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

Optimal. Leaf size=35 \[ \frac {1}{2} \tanh ^{-1}\left (\sqrt {1-\frac {1}{x^2}}\right )-\frac {1}{2} \sqrt {1-\frac {1}{x^2}} x^2 \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6175, 6178, 266, 47, 63, 206} \[ \frac {1}{2} \tanh ^{-1}\left (\sqrt {1-\frac {1}{x^2}}\right )-\frac {1}{2} \sqrt {1-\frac {1}{x^2}} x^2 \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCoth[x]*(1 - x),x]

[Out]

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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])

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 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]

Rubi steps

\begin {align*} \int e^{\coth ^{-1}(x)} (1-x) \, dx &=-\int e^{\coth ^{-1}(x)} \left (1-\frac {1}{x}\right ) x \, dx\\ &=\operatorname {Subst}\left (\int \frac {\sqrt {1-x^2}}{x^3} \, dx,x,\frac {1}{x}\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\sqrt {1-x}}{x^2} \, dx,x,\frac {1}{x^2}\right )\\ &=-\frac {1}{2} \sqrt {1-\frac {1}{x^2}} x^2-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x} x} \, dx,x,\frac {1}{x^2}\right )\\ &=-\frac {1}{2} \sqrt {1-\frac {1}{x^2}} x^2+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {1-\frac {1}{x^2}}\right )\\ &=-\frac {1}{2} \sqrt {1-\frac {1}{x^2}} x^2+\frac {1}{2} \tanh ^{-1}\left (\sqrt {1-\frac {1}{x^2}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 39, normalized size = 1.11 \[ \frac {1}{2} \log \left (\left (\sqrt {1-\frac {1}{x^2}}+1\right ) x\right )-\frac {1}{2} \sqrt {1-\frac {1}{x^2}} x^2 \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcCoth[x]*(1 - x),x]

[Out]

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

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fricas [A]  time = 0.45, size = 51, normalized size = 1.46 \[ -\frac {1}{2} \, {\left (x^{2} + x\right )} \sqrt {\frac {x - 1}{x + 1}} + \frac {1}{2} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) - \frac {1}{2} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} - 1\right ) \]

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]

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

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giac [B]  time = 0.14, size = 110, normalized size = 3.14 \[ -\frac {\sqrt {\frac {x - 1}{x + 1}} + \frac {1}{\sqrt {\frac {x - 1}{x + 1}}}}{{\left (\sqrt {\frac {x - 1}{x + 1}} + \frac {1}{\sqrt {\frac {x - 1}{x + 1}}}\right )}^{2} - 4} + \frac {1}{4} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} + \frac {1}{\sqrt {\frac {x - 1}{x + 1}}} + 2\right ) - \frac {1}{4} \, \log \left ({\left | \sqrt {\frac {x - 1}{x + 1}} + \frac {1}{\sqrt {\frac {x - 1}{x + 1}}} - 2 \right |}\right ) \]

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]

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

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maple [A]  time = 0.04, size = 48, normalized size = 1.37 \[ -\frac {\left (-1+x \right ) \left (x \sqrt {x^{2}-1}-\ln \left (x +\sqrt {x^{2}-1}\right )\right )}{2 \sqrt {\frac {-1+x}{1+x}}\, \sqrt {\left (1+x \right ) \left (-1+x \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.32, size = 83, normalized size = 2.37 \[ \frac {\left (\frac {x - 1}{x + 1}\right )^{\frac {3}{2}} + \sqrt {\frac {x - 1}{x + 1}}}{\frac {2 \, {\left (x - 1\right )}}{x + 1} - \frac {{\left (x - 1\right )}^{2}}{{\left (x + 1\right )}^{2}} - 1} + \frac {1}{2} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) - \frac {1}{2} \, \log \left (\sqrt {\frac {x - 1}{x + 1}} - 1\right ) \]

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]

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

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mupad [B]  time = 0.03, size = 63, normalized size = 1.80 \[ \mathrm {atanh}\left (\sqrt {\frac {x-1}{x+1}}\right )-\frac {\sqrt {\frac {x-1}{x+1}}+{\left (\frac {x-1}{x+1}\right )}^{3/2}}{\frac {{\left (x-1\right )}^2}{{\left (x+1\right )}^2}-\frac {2\,\left (x-1\right )}{x+1}+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {x}{\sqrt {\frac {x}{x + 1} - \frac {1}{x + 1}}}\, dx - \int \left (- \frac {1}{\sqrt {\frac {x}{x + 1} - \frac {1}{x + 1}}}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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