3.1.29 \(\int \frac {\tan ^{-1}(x)}{x^2 \sqrt {1-x^2}} \, dx\) [29]

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 = 0.05, 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 = {270, 5096, 457, 85, 65, 212} \begin {gather*} -\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 {gather*}

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 65

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 85

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 212

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

Rule 270

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 457

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 5096

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} \text {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} \text {Subst}\left (\int \frac {1}{\sqrt {1-x} x} \, dx,x,x^2\right )-\text {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 \text {Subst}\left (\int \frac {1}{2-x^2} \, dx,x,\sqrt {1-x^2}\right )-\text {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 = 0.04, size = 77, normalized size = 1.35 \begin {gather*} -\frac {\sqrt {1-x^2} \tan ^{-1}(x)}{x}+\log (x)-\frac {\log \left (1+x^2\right )}{\sqrt {2}}+\frac {\log \left (3-x^2+2 \sqrt {2-2 x^2}\right )}{\sqrt {2}}-\log \left (1+\sqrt {1-x^2}\right ) \end {gather*}

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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Mathics [F(-1)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[ArcTan[x]/(x^2*Sqrt[1 - x^2]),x]')

[Out]

Timed out

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Maple [F]
time = 0.12, size = 0, normalized size = 0.00 \[\int \frac {\arctan \left (x \right )}{x^{2} \sqrt {-x^{2}+1}}\, dx\]

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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [A]
time = 0.36, size = 81, normalized size = 1.42 \begin {gather*} \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 \left (x\right )}{2 \, x} \end {gather*}

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

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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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 104 vs. \(2 (48) = 96\).
time = 0.01, size = 130, normalized size = 2.28 \begin {gather*} \frac {\ln \left (-\sqrt {-x^{2}+1}+1\right )}{2}-\frac {\ln \left (\sqrt {-x^{2}+1}+1\right )}{2}-\frac {\ln \left (\frac {-2 \sqrt {-x^{2}+1}+2 \sqrt {2}}{2 \sqrt {-x^{2}+1}+2 \sqrt {2}}\right )}{\sqrt {2}}+\left (-\frac {x}{-2 \sqrt {-x^{2}+1}+2}+\frac {-2 \sqrt {-x^{2}+1}+2}{4 x}\right ) \arctan x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

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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