3.294 \(\int \frac {e^{\coth ^{-1}(x)}}{(1-x)^2} \, dx\)

Optimal. Leaf size=24 \[ -\frac {\left (1-\frac {1}{x^2}\right )^{3/2}}{3 \left (1-\frac {1}{x}\right )^3} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6175, 6178, 651} \[ -\frac {\left (1-\frac {1}{x^2}\right )^{3/2}}{3 \left (1-\frac {1}{x}\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCoth[x]/(1 - x)^2,x]

[Out]

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

Rule 651

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^m*(a + c*x^2)^(p + 1))
/(2*c*d*(p + 1)), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p
+ 2, 0]

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

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Mathematica [A]  time = 0.02, size = 24, normalized size = 1.00 \[ -\frac {\sqrt {1-\frac {1}{x^2}} x (x+1)}{3 (x-1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcCoth[x]/(1 - x)^2,x]

[Out]

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

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fricas [A]  time = 0.51, size = 31, normalized size = 1.29 \[ -\frac {{\left (x^{2} + 2 \, x + 1\right )} \sqrt {\frac {x - 1}{x + 1}}}{3 \, {\left (x^{2} - 2 \, x + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.14, size = 21, normalized size = 0.88 \[ -\frac {x + 1}{3 \, {\left (x - 1\right )} \sqrt {\frac {x - 1}{x + 1}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(x + 1)/((x - 1)*sqrt((x - 1)/(x + 1)))

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maple [A]  time = 0.03, size = 22, normalized size = 0.92 \[ -\frac {1+x}{3 \left (-1+x \right ) \sqrt {\frac {-1+x}{1+x}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 13, normalized size = 0.54 \[ -\frac {1}{3 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/((x - 1)/(x + 1))^(3/2)

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mupad [B]  time = 0.02, size = 13, normalized size = 0.54 \[ -\frac {1}{3\,{\left (\frac {x-1}{x+1}\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(3*((x - 1)/(x + 1))^(3/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {\frac {x - 1}{x + 1}} \left (x - 1\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sqrt((x - 1)/(x + 1))*(x - 1)**2), x)

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