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]

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

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Rubi [A]  time = 0.05, antiderivative size = 29, normalized size of antiderivative = 1.00, 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 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])

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 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]

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*}

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Mathematica [A]  time = 0.03, 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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fricas [A]  time = 0.44, size = 47, normalized size = 1.62 \[ -\frac {x \log \left (-x + \sqrt {x^{2} + 1} + 1\right ) - x \log \left (-x + \sqrt {x^{2} + 1} - 1\right ) + \sqrt {x^{2} + 1} \arctan \relax (x)}{x} \]

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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giac [B]  time = 1.10, size = 54, normalized size = 1.86 \[ \frac {2 \, \arctan \relax (x)}{{\left (x - \sqrt {x^{2} + 1}\right )}^{2} - 1} + \arctan \relax (x) - \log \left ({\left | -x + \sqrt {x^{2} + 1} + 1 \right |}\right ) + \log \left ({\left | -x + \sqrt {x^{2} + 1} - 1 \right |}\right ) \]

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))

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maple [C]  time = 0.17, size = 56, normalized size = 1.93 \[ -\ln \left (1+\frac {i x +1}{\sqrt {x^{2}+1}}\right )+\ln \left (\frac {i x +1}{\sqrt {x^{2}+1}}-1\right )-\frac {\sqrt {\left (x -i\right ) \left (x +i\right )}\, \arctan \relax (x )}{x} \]

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

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maxima [A]  time = 0.95, size = 22, normalized size = 0.76 \[ -\frac {\sqrt {x^{2} + 1} \arctan \relax (x)}{x} - \operatorname {arsinh}\left (\frac {1}{{\left | x \right |}}\right ) \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {\mathrm {atan}\relax (x)}{x^2\,\sqrt {x^2+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(x)/(x^2*(x^2 + 1)^(1/2)),x)

[Out]

int(atan(x)/(x^2*(x^2 + 1)^(1/2)), x)

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sympy [A]  time = 7.26, size = 19, normalized size = 0.66 \[ - \operatorname {asinh}{\left (\frac {1}{x} \right )} - \frac {\sqrt {x^{2} + 1} \operatorname {atan}{\relax (x )}}{x} \]

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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