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

Optimal. Leaf size=29 \[ -\frac{\sqrt{x^2+1} \tan ^{-1}(x)}{x}-\tanh ^{-1}\left (\sqrt{x^2+1}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0478196, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {4944, 266, 63, 207} \[ -\frac{\sqrt{x^2+1} \tan ^{-1}(x)}{x}-\tanh ^{-1}\left (\sqrt{x^2+1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[x]/(x^2*Sqrt[1 + x^2]),x]

[Out]

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

Rule 4944

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp
[((f*x)^(m + 1)*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^p)/(d*f*(m + 1)), x] - Dist[(b*c*p)/(f*(m + 1)), Int[(
f*x)^(m + 1)*(d + e*x^2)^q*(a + b*ArcTan[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[e,
c^2*d] && EqQ[m + 2*q + 3, 0] && GtQ[p, 0] && NeQ[m, -1]

Rule 266

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0248081, size = 33, normalized size = 1.14 \[ -\log \left (\sqrt{x^2+1}+1\right )-\frac{\sqrt{x^2+1} \tan ^{-1}(x)}{x}+\log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.099, size = 56, normalized size = 1.9 \begin{align*} -{\frac{\arctan \left ( x \right ) }{x}\sqrt{ \left ( x-i \right ) \left ( x+i \right ) }}-\ln \left ({(1+ix){\frac{1}{\sqrt{{x}^{2}+1}}}}+1 \right ) +\ln \left ({(1+ix){\frac{1}{\sqrt{{x}^{2}+1}}}}-1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.4316, size = 30, normalized size = 1.03 \begin{align*} -\frac{\sqrt{x^{2} + 1} \arctan \left (x\right )}{x} - \operatorname{arsinh}\left (\frac{1}{{\left | x \right |}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.0765, size = 127, normalized size = 4.38 \begin{align*} -\frac{x \log \left (-x + \sqrt{x^{2} + 1} + 1\right ) - x \log \left (-x + \sqrt{x^{2} + 1} - 1\right ) + \sqrt{x^{2} + 1} \arctan \left (x\right )}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 21.4801, size = 19, normalized size = 0.66 \begin{align*} - \operatorname{asinh}{\left (\frac{1}{x} \right )} - \frac{\sqrt{x^{2} + 1} \operatorname{atan}{\left (x \right )}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.08828, size = 73, normalized size = 2.52 \begin{align*} \frac{2 \, \arctan \left (x\right )}{{\left (x - \sqrt{x^{2} + 1}\right )}^{2} - 1} + \arctan \left (x\right ) - \log \left ({\left | -x + \sqrt{x^{2} + 1} + 1 \right |}\right ) + \log \left ({\left | -x + \sqrt{x^{2} + 1} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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