3.700 \(\int \frac{x \tan ^{-1}(\sqrt{1+x^2})}{\sqrt{1+x^2}} \, dx\)

Optimal. Leaf size=31 \[ \sqrt{x^2+1} \tan ^{-1}\left (\sqrt{x^2+1}\right )-\frac{1}{2} \log \left (x^2+2\right ) \]

[Out]

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

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Rubi [A]  time = 0.0380929, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {261, 5207, 260} \[ \sqrt{x^2+1} \tan ^{-1}\left (\sqrt{x^2+1}\right )-\frac{1}{2} \log \left (x^2+2\right ) \]

Antiderivative was successfully verified.

[In]

Int[(x*ArcTan[Sqrt[1 + x^2]])/Sqrt[1 + x^2],x]

[Out]

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

Rule 261

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

Rule 5207

Int[((a_.) + ArcTan[u_]*(b_.))*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[a + b*ArcTan[u], w, x] - Dist
[b, Int[SimplifyIntegrand[(w*D[u, x])/(1 + u^2), x], x], x] /; InverseFunctionFreeQ[w, x]] /; FreeQ[{a, b}, x]
&& InverseFunctionFreeQ[u, x] &&  !MatchQ[v, ((c_.) + (d_.)*x)^(m_.) /; FreeQ[{c, d, m}, x]] && FalseQ[Functi
onOfLinear[v*(a + b*ArcTan[u]), x]]

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

Mathematica [A]  time = 0.0272257, size = 31, normalized size = 1. \[ \sqrt{x^2+1} \tan ^{-1}\left (\sqrt{x^2+1}\right )-\frac{1}{2} \log \left (x^2+2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(x*ArcTan[Sqrt[1 + x^2]])/Sqrt[1 + x^2],x]

[Out]

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

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Maple [A]  time = 0.007, size = 26, normalized size = 0.8 \begin{align*} -{\frac{\ln \left ({x}^{2}+2 \right ) }{2}}+\arctan \left ( \sqrt{{x}^{2}+1} \right ) \sqrt{{x}^{2}+1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.942101, size = 34, normalized size = 1.1 \begin{align*} \sqrt{x^{2} + 1} \arctan \left (\sqrt{x^{2} + 1}\right ) - \frac{1}{2} \, \log \left (x^{2} + 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.52711, size = 76, normalized size = 2.45 \begin{align*} \sqrt{x^{2} + 1} \arctan \left (\sqrt{x^{2} + 1}\right ) - \frac{1}{2} \, \log \left (x^{2} + 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 3.32846, size = 26, normalized size = 0.84 \begin{align*} \sqrt{x^{2} + 1} \operatorname{atan}{\left (\sqrt{x^{2} + 1} \right )} - \frac{\log{\left (x^{2} + 2 \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.07154, size = 34, normalized size = 1.1 \begin{align*} \sqrt{x^{2} + 1} \arctan \left (\sqrt{x^{2} + 1}\right ) - \frac{1}{2} \, \log \left (x^{2} + 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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