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

Optimal. Leaf size=147 \[ \sqrt {x-\sqrt {x^2-1}} \left (\sqrt {\sqrt {x^2-1}+x} \left (-\frac {\tan ^{-1}\left (\frac {\frac {\sqrt {x^2-1}}{\sqrt {2}}+\frac {x}{\sqrt {2}}-\frac {1}{\sqrt {2}}}{\sqrt {\sqrt {x^2-1}+x}}\right )}{\sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {\frac {\sqrt {x^2-1}}{\sqrt {2}}+\frac {x}{\sqrt {2}}+\frac {1}{\sqrt {2}}}{\sqrt {\sqrt {x^2-1}+x}}\right )}{\sqrt {2}}\right )-\frac {1}{x}\right ) \]

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Rubi [A]  time = 0.14, antiderivative size = 200, normalized size of antiderivative = 1.36, number of steps used = 12, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2119, 457, 329, 297, 1162, 617, 204, 1165, 628} \begin {gather*} -\frac {2 \left (x-\sqrt {x^2-1}\right )^{3/2}}{\left (x-\sqrt {x^2-1}\right )^2+1}+\frac {\log \left (-\sqrt {x^2-1}-\sqrt {2} \sqrt {x-\sqrt {x^2-1}}+x+1\right )}{2 \sqrt {2}}-\frac {\log \left (-\sqrt {x^2-1}+\sqrt {2} \sqrt {x-\sqrt {x^2-1}}+x+1\right )}{2 \sqrt {2}}-\frac {\tan ^{-1}\left (1-\sqrt {2} \sqrt {x-\sqrt {x^2-1}}\right )}{\sqrt {2}}+\frac {\tan ^{-1}\left (\sqrt {2} \sqrt {x-\sqrt {x^2-1}}+1\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[x - Sqrt[-1 + x^2]]/x^2,x]

[Out]

(-2*(x - Sqrt[-1 + x^2])^(3/2))/(1 + (x - Sqrt[-1 + x^2])^2) - ArcTan[1 - Sqrt[2]*Sqrt[x - Sqrt[-1 + x^2]]]/Sq
rt[2] + ArcTan[1 + Sqrt[2]*Sqrt[x - Sqrt[-1 + x^2]]]/Sqrt[2] + Log[1 + x - Sqrt[-1 + x^2] - Sqrt[2]*Sqrt[x - S
qrt[-1 + x^2]]]/(2*Sqrt[2]) - Log[1 + x - Sqrt[-1 + x^2] + Sqrt[2]*Sqrt[x - Sqrt[-1 + x^2]]]/(2*Sqrt[2])

Rule 204

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

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 329

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

Rule 457

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 2119

Int[((g_.) + (h_.)*(x_))^(m_.)*((e_.)*(x_) + (f_.)*Sqrt[(a_.) + (c_.)*(x_)^2])^(n_.), x_Symbol] :> Dist[1/(2^(
m + 1)*e^(m + 1)), Subst[Int[x^(n - m - 2)*(a*f^2 + x^2)*(-(a*f^2*h) + 2*e*g*x + h*x^2)^m, x], x, e*x + f*Sqrt
[a + c*x^2]], x] /; FreeQ[{a, c, e, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[m]

Rubi steps

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

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Mathematica [C]  time = 0.23, size = 102, normalized size = 0.69 \begin {gather*} \frac {4 \sqrt {x^2-1} \left (x-\sqrt {x^2-1}\right )^{5/2} \left (\, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\left (x-\sqrt {x^2-1}\right )^2\right )-2 \, _2F_1\left (\frac {3}{4},2;\frac {7}{4};-\left (x-\sqrt {x^2-1}\right )^2\right )\right )}{-3 x^2+3 \sqrt {x^2-1} x+3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[-1 + x^2]*(x - Sqrt[-1 + x^2])^(5/2)*(Hypergeometric2F1[3/4, 1, 7/4, -(x - Sqrt[-1 + x^2])^2] - 2*Hype
rgeometric2F1[3/4, 2, 7/4, -(x - Sqrt[-1 + x^2])^2]))/(3 - 3*x^2 + 3*x*Sqrt[-1 + x^2])

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IntegrateAlgebraic [A]  time = 0.24, size = 135, normalized size = 0.92 \begin {gather*} -\frac {\sqrt {x-\sqrt {-1+x^2}}}{x}+\frac {\tan ^{-1}\left (\frac {-\frac {1}{\sqrt {2}}+\frac {x}{\sqrt {2}}-\frac {\sqrt {-1+x^2}}{\sqrt {2}}}{\sqrt {x-\sqrt {-1+x^2}}}\right )}{\sqrt {2}}+\frac {\tanh ^{-1}\left (\frac {-\frac {1}{\sqrt {2}}-\frac {x}{\sqrt {2}}+\frac {\sqrt {-1+x^2}}{\sqrt {2}}}{\sqrt {x-\sqrt {-1+x^2}}}\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[x - Sqrt[-1 + x^2]]/x^2,x]

[Out]

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

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fricas [B]  time = 0.50, size = 226, normalized size = 1.54 \begin {gather*} -\frac {4 \, \sqrt {2} x \arctan \left (\sqrt {2} \sqrt {\sqrt {2} \sqrt {x - \sqrt {x^{2} - 1}} + x - \sqrt {x^{2} - 1} + 1} - \sqrt {2} \sqrt {x - \sqrt {x^{2} - 1}} - 1\right ) + 4 \, \sqrt {2} x \arctan \left (\frac {1}{2} \, \sqrt {2} \sqrt {-4 \, \sqrt {2} \sqrt {x - \sqrt {x^{2} - 1}} + 4 \, x - 4 \, \sqrt {x^{2} - 1} + 4} - \sqrt {2} \sqrt {x - \sqrt {x^{2} - 1}} + 1\right ) + \sqrt {2} x \log \left (4 \, \sqrt {2} \sqrt {x - \sqrt {x^{2} - 1}} + 4 \, x - 4 \, \sqrt {x^{2} - 1} + 4\right ) - \sqrt {2} x \log \left (-4 \, \sqrt {2} \sqrt {x - \sqrt {x^{2} - 1}} + 4 \, x - 4 \, \sqrt {x^{2} - 1} + 4\right ) + 4 \, \sqrt {x - \sqrt {x^{2} - 1}}}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(4*sqrt(2)*x*arctan(sqrt(2)*sqrt(sqrt(2)*sqrt(x - sqrt(x^2 - 1)) + x - sqrt(x^2 - 1) + 1) - sqrt(2)*sqrt(
x - sqrt(x^2 - 1)) - 1) + 4*sqrt(2)*x*arctan(1/2*sqrt(2)*sqrt(-4*sqrt(2)*sqrt(x - sqrt(x^2 - 1)) + 4*x - 4*sqr
t(x^2 - 1) + 4) - sqrt(2)*sqrt(x - sqrt(x^2 - 1)) + 1) + sqrt(2)*x*log(4*sqrt(2)*sqrt(x - sqrt(x^2 - 1)) + 4*x
- 4*sqrt(x^2 - 1) + 4) - sqrt(2)*x*log(-4*sqrt(2)*sqrt(x - sqrt(x^2 - 1)) + 4*x - 4*sqrt(x^2 - 1) + 4) + 4*sq
rt(x - sqrt(x^2 - 1)))/x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x - sqrt(x^2 - 1))/x^2, x)

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\sqrt {x -\sqrt {x^{2}-1}}}{x^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x - sqrt(x^2 - 1))/x^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {x-\sqrt {x^2-1}}}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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