### 3.289 $$\int \frac{e^{\coth ^{-1}(x)} x}{1-x} \, dx$$

Optimal. Leaf size=47 $\frac{2 \left (\frac{1}{x}+1\right )}{\sqrt{1-\frac{1}{x^2}}}-\sqrt{1-\frac{1}{x^2}} x-2 \tanh ^{-1}\left (\sqrt{1-\frac{1}{x^2}}\right )$

[Out]

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

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Rubi [A]  time = 0.135316, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 13, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.615, Rules used = {6175, 6177, 852, 1805, 807, 266, 63, 206} $\frac{2 \left (\frac{1}{x}+1\right )}{\sqrt{1-\frac{1}{x^2}}}-\sqrt{1-\frac{1}{x^2}} x-2 \tanh ^{-1}\left (\sqrt{1-\frac{1}{x^2}}\right )$

Antiderivative was successfully veriﬁed.

[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 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 6177

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]

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 807

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

Mathematica [A]  time = 0.0521181, size = 41, normalized size = 0.87 $-\frac{\sqrt{1-\frac{1}{x^2}} (x-3) x}{x-1}-2 \log \left (\left (\sqrt{1-\frac{1}{x^2}}+1\right ) x\right )$

Warning: Unable to verify antiderivative.

[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]

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Maple [B]  time = 0.118, size = 106, normalized size = 2.3 \begin{align*}{\frac{1}{-1+x} \left ( \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 ) \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)*x/(1-x),x)

[Out]

((x^2-1)^(3/2)-2*x^2*(x^2-1)^(1/2)-2*ln(x+(x^2-1)^(1/2))*x^2+4*x*(x^2-1)^(1/2)+4*ln(x+(x^2-1)^(1/2))*x-2*(x^2-
1)^(1/2)-2*ln(x+(x^2-1)^(1/2)))/(-1+x)/((1+x)*(-1+x))^(1/2)/((-1+x)/(1+x))^(1/2)

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Maxima [A]  time = 1.03955, size = 100, normalized size = 2.13 \begin{align*} \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 ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.67737, size = 184, normalized size = 3.91 \begin{align*} -\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} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{x}{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)*x/(1-x),x)

[Out]

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

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Giac [B]  time = 1.15349, size = 113, normalized size = 2.4 \begin{align*} \frac{2 \,{\left (\frac{2 \,{\left (x - 1\right )}}{x + 1} - 1\right )}}{\frac{{\left (x - 1\right )} \sqrt{\frac{x - 1}{x + 1}}}{x + 1} - \sqrt{\frac{x - 1}{x + 1}}} - 2 \, \log \left (\sqrt{\frac{x - 1}{x + 1}} + 1\right ) + 2 \, \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)*x/(1-x),x, algorithm="giac")

[Out]

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