3.3.85 \(\int e^{\coth ^{-1}(x)} (1-x)^2 x \, dx\) [285]

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

[Out]

-1/3*(1-1/x^2)^(3/2)*x^3+1/4*(1-1/x^2)^(3/2)*x^4-1/8*arctanh((1-1/x^2)^(1/2))+1/8*x^2*(1-1/x^2)^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 71, normalized size of antiderivative = 1.00, 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 = {6310, 6313, 849, 821, 272, 43, 65, 212} \begin {gather*} \frac {1}{8} \sqrt {1-\frac {1}{x^2}} x^2-\frac {1}{8} \tanh ^{-1}\left (\sqrt {1-\frac {1}{x^2}}\right )+\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 \end {gather*}

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 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

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 821

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 849

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 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 e^{\coth ^{-1}(x)} (1-x)^2 x \, dx &=\int e^{\coth ^{-1}(x)} \left (1-\frac {1}{x}\right )^2 x^3 \, dx\\ &=-\text {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} \text {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} \text {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} \text {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} \text {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} \text {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*}

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Mathematica [A]
time = 0.02, size = 52, normalized size = 0.73 \begin {gather*} \frac {1}{24} \sqrt {1-\frac {1}{x^2}} x \left (8-3 x-8 x^2+6 x^3\right )-\frac {1}{8} \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)^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 = 0.12, size = 70, normalized size = 0.99

method	result	size
default	\(-\frac {\left (-1+x \right ) \left (-6 x \left (x^{2}-1\right )^{\frac {3}{2}}+8 \left (\left (1+x \right ) \left (-1+x \right )\right )^{\frac {3}{2}}-3 x \sqrt {x^{2}-1}+3 \ln \left (x +\sqrt {x^{2}-1}\right )\right )}{24 \sqrt {\frac {-1+x}{1+x}}\, \sqrt {\left (1+x \right ) \left (-1+x \right )}}\)	\(70\)
risch	\(\frac {\left (6 x^{3}-8 x^{2}-3 x +8\right ) \left (-1+x \right )}{24 \sqrt {\frac {-1+x}{1+x}}}-\frac {\ln \left (x +\sqrt {x^{2}-1}\right ) \sqrt {\left (1+x \right ) \left (-1+x \right )}}{8 \sqrt {\frac {-1+x}{1+x}}\, \left (1+x \right )}\)	\(70\)
trager	\(\frac {\left (1+x \right ) \left (6 x^{3}-8 x^{2}-3 x +8\right ) \sqrt {-\frac {1-x}{1+x}}}{24}-\frac {\ln \left (\sqrt {-\frac {1-x}{1+x}}\, x +\sqrt {-\frac {1-x}{1+x}}+x \right )}{8}\)	\(71\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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] Leaf count of result is larger than twice the leaf count of optimal. 138 vs. \(2 (55) = 110\).
time = 0.26, size = 138, normalized size = 1.94 \begin {gather*} -\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 {gather*}

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 = 0.34, size = 66, normalized size = 0.93 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \left (x - 1\right )^{2}}{\sqrt {\frac {x - 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)**2*x,x)

[Out]

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

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Giac [A]
time = 0.42, size = 72, normalized size = 1.01 \begin {gather*} \frac {1}{24} \, {\left ({\left (2 \, x {\left (\frac {3 \, x}{\mathrm {sgn}\left (x + 1\right )} - \frac {4}{\mathrm {sgn}\left (x + 1\right )}\right )} - \frac {3}{\mathrm {sgn}\left (x + 1\right )}\right )} x + \frac {8}{\mathrm {sgn}\left (x + 1\right )}\right )} \sqrt {x^{2} - 1} + \frac {\log \left ({\left | -x + \sqrt {x^{2} - 1} \right |}\right )}{8 \, \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)^2*x,x, algorithm="giac")

[Out]

1/24*((2*x*(3*x/sgn(x + 1) - 4/sgn(x + 1)) - 3/sgn(x + 1))*x + 8/sgn(x + 1))*sqrt(x^2 - 1) + 1/8*log(abs(-x +
sqrt(x^2 - 1)))/sgn(x + 1)

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Mupad [B]
time = 1.19, size = 118, normalized size = 1.66 \begin {gather*} \frac {\frac {\sqrt {\frac {x-1}{x+1}}}{4}-\frac {11\,{\left (\frac {x-1}{x+1}\right )}^{3/2}}{12}+\frac {53\,{\left (\frac {x-1}{x+1}\right )}^{5/2}}{12}+\frac {{\left (\frac {x-1}{x+1}\right )}^{7/2}}{4}}{\frac {6\,{\left (x-1\right )}^2}{{\left (x+1\right )}^2}-\frac {4\,\left (x-1\right )}{x+1}-\frac {4\,{\left (x-1\right )}^3}{{\left (x+1\right )}^3}+\frac {{\left (x-1\right )}^4}{{\left (x+1\right )}^4}+1}-\frac {\mathrm {atanh}\left (\sqrt {\frac {x-1}{x+1}}\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((x - 1)/(x + 1))^(1/2)/4 - (11*((x - 1)/(x + 1))^(3/2))/12 + (53*((x - 1)/(x + 1))^(5/2))/12 + ((x - 1)/(x +
1))^(7/2)/4)/((6*(x - 1)^2)/(x + 1)^2 - (4*(x - 1))/(x + 1) - (4*(x - 1)^3)/(x + 1)^3 + (x - 1)^4/(x + 1)^4 +
1) - atanh(((x - 1)/(x + 1))^(1/2))/4

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