3.3.90 \(\int \frac {e^{\coth ^{-1}(x)}}{1-x} \, dx\) [290]

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

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 33, normalized size of antiderivative = 1.00, 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 = {6310, 6313, 866, 1819, 272, 65, 212} \begin {gather*} \frac {2 \left (\frac {1}{x}+1\right )}{\sqrt {1-\frac {1}{x^2}}}-\tanh ^{-1}\left (\sqrt {1-\frac {1}{x^2}}\right ) \end {gather*}

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

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Mathematica [A]
time = 0.02, size = 38, normalized size = 1.15 \begin {gather*} \frac {2 \sqrt {1-\frac {1}{x^2}} x}{-1+x}-\log \left (\left (1+\sqrt {1-\frac {1}{x^2}}\right ) x\right ) \end {gather*}

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]

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(105\) vs. \(2(29)=58\).
time = 0.12, size = 106, normalized size = 3.21

method	result	size
risch	\(\frac {2}{\sqrt {\frac {-1+x}{1+x}}}-\frac {\ln \left (x +\sqrt {x^{2}-1}\right ) \sqrt {\left (1+x \right ) \left (-1+x \right )}}{\sqrt {\frac {-1+x}{1+x}}\, \left (1+x \right )}\)	\(52\)
trager	\(\frac {2 \sqrt {-\frac {1-x}{1+x}}\, \left (1+x \right )}{-1+x}-\ln \left (\sqrt {-\frac {1-x}{1+x}}\, x +\sqrt {-\frac {1-x}{1+x}}+x \right )\)	\(61\)
default	\(\frac {\left (x^{2}-1\right )^{\frac {3}{2}}-\sqrt {x^{2}-1}\, x^{2}-\ln \left (x +\sqrt {x^{2}-1}\right ) x^{2}+2 x \sqrt {x^{2}-1}+2 \ln \left (x +\sqrt {x^{2}-1}\right ) x -\sqrt {x^{2}-1}-\ln \left (x +\sqrt {x^{2}-1}\right )}{\left (-1+x \right ) \sqrt {\left (1+x \right ) \left (-1+x \right )}\, \sqrt {\frac {-1+x}{1+x}}}\)	\(106\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 44, normalized size = 1.33 \begin {gather*} \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 {gather*}

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)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (29) = 58\).
time = 0.38, size = 61, normalized size = 1.85 \begin {gather*} -\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 {gather*}

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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {1}{x \sqrt {\frac {x}{x + 1} - \frac {1}{x + 1}} - \sqrt {\frac {x}{x + 1} - \frac {1}{x + 1}}}\, dx \end {gather*}

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)

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Giac [A]
time = 0.41, size = 49, normalized size = 1.48 \begin {gather*} \frac {\log \left ({\left | -x + \sqrt {x^{2} - 1} \right |}\right )}{\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 ) \end {gather*}

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]

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

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Mupad [B]
time = 1.19, size = 28, normalized size = 0.85 \begin {gather*} \frac {2}{\sqrt {\frac {x-1}{x+1}}}-2\,\mathrm {atanh}\left (\sqrt {\frac {x-1}{x+1}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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