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

Optimal. Leaf size=58 \[ -\frac{1}{2} \log ^2\left (\sqrt{x^2+1}+x\right )+\frac{1}{2} \log \left (x^2+1\right )+\sqrt{x^2+1} \log \left (\sqrt{x^2+1}+x\right ) \tan ^{-1}(x)-x \tan ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.141765, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36, Rules used = {4930, 215, 261, 2554, 8, 5212, 6686, 4846, 260} \[ -\frac{1}{2} \log ^2\left (\sqrt{x^2+1}+x\right )+\frac{1}{2} \log \left (x^2+1\right )+\sqrt{x^2+1} \log \left (\sqrt{x^2+1}+x\right ) \tan ^{-1}(x)-x \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 5212

Int[ArcTan[v_]*Log[w_]*(u_), x_Symbol] :> With[{z = IntHide[u, x]}, Dist[ArcTan[v]*Log[w], z, x] + (-Int[Simpl
ifyIntegrand[(z*Log[w]*D[v, x])/(1 + v^2), x], x] - Int[SimplifyIntegrand[(z*ArcTan[v]*D[w, x])/w, x], x]) /;
InverseFunctionFreeQ[z, x]] /; InverseFunctionFreeQ[v, x] && InverseFunctionFreeQ[w, x]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 4846

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

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

Mathematica [A]  time = 0.0359266, size = 58, normalized size = 1. \[ -\frac{1}{2} \log ^2\left (\sqrt{x^2+1}+x\right )+\frac{1}{2} \log \left (x^2+1\right )+\sqrt{x^2+1} \log \left (\sqrt{x^2+1}+x\right ) \tan ^{-1}(x)-x \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.09, size = 0, normalized size = 0. \begin{align*} \int{x\arctan \left ( x \right ) \ln \left ( x+\sqrt{{x}^{2}+1} \right ){\frac{1}{\sqrt{{x}^{2}+1}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \arctan \left (x\right ) \log \left (x + \sqrt{x^{2} + 1}\right )}{\sqrt{x^{2} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.14889, size = 151, normalized size = 2.6 \begin{align*} \sqrt{x^{2} + 1} \arctan \left (x\right ) \log \left (x + \sqrt{x^{2} + 1}\right ) - x \arctan \left (x\right ) - \frac{1}{2} \, \log \left (x + \sqrt{x^{2} + 1}\right )^{2} + \frac{1}{2} \, \log \left (x^{2} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \arctan \left (x\right ) \log \left (x + \sqrt{x^{2} + 1}\right )}{\sqrt{x^{2} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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