3.10.68 \(\int \frac {x}{x-\sqrt {1-x^2}} \, dx\) [968]

Optimal. Leaf size=65 \[ \frac {x}{2}+\frac {\sqrt {1-x^2}}{2}-\frac {\tanh ^{-1}\left (\sqrt {2} x\right )}{2 \sqrt {2}}-\frac {\tanh ^{-1}\left (\sqrt {2} \sqrt {1-x^2}\right )}{2 \sqrt {2}} \]

[Out]

1/2*x-1/4*arctanh(x*2^(1/2))*2^(1/2)-1/4*arctanh(2^(1/2)*(-x^2+1)^(1/2))*2^(1/2)+1/2*(-x^2+1)^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {2132, 327, 212, 455, 52, 65, 213} \begin {gather*} \frac {\sqrt {1-x^2}}{2}-\frac {\tanh ^{-1}\left (\sqrt {2} \sqrt {1-x^2}\right )}{2 \sqrt {2}}+\frac {x}{2}-\frac {\tanh ^{-1}\left (\sqrt {2} x\right )}{2 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/(x - Sqrt[1 - x^2]),x]

[Out]

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

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, 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 213

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

Rule 327

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

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 2132

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

Rubi steps

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

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Mathematica [A]
time = 0.12, size = 49, normalized size = 0.75 \begin {gather*} \frac {1}{2} \left (x+\sqrt {1-x^2}+\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} x}{-1-x+\sqrt {1-x^2}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/(x - Sqrt[1 - x^2]),x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(174\) vs. \(2(45)=90\).
time = 0.14, size = 175, normalized size = 2.69

method	result	size
trager	\(\frac {x}{2}+\frac {\sqrt {-x^{2}+1}}{2}-\frac {\RootOf \left (\textit {\_Z}^{2}-2\right ) \ln \left (-\frac {\RootOf \left (\textit {\_Z}^{2}-2\right )+2 \sqrt {-x^{2}+1}}{\RootOf \left (\textit {\_Z}^{2}-2\right ) x -1}\right )}{4}\)	\(57\)
default	\(\frac {x}{2}-\frac {\arctanh \left (\sqrt {2}\, x \right ) \sqrt {2}}{4}+\frac {\sqrt {-4 \left (x +\frac {\sqrt {2}}{2}\right )^{2}+4 \left (x +\frac {\sqrt {2}}{2}\right ) \sqrt {2}+2}}{8}-\frac {\sqrt {2}\, \arctanh \left (\frac {\left (\left (x +\frac {\sqrt {2}}{2}\right ) \sqrt {2}+1\right ) \sqrt {2}}{\sqrt {-4 \left (x +\frac {\sqrt {2}}{2}\right )^{2}+4 \left (x +\frac {\sqrt {2}}{2}\right ) \sqrt {2}+2}}\right )}{8}+\frac {\sqrt {-4 \left (x -\frac {\sqrt {2}}{2}\right )^{2}-4 \left (x -\frac {\sqrt {2}}{2}\right ) \sqrt {2}+2}}{8}-\frac {\sqrt {2}\, \arctanh \left (\frac {\left (1-\left (x -\frac {\sqrt {2}}{2}\right ) \sqrt {2}\right ) \sqrt {2}}{\sqrt {-4 \left (x -\frac {\sqrt {2}}{2}\right )^{2}-4 \left (x -\frac {\sqrt {2}}{2}\right ) \sqrt {2}+2}}\right )}{8}\)	\(175\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x-1/4*arctanh(2^(1/2)*x)*2^(1/2)+1/8*(-4*(x+1/2*2^(1/2))^2+4*(x+1/2*2^(1/2))*2^(1/2)+2)^(1/2)-1/8*2^(1/2)*
arctanh(((x+1/2*2^(1/2))*2^(1/2)+1)*2^(1/2)/(-4*(x+1/2*2^(1/2))^2+4*(x+1/2*2^(1/2))*2^(1/2)+2)^(1/2))+1/8*(-4*
(x-1/2*2^(1/2))^2-4*(x-1/2*2^(1/2))*2^(1/2)+2)^(1/2)-1/8*2^(1/2)*arctanh((1-(x-1/2*2^(1/2))*2^(1/2))*2^(1/2)/(
-4*(x-1/2*2^(1/2))^2-4*(x-1/2*2^(1/2))*2^(1/2)+2)^(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(x/(x-(-x^2+1)^(1/2)),x, algorithm="maxima")

[Out]

integrate(x/(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. 97 vs. \(2 (45) = 90\).
time = 0.33, size = 97, normalized size = 1.49 \begin {gather*} \frac {1}{8} \, \sqrt {2} \log \left (\frac {6 \, x^{2} - 2 \, \sqrt {2} {\left (2 \, x^{2} - 3\right )} + 2 \, \sqrt {-x^{2} + 1} {\left (3 \, \sqrt {2} - 4\right )} - 9}{2 \, x^{2} - 1}\right ) + \frac {1}{8} \, \sqrt {2} \log \left (\frac {2 \, x^{2} - 2 \, \sqrt {2} x + 1}{2 \, x^{2} - 1}\right ) + \frac {1}{2} \, x + \frac {1}{2} \, \sqrt {-x^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(2)*log((6*x^2 - 2*sqrt(2)*(2*x^2 - 3) + 2*sqrt(-x^2 + 1)*(3*sqrt(2) - 4) - 9)/(2*x^2 - 1)) + 1/8*sqrt
(2)*log((2*x^2 - 2*sqrt(2)*x + 1)/(2*x^2 - 1)) + 1/2*x + 1/2*sqrt(-x^2 + 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/(x - sqrt(1 - x**2)), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 105 vs. \(2 (45) = 90\).
time = 5.39, size = 105, normalized size = 1.62 \begin {gather*} \frac {1}{8} \, \sqrt {2} \log \left (\frac {{\left | 4 \, x - 2 \, \sqrt {2} \right |}}{{\left | 4 \, x + 2 \, \sqrt {2} \right |}}\right ) - \frac {1}{8} \, \sqrt {2} \log \left (\frac {{\left | -4 \, \sqrt {2} + \frac {2 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{x^{2}} - 6 \right |}}{{\left | 4 \, \sqrt {2} + \frac {2 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{x^{2}} - 6 \right |}}\right ) + \frac {1}{2} \, x + \frac {1}{2} \, \sqrt {-x^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(2)*log(abs(4*x - 2*sqrt(2))/abs(4*x + 2*sqrt(2))) - 1/8*sqrt(2)*log(abs(-4*sqrt(2) + 2*(sqrt(-x^2 + 1
) - 1)^2/x^2 - 6)/abs(4*sqrt(2) + 2*(sqrt(-x^2 + 1) - 1)^2/x^2 - 6)) + 1/2*x + 1/2*sqrt(-x^2 + 1)

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Mupad [B]
time = 3.50, size = 127, normalized size = 1.95 \begin {gather*} \frac {x}{2}-\frac {\sqrt {2}\,\ln \left (\frac {\sqrt {2}\,\left (\frac {\sqrt {2}\,x}{2}-1\right )\,1{}\mathrm {i}-\sqrt {1-x^2}\,1{}\mathrm {i}}{x-\frac {\sqrt {2}}{2}}\right )}{8}-\frac {\sqrt {2}\,\ln \left (\frac {\sqrt {2}\,\left (\frac {\sqrt {2}\,x}{2}+1\right )\,1{}\mathrm {i}+\sqrt {1-x^2}\,1{}\mathrm {i}}{x+\frac {\sqrt {2}}{2}}\right )}{8}+\frac {\sqrt {2}\,\ln \left (x-\frac {\sqrt {2}}{2}\right )}{8}-\frac {\sqrt {2}\,\ln \left (x+\frac {\sqrt {2}}{2}\right )}{8}+\frac {\sqrt {1-x^2}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(x - (1 - x^2)^(1/2)),x)

[Out]

x/2 - (2^(1/2)*log((2^(1/2)*((2^(1/2)*x)/2 - 1)*1i - (1 - x^2)^(1/2)*1i)/(x - 2^(1/2)/2)))/8 - (2^(1/2)*log((2
^(1/2)*((2^(1/2)*x)/2 + 1)*1i + (1 - x^2)^(1/2)*1i)/(x + 2^(1/2)/2)))/8 + (2^(1/2)*log(x - 2^(1/2)/2))/8 - (2^
(1/2)*log(x + 2^(1/2)/2))/8 + (1 - x^2)^(1/2)/2

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