\(\int \frac {e^{\coth ^{-1}(x)} x}{1-x} \, dx\) [289]

Optimal result

Integrand size = 13, antiderivative size = 47 \[ \int \frac {e^{\coth ^{-1}(x)} x}{1-x} \, dx=\frac {2 \left (1+\frac {1}{x}\right )}{\sqrt {1-\frac {1}{x^2}}}-\sqrt {1-\frac {1}{x^2}} x-2 \text {arctanh}\left (\sqrt {1-\frac {1}{x^2}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {6310, 6312, 866, 1819, 821, 272, 65, 212} \[ \int \frac {e^{\coth ^{-1}(x)} x}{1-x} \, dx=-2 \text {arctanh}\left (\sqrt {1-\frac {1}{x^2}}\right )+\frac {2 \left (\frac {1}{x}+1\right )}{\sqrt {1-\frac {1}{x^2}}}-\sqrt {1-\frac {1}{x^2}} x \]

[In]

Int[(E^ArcCoth[x]*x)/(1 - x),x]

[Out]

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

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 821

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(e*f - d*g
))*(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 + a*e^2))), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e
^2), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0
] && EqQ[Simplify[m + 2*p + 3], 0]

Rule 866

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[d^(2*m)/a
^m, Int[(f + g*x)^n*((a + c*x^2)^(m + p)/(d - e*x)^m), x], x] /; FreeQ[{a, c, d, e, f, g, n, p}, x] && NeQ[e*f
- d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[f, 0] && ILtQ[m, -1] &&  !(IGtQ[n, 0] && ILtQ[m +
n, 0] &&  !GtQ[p, 1])

Rule 1819

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 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 6312

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

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.87 \[ \int \frac {e^{\coth ^{-1}(x)} x}{1-x} \, dx=-\frac {\sqrt {1-\frac {1}{x^2}} (-3+x) x}{-1+x}-2 \log \left (\left (1+\sqrt {1-\frac {1}{x^2}}\right ) x\right ) \]

[In]

Integrate[(E^ArcCoth[x]*x)/(1 - x),x]

[Out]

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

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.36

method	result	size
trager	\(-\frac {\left (1+x \right ) \left (-3+x \right ) \sqrt {-\frac {1-x}{1+x}}}{x -1}-2 \ln \left (\sqrt {-\frac {1-x}{1+x}}\, x +\sqrt {-\frac {1-x}{1+x}}+x \right )\)	\(64\)
risch	\(-\frac {x^{2}-2 x -3}{\sqrt {\frac {x -1}{1+x}}\, \left (1+x \right )}-\frac {2 \ln \left (x +\sqrt {x^{2}-1}\right ) \sqrt {\left (x -1\right ) \left (1+x \right )}}{\sqrt {\frac {x -1}{1+x}}\, \left (1+x \right )}\)	\(65\)
default	\(\frac {\left (x^{2}-1\right )^{\frac {3}{2}}-2 x^{2} \sqrt {x^{2}-1}-2 \ln \left (x +\sqrt {x^{2}-1}\right ) x^{2}+4 x \sqrt {x^{2}-1}+4 \ln \left (x +\sqrt {x^{2}-1}\right ) x -2 \sqrt {x^{2}-1}-2 \ln \left (x +\sqrt {x^{2}-1}\right )}{\left (x -1\right ) \sqrt {\left (x -1\right ) \left (1+x \right )}\, \sqrt {\frac {x -1}{1+x}}}\)	\(106\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.40 \[ \int \frac {e^{\coth ^{-1}(x)} x}{1-x} \, dx=-\frac {2 \, {\left (x - 1\right )} \log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) - 2 \, {\left (x - 1\right )} \log \left (\sqrt {\frac {x - 1}{x + 1}} - 1\right ) + {\left (x^{2} - 2 \, x - 3\right )} \sqrt {\frac {x - 1}{x + 1}}}{x - 1} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.57 \[ \int \frac {e^{\coth ^{-1}(x)} x}{1-x} \, dx=\frac {2 \, {\left (\frac {2 \, {\left (x - 1\right )}}{x + 1} - 1\right )}}{\left (\frac {x - 1}{x + 1}\right )^{\frac {3}{2}} - \sqrt {\frac {x - 1}{x + 1}}} - 2 \, \log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) + 2 \, \log \left (\sqrt {\frac {x - 1}{x + 1}} - 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.38 \[ \int \frac {e^{\coth ^{-1}(x)} x}{1-x} \, dx=\frac {2 \, \log \left ({\left | -x + \sqrt {x^{2} - 1} \right |}\right )}{\mathrm {sgn}\left (x + 1\right )} - \frac {\sqrt {x^{2} - 1}}{\mathrm {sgn}\left (x + 1\right )} - \frac {4}{{\left (x - \sqrt {x^{2} - 1} - 1\right )} \mathrm {sgn}\left (x + 1\right )} - 2 \, \mathrm {sgn}\left (x + 1\right ) \]

[In]

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

[Out]

2*log(abs(-x + sqrt(x^2 - 1)))/sgn(x + 1) - sqrt(x^2 - 1)/sgn(x + 1) - 4/((x - sqrt(x^2 - 1) - 1)*sgn(x + 1))
- 2*sgn(x + 1)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.91 \[ \int \frac {e^{\coth ^{-1}(x)} x}{1-x} \, dx=-\frac {2\,x+8\,\mathrm {atanh}\left (\sqrt {\frac {x-1}{x+1}}\right )\,\sqrt {\frac {x-1}{x+1}}-6}{2\,\sqrt {\frac {x-1}{x+1}}} \]

[In]

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

[Out]

-(2*x + 8*atanh(((x - 1)/(x + 1))^(1/2))*((x - 1)/(x + 1))^(1/2) - 6)/(2*((x - 1)/(x + 1))^(1/2))