3.8.1 \(\int \frac {\sin ^{-1}(x)}{(1-x)^{5/2}} \, dx\) [701]

Optimal. Leaf size=57 \[ -\frac {\sqrt {1+x}}{3 (1-x)}+\frac {2 \sin ^{-1}(x)}{3 (1-x)^{3/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {1+x}}{\sqrt {2}}\right )}{3 \sqrt {2}} \]

[Out]

2/3*arcsin(x)/(1-x)^(3/2)-1/6*arctanh(1/2*2^(1/2)*(1+x)^(1/2))*2^(1/2)-1/3*(1+x)^(1/2)/(1-x)

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Rubi [A]
time = 0.02, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {4827, 641, 44, 65, 212} \begin {gather*} \frac {2 \text {ArcSin}(x)}{3 (1-x)^{3/2}}-\frac {\sqrt {x+1}}{3 (1-x)}-\frac {\tanh ^{-1}\left (\frac {\sqrt {x+1}}{\sqrt {2}}\right )}{3 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcSin[x]/(1 - x)^(5/2),x]

[Out]

-1/3*Sqrt[1 + x]/(1 - x) + (2*ArcSin[x])/(3*(1 - x)^(3/2)) - ArcTanh[Sqrt[1 + x]/Sqrt[2]]/(3*Sqrt[2])

Rule 44

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

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 641

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c/e)*x)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 4827

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

Rubi steps

\begin {align*} \int \frac {\sin ^{-1}(x)}{(1-x)^{5/2}} \, dx &=\frac {2 \sin ^{-1}(x)}{3 (1-x)^{3/2}}-\frac {2}{3} \int \frac {1}{(1-x)^{3/2} \sqrt {1-x^2}} \, dx\\ &=\frac {2 \sin ^{-1}(x)}{3 (1-x)^{3/2}}-\frac {2}{3} \int \frac {1}{(1-x)^2 \sqrt {1+x}} \, dx\\ &=-\frac {\sqrt {1+x}}{3 (1-x)}+\frac {2 \sin ^{-1}(x)}{3 (1-x)^{3/2}}-\frac {1}{6} \int \frac {1}{(1-x) \sqrt {1+x}} \, dx\\ &=-\frac {\sqrt {1+x}}{3 (1-x)}+\frac {2 \sin ^{-1}(x)}{3 (1-x)^{3/2}}-\frac {1}{3} \text {Subst}\left (\int \frac {1}{2-x^2} \, dx,x,\sqrt {1+x}\right )\\ &=-\frac {\sqrt {1+x}}{3 (1-x)}+\frac {2 \sin ^{-1}(x)}{3 (1-x)^{3/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {1+x}}{\sqrt {2}}\right )}{3 \sqrt {2}}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 61, normalized size = 1.07 \begin {gather*} \frac {1}{6} \left (-\frac {2 \left (\sqrt {1-x^2}-2 \sin ^{-1}(x)\right )}{(1-x)^{3/2}}-\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {1-x^2}}{\sqrt {2-2 x}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[x]/(1 - x)^(5/2),x]

[Out]

((-2*(Sqrt[1 - x^2] - 2*ArcSin[x]))/(1 - x)^(3/2) - Sqrt[2]*ArcTanh[Sqrt[1 - x^2]/Sqrt[2 - 2*x]])/6

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Maple [A]
time = 0.07, size = 70, normalized size = 1.23

method	result	size
derivativedivides	\(\frac {2 \arcsin \left (x \right )}{3 \left (1-x \right )^{\frac {3}{2}}}-\frac {\sqrt {1+x}\, \left (\sqrt {2}\, \arctanh \left (\frac {\sqrt {2}}{\sqrt {1+x}}\right ) \left (1-x \right )+2 \sqrt {1+x}\right )}{6 \sqrt {1-x}\, \sqrt {-\left (1-x \right )^{2}+2-2 x}}\)	\(70\)
default	\(\frac {2 \arcsin \left (x \right )}{3 \left (1-x \right )^{\frac {3}{2}}}-\frac {\sqrt {1+x}\, \left (\sqrt {2}\, \arctanh \left (\frac {\sqrt {2}}{\sqrt {1+x}}\right ) \left (1-x \right )+2 \sqrt {1+x}\right )}{6 \sqrt {1-x}\, \sqrt {-\left (1-x \right )^{2}+2-2 x}}\)	\(70\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsin(x)/(1-x)^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/3*arcsin(x)/(1-x)^(3/2)-1/6/(1-x)^(1/2)*(1+x)^(1/2)*(2^(1/2)*arctanh(2^(1/2)/(1+x)^(1/2))*(1-x)+2*(1+x)^(1/2
))/(-(1-x)^2+2-2*x)^(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)/(1-x)^(5/2),x, algorithm="maxima")

[Out]

-2/3*(3*(x - 1)*sqrt(-x + 1)*integrate(1/3*sqrt(x + 1)*x^2/(x^5 - x^4 - x^3 + x^2 + (x^3 - x^2 - x + 1)*e^(log
(x + 1) + log(-x + 1))), x) + arctan2(x, sqrt(x + 1)*sqrt(-x + 1)))/((x - 1)*sqrt(-x + 1))

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 90 vs. \(2 (40) = 80\).
time = 1.14, size = 90, normalized size = 1.58 \begin {gather*} \frac {\sqrt {2} {\left (x^{2} - 2 \, x + 1\right )} \log \left (-\frac {x^{2} + 2 \, \sqrt {2} \sqrt {-x^{2} + 1} \sqrt {-x + 1} + 2 \, x - 3}{x^{2} - 2 \, x + 1}\right ) - 4 \, \sqrt {-x + 1} {\left (\sqrt {-x^{2} + 1} - 2 \, \arcsin \left (x\right )\right )}}{12 \, {\left (x^{2} - 2 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(x)/(1-x)^(5/2),x, algorithm="fricas")

[Out]

1/12*(sqrt(2)*(x^2 - 2*x + 1)*log(-(x^2 + 2*sqrt(2)*sqrt(-x^2 + 1)*sqrt(-x + 1) + 2*x - 3)/(x^2 - 2*x + 1)) -
4*sqrt(-x + 1)*(sqrt(-x^2 + 1) - 2*arcsin(x)))/(x^2 - 2*x + 1)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {asin}{\left (x \right )}}{\left (1 - x\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asin(x)/(1-x)**(5/2),x)

[Out]

Integral(asin(x)/(1 - x)**(5/2), x)

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Giac [A]
time = 1.11, size = 58, normalized size = 1.02 \begin {gather*} \frac {1}{12} \, \sqrt {2} \log \left (\frac {\sqrt {2} - \sqrt {x + 1}}{\sqrt {2} + \sqrt {x + 1}}\right ) + \frac {\sqrt {x + 1}}{3 \, {\left (x - 1\right )}} - \frac {2 \, \arcsin \left (x\right )}{3 \, {\left (x - 1\right )} \sqrt {-x + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(x)/(1-x)^(5/2),x, algorithm="giac")

[Out]

1/12*sqrt(2)*log((sqrt(2) - sqrt(x + 1))/(sqrt(2) + sqrt(x + 1))) + 1/3*sqrt(x + 1)/(x - 1) - 2/3*arcsin(x)/((
x - 1)*sqrt(-x + 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 )}{{\left (1-x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(x)/(1 - x)^(5/2),x)

[Out]

int(asin(x)/(1 - x)^(5/2), x)

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