3.1.19 \(\int \tan ^{-1}(x) \, dx\) [19]

Optimal. Leaf size=15 \[ x \tan ^{-1}(x)-\frac {1}{2} \log \left (1+x^2\right ) \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {4930, 266} \begin {gather*} x \tan ^{-1}(x)-\frac {1}{2} \log \left (x^2+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcTan[x],x]

[Out]

x*ArcTan[x] - Log[1 + x^2]/2

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 4930

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTan[c*x^n])^p, x] - Dist[b*c
*n*p, Int[x^n*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0
] && (EqQ[n, 1] || EqQ[p, 1])

Rubi steps

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

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Mathematica [A]
time = 0.00, size = 15, normalized size = 1.00 \begin {gather*} x \tan ^{-1}(x)-\frac {1}{2} \log \left (1+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[x],x]

[Out]

x*ArcTan[x] - Log[1 + x^2]/2

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Mathics [A]
time = 1.80, size = 13, normalized size = 0.87 \begin {gather*} x \text {ArcTan}\left [x\right ]-\frac {\text {Log}\left [1+x^2\right ]}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[ArcTan[x],x]')

[Out]

x ArcTan[x] - Log[1 + x ^ 2] / 2

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Maple [A]
time = 0.03, size = 14, normalized size = 0.93

method	result	size
lookup	\(x \arctan \left (x \right )-\frac {\ln \left (x^{2}+1\right )}{2}\)	\(14\)
default	\(x \arctan \left (x \right )-\frac {\ln \left (x^{2}+1\right )}{2}\)	\(14\)
meijerg	\(\frac {x^{2} \arctan \left (\sqrt {x^{2}}\right )}{\sqrt {x^{2}}}-\frac {\ln \left (x^{2}+1\right )}{2}\)	\(25\)
risch	\(-\frac {i x \ln \left (i x +1\right )}{2}+\frac {i x \ln \left (-i x +1\right )}{2}-\frac {\ln \left (x^{2}+1\right )}{2}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(x),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 13, normalized size = 0.87 \begin {gather*} x \arctan \left (x\right ) - \frac {1}{2} \, \log \left (x^{2} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(x),x, algorithm="maxima")

[Out]

x*arctan(x) - 1/2*log(x^2 + 1)

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Fricas [A]
time = 0.34, size = 13, normalized size = 0.87 \begin {gather*} x \arctan \left (x\right ) - \frac {1}{2} \, \log \left (x^{2} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(x),x, algorithm="fricas")

[Out]

x*arctan(x) - 1/2*log(x^2 + 1)

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Sympy [A]
time = 0.08, size = 12, normalized size = 0.80 \begin {gather*} x \operatorname {atan}{\left (x \right )} - \frac {\log {\left (x^{2} + 1 \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atan(x),x)

[Out]

x*atan(x) - log(x**2 + 1)/2

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Giac [A]
time = 0.00, size = 15, normalized size = 1.00 \begin {gather*} x \arctan x-\frac {\ln \left (x^{2}+1\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(x),x)

[Out]

x*arctan(x) - 1/2*log(x^2 + 1)

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Mupad [B]
time = 0.16, size = 13, normalized size = 0.87 \begin {gather*} x\,\mathrm {atan}\left (x\right )-\frac {\ln \left (x^2+1\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(x),x)

[Out]

x*atan(x) - log(x^2 + 1)/2

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