\(\int \frac {1}{x (1+x^2)^2} \, dx\) [123]

Optimal result

Integrand size = 11, antiderivative size = 24 \[ \int \frac {1}{x \left (1+x^2\right )^2} \, dx=\frac {1}{2 \left (1+x^2\right )}+\log (x)-\frac {1}{2} \log \left (1+x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {272, 46} \[ \int \frac {1}{x \left (1+x^2\right )^2} \, dx=\frac {1}{2 \left (x^2+1\right )}-\frac {1}{2} \log \left (x^2+1\right )+\log (x) \]

[In]

Int[1/(x*(1 + x^2)^2),x]

[Out]

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

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \left (1+x^2\right )^2} \, dx=\frac {1}{2 \left (1+x^2\right )}+\log (x)-\frac {1}{2} \log \left (1+x^2\right ) \]

[In]

Integrate[1/(x*(1 + x^2)^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88

method	result	size
default	\(\frac {1}{2 x^{2}+2}+\ln \left (x \right )-\frac {\ln \left (x^{2}+1\right )}{2}\)	\(21\)
norman	\(\frac {1}{2 x^{2}+2}+\ln \left (x \right )-\frac {\ln \left (x^{2}+1\right )}{2}\)	\(21\)
risch	\(\frac {1}{2 x^{2}+2}+\ln \left (x \right )-\frac {\ln \left (x^{2}+1\right )}{2}\)	\(21\)
meijerg	\(\frac {1}{2}+\ln \left (x \right )-\frac {x^{2}}{2 x^{2}+2}-\frac {\ln \left (x^{2}+1\right )}{2}\)	\(27\)
parallelrisch	\(\frac {2 x^{2} \ln \left (x \right )-\ln \left (x^{2}+1\right ) x^{2}+1+2 \ln \left (x \right )-\ln \left (x^{2}+1\right )}{2 x^{2}+2}\)	\(42\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {1}{x \left (1+x^2\right )^2} \, dx=-\frac {{\left (x^{2} + 1\right )} \log \left (x^{2} + 1\right ) - 2 \, {\left (x^{2} + 1\right )} \log \left (x\right ) - 1}{2 \, {\left (x^{2} + 1\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {1}{x \left (1+x^2\right )^2} \, dx=\log {\left (x \right )} - \frac {\log {\left (x^{2} + 1 \right )}}{2} + \frac {1}{2 x^{2} + 2} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \left (1+x^2\right )^2} \, dx=\frac {1}{2 \, {\left (x^{2} + 1\right )}} - \frac {1}{2} \, \log \left (x^{2} + 1\right ) + \frac {1}{2} \, \log \left (x^{2}\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {1}{x \left (1+x^2\right )^2} \, dx=\frac {x^{2} + 2}{2 \, {\left (x^{2} + 1\right )}} - \frac {1}{2} \, \log \left (x^{2} + 1\right ) + \frac {1}{2} \, \log \left (x^{2}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x \left (1+x^2\right )^2} \, dx=\ln \left (x\right )-\frac {\ln \left (x^2+1\right )}{2}+\frac {1}{2\,\left (x^2+1\right )} \]

[In]

int(1/(x*(x^2 + 1)^2),x)

[Out]

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