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

Optimal. Leaf size=55 \[ -\frac {4 \left (\frac {1}{x}+1\right )}{3 \left (1-\frac {1}{x^2}\right )^{3/2}}-\frac {\frac {5}{x}+3}{3 \sqrt {1-\frac {1}{x^2}}}+\tanh ^{-1}\left (\sqrt {1-\frac {1}{x^2}}\right ) \]

[Out]

-4/3*(1/x+1)/(1-1/x^2)^(3/2)+arctanh((1-1/x^2)^(1/2))+1/3*(-3-5/x)/(1-1/x^2)^(1/2)

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Rubi [A]  time = 0.15, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.692, Rules used = {6175, 6178, 852, 1805, 823, 12, 266, 63, 206} \[ -\frac {4 \left (\frac {1}{x}+1\right )}{3 \left (1-\frac {1}{x^2}\right )^{3/2}}-\frac {\frac {5}{x}+3}{3 \sqrt {1-\frac {1}{x^2}}}+\tanh ^{-1}\left (\sqrt {1-\frac {1}{x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcCoth[x]*x)/(1 - x)^2,x]

[Out]

(-4*(1 + x^(-1)))/(3*(1 - x^(-2))^(3/2)) - (3 + 5/x)/(3*Sqrt[1 - x^(-2)]) + ArcTanh[Sqrt[1 - x^(-2)]]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 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 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)} x}{(1-x)^2} \, dx &=\int \frac {e^{\coth ^{-1}(x)}}{\left (1-\frac {1}{x}\right )^2 x} \, dx\\ &=-\operatorname {Subst}\left (\int \frac {\sqrt {1-x^2}}{(1-x)^3 x} \, dx,x,\frac {1}{x}\right )\\ &=-\operatorname {Subst}\left (\int \frac {(1+x)^3}{x \left (1-x^2\right )^{5/2}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {4 \left (1+\frac {1}{x}\right )}{3 \left (1-\frac {1}{x^2}\right )^{3/2}}+\frac {1}{3} \operatorname {Subst}\left (\int \frac {-3-5 x}{x \left (1-x^2\right )^{3/2}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {4 \left (1+\frac {1}{x}\right )}{3 \left (1-\frac {1}{x^2}\right )^{3/2}}-\frac {3+\frac {5}{x}}{3 \sqrt {1-\frac {1}{x^2}}}+\frac {1}{3} \operatorname {Subst}\left (\int -\frac {3}{x \sqrt {1-x^2}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {4 \left (1+\frac {1}{x}\right )}{3 \left (1-\frac {1}{x^2}\right )^{3/2}}-\frac {3+\frac {5}{x}}{3 \sqrt {1-\frac {1}{x^2}}}-\operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-x^2}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {4 \left (1+\frac {1}{x}\right )}{3 \left (1-\frac {1}{x^2}\right )^{3/2}}-\frac {3+\frac {5}{x}}{3 \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 {4 \left (1+\frac {1}{x}\right )}{3 \left (1-\frac {1}{x^2}\right )^{3/2}}-\frac {3+\frac {5}{x}}{3 \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 {4 \left (1+\frac {1}{x}\right )}{3 \left (1-\frac {1}{x^2}\right )^{3/2}}-\frac {3+\frac {5}{x}}{3 \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 = 0.07, size = 43, normalized size = 0.78 \[ \frac {\sqrt {1-\frac {1}{x^2}} (5-7 x) x}{3 (x-1)^2}+\log \left (\left (\sqrt {1-\frac {1}{x^2}}+1\right ) x\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(E^ArcCoth[x]*x)/(1 - x)^2,x]

[Out]

(Sqrt[1 - x^(-2)]*(5 - 7*x)*x)/(3*(-1 + x)^2) + Log[(1 + Sqrt[1 - x^(-2)])*x]

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fricas [A]  time = 0.61, size = 84, normalized size = 1.53 \[ \frac {3 \, {\left (x^{2} - 2 \, x + 1\right )} \log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) - 3 \, {\left (x^{2} - 2 \, x + 1\right )} \log \left (\sqrt {\frac {x - 1}{x + 1}} - 1\right ) - {\left (7 \, x^{2} + 2 \, x - 5\right )} \sqrt {\frac {x - 1}{x + 1}}}{3 \, {\left (x^{2} - 2 \, x + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*(x^2 - 2*x + 1)*log(sqrt((x - 1)/(x + 1)) + 1) - 3*(x^2 - 2*x + 1)*log(sqrt((x - 1)/(x + 1)) - 1) - (7*
x^2 + 2*x - 5)*sqrt((x - 1)/(x + 1)))/(x^2 - 2*x + 1)

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giac [A]  time = 0.15, size = 65, normalized size = 1.18 \[ -\frac {{\left (x + 1\right )} {\left (\frac {6 \, {\left (x - 1\right )}}{x + 1} + 1\right )}}{3 \, {\left (x - 1\right )} \sqrt {\frac {x - 1}{x + 1}}} + \log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) - \log \left ({\left | \sqrt {\frac {x - 1}{x + 1}} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(x + 1)*(6*(x - 1)/(x + 1) + 1)/((x - 1)*sqrt((x - 1)/(x + 1))) + log(sqrt((x - 1)/(x + 1)) + 1) - log(ab
s(sqrt((x - 1)/(x + 1)) - 1))

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maple [B]  time = 0.05, size = 146, normalized size = 2.65 \[ -\frac {3 x \left (x^{2}-1\right )^{\frac {3}{2}}-3 x^{3} \sqrt {x^{2}-1}-3 \ln \left (x +\sqrt {x^{2}-1}\right ) x^{3}-2 \left (x^{2}-1\right )^{\frac {3}{2}}+9 x^{2} \sqrt {x^{2}-1}+9 \ln \left (x +\sqrt {x^{2}-1}\right ) x^{2}-9 x \sqrt {x^{2}-1}-9 \ln \left (x +\sqrt {x^{2}-1}\right ) x +3 \sqrt {x^{2}-1}+3 \ln \left (x +\sqrt {x^{2}-1}\right )}{3 \left (-1+x \right )^{2} \sqrt {\left (1+x \right ) \left (-1+x \right )}\, \sqrt {\frac {-1+x}{1+x}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(3*x*(x^2-1)^(3/2)-3*x^3*(x^2-1)^(1/2)-3*ln(x+(x^2-1)^(1/2))*x^3-2*(x^2-1)^(3/2)+9*x^2*(x^2-1)^(1/2)+9*ln
(x+(x^2-1)^(1/2))*x^2-9*x*(x^2-1)^(1/2)-9*ln(x+(x^2-1)^(1/2))*x+3*(x^2-1)^(1/2)+3*ln(x+(x^2-1)^(1/2)))/(-1+x)^
2/((1+x)*(-1+x))^(1/2)/((-1+x)/(1+x))^(1/2)

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maxima [A]  time = 0.32, size = 56, normalized size = 1.02 \[ -\frac {\frac {6 \, {\left (x - 1\right )}}{x + 1} + 1}{3 \, \left (\frac {x - 1}{x + 1}\right )^{\frac {3}{2}}} + \log \left (\sqrt {\frac {x - 1}{x + 1}} + 1\right ) - \log \left (\sqrt {\frac {x - 1}{x + 1}} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(6*(x - 1)/(x + 1) + 1)/((x - 1)/(x + 1))^(3/2) + log(sqrt((x - 1)/(x + 1)) + 1) - log(sqrt((x - 1)/(x +
1)) - 1)

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mupad [B]  time = 0.04, size = 40, normalized size = 0.73 \[ 2\,\mathrm {atanh}\left (\sqrt {\frac {x-1}{x+1}}\right )-\frac {\frac {2\,\left (x-1\right )}{x+1}+\frac {1}{3}}{{\left (\frac {x-1}{x+1}\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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