3.1.50 \(\int \frac {1}{-1+\sqrt {x}} \, dx\) [50]

Optimal. Leaf size=20 \[ 2 \sqrt {x}+2 \log \left (1-\sqrt {x}\right ) \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {196, 45} \begin {gather*} 2 \sqrt {x}+2 \log \left (1-\sqrt {x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + Sqrt[x])^(-1),x]

[Out]

2*Sqrt[x] + 2*Log[1 - Sqrt[x]]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 196

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

Rubi steps

\begin {gather*} \begin {aligned} \text {Integral} &=2 \text {Subst}\left (\int \frac {x}{-1+x} \, dx,x,\sqrt {x}\right )\\ &=2 \text {Subst}\left (\int \left (1+\frac {1}{-1+x}\right ) \, dx,x,\sqrt {x}\right )\\ &=2 \sqrt {x}+2 \log \left (1-\sqrt {x}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.01, size = 18, normalized size = 0.90 \begin {gather*} 2 \sqrt {x}+2 \log \left (-1+\sqrt {x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + Sqrt[x])^(-1),x]

[Out]

2*Sqrt[x] + 2*Log[-1 + Sqrt[x]]

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Maple [A]
time = 0.06, size = 25, normalized size = 1.25

method	result	size
derivativedivides	\(2 \sqrt {x}+2 \ln \left (\sqrt {x}-1\right )\)	\(15\)
meijerg	\(2 \sqrt {x}+2 \ln \left (1-\sqrt {x}\right )\)	\(17\)
trager	\(2 \sqrt {x}+\ln \left (2 \sqrt {x}-1-x \right )\)	\(18\)
default	\(\ln \left (x -1\right )+2 \sqrt {x}+\ln \left (\sqrt {x}-1\right )-\ln \left (1+\sqrt {x}\right )\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.35, size = 15, normalized size = 0.75 \begin {gather*} 2 \, \sqrt {x} + 2 \, \log \left (\sqrt {x} - 1\right ) - 2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(x) + 2*log(sqrt(x) - 1) - 2

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Fricas [A]
time = 0.63, size = 14, normalized size = 0.70 \begin {gather*} 2 \, \sqrt {x} + 2 \, \log \left (\sqrt {x} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(x) + 2*log(sqrt(x) - 1)

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Sympy [A]
time = 0.05, size = 15, normalized size = 0.75 \begin {gather*} 2 \sqrt {x} + 2 \log {\left (\sqrt {x} - 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(x) + 2*log(sqrt(x) - 1)

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Giac [A]
time = 0.40, size = 15, normalized size = 0.75 \begin {gather*} 2 \, \sqrt {x} + 2 \, \log \left ({\left | \sqrt {x} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(x) + 2*log(abs(sqrt(x) - 1))

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Mupad [B]
time = 0.06, size = 14, normalized size = 0.70 \begin {gather*} 2\,\ln \left (\sqrt {x}-1\right )+2\,\sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(1/2) - 1),x)

[Out]

2*log(x^(1/2) - 1) + 2*x^(1/2)

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Chatgpt [A]
time = 1.00, size = 14, normalized size = 0.70 \begin {gather*} 2 \sqrt {x}+2 \ln \left (\sqrt {x}-1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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