3.7.64 \(\int \frac {x^3 \sin ^{-1}(x)}{(1-x^2)^{3/2}} \, dx\) [664]

Optimal. Leaf size=36 \[ -x+\frac {\sin ^{-1}(x)}{\sqrt {1-x^2}}+\sqrt {1-x^2} \sin ^{-1}(x)-\tanh ^{-1}(x) \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {272, 45, 4779, 396, 212} \begin {gather*} \sqrt {1-x^2} \sin ^{-1}(x)+\frac {\sin ^{-1}(x)}{\sqrt {1-x^2}}-x-\tanh ^{-1}(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*ArcSin[x])/(1 - x^2)^(3/2),x]

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 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 396

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

Rule 4779

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> With[{u = IntHide[x^
m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSin[c*x], u, x] - Dist[b*c*Simp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2]], Int[Si
mplifyIntegrand[u/Sqrt[d + e*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IntegerQ[p
- 1/2] && NeQ[p, -2^(-1)] && (IGtQ[(m + 1)/2, 0] || ILtQ[(m + 2*p + 3)/2, 0])

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 40, normalized size = 1.11 \begin {gather*} \frac {1}{2} \left (-2 x-\frac {2 \left (-2+x^2\right ) \sin ^{-1}(x)}{\sqrt {1-x^2}}+\log (1-x)-\log (1+x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathics [A]
time = 9.97, size = 45, normalized size = 1.25 \begin {gather*} \frac {-\text {ArcSin}\left [x\right ] \left (-2+x^2\right )+\frac {\left (-2 x+\text {Log}\left [-1+x\right ]-\text {Log}\left [1+x\right ]\right ) \sqrt {1-x^2}}{2}}{\sqrt {1-x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[(ArcSin[x]*x^3)/(1 - x^2)^(3/2),x]')

[Out]

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

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Maple [A]
time = 0.42, size = 61, normalized size = 1.69

method	result	size
default	\(-x +\arcsin \left (x \right ) \sqrt {-x^{2}+1}-\frac {\sqrt {-x^{2}+1}\, \arcsin \left (x \right )}{x^{2}-1}-\ln \left (\frac {1}{\sqrt {-x^{2}+1}}+\frac {x}{\sqrt {-x^{2}+1}}\right )\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arcsin(x)/(-x^2+1)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.33, size = 45, normalized size = 1.25 \begin {gather*} -{\left (\frac {x^{2}}{\sqrt {-x^{2} + 1}} - \frac {2}{\sqrt {-x^{2} + 1}}\right )} \arcsin \left (x\right ) - x - \frac {1}{2} \, \log \left (x + 1\right ) + \frac {1}{2} \, \log \left (x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.33, size = 57, normalized size = 1.58 \begin {gather*} -\frac {2 \, x^{3} - 2 \, {\left (x^{2} - 2\right )} \sqrt {-x^{2} + 1} \arcsin \left (x\right ) + {\left (x^{2} - 1\right )} \log \left (x + 1\right ) - {\left (x^{2} - 1\right )} \log \left (x - 1\right ) - 2 \, x}{2 \, {\left (x^{2} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*x^3 - 2*(x^2 - 2)*sqrt(-x^2 + 1)*arcsin(x) + (x^2 - 1)*log(x + 1) - (x^2 - 1)*log(x - 1) - 2*x)/(x^2 -
1)

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Sympy [A]
time = 8.57, size = 37, normalized size = 1.03 \begin {gather*} - x - \left (- \sqrt {1 - x^{2}} - \frac {1}{\sqrt {1 - x^{2}}}\right ) \operatorname {asin}{\left (x \right )} + \frac {\log {\left (x - 1 \right )}}{2} - \frac {\log {\left (x + 1 \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x - (-sqrt(1 - x**2) - 1/sqrt(1 - x**2))*asin(x) + log(x - 1)/2 - log(x + 1)/2

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Giac [A]
time = 0.01, size = 45, normalized size = 1.25 \begin {gather*} -x+\frac {\ln \left |x-1\right |}{2}-\frac {\ln \left |x+1\right |}{2}+\left (\sqrt {-x^{2}+1}+\frac 1{\sqrt {-x^{2}+1}}\right ) \arcsin x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsin(x)/(-x^2+1)^(3/2),x)

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {x^3\,\mathrm {asin}\left (x\right )}{{\left (1-x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*asin(x))/(1 - x^2)^(3/2),x)

[Out]

int((x^3*asin(x))/(1 - x^2)^(3/2), x)

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