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

Optimal. Leaf size=57 \[ -\frac {\sqrt {1-x^2} \tan ^{-1}(x)}{x}-\tanh ^{-1}\left (\sqrt {1-x^2}\right )+\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {1-x^2}}{\sqrt {2}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {264, 4976, 446, 83, 63, 206} \[ -\frac {\sqrt {1-x^2} \tan ^{-1}(x)}{x}-\tanh ^{-1}\left (\sqrt {1-x^2}\right )+\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {1-x^2}}{\sqrt {2}}\right ) \]

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]] + Sqrt[2]*ArcTanh[Sqrt[1 - x^2]/Sqrt[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 83

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[(b*e - a*f)/(b*c
- a*d), Int[(e + f*x)^(p - 1)/(a + b*x), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[(e + f*x)^(p - 1)/(c + d*
x), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[0, p, 1]

Rule 206

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

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 4976

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^q, x]}, Dist[a + b*ArcTan[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/(1 + c^
2*x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && ((IGtQ[q, 0] &&  !(ILtQ[(m - 1)/2, 0] && GtQ[m +
2*q + 3, 0])) || (IGtQ[(m + 1)/2, 0] &&  !(ILtQ[q, 0] && GtQ[m + 2*q + 3, 0])) || (ILtQ[(m + 2*q + 1)/2, 0] &&
!ILtQ[(m - 1)/2, 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 {\sqrt {1-x^2}}{x \left (1+x^2\right )} \, dx\\ &=-\frac {\sqrt {1-x^2} \tan ^{-1}(x)}{x}+\frac {1}{2} \operatorname {Subst}\left (\int \frac {\sqrt {1-x}}{x (1+x)} \, dx,x,x^2\right )\\ &=-\frac {\sqrt {1-x^2} \tan ^{-1}(x)}{x}+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x} x} \, dx,x,x^2\right )-\operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x} (1+x)} \, dx,x,x^2\right )\\ &=-\frac {\sqrt {1-x^2} \tan ^{-1}(x)}{x}+2 \operatorname {Subst}\left (\int \frac {1}{2-x^2} \, dx,x,\sqrt {1-x^2}\right )-\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 )+\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {1-x^2}}{\sqrt {2}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 77, normalized size = 1.35 \[ -\frac {\log \left (x^2+1\right )}{\sqrt {2}}+\frac {\log \left (-x^2+2 \sqrt {2-2 x^2}+3\right )}{\sqrt {2}}-\log \left (\sqrt {1-x^2}+1\right )-\frac {\sqrt {1-x^2} \tan ^{-1}(x)}{x}+\log (x) \]

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 + x^2]/Sqrt[2] + Log[3 - x^2 + 2*Sqrt[2 - 2*x^2]]/Sqrt[2] - Lo
g[1 + Sqrt[1 - x^2]]

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fricas [A]  time = 0.47, size = 81, normalized size = 1.42 \[ \frac {\sqrt {2} x \log \left (\frac {x^{2} - 2 \, \sqrt {2} \sqrt {-x^{2} + 1} - 3}{x^{2} + 1}\right ) - x \log \left (\sqrt {-x^{2} + 1} + 1\right ) + x \log \left (\sqrt {-x^{2} + 1} - 1\right ) - 2 \, \sqrt {-x^{2} + 1} \arctan \relax (x)}{2 \, x} \]

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]

1/2*(sqrt(2)*x*log((x^2 - 2*sqrt(2)*sqrt(-x^2 + 1) - 3)/(x^2 + 1)) - x*log(sqrt(-x^2 + 1) + 1) + x*log(sqrt(-x
^2 + 1) - 1) - 2*sqrt(-x^2 + 1)*arctan(x))/x

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giac [B]  time = 1.09, size = 104, normalized size = 1.82 \[ \frac {1}{2} \, {\left (\frac {x}{\sqrt {-x^{2} + 1} - 1} - \frac {\sqrt {-x^{2} + 1} - 1}{x}\right )} \arctan \relax (x) - \frac {1}{2} \, \sqrt {2} \log \left (\frac {\sqrt {2} - \sqrt {-x^{2} + 1}}{\sqrt {2} + \sqrt {-x^{2} + 1}}\right ) - \frac {1}{2} \, \log \left (\sqrt {-x^{2} + 1} + 1\right ) + \frac {1}{2} \, \log \left (-\sqrt {-x^{2} + 1} + 1\right ) \]

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]

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

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maple [F]  time = 0.34, size = 0, normalized size = 0.00 \[ \int \frac {\arctan \relax (x )}{\sqrt {-x^{2}+1}\, x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\arctan \relax (x)}{\sqrt {-x^{2} + 1} x^{2}}\,{d x} \]

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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {atan}\relax (x)}{x^2\,\sqrt {1-x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(x)/(x^2*(1 - x^2)^(1/2)),x)

[Out]

int(atan(x)/(x^2*(1 - x^2)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atan}{\relax (x )}}{x^{2} \sqrt {- \left (x - 1\right ) \left (x + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(atan(x)/(x**2*sqrt(-(x - 1)*(x + 1))), x)

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