3.979 \(\int \frac {1}{\sqrt {\frac {-1-x}{x}}} \, dx\)

Optimal. Leaf size=29 \[ \tan ^{-1}\left (\sqrt {-\frac {x+1}{x}}\right )-x \sqrt {-\frac {x+1}{x}} \]

[Out]

arctan(((-1-x)/x)^(1/2))-x*((-1-x)/x)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {1972, 242, 51, 63, 204} \[ \tan ^{-1}\left (\sqrt {-\frac {x+1}{x}}\right )-x \sqrt {-\frac {x+1}{x}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[(-1 - x)/x],x]

[Out]

-(x*Sqrt[-((1 + x)/x)]) + ArcTan[Sqrt[-((1 + x)/x)]]

Rule 51

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] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 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 242

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^2, x], x, 1/x] /; FreeQ[{a, b, p},
x] && ILtQ[n, 0]

Rule 1972

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && BinomialQ[u, x] &&  !BinomialMatchQ[
u, x]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 43, normalized size = 1.48 \[ \frac {\sqrt {x} (x+1)-\sqrt {x+1} \sinh ^{-1}\left (\sqrt {x}\right )}{\sqrt {x} \sqrt {-\frac {x+1}{x}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[(-1 - x)/x],x]

[Out]

(Sqrt[x]*(1 + x) - Sqrt[1 + x]*ArcSinh[Sqrt[x]])/(Sqrt[x]*Sqrt[-((1 + x)/x)])

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fricas [A]  time = 0.44, size = 25, normalized size = 0.86 \[ -x \sqrt {-\frac {x + 1}{x}} + \arctan \left (\sqrt {-\frac {x + 1}{x}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x*sqrt(-(x + 1)/x) + arctan(sqrt(-(x + 1)/x))

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giac [A]  time = 0.46, size = 35, normalized size = 1.21 \[ \frac {1}{4} \, \pi \mathrm {sgn}\relax (x) - \frac {\arcsin \left (2 \, x + 1\right )}{2 \, \mathrm {sgn}\relax (x)} - \frac {\sqrt {-x^{2} - x}}{\mathrm {sgn}\relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*pi*sgn(x) - 1/2*arcsin(2*x + 1)/sgn(x) - sqrt(-x^2 - x)/sgn(x)

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maple [A]  time = 0.01, size = 44, normalized size = 1.52 \[ \frac {\left (x +1\right ) \left (\arcsin \left (2 x +1\right )+2 \sqrt {-x^{2}-x}\right )}{2 \sqrt {-\frac {x +1}{x}}\, \sqrt {-\left (x +1\right ) x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.97, size = 35, normalized size = 1.21 \[ -\frac {\sqrt {-\frac {x + 1}{x}}}{\frac {x + 1}{x} - 1} + \arctan \left (\sqrt {-\frac {x + 1}{x}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-(x + 1)/x)/((x + 1)/x - 1) + arctan(sqrt(-(x + 1)/x))

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mupad [B]  time = 3.37, size = 23, normalized size = 0.79 \[ \mathrm {atan}\left (\sqrt {-\frac {1}{x}-1}\right )-x\,\sqrt {-\frac {1}{x}-1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-(x + 1)/x)^(1/2),x)

[Out]

atan((- 1/x - 1)^(1/2)) - x*(- 1/x - 1)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt((-x - 1)/x), x)

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