3.86 \(\int \frac {\sin ^{-1}(x)}{x^2} \, dx\)

Optimal. Leaf size=22 \[ -\tanh ^{-1}\left (\sqrt {1-x^2}\right )-\frac {\sin ^{-1}(x)}{x} \]

[Out]

-arcsin(x)/x-arctanh((-x^2+1)^(1/2))

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Rubi [A]  time = 0.02, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {4627, 266, 63, 206} \[ -\tanh ^{-1}\left (\sqrt {1-x^2}\right )-\frac {\sin ^{-1}(x)}{x} \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[x]/x^2,x]

[Out]

-(ArcSin[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 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 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 4627

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcSi
n[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSin[c*x])^(n - 1))/Sqrt[1
- c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 22, normalized size = 1.00 \[ -\tanh ^{-1}\left (\sqrt {1-x^2}\right )-\frac {\sin ^{-1}(x)}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[x]/x^2,x]

[Out]

-(ArcSin[x]/x) - ArcTanh[Sqrt[1 - x^2]]

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fricas [A]  time = 0.45, size = 39, normalized size = 1.77 \[ -\frac {x \log \left (\sqrt {-x^{2} + 1} + 1\right ) - x \log \left (\sqrt {-x^{2} + 1} - 1\right ) + 2 \, \arcsin \relax (x)}{2 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(x)/x^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.02, size = 38, normalized size = 1.73 \[ -\frac {\arcsin \relax (x)}{x} - \frac {1}{2} \, \log \left (\sqrt {-x^{2} + 1} + 1\right ) + \frac {1}{2} \, \log \left (-\sqrt {-x^{2} + 1} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(x)/x^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.00, size = 21, normalized size = 0.95 \[ -\arctanh \left (\frac {1}{\sqrt {-x^{2}+1}}\right )-\frac {\arcsin \relax (x )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsin(x)/x^2,x)

[Out]

-arcsin(x)/x-arctanh(1/(-x^2+1)^(1/2))

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maxima [A]  time = 1.23, size = 33, normalized size = 1.50 \[ -\frac {\arcsin \relax (x)}{x} - \log \left (\frac {2 \, \sqrt {-x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(x)/x^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.02, size = 20, normalized size = 0.91 \[ -\mathrm {atanh}\left (\frac {1}{\sqrt {1-x^2}}\right )-\frac {\mathrm {asin}\relax (x)}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(x)/x^2,x)

[Out]

- atanh(1/(1 - x^2)^(1/2)) - asin(x)/x

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sympy [A]  time = 2.02, size = 22, normalized size = 1.00 \[ \begin {cases} - \operatorname {acosh}{\left (\frac {1}{x} \right )} & \text {for}\: \frac {1}{\left |{x^{2}}\right |} > 1 \\i \operatorname {asin}{\left (\frac {1}{x} \right )} & \text {otherwise} \end {cases} - \frac {\operatorname {asin}{\relax (x )}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asin(x)/x**2,x)

[Out]

Piecewise((-acosh(1/x), 1/Abs(x**2) > 1), (I*asin(1/x), True)) - asin(x)/x

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