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

Optimal. Leaf size=71 \[ \frac{1}{4} \left (1-\frac{1}{x^2}\right )^{3/2} x^4-\frac{1}{3} \left (1-\frac{1}{x^2}\right )^{3/2} x^3+\frac{1}{8} \sqrt{1-\frac{1}{x^2}} x^2-\frac{1}{8} \tanh ^{-1}\left (\sqrt{1-\frac{1}{x^2}}\right ) \]

[Out]

(Sqrt[1 - x^(-2)]*x^2)/8 - ((1 - x^(-2))^(3/2)*x^3)/3 + ((1 - x^(-2))^(3/2)*x^4)/4 - ArcTanh[Sqrt[1 - x^(-2)]]
/8

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Rubi [A]  time = 0.116344, antiderivative size = 71, 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, 6178, 835, 807, 266, 47, 63, 206} \[ \frac{1}{4} \left (1-\frac{1}{x^2}\right )^{3/2} x^4-\frac{1}{3} \left (1-\frac{1}{x^2}\right )^{3/2} x^3+\frac{1}{8} \sqrt{1-\frac{1}{x^2}} x^2-\frac{1}{8} \tanh ^{-1}\left (\sqrt{1-\frac{1}{x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCoth[x]*(1 - x)^2*x,x]

[Out]

(Sqrt[1 - x^(-2)]*x^2)/8 - ((1 - x^(-2))^(3/2)*x^3)/3 + ((1 - x^(-2))^(3/2)*x^4)/4 - ArcTanh[Sqrt[1 - x^(-2)]]
/8

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]

Rule 835

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))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

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 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])

Rubi steps

\begin{align*} \int e^{\coth ^{-1}(x)} (1-x)^2 x \, dx &=\int e^{\coth ^{-1}(x)} \left (1-\frac{1}{x}\right )^2 x^3 \, dx\\ &=-\operatorname{Subst}\left (\int \frac{(1-x) \sqrt{1-x^2}}{x^5} \, dx,x,\frac{1}{x}\right )\\ &=\frac{1}{4} \left (1-\frac{1}{x^2}\right )^{3/2} x^4+\frac{1}{4} \operatorname{Subst}\left (\int \frac{(4-x) \sqrt{1-x^2}}{x^4} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{1}{3} \left (1-\frac{1}{x^2}\right )^{3/2} x^3+\frac{1}{4} \left (1-\frac{1}{x^2}\right )^{3/2} x^4-\frac{1}{4} \operatorname{Subst}\left (\int \frac{\sqrt{1-x^2}}{x^3} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{1}{3} \left (1-\frac{1}{x^2}\right )^{3/2} x^3+\frac{1}{4} \left (1-\frac{1}{x^2}\right )^{3/2} x^4-\frac{1}{8} \operatorname{Subst}\left (\int \frac{\sqrt{1-x}}{x^2} \, dx,x,\frac{1}{x^2}\right )\\ &=\frac{1}{8} \sqrt{1-\frac{1}{x^2}} x^2-\frac{1}{3} \left (1-\frac{1}{x^2}\right )^{3/2} x^3+\frac{1}{4} \left (1-\frac{1}{x^2}\right )^{3/2} x^4+\frac{1}{16} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x} x} \, dx,x,\frac{1}{x^2}\right )\\ &=\frac{1}{8} \sqrt{1-\frac{1}{x^2}} x^2-\frac{1}{3} \left (1-\frac{1}{x^2}\right )^{3/2} x^3+\frac{1}{4} \left (1-\frac{1}{x^2}\right )^{3/2} x^4-\frac{1}{8} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sqrt{1-\frac{1}{x^2}}\right )\\ &=\frac{1}{8} \sqrt{1-\frac{1}{x^2}} x^2-\frac{1}{3} \left (1-\frac{1}{x^2}\right )^{3/2} x^3+\frac{1}{4} \left (1-\frac{1}{x^2}\right )^{3/2} x^4-\frac{1}{8} \tanh ^{-1}\left (\sqrt{1-\frac{1}{x^2}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0391362, size = 52, normalized size = 0.73 \[ \frac{1}{24} \sqrt{1-\frac{1}{x^2}} x \left (6 x^3-8 x^2-3 x+8\right )-\frac{1}{8} \log \left (\left (\sqrt{1-\frac{1}{x^2}}+1\right ) x\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcCoth[x]*(1 - x)^2*x,x]

[Out]

(Sqrt[1 - x^(-2)]*x*(8 - 3*x - 8*x^2 + 6*x^3))/24 - Log[(1 + Sqrt[1 - x^(-2)])*x]/8

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Maple [A]  time = 0.118, size = 70, normalized size = 1. \begin{align*}{\frac{-1+x}{24} \left ( 6\,x \left ({x}^{2}-1 \right ) ^{3/2}-8\, \left ( \left ( 1+x \right ) \left ( -1+x \right ) \right ) ^{3/2}+3\,x\sqrt{{x}^{2}-1}-3\,\ln \left ( x+\sqrt{{x}^{2}-1} \right ) \right ){\frac{1}{\sqrt{{\frac{-1+x}{1+x}}}}}{\frac{1}{\sqrt{ \left ( 1+x \right ) \left ( -1+x \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(-1+x)*(6*x*(x^2-1)^(3/2)-8*((1+x)*(-1+x))^(3/2)+3*x*(x^2-1)^(1/2)-3*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 = 1.07647, size = 186, normalized size = 2.62 \begin{align*} -\frac{3 \, \left (\frac{x - 1}{x + 1}\right )^{\frac{7}{2}} + 53 \, \left (\frac{x - 1}{x + 1}\right )^{\frac{5}{2}} - 11 \, \left (\frac{x - 1}{x + 1}\right )^{\frac{3}{2}} + 3 \, \sqrt{\frac{x - 1}{x + 1}}}{12 \,{\left (\frac{4 \,{\left (x - 1\right )}}{x + 1} - \frac{6 \,{\left (x - 1\right )}^{2}}{{\left (x + 1\right )}^{2}} + \frac{4 \,{\left (x - 1\right )}^{3}}{{\left (x + 1\right )}^{3}} - \frac{{\left (x - 1\right )}^{4}}{{\left (x + 1\right )}^{4}} - 1\right )}} - \frac{1}{8} \, \log \left (\sqrt{\frac{x - 1}{x + 1}} + 1\right ) + \frac{1}{8} \, \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)*(1-x)^2*x,x, algorithm="maxima")

[Out]

-1/12*(3*((x - 1)/(x + 1))^(7/2) + 53*((x - 1)/(x + 1))^(5/2) - 11*((x - 1)/(x + 1))^(3/2) + 3*sqrt((x - 1)/(x
+ 1)))/(4*(x - 1)/(x + 1) - 6*(x - 1)^2/(x + 1)^2 + 4*(x - 1)^3/(x + 1)^3 - (x - 1)^4/(x + 1)^4 - 1) - 1/8*lo
g(sqrt((x - 1)/(x + 1)) + 1) + 1/8*log(sqrt((x - 1)/(x + 1)) - 1)

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Fricas [A]  time = 2.00583, size = 185, normalized size = 2.61 \begin{align*} \frac{1}{24} \,{\left (6 \, x^{4} - 2 \, x^{3} - 11 \, x^{2} + 5 \, x + 8\right )} \sqrt{\frac{x - 1}{x + 1}} - \frac{1}{8} \, \log \left (\sqrt{\frac{x - 1}{x + 1}} + 1\right ) + \frac{1}{8} \, \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)*(1-x)^2*x,x, algorithm="fricas")

[Out]

1/24*(6*x^4 - 2*x^3 - 11*x^2 + 5*x + 8)*sqrt((x - 1)/(x + 1)) - 1/8*log(sqrt((x - 1)/(x + 1)) + 1) + 1/8*log(s
qrt((x - 1)/(x + 1)) - 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \left (x - 1\right )^{2}}{\sqrt{\frac{x - 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)*(1-x)**2*x,x)

[Out]

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

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Giac [B]  time = 1.17352, size = 176, normalized size = 2.48 \begin{align*} -\frac{\frac{11 \,{\left (x - 1\right )} \sqrt{\frac{x - 1}{x + 1}}}{x + 1} - \frac{53 \,{\left (x - 1\right )}^{2} \sqrt{\frac{x - 1}{x + 1}}}{{\left (x + 1\right )}^{2}} - \frac{3 \,{\left (x - 1\right )}^{3} \sqrt{\frac{x - 1}{x + 1}}}{{\left (x + 1\right )}^{3}} - 3 \, \sqrt{\frac{x - 1}{x + 1}}}{12 \,{\left (\frac{x - 1}{x + 1} - 1\right )}^{4}} - \frac{1}{8} \, \log \left (\sqrt{\frac{x - 1}{x + 1}} + 1\right ) + \frac{1}{8} \, \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)*(1-x)^2*x,x, algorithm="giac")

[Out]

-1/12*(11*(x - 1)*sqrt((x - 1)/(x + 1))/(x + 1) - 53*(x - 1)^2*sqrt((x - 1)/(x + 1))/(x + 1)^2 - 3*(x - 1)^3*s
qrt((x - 1)/(x + 1))/(x + 1)^3 - 3*sqrt((x - 1)/(x + 1)))/((x - 1)/(x + 1) - 1)^4 - 1/8*log(sqrt((x - 1)/(x +
1)) + 1) + 1/8*log(abs(sqrt((x - 1)/(x + 1)) - 1))