3.828 \(\int \frac {-1+\sqrt {1-x^2}}{\sqrt {1-x^2} (2+x-2 \sqrt {1-x^2})^2} \, dx\)

Optimal. Leaf size=31 \[ \frac {\sqrt {1-x^2}}{5 x+4}+\frac {3}{5 (5 x+4)} \]

[Out]

3/5/(4+5*x)+(-x^2+1)^(1/2)/(4+5*x)

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Rubi [A]  time = 0.65, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 31, number of rules used = 13, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.302, Rules used = {6742, 277, 216, 266, 50, 63, 206, 733, 844, 725, 735, 264, 731} \[ \frac {\sqrt {1-x^2}}{5 x+4}+\frac {3}{5 (5 x+4)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

3/(5*(4 + 5*x)) + Sqrt[1 - x^2]/(4 + 5*x)

Rule 50

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 63

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 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 264

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

Rule 266

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 277

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 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 731

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1)*(a + c*x^2)^(p
+ 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d)/(c*d^2 + a*e^2), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x]
/; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 3, 0]

Rule 733

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a + c*x^2)^p)/(
e*(m + 1)), x] - Dist[(2*c*p)/(e*(m + 1)), Int[x*(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1), x], x] /; FreeQ[{a, c,
d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] || LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m +
2*p + 1, 0] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 735

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a + c*x^2)^p)/(
e*(m + 2*p + 1)), x] + Dist[(2*p)/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[a*e - c*d*x, x]*(a + c*x^2)^(p - 1),
x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !Ration
alQ[m] || LtQ[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 6742

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

Rubi steps

\begin {align*} \int \frac {-1+\sqrt {1-x^2}}{\sqrt {1-x^2} \left (2+x-2 \sqrt {1-x^2}\right )^2} \, dx &=\int \left (\frac {1}{\left (-2-x+2 \sqrt {1-x^2}\right )^2}-\frac {1}{\sqrt {1-x^2} \left (-2-x+2 \sqrt {1-x^2}\right )^2}\right ) \, dx\\ &=\int \frac {1}{\left (-2-x+2 \sqrt {1-x^2}\right )^2} \, dx-\int \frac {1}{\sqrt {1-x^2} \left (-2-x+2 \sqrt {1-x^2}\right )^2} \, dx\\ &=-\int \left (\frac {1}{2 x^2}-\frac {1}{x}+\frac {15}{2 (4+5 x)^2}+\frac {5}{4+5 x}+\frac {1}{2 x^2 \sqrt {1-x^2}}-\frac {1}{x \sqrt {1-x^2}}+\frac {9}{2 (4+5 x)^2 \sqrt {1-x^2}}+\frac {5}{(4+5 x) \sqrt {1-x^2}}\right ) \, dx+\int \left (\frac {1}{2 x^2}-\frac {1}{x}+\frac {9}{2 (4+5 x)^2}+\frac {5}{4+5 x}+\frac {\sqrt {1-x^2}}{2 x^2}-\frac {\sqrt {1-x^2}}{x}+\frac {15 \sqrt {1-x^2}}{2 (4+5 x)^2}+\frac {5 \sqrt {1-x^2}}{4+5 x}\right ) \, dx\\ &=\frac {3}{5 (4+5 x)}-\frac {1}{2} \int \frac {1}{x^2 \sqrt {1-x^2}} \, dx+\frac {1}{2} \int \frac {\sqrt {1-x^2}}{x^2} \, dx-\frac {9}{2} \int \frac {1}{(4+5 x)^2 \sqrt {1-x^2}} \, dx-5 \int \frac {1}{(4+5 x) \sqrt {1-x^2}} \, dx+5 \int \frac {\sqrt {1-x^2}}{4+5 x} \, dx+\frac {15}{2} \int \frac {\sqrt {1-x^2}}{(4+5 x)^2} \, dx+\int \frac {1}{x \sqrt {1-x^2}} \, dx-\int \frac {\sqrt {1-x^2}}{x} \, dx\\ &=\frac {3}{5 (4+5 x)}+\sqrt {1-x^2}+\frac {\sqrt {1-x^2}}{4+5 x}-\frac {1}{2} \int \frac {1}{\sqrt {1-x^2}} \, dx+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x} x} \, dx,x,x^2\right )-\frac {1}{2} \operatorname {Subst}\left (\int \frac {\sqrt {1-x}}{x} \, dx,x,x^2\right )-\frac {3}{2} \int \frac {x}{(4+5 x) \sqrt {1-x^2}} \, dx+2 \int \frac {1}{(4+5 x) \sqrt {1-x^2}} \, dx+5 \operatorname {Subst}\left (\int \frac {1}{9-x^2} \, dx,x,\frac {5+4 x}{\sqrt {1-x^2}}\right )+\int \frac {5+4 x}{(4+5 x) \sqrt {1-x^2}} \, dx\\ &=\frac {3}{5 (4+5 x)}+\frac {\sqrt {1-x^2}}{4+5 x}-\frac {1}{2} \sin ^{-1}(x)+\frac {5}{3} \tanh ^{-1}\left (\frac {5+4 x}{3 \sqrt {1-x^2}}\right )-\frac {3}{10} \int \frac {1}{\sqrt {1-x^2}} \, dx-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x} x} \, dx,x,x^2\right )+\frac {4}{5} \int \frac {1}{\sqrt {1-x^2}} \, dx+\frac {6}{5} \int \frac {1}{(4+5 x) \sqrt {1-x^2}} \, dx+\frac {9}{5} \int \frac {1}{(4+5 x) \sqrt {1-x^2}} \, dx-2 \operatorname {Subst}\left (\int \frac {1}{9-x^2} \, dx,x,\frac {5+4 x}{\sqrt {1-x^2}}\right )-\operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {1-x^2}\right )\\ &=\frac {3}{5 (4+5 x)}+\frac {\sqrt {1-x^2}}{4+5 x}+\tanh ^{-1}\left (\frac {5+4 x}{3 \sqrt {1-x^2}}\right )-\tanh ^{-1}\left (\sqrt {1-x^2}\right )-\frac {6}{5} \operatorname {Subst}\left (\int \frac {1}{9-x^2} \, dx,x,\frac {5+4 x}{\sqrt {1-x^2}}\right )-\frac {9}{5} \operatorname {Subst}\left (\int \frac {1}{9-x^2} \, dx,x,\frac {5+4 x}{\sqrt {1-x^2}}\right )+\operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {1-x^2}\right )\\ &=\frac {3}{5 (4+5 x)}+\frac {\sqrt {1-x^2}}{4+5 x}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 23, normalized size = 0.74 \[ \frac {5 \sqrt {1-x^2}+3}{25 x+20} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3 + 5*Sqrt[1 - x^2])/(20 + 25*x)

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fricas [A]  time = 0.41, size = 25, normalized size = 0.81 \[ \frac {25 \, x + 20 \, \sqrt {-x^{2} + 1} + 32}{20 \, {\left (5 \, x + 4\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*(25*x + 20*sqrt(-x^2 + 1) + 32)/(5*x + 4)

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giac [B]  time = 0.62, size = 68, normalized size = 2.19 \[ \frac {\frac {5 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}{x} - 4}{4 \, {\left (\frac {5 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}{x} - \frac {2 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{x^{2}} - 2\right )}} + \frac {3}{5 \, {\left (5 \, x + 4\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(5*(sqrt(-x^2 + 1) - 1)/x - 4)/(5*(sqrt(-x^2 + 1) - 1)/x - 2*(sqrt(-x^2 + 1) - 1)^2/x^2 - 2) + 3/5/(5*x +
4)

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maple [A]  time = 0.01, size = 32, normalized size = 1.03 \[ \frac {\sqrt {\frac {8 x}{5}-\left (x +\frac {4}{5}\right )^{2}+\frac {41}{25}}}{5 x +4}+\frac {3}{5 \left (5 x +4\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5/(x+4/5)*(8/5*x-(x+4/5)^2+41/25)^(1/2)+3/5/(5*x+4)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{56} \, \sqrt {7} \log \left (\frac {3 \, x - 2 \, \sqrt {7} - 2}{3 \, x + 2 \, \sqrt {7} - 2}\right ) - \int -\frac {100 \, x^{7} + 285 \, x^{6} + 264 \, x^{5} + 80 \, x^{4}}{8 \, {\left (21 \, x^{9} + 278 \, x^{8} + 283 \, x^{7} - 2022 \, x^{6} - 3632 \, x^{5} + 2256 \, x^{4} + 7424 \, x^{3} + 1536 \, x^{2} - 8 \, {\left (9 \, x^{8} + 12 \, x^{7} - 101 \, x^{6} - 172 \, x^{5} + 284 \, x^{4} + 672 \, x^{3} + 64 \, x^{2} - 512 \, x - 256\right )} \sqrt {x + 1} \sqrt {-x + 1} - 4096 \, x - 2048\right )}}\,{d x} - \frac {1}{24} \, \log \left (x + 2\right ) + \frac {1}{16} \, \log \left (x + 1\right ) - \frac {1}{48} \, \log \left (x - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/56*sqrt(7)*log((3*x - 2*sqrt(7) - 2)/(3*x + 2*sqrt(7) - 2)) - integrate(-1/8*(100*x^7 + 285*x^6 + 264*x^5 +
80*x^4)/(21*x^9 + 278*x^8 + 283*x^7 - 2022*x^6 - 3632*x^5 + 2256*x^4 + 7424*x^3 + 1536*x^2 - 8*(9*x^8 + 12*x^
7 - 101*x^6 - 172*x^5 + 284*x^4 + 672*x^3 + 64*x^2 - 512*x - 256)*sqrt(x + 1)*sqrt(-x + 1) - 4096*x - 2048), x
) - 1/24*log(x + 2) + 1/16*log(x + 1) - 1/48*log(x - 1)

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mupad [B]  time = 3.32, size = 19, normalized size = 0.61 \[ \frac {\sqrt {1-x^2}+\frac {3}{5}}{5\,x+4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((1 - x^2)^(1/2) + 3/5)/(5*x + 4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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