\(\int \frac {e^{\coth ^{-1}(x)}}{1+x} \, dx\) [288]

Optimal result

Integrand size = 10, antiderivative size = 22 \[ \int \frac {e^{\coth ^{-1}(x)}}{1+x} \, dx=\text {arctanh}\left (\sqrt {1+\frac {1}{x}} \sqrt {\frac {-1+x}{x}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6310, 6315, 94, 212} \[ \int \frac {e^{\coth ^{-1}(x)}}{1+x} \, dx=\text {arctanh}\left (\sqrt {\frac {1}{x}+1} \sqrt {\frac {x-1}{x}}\right ) \]

[In]

Int[E^ArcCoth[x]/(1 + x),x]

[Out]

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

Rule 94

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))), x_Symbol] :> Dist[b*f, Subst[I
nt[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sqrt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
EqQ[2*b*d*e - f*(b*c + a*d), 0]

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 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 6315

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

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {e^{\coth ^{-1}(x)}}{1+x} \, dx=\log \left (x \left (1+\sqrt {\frac {-1+x^2}{x^2}}\right )\right ) \]

[In]

Integrate[E^ArcCoth[x]/(1 + x),x]

[Out]

Log[x*(1 + Sqrt[(-1 + x^2)/x^2])]

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.59

method	result	size
default	\(\frac {\left (x -1\right ) \ln \left (x +\sqrt {x^{2}-1}\right )}{\sqrt {\frac {x -1}{1+x}}\, \sqrt {\left (x -1\right ) \left (1+x \right )}}\)	\(35\)
trager	\(-\ln \left (-\sqrt {-\frac {1-x}{1+x}}\, x -\sqrt {-\frac {1-x}{1+x}}+x \right )\)	\(39\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {e^{\coth ^{-1}(x)}}{1+x} \, dx=\log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) - \log \left (\sqrt {\frac {x - 1}{x + 1}} - 1\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 2.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {e^{\coth ^{-1}(x)}}{1+x} \, dx=- \log {\left (\sqrt {1 - \frac {2}{x + 1}} - 1 \right )} + \log {\left (\sqrt {1 - \frac {2}{x + 1}} + 1 \right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {e^{\coth ^{-1}(x)}}{1+x} \, dx=\log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) - \log \left (\sqrt {\frac {x - 1}{x + 1}} - 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {e^{\coth ^{-1}(x)}}{1+x} \, dx=-\frac {\log \left ({\left | -x + \sqrt {x^{2} - 1} \right |}\right )}{\mathrm {sgn}\left (x + 1\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.64 \[ \int \frac {e^{\coth ^{-1}(x)}}{1+x} \, dx=2\,\mathrm {atanh}\left (\sqrt {\frac {x-1}{x+1}}\right ) \]

[In]

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

[Out]

2*atanh(((x - 1)/(x + 1))^(1/2))