3.1.49 \(\int \sin ^{-1}(\frac {x}{\sqrt {1-x^2}}) \, dx\) [49]

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

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {4924, 455, 65, 209} \begin {gather*} x \text {ArcSin}\left (\frac {x}{\sqrt {1-x^2}}\right )+\text {ArcTan}\left (\sqrt {1-2 x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcSin[x/Sqrt[1 - x^2]],x]

[Out]

x*ArcSin[x/Sqrt[1 - x^2]] + ArcTan[Sqrt[1 - 2*x^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 209

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

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rule 4924

Int[ArcSin[u_], x_Symbol] :> Simp[x*ArcSin[u], x] - Int[SimplifyIntegrand[x*(D[u, x]/Sqrt[1 - u^2]), x], x] /;
InverseFunctionFreeQ[u, x] &&  !FunctionOfExponentialQ[u, x]

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 29, normalized size = 1.00 \begin {gather*} x \sin ^{-1}\left (\frac {x}{\sqrt {1-x^2}}\right )+\tan ^{-1}\left (\sqrt {1-2 x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[x/Sqrt[1 - x^2]],x]

[Out]

x*ArcSin[x/Sqrt[1 - x^2]] + ArcTan[Sqrt[1 - 2*x^2]]

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(137\) vs. \(2(25)=50\).
time = 0.07, size = 138, normalized size = 4.76

method	result	size
default	\(x \arcsin \left (\frac {x}{\sqrt {-x^{2}+1}}\right )+\frac {\sqrt {\frac {2 x^{2}-1}{x^{2}-1}}\, \left (\sqrt {-2 x^{2}+1}+\arctan \left (\frac {2 x -1}{\sqrt {-2 x^{2}+1}}\right )-\arctan \left (\frac {1+2 x}{\sqrt {-2 x^{2}+1}}\right )\right ) \sqrt {-x^{2}+1}}{\sqrt {-2 x^{2}+1}\, \left (2+\sqrt {2}\right ) \left (-2+\sqrt {2}\right )}+\frac {\sqrt {\frac {2 x^{2}-1}{x^{2}-1}}\, \sqrt {-x^{2}+1}}{2}\)	\(138\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsin(x/(-x^2+1)^(1/2)),x,method=_RETURNVERBOSE)

[Out]

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

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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(arcsin(x/(-x^2+1)^(1/2)),x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (25) = 50\).
time = 1.20, size = 60, normalized size = 2.07 \begin {gather*} -x \arcsin \left (\frac {\sqrt {-x^{2} + 1} x}{x^{2} - 1}\right ) + \arctan \left (\frac {x^{2} + \sqrt {-x^{2} + 1} \sqrt {\frac {2 \, x^{2} - 1}{x^{2} - 1}} - 1}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asin(x/sqrt(1 - x**2)), x)

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Giac [A]
time = 0.61, size = 34, normalized size = 1.17 \begin {gather*} x \arcsin \left (\frac {x}{\sqrt {-x^{2} + 1}}\right ) + \frac {\arctan \left (\sqrt {-2 \, x^{2} + 1}\right )}{\mathrm {sgn}\left (x^{2} - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \mathrm {asin}\left (\frac {x}{\sqrt {1-x^2}}\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(x/(1 - x^2)^(1/2)),x)

[Out]

int(asin(x/(1 - x^2)^(1/2)), x)

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