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

Optimal. Leaf size=65 \[ \frac{1}{5} \left (5-\sqrt{5}\right ) \log \left (2 \sqrt{1-x}-\sqrt{5}+1\right )+\frac{1}{5} \left (5+\sqrt{5}\right ) \log \left (2 \sqrt{1-x}+\sqrt{5}+1\right ) \]

[Out]

((5 - Sqrt[5])*Log[1 - Sqrt[5] + 2*Sqrt[1 - x]])/5 + ((5 + Sqrt[5])*Log[1 + Sqrt[5] + 2*Sqrt[1 - x]])/5

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Rubi [A]  time = 0.0372839, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {632, 31} \[ \frac{1}{5} \left (5-\sqrt{5}\right ) \log \left (2 \sqrt{1-x}-\sqrt{5}+1\right )+\frac{1}{5} \left (5+\sqrt{5}\right ) \log \left (2 \sqrt{1-x}+\sqrt{5}+1\right ) \]

Antiderivative was successfully verified.

[In]

Int[(-Sqrt[1 - x] + x)^(-1),x]

[Out]

((5 - Sqrt[5])*Log[1 - Sqrt[5] + 2*Sqrt[1 - x]])/5 + ((5 + Sqrt[5])*Log[1 + Sqrt[5] + 2*Sqrt[1 - x]])/5

Rule 632

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{1}{-\sqrt{1-x}+x} \, dx &=2 \operatorname{Subst}\left (\int \frac{x}{-1+x+x^2} \, dx,x,\sqrt{1-x}\right )\\ &=\frac{1}{5} \left (5-\sqrt{5}\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{2}-\frac{\sqrt{5}}{2}+x} \, dx,x,\sqrt{1-x}\right )+\frac{1}{5} \left (5+\sqrt{5}\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{2}+\frac{\sqrt{5}}{2}+x} \, dx,x,\sqrt{1-x}\right )\\ &=\frac{1}{5} \left (5-\sqrt{5}\right ) \log \left (1-\sqrt{5}+2 \sqrt{1-x}\right )+\frac{1}{5} \left (5+\sqrt{5}\right ) \log \left (1+\sqrt{5}+2 \sqrt{1-x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0274428, size = 62, normalized size = 0.95 \[ \frac{1}{5} \left (\left (5+\sqrt{5}\right ) \log \left (2 \sqrt{1-x}+\sqrt{5}+1\right )-\left (\sqrt{5}-5\right ) \log \left (2 \sqrt{1-x}-\sqrt{5}+1\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(-Sqrt[1 - x] + x)^(-1),x]

[Out]

(-((-5 + Sqrt[5])*Log[1 - Sqrt[5] + 2*Sqrt[1 - x]]) + (5 + Sqrt[5])*Log[1 + Sqrt[5] + 2*Sqrt[1 - x]])/5

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Maple [B]  time = 0.004, size = 101, normalized size = 1.6 \begin{align*}{\frac{\ln \left ({x}^{2}+x-1 \right ) }{2}}+{\frac{\sqrt{5}}{5}{\it Artanh} \left ({\frac{ \left ( 1+2\,x \right ) \sqrt{5}}{5}} \right ) }+{\frac{1}{2}\ln \left ( -x+\sqrt{1-x} \right ) }+{\frac{\sqrt{5}}{5}{\it Artanh} \left ({\frac{\sqrt{5}}{5} \left ( 2\,\sqrt{1-x}+1 \right ) } \right ) }-{\frac{1}{2}\ln \left ( -x-\sqrt{1-x} \right ) }+{\frac{\sqrt{5}}{5}{\it Artanh} \left ({\frac{\sqrt{5}}{5} \left ( 2\,\sqrt{1-x}-1 \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*ln(x^2+x-1)+1/5*5^(1/2)*arctanh(1/5*(1+2*x)*5^(1/2))+1/2*ln(-x+(1-x)^(1/2))+1/5*5^(1/2)*arctanh(1/5*(2*(1-
x)^(1/2)+1)*5^(1/2))-1/2*ln(-x-(1-x)^(1/2))+1/5*5^(1/2)*arctanh(1/5*(2*(1-x)^(1/2)-1)*5^(1/2))

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Maxima [A]  time = 1.47626, size = 69, normalized size = 1.06 \begin{align*} -\frac{1}{5} \, \sqrt{5} \log \left (-\frac{\sqrt{5} - 2 \, \sqrt{-x + 1} - 1}{\sqrt{5} + 2 \, \sqrt{-x + 1} + 1}\right ) + \log \left (-x + \sqrt{-x + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*sqrt(5)*log(-(sqrt(5) - 2*sqrt(-x + 1) - 1)/(sqrt(5) + 2*sqrt(-x + 1) + 1)) + log(-x + sqrt(-x + 1))

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Fricas [A]  time = 1.43954, size = 178, normalized size = 2.74 \begin{align*} \frac{1}{5} \, \sqrt{5} \log \left (\frac{2 \, x^{2} + \sqrt{5}{\left (x - 2\right )} -{\left (\sqrt{5}{\left (2 \, x + 1\right )} + 5\right )} \sqrt{-x + 1} - 3 \, x - 2}{x^{2} + x - 1}\right ) + \log \left (-x + \sqrt{-x + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*sqrt(5)*log((2*x^2 + sqrt(5)*(x - 2) - (sqrt(5)*(2*x + 1) + 5)*sqrt(-x + 1) - 3*x - 2)/(x^2 + x - 1)) + lo
g(-x + sqrt(-x + 1))

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Sympy [A]  time = 1.15198, size = 92, normalized size = 1.42 \begin{align*} - 4 \left (\begin{cases} - \frac{\sqrt{5} \operatorname{acoth}{\left (\frac{2 \sqrt{5} \left (\sqrt{1 - x} + \frac{1}{2}\right )}{5} \right )}}{10} & \text{for}\: \left (\sqrt{1 - x} + \frac{1}{2}\right )^{2} > \frac{5}{4} \\- \frac{\sqrt{5} \operatorname{atanh}{\left (\frac{2 \sqrt{5} \left (\sqrt{1 - x} + \frac{1}{2}\right )}{5} \right )}}{10} & \text{for}\: \left (\sqrt{1 - x} + \frac{1}{2}\right )^{2} < \frac{5}{4} \end{cases}\right ) + \log{\left (x - \sqrt{1 - x} \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*Piecewise((-sqrt(5)*acoth(2*sqrt(5)*(sqrt(1 - x) + 1/2)/5)/10, (sqrt(1 - x) + 1/2)**2 > 5/4), (-sqrt(5)*ata
nh(2*sqrt(5)*(sqrt(1 - x) + 1/2)/5)/10, (sqrt(1 - x) + 1/2)**2 < 5/4)) + log(x - sqrt(1 - x))

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Giac [A]  time = 1.13885, size = 73, normalized size = 1.12 \begin{align*} -\frac{1}{5} \, \sqrt{5} \log \left (\frac{{\left | -\sqrt{5} + 2 \, \sqrt{-x + 1} + 1 \right |}}{\sqrt{5} + 2 \, \sqrt{-x + 1} + 1}\right ) + \log \left ({\left | -x + \sqrt{-x + 1} \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*sqrt(5)*log(abs(-sqrt(5) + 2*sqrt(-x + 1) + 1)/(sqrt(5) + 2*sqrt(-x + 1) + 1)) + log(abs(-x + sqrt(-x + 1
)))