\(\int e^{\coth ^{-1}(x)} \sqrt {1-x} \, dx\) [326]

Optimal result

Integrand size = 14, antiderivative size = 20 \[ \int e^{\coth ^{-1}(x)} \sqrt {1-x} \, dx=\frac {2}{3} e^{\coth ^{-1}(x)} \sqrt {1-x} (1+x) \]

[Out]

2/3/((-1+x)/(1+x))^(1/2)*(1+x)*(1-x)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {6309} \[ \int e^{\coth ^{-1}(x)} \sqrt {1-x} \, dx=\frac {2}{3} \sqrt {1-x} (x+1) e^{\coth ^{-1}(x)} \]

[In]

Int[E^ArcCoth[x]*Sqrt[1 - x],x]

[Out]

(2*E^ArcCoth[x]*Sqrt[1 - x]*(1 + x))/3

Rule 6309

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> Simp[(1 + a*x)*(c + d*x)^p*(E^(n*Arc
Coth[a*x])/(a*(p + 1))), x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a*c + d, 0] && EqQ[p, n/2] &&  !IntegerQ[n/2]

Rubi steps \begin{align*} \text {integral}& = \frac {2}{3} e^{\coth ^{-1}(x)} \sqrt {1-x} (1+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.70 \[ \int e^{\coth ^{-1}(x)} \sqrt {1-x} \, dx=\frac {2 \left (1+\frac {1}{x}\right )^{3/2} \sqrt {1-x} x}{3 \sqrt {1-\frac {1}{x}}} \]

[In]

Integrate[E^ArcCoth[x]*Sqrt[1 - x],x]

[Out]

(2*(1 + x^(-1))^(3/2)*Sqrt[1 - x]*x)/(3*Sqrt[1 - x^(-1)])

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20

method	result	size
gosper	\(\frac {2 \left (1+x \right ) \sqrt {1-x}}{3 \sqrt {\frac {x -1}{1+x}}}\)	\(24\)
default	\(\frac {2 \left (1+x \right ) \sqrt {1-x}}{3 \sqrt {\frac {x -1}{1+x}}}\)	\(24\)
risch	\(-\frac {2 \left (1+x \right ) \sqrt {\frac {\left (1+x \right ) \left (1-x \right )}{x -1}}\, \left (x -1\right )}{3 \sqrt {\frac {x -1}{1+x}}\, \sqrt {1-x}\, \sqrt {-1-x}}\)	\(50\)

[In]

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

[Out]

2/3/((x-1)/(1+x))^(1/2)*(1+x)*(1-x)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.65 \[ \int e^{\coth ^{-1}(x)} \sqrt {1-x} \, dx=\frac {2 \, {\left (x^{2} + 2 \, x + 1\right )} \sqrt {-x + 1} \sqrt {\frac {x - 1}{x + 1}}}{3 \, {\left (x - 1\right )}} \]

[In]

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

[Out]

2/3*(x^2 + 2*x + 1)*sqrt(-x + 1)*sqrt((x - 1)/(x + 1))/(x - 1)

Sympy [F]

\[ \int e^{\coth ^{-1}(x)} \sqrt {1-x} \, dx=\int \frac {\sqrt {1 - x}}{\sqrt {\frac {x - 1}{x + 1}}}\, dx \]

[In]

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

[Out]

Integral(sqrt(1 - x)/sqrt((x - 1)/(x + 1)), x)

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.20 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.60 \[ \int e^{\coth ^{-1}(x)} \sqrt {1-x} \, dx=-\frac {2}{3} \, \sqrt {x + 1} {\left (-i \, x - i\right )} \]

[In]

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

[Out]

-2/3*sqrt(x + 1)*(-I*x - I)

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.65 \[ \int e^{\coth ^{-1}(x)} \sqrt {1-x} \, dx=-\frac {4}{3} i \, \sqrt {2} \mathrm {sgn}\left (x + 1\right ) - \frac {2 \, {\left ({\left (-x - 1\right )}^{\frac {3}{2}} + 2 i \, \sqrt {2}\right )} \mathrm {sgn}\left (x\right )}{3 \, \mathrm {sgn}\left (x + 1\right )} \]

[In]

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

[Out]

-4/3*I*sqrt(2)*sgn(x + 1) - 2/3*((-x - 1)^(3/2) + 2*I*sqrt(2))*sgn(x)/sgn(x + 1)

Mupad [B] (verification not implemented)

Time = 4.52 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25 \[ \int e^{\coth ^{-1}(x)} \sqrt {1-x} \, dx=-\frac {2\,\sqrt {\frac {x-1}{x+1}}\,{\left (x+1\right )}^2}{3\,\sqrt {1-x}} \]

[In]

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

[Out]

-(2*((x - 1)/(x + 1))^(1/2)*(x + 1)^2)/(3*(1 - x)^(1/2))