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

Optimal. Leaf size=17 \[ \sqrt{x^2+1} \tan ^{-1}(x)-\sinh ^{-1}(x) \]

[Out]

-ArcSinh[x] + Sqrt[1 + x^2]*ArcTan[x]

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Rubi [A]  time = 0.0280095, antiderivative size = 17, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {4930, 215} \[ \sqrt{x^2+1} \tan ^{-1}(x)-\sinh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(x*ArcTan[x])/Sqrt[1 + x^2],x]

[Out]

-ArcSinh[x] + Sqrt[1 + x^2]*ArcTan[x]

Rule 4930

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[((d + e*x^2)^
(q + 1)*(a + b*ArcTan[c*x])^p)/(2*e*(q + 1)), x] - Dist[(b*p)/(2*c*(q + 1)), Int[(d + e*x^2)^q*(a + b*ArcTan[c
*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, q}, x] && EqQ[e, c^2*d] && GtQ[p, 0] && NeQ[q, -1]

Rule 215

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

Rubi steps

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

Mathematica [A]  time = 0.0147078, size = 17, normalized size = 1. \[ \sqrt{x^2+1} \tan ^{-1}(x)-\sinh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcSinh[x] + Sqrt[1 + x^2]*ArcTan[x]

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Maple [C]  time = 0.104, size = 54, normalized size = 3.2 \begin{align*} \sqrt{ \left ( x-i \right ) \left ( x+i \right ) }\arctan \left ( x \right ) -\ln \left ({(1+ix){\frac{1}{\sqrt{{x}^{2}+1}}}}+i \right ) +\ln \left ({(1+ix){\frac{1}{\sqrt{{x}^{2}+1}}}}-i \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.42233, size = 20, normalized size = 1.18 \begin{align*} \sqrt{x^{2} + 1} \arctan \left (x\right ) - \operatorname{arsinh}\left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(x^2 + 1)*arctan(x) - arcsinh(x)

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Fricas [A]  time = 2.10871, size = 69, normalized size = 4.06 \begin{align*} \sqrt{x^{2} + 1} \arctan \left (x\right ) + \log \left (-x + \sqrt{x^{2} + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 2.71983, size = 29, normalized size = 1.71 \begin{align*} \frac{x^{2} \operatorname{atan}{\left (x \right )}}{\sqrt{x^{2} + 1}} - \operatorname{asinh}{\left (x \right )} + \frac{\operatorname{atan}{\left (x \right )}}{\sqrt{x^{2} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.10476, size = 31, normalized size = 1.82 \begin{align*} \sqrt{x^{2} + 1} \arctan \left (x\right ) + \log \left (-x + \sqrt{x^{2} + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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