3.1.14 \(\int \frac {\sin ^{-1}(x)}{1+\sqrt {1-x^2}} \, dx\) [14]

Optimal. Leaf size=45 \[ -\frac {x \sin ^{-1}(x)}{1+\sqrt {1-x^2}}+\frac {1}{2} \sin ^{-1}(x)^2-\log \left (1+\sqrt {1-x^2}\right ) \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 51, normalized size of antiderivative = 1.13, number of steps used = 9, number of rules used = 11, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.611, Rules used = {6874, 283, 222, 4875, 4723, 272, 65, 212, 4781, 29, 4737} \begin {gather*} \frac {\sqrt {1-x^2} \text {ArcSin}(x)}{x}+\frac {\text {ArcSin}(x)^2}{2}-\frac {\text {ArcSin}(x)}{x}-\tanh ^{-1}\left (\sqrt {1-x^2}\right )-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcSin[x]/(1 + Sqrt[1 - x^2]),x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

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 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 222

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

Rule 272

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 283

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

Rule 4723

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]

Rule 4737

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(1/(b*c*(n + 1)))*Si
mp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSin[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c
^2*d + e, 0] && NeQ[n, -1]

Rule 4781

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(f
*x)^(m + 1)*Sqrt[d + e*x^2]*((a + b*ArcSin[c*x])^n/(f*(m + 1))), x] + (-Dist[b*c*(n/(f*(m + 1)))*Simp[Sqrt[d +
e*x^2]/Sqrt[1 - c^2*x^2]], Int[(f*x)^(m + 1)*(a + b*ArcSin[c*x])^(n - 1), x], x] + Dist[(c^2/(f^2*(m + 1)))*S
imp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2]], Int[(f*x)^(m + 2)*((a + b*ArcSin[c*x])^n/Sqrt[1 - c^2*x^2]), x], x]) /
; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[m, -1]

Rule 4875

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(Px_.)*((f_) + (g_.)*((d_) + (e_.)*(x_)^2)^(p_))^(m_.), x_Symbol]
:> With[{u = ExpandIntegrand[Px*(f + g*(d + e*x^2)^p)^m*(a + b*ArcSin[c*x])^n, x]}, Int[u, x] /; SumQ[u]] /;
FreeQ[{a, b, c, d, e, f, g}, x] && PolynomialQ[Px, x] && EqQ[c^2*d + e, 0] && IGtQ[p + 1/2, 0] && IntegersQ[m,
n]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 44, normalized size = 0.98 \begin {gather*} \frac {\left (-1+\sqrt {1-x^2}\right ) \sin ^{-1}(x)}{x}+\frac {1}{2} \sin ^{-1}(x)^2-\log \left (1+\sqrt {1-x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[x]/(1 + Sqrt[1 - x^2]),x]

[Out]

((-1 + Sqrt[1 - x^2])*ArcSin[x])/x + ArcSin[x]^2/2 - Log[1 + Sqrt[1 - x^2]]

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\arcsin \left (x \right )}{1+\sqrt {-x^{2}+1}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [A]
time = 0.77, size = 63, normalized size = 1.40 \begin {gather*} \frac {x \arcsin \left (x\right )^{2} - 2 \, x \log \left (x\right ) - x \log \left (\sqrt {-x^{2} + 1} + 1\right ) + x \log \left (\sqrt {-x^{2} + 1} - 1\right ) + 2 \, \sqrt {-x^{2} + 1} \arcsin \left (x\right ) - 2 \, \arcsin \left (x\right )}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.47, size = 57, normalized size = 1.27 \begin {gather*} \frac {1}{2} \, \arcsin \left (x\right )^{2} - \frac {x \arcsin \left (x\right )}{\sqrt {-x^{2} + 1} + 1} - 2 \, \log \left (2\right ) + \log \left (2 \, \sqrt {-x^{2} + 1} + 2\right ) - 2 \, \log \left (\sqrt {-x^{2} + 1} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*arcsin(x)^2 - x*arcsin(x)/(sqrt(-x^2 + 1) + 1) - 2*log(2) + log(2*sqrt(-x^2 + 1) + 2) - 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 {asin}\left (x\right )}{\sqrt {1-x^2}+1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(x)/((1 - x^2)^(1/2) + 1),x)

[Out]

int(asin(x)/((1 - x^2)^(1/2) + 1), x)

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