3.1.23 \(\int \frac {x \tan ^{-1}(x) \log (x+\sqrt {1+x^2})}{\sqrt {1+x^2}} \, dx\) [23]

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

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {5050, 221, 267, 2634, 8, 5320, 6818, 4930, 266} \begin {gather*} \sqrt {x^2+1} \text {ArcTan}(x) \log \left (\sqrt {x^2+1}+x\right )-x \text {ArcTan}(x)-\frac {1}{2} \log ^2\left (\sqrt {x^2+1}+x\right )+\frac {1}{2} \log \left (x^2+1\right ) \end {gather*}

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 8

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

Rule 221

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 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 267

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 2634

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 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])

Rule 5050

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 5320

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 6818

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]

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*}

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Mathematica [A]
time = 0.03, size = 58, normalized size = 1.00 \begin {gather*} -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 {gather*}

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 = 0.05, size = 0, normalized size = 0.00 \[\int \frac {x \arctan \left (x \right ) \ln \left (x +\sqrt {x^{2}+1}\right )}{\sqrt {x^{2}+1}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.47, size = 48, normalized size = 0.83 \begin {gather*} \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 {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x\,\mathrm {atan}\left (x\right )\,\ln \left (x+\sqrt {x^2+1}\right )}{\sqrt {x^2+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*atan(x)*log(x + (x^2 + 1)^(1/2)))/(x^2 + 1)^(1/2),x)

[Out]

int((x*atan(x)*log(x + (x^2 + 1)^(1/2)))/(x^2 + 1)^(1/2), x)

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