\(\int e^{\coth ^{-1}(x)} (1-x) \, dx\) [282]

Optimal result

Integrand size = 10, antiderivative size = 35 \[ \int e^{\coth ^{-1}(x)} (1-x) \, dx=-\frac {1}{2} \sqrt {1-\frac {1}{x^2}} x^2+\frac {1}{2} \text {arctanh}\left (\sqrt {1-\frac {1}{x^2}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6310, 6313, 272, 43, 65, 212} \[ \int e^{\coth ^{-1}(x)} (1-x) \, dx=\frac {1}{2} \text {arctanh}\left (\sqrt {1-\frac {1}{x^2}}\right )-\frac {1}{2} \sqrt {1-\frac {1}{x^2}} x^2 \]

[In]

Int[E^ArcCoth[x]*(1 - x),x]

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 65

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 212

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

Rule 272

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

Rule 6310

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

Rule 6313

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[-c^n, Subst[Int[(c +
d*x)^(p - n)*((1 - x^2/a^2)^(n/2)/x^(m + 2)), x], x, 1/x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[c + a*d, 0] &&
IntegerQ[(n - 1)/2] && IntegerQ[m] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p, n/2 + 1] || LtQ[-5, m, -1]) && In
tegerQ[2*p]

Rubi steps \begin{align*} \text {integral}& = -\int e^{\coth ^{-1}(x)} \left (1-\frac {1}{x}\right ) x \, dx \\ & = \text {Subst}\left (\int \frac {\sqrt {1-x^2}}{x^3} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {\sqrt {1-x}}{x^2} \, dx,x,\frac {1}{x^2}\right ) \\ & = -\frac {1}{2} \sqrt {1-\frac {1}{x^2}} x^2-\frac {1}{4} \text {Subst}\left (\int \frac {1}{\sqrt {1-x} x} \, dx,x,\frac {1}{x^2}\right ) \\ & = -\frac {1}{2} \sqrt {1-\frac {1}{x^2}} x^2+\frac {1}{2} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {1-\frac {1}{x^2}}\right ) \\ & = -\frac {1}{2} \sqrt {1-\frac {1}{x^2}} x^2+\frac {1}{2} \text {arctanh}\left (\sqrt {1-\frac {1}{x^2}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.11 \[ \int e^{\coth ^{-1}(x)} (1-x) \, dx=-\frac {1}{2} \sqrt {1-\frac {1}{x^2}} x^2+\frac {1}{2} \log \left (\left (1+\sqrt {1-\frac {1}{x^2}}\right ) x\right ) \]

[In]

Integrate[E^ArcCoth[x]*(1 - x),x]

[Out]

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

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.37

method	result	size
default	\(-\frac {\left (x -1\right ) \left (x \sqrt {x^{2}-1}-\ln \left (x +\sqrt {x^{2}-1}\right )\right )}{2 \sqrt {\frac {x -1}{1+x}}\, \sqrt {\left (x -1\right ) \left (1+x \right )}}\)	\(48\)
risch	\(-\frac {x \left (x -1\right )}{2 \sqrt {\frac {x -1}{1+x}}}+\frac {\ln \left (x +\sqrt {x^{2}-1}\right ) \sqrt {\left (x -1\right ) \left (1+x \right )}}{2 \sqrt {\frac {x -1}{1+x}}\, \left (1+x \right )}\)	\(56\)
trager	\(-\frac {\left (1+x \right ) \sqrt {-\frac {1-x}{1+x}}\, x}{2}+\frac {\ln \left (\sqrt {-\frac {1-x}{1+x}}\, x +\sqrt {-\frac {1-x}{1+x}}+x \right )}{2}\)	\(57\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.46 \[ \int e^{\coth ^{-1}(x)} (1-x) \, dx=-\frac {1}{2} \, {\left (x^{2} + x\right )} \sqrt {\frac {x - 1}{x + 1}} + \frac {1}{2} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) - \frac {1}{2} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} - 1\right ) \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 83 vs. \(2 (27) = 54\).

Time = 0.21 (sec) , antiderivative size = 83, normalized size of antiderivative = 2.37 \[ \int e^{\coth ^{-1}(x)} (1-x) \, dx=\frac {\left (\frac {x - 1}{x + 1}\right )^{\frac {3}{2}} + \sqrt {\frac {x - 1}{x + 1}}}{\frac {2 \, {\left (x - 1\right )}}{x + 1} - \frac {{\left (x - 1\right )}^{2}}{{\left (x + 1\right )}^{2}} - 1} + \frac {1}{2} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) - \frac {1}{2} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} - 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.09 \[ \int e^{\coth ^{-1}(x)} (1-x) \, dx=-\frac {\sqrt {x^{2} - 1} x}{2 \, \mathrm {sgn}\left (x + 1\right )} - \frac {\log \left ({\left | -x + \sqrt {x^{2} - 1} \right |}\right )}{2 \, \mathrm {sgn}\left (x + 1\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.01 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.80 \[ \int e^{\coth ^{-1}(x)} (1-x) \, dx=\mathrm {atanh}\left (\sqrt {\frac {x-1}{x+1}}\right )-\frac {\sqrt {\frac {x-1}{x+1}}+{\left (\frac {x-1}{x+1}\right )}^{3/2}}{\frac {{\left (x-1\right )}^2}{{\left (x+1\right )}^2}-\frac {2\,\left (x-1\right )}{x+1}+1} \]

[In]

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

[Out]

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