\(\int \frac {1}{(1+x^2) \cot ^{-1}(x)} \, dx\) [51]

Optimal result

Integrand size = 12, antiderivative size = 5 \[ \int \frac {1}{\left (1+x^2\right ) \cot ^{-1}(x)} \, dx=-\log \left (\cot ^{-1}(x)\right ) \]

[Out]

-ln(arccot(x))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 5, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {5003} \[ \int \frac {1}{\left (1+x^2\right ) \cot ^{-1}(x)} \, dx=-\log \left (\cot ^{-1}(x)\right ) \]

[In]

Int[1/((1 + x^2)*ArcCot[x]),x]

[Out]

-Log[ArcCot[x]]

Rule 5003

Int[1/(((a_.) + ArcCot[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Simp[-Log[RemoveContent[a + b*A
rcCot[c*x], x]]/(b*c*d), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d]

Rubi steps \begin{align*} \text {integral}& = -\log \left (\cot ^{-1}(x)\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 5, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (1+x^2\right ) \cot ^{-1}(x)} \, dx=-\log \left (\cot ^{-1}(x)\right ) \]

[In]

Integrate[1/((1 + x^2)*ArcCot[x]),x]

[Out]

-Log[ArcCot[x]]

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 6, normalized size of antiderivative = 1.20

method	result	size
derivativedivides	\(-\ln \left (\operatorname {arccot}\left (x \right )\right )\)	\(6\)
default	\(-\ln \left (\operatorname {arccot}\left (x \right )\right )\)	\(6\)
parallelrisch	\(-\ln \left (\operatorname {arccot}\left (x \right )\right )\)	\(6\)
risch	\(-\ln \left (\ln \left (i x +1\right )+i \left (i \ln \left (-i x +1\right )-\pi \right )\right )\)	\(29\)

[In]

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

[Out]

-ln(arccot(x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 5, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (1+x^2\right ) \cot ^{-1}(x)} \, dx=-\log \left (\operatorname {arccot}\left (x\right )\right ) \]

[In]

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

[Out]

-log(arccot(x))

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 5, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (1+x^2\right ) \cot ^{-1}(x)} \, dx=- \log {\left (\operatorname {acot}{\left (x \right )} \right )} \]

[In]

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

[Out]

-log(acot(x))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 5, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (1+x^2\right ) \cot ^{-1}(x)} \, dx=-\log \left (\operatorname {arccot}\left (x\right )\right ) \]

[In]

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

[Out]

-log(arccot(x))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.60 \[ \int \frac {1}{\left (1+x^2\right ) \cot ^{-1}(x)} \, dx=-\log \left ({\left | \arctan \left (\frac {1}{x}\right ) \right |}\right ) \]

[In]

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

[Out]

-log(abs(arctan(1/x)))

Mupad [B] (verification not implemented)

Time = 0.71 (sec) , antiderivative size = 5, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (1+x^2\right ) \cot ^{-1}(x)} \, dx=-\ln \left (\mathrm {acot}\left (x\right )\right ) \]

[In]

int(1/(acot(x)*(x^2 + 1)),x)

[Out]

-log(acot(x))