3.8.41 \(\int \sqrt {-\frac {x}{1+x}} \, dx\) [741]

Optimal. Leaf size=32 \[ \sqrt {-\frac {x}{1+x}} (1+x)-\tan ^{-1}\left (\sqrt {-\frac {x}{1+x}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1979, 294, 210} \begin {gather*} \sqrt {-\frac {x}{x+1}} (x+1)-\text {ArcTan}\left (\sqrt {-\frac {x}{x+1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[-(x/(1 + x))],x]

[Out]

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

Rule 210

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

Rule 294

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*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 1979

Int[(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> With[{q = Denominator[p]
}, Dist[q*e*((b*c - a*d)/n), Subst[Int[x^(q*(p + 1) - 1)*(((-a)*e + c*x^q)^(1/n - 1)/(b*e - d*x^q)^(1/n + 1)),
x], x, (e*((a + b*x^n)/(c + d*x^n)))^(1/q)], x]] /; FreeQ[{a, b, c, d, e}, x] && FractionQ[p] && IntegerQ[1/n
]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 49, normalized size = 1.53 \begin {gather*} \frac {\sqrt {-\frac {x}{1+x}} \left (\sqrt {x} (1+x)-\sqrt {1+x} \tanh ^{-1}\left (\sqrt {\frac {x}{1+x}}\right )\right )}{\sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[-(x/(1 + x))],x]

[Out]

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

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Maple [A]
time = 0.11, size = 46, normalized size = 1.44

method	result	size
risch	\(\left (1+x \right ) \sqrt {-\frac {x}{1+x}}-\frac {\arcsin \left (2 x +1\right ) \sqrt {-\frac {x}{1+x}}\, \sqrt {-x \left (1+x \right )}}{2 x}\)	\(45\)
default	\(\frac {\sqrt {-\frac {x}{1+x}}\, \left (1+x \right ) \left (2 \sqrt {x^{2}+x}-\ln \left (x +\frac {1}{2}+\sqrt {x^{2}+x}\right )\right )}{2 \sqrt {x \left (1+x \right )}}\)	\(46\)
trager	\(2 \left (\frac {1}{2}+\frac {x}{2}\right ) \sqrt {-\frac {x}{1+x}}-\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (2 x \RootOf \left (\textit {\_Z}^{2}+1\right )+2 \sqrt {-\frac {x}{1+x}}\, x +\RootOf \left (\textit {\_Z}^{2}+1\right )+2 \sqrt {-\frac {x}{1+x}}\right )}{2}\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.49, size = 37, normalized size = 1.16 \begin {gather*} -\frac {\sqrt {-\frac {x}{x + 1}}}{\frac {x}{x + 1} - 1} - \arctan \left (\sqrt {-\frac {x}{x + 1}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.37, size = 28, normalized size = 0.88 \begin {gather*} {\left (x + 1\right )} \sqrt {-\frac {x}{x + 1}} - \arctan \left (\sqrt {-\frac {x}{x + 1}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(-x/(x + 1)), x)

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Giac [A]
time = 4.59, size = 36, normalized size = 1.12 \begin {gather*} \frac {1}{4} \, \pi \mathrm {sgn}\left (x + 1\right ) + \frac {1}{2} \, \arcsin \left (2 \, x + 1\right ) \mathrm {sgn}\left (x + 1\right ) + \sqrt {-x^{2} - x} \mathrm {sgn}\left (x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 3.13, size = 37, normalized size = 1.16 \begin {gather*} -\mathrm {atan}\left (\sqrt {-\frac {x}{x+1}}\right )-\frac {\sqrt {-\frac {x}{x+1}}}{\frac {x}{x+1}-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x/(x + 1))^(1/2),x)

[Out]

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

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