \frac {1}{4} \, \log \left (x^{2} + 1\right ) - \frac {1}{2} \, \log \left ({\left | x + 1 \right |}\right ) + \log \left ({\left | x \right |}\right ) \]

[In]

integrate(1/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)) + log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x (1+x) \left (1+x^2\right )} \, dx=\ln \left (x\right )-\frac {\ln \left (x+1\right )}{2}+\ln \left (x-\mathrm {i}\right )\,\left (-\frac {1}{4}+\frac {1}{4}{}\mathrm {i}\right )+\ln \left (x+1{}\mathrm {i}\right )\,\left (-\frac {1}{4}-\frac {1}{4}{}\mathrm {i}\right ) \]

[In]

int(1/(x*(x^2 + 1)*(x + 1)),x)

[Out]

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