3.290 \(\int \frac{e^{\coth ^{-1}(x)}}{1-x} \, dx\)

Optimal. Leaf size=33 \[ \frac{2 \left (\frac{1}{x}+1\right )}{\sqrt{1-\frac{1}{x^2}}}-\tanh ^{-1}\left (\sqrt{1-\frac{1}{x^2}}\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.131617, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.583, Rules used = {6175, 6178, 852, 1805, 266, 63, 206} \[ \frac{2 \left (\frac{1}{x}+1\right )}{\sqrt{1-\frac{1}{x^2}}}-\tanh ^{-1}\left (\sqrt{1-\frac{1}{x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCoth[x]/(1 - x),x]

[Out]

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

Rule 6175

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 6178

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]

Rule 852

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 1805

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 266

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 63

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 206

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

Rubi steps

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

Mathematica [A]  time = 0.0227887, size = 38, normalized size = 1.15 \[ \frac{2 \sqrt{1-\frac{1}{x^2}} x}{x-1}-\log \left (\left (\sqrt{1-\frac{1}{x^2}}+1\right ) x\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcCoth[x]/(1 - x),x]

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.12, size = 106, normalized size = 3.2 \begin{align*}{\frac{1}{-1+x} \left ( \left ({x}^{2}-1 \right ) ^{{\frac{3}{2}}}-{x}^{2}\sqrt{{x}^{2}-1}-\ln \left ( x+\sqrt{{x}^{2}-1} \right ){x}^{2}+2\,x\sqrt{{x}^{2}-1}+2\,\ln \left ( x+\sqrt{{x}^{2}-1} \right ) x-\sqrt{{x}^{2}-1}-\ln \left ( x+\sqrt{{x}^{2}-1} \right ) \right ){\frac{1}{\sqrt{ \left ( 1+x \right ) \left ( -1+x \right ) }}}{\frac{1}{\sqrt{{\frac{-1+x}{1+x}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.00447, size = 59, normalized size = 1.79 \begin{align*} \frac{2}{\sqrt{\frac{x - 1}{x + 1}}} - \log \left (\sqrt{\frac{x - 1}{x + 1}} + 1\right ) + \log \left (\sqrt{\frac{x - 1}{x + 1}} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 1.58125, size = 170, normalized size = 5.15 \begin{align*} -\frac{{\left (x - 1\right )} \log \left (\sqrt{\frac{x - 1}{x + 1}} + 1\right ) -{\left (x - 1\right )} \log \left (\sqrt{\frac{x - 1}{x + 1}} - 1\right ) - 2 \,{\left (x + 1\right )} \sqrt{\frac{x - 1}{x + 1}}}{x - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{1}{x \sqrt{\frac{x}{x + 1} - \frac{1}{x + 1}} - \sqrt{\frac{x}{x + 1} - \frac{1}{x + 1}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.16523, size = 61, normalized size = 1.85 \begin{align*} \frac{2}{\sqrt{\frac{x - 1}{x + 1}}} - \log \left (\sqrt{\frac{x - 1}{x + 1}} + 1\right ) + \log \left ({\left | \sqrt{\frac{x - 1}{x + 1}} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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