3.16 \(\int \frac {\sin ^{-1}(x)}{(1+x^2)^{3/2}} \, dx\)

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

[Out]

-1/2*arcsin(x^2)+x*arcsin(x)/(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 = 3, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {191, 4665, 275, 216} \[ \frac {x \sin ^{-1}(x)}{\sqrt {x^2+1}}-\frac {1}{2} \sin ^{-1}\left (x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[x]/(1 + x^2)^(3/2),x]

[Out]

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

Rule 191

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

Rule 216

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 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 4665

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(d + e*x^2)
^p, x]}, Dist[a + b*ArcSin[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/Sqrt[1 - c^2*x^2], x], x], x]] /; F
reeQ[{a, b, c, d, e}, x] && NeQ[c^2*d + e, 0] && (IGtQ[p, 0] || ILtQ[p + 1/2, 0])

Rubi steps

\begin {align*} \int \frac {\sin ^{-1}(x)}{\left (1+x^2\right )^{3/2}} \, dx &=\frac {x \sin ^{-1}(x)}{\sqrt {1+x^2}}-\int \frac {x}{\sqrt {1-x^4}} \, dx\\ &=\frac {x \sin ^{-1}(x)}{\sqrt {1+x^2}}-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2}} \, dx,x,x^2\right )\\ &=\frac {x \sin ^{-1}(x)}{\sqrt {1+x^2}}-\frac {1}{2} \sin ^{-1}\left (x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 22, normalized size = 1.00 \[ \frac {x \sin ^{-1}(x)}{\sqrt {x^2+1}}-\frac {1}{2} \sin ^{-1}\left (x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[x]/(1 + x^2)^(3/2),x]

[Out]

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

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fricas [B]  time = 0.45, size = 56, normalized size = 2.55 \[ \frac {2 \, \sqrt {x^{2} + 1} x \arcsin \relax (x) + {\left (x^{2} + 1\right )} \arctan \left (\frac {\sqrt {x^{2} + 1} \sqrt {-x^{2} + 1} x^{2}}{x^{4} - 1}\right )}{2 \, {\left (x^{2} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(x)/(x^2+1)^(3/2),x, algorithm="fricas")

[Out]

1/2*(2*sqrt(x^2 + 1)*x*arcsin(x) + (x^2 + 1)*arctan(sqrt(x^2 + 1)*sqrt(-x^2 + 1)*x^2/(x^4 - 1)))/(x^2 + 1)

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giac [A]  time = 1.06, size = 18, normalized size = 0.82 \[ \frac {x \arcsin \relax (x)}{\sqrt {x^{2} + 1}} - \frac {1}{2} \, \arcsin \left (x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(x)/(x^2+1)^(3/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.19, size = 0, normalized size = 0.00 \[ \int \frac {\arcsin \relax (x )}{\left (x^{2}+1\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsin(x)/(x^2+1)^(3/2),x)

[Out]

int(arcsin(x)/(x^2+1)^(3/2),x)

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maxima [A]  time = 0.96, size = 18, normalized size = 0.82 \[ \frac {x \arcsin \relax (x)}{\sqrt {x^{2} + 1}} - \frac {1}{2} \, \arcsin \left (x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(x)/(x^2+1)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \[ \int \frac {\mathrm {asin}\relax (x)}{{\left (x^2+1\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(x)/(x^2 + 1)^(3/2),x)

[Out]

int(asin(x)/(x^2 + 1)^(3/2), x)

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sympy [C]  time = 14.50, size = 78, normalized size = 3.55 \[ \frac {x \operatorname {asin}{\relax (x )}}{\sqrt {x^{2} + 1}} + \frac {i {G_{6, 6}^{6, 2}\left (\begin {matrix} \frac {1}{4}, \frac {3}{4} & \frac {1}{2}, \frac {1}{2}, 1, 1 \\0, \frac {1}{4}, \frac {1}{2}, \frac {3}{4}, 1, 0 & \end {matrix} \middle | {\frac {1}{x^{4}}} \right )}}{8 \pi ^{\frac {3}{2}}} - \frac {{G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4}, 0, \frac {1}{4}, \frac {1}{2}, 1 & \\- \frac {1}{4}, \frac {1}{4} & - \frac {1}{2}, 0, 0, 0 \end {matrix} \middle | {\frac {e^{- 2 i \pi }}{x^{4}}} \right )}}{8 \pi ^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asin(x)/(x**2+1)**(3/2),x)

[Out]

x*asin(x)/sqrt(x**2 + 1) + I*meijerg(((1/4, 3/4), (1/2, 1/2, 1, 1)), ((0, 1/4, 1/2, 3/4, 1, 0), ()), x**(-4))/
(8*pi**(3/2)) - meijerg(((-1/2, -1/4, 0, 1/4, 1/2, 1), ()), ((-1/4, 1/4), (-1/2, 0, 0, 0)), exp_polar(-2*I*pi)
/x**4)/(8*pi**(3/2))

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