3.4.25 \(\int e^{\coth ^{-1}(x)} \sqrt {1-x} x \, dx\) [325]

Optimal. Leaf size=71 \[ -\frac {4 \left (1+\frac {1}{x}\right )^{3/2} \sqrt {1-x} x}{15 \sqrt {1-\frac {1}{x}}}+\frac {2 \left (1+\frac {1}{x}\right )^{3/2} \sqrt {1-x} x^2}{5 \sqrt {1-\frac {1}{x}}} \]

[Out]

-4/15*(1+1/x)^(3/2)*x*(1-x)^(1/2)/(1-1/x)^(1/2)+2/5*(1+1/x)^(3/2)*x^2*(1-x)^(1/2)/(1-1/x)^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {6311, 6316, 47, 37} \begin {gather*} \frac {2 \left (\frac {1}{x}+1\right )^{3/2} \sqrt {1-x} x^2}{5 \sqrt {1-\frac {1}{x}}}-\frac {4 \left (\frac {1}{x}+1\right )^{3/2} \sqrt {1-x} x}{15 \sqrt {1-\frac {1}{x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^ArcCoth[x]*Sqrt[1 - x]*x,x]

[Out]

(-4*(1 + x^(-1))^(3/2)*Sqrt[1 - x]*x)/(15*Sqrt[1 - x^(-1)]) + (2*(1 + x^(-1))^(3/2)*Sqrt[1 - x]*x^2)/(5*Sqrt[1
- x^(-1)])

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 47

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

Rule 6311

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_), x_Symbol] :> Dist[(c + d*x)^p/(x^p*(1 + c/(d
*x))^p), Int[u*x^p*(1 + c/(d*x))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c^2 - d^
2, 0] &&  !IntegerQ[n/2] &&  !IntegerQ[p]

Rule 6316

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_), x_Symbol] :> Dist[(-c^p)*x^m*(1/x)^m, S
ubst[Int[(1 + d*(x/c))^p*((1 + x/a)^(n/2)/(x^(m + 2)*(1 - x/a)^(n/2))), x], x, 1/x], x] /; FreeQ[{a, c, d, m,
n, p}, x] && EqQ[c^2 - a^2*d^2, 0] &&  !IntegerQ[n/2] && (IntegerQ[p] || GtQ[c, 0]) &&  !IntegerQ[m]

Rubi steps

\begin {align*} \int e^{\coth ^{-1}(x)} \sqrt {1-x} x \, dx &=\frac {\sqrt {1-x} \int e^{\coth ^{-1}(x)} \sqrt {1-\frac {1}{x}} x^{3/2} \, dx}{\sqrt {1-\frac {1}{x}} \sqrt {x}}\\ &=-\frac {\left (\sqrt {1-x} \sqrt {\frac {1}{x}}\right ) \text {Subst}\left (\int \frac {\sqrt {1+x}}{x^{7/2}} \, dx,x,\frac {1}{x}\right )}{\sqrt {1-\frac {1}{x}}}\\ &=\frac {2 \left (1+\frac {1}{x}\right )^{3/2} \sqrt {1-x} x^2}{5 \sqrt {1-\frac {1}{x}}}+\frac {\left (2 \sqrt {1-x} \sqrt {\frac {1}{x}}\right ) \text {Subst}\left (\int \frac {\sqrt {1+x}}{x^{5/2}} \, dx,x,\frac {1}{x}\right )}{5 \sqrt {1-\frac {1}{x}}}\\ &=-\frac {4 \left (1+\frac {1}{x}\right )^{3/2} \sqrt {1-x} x}{15 \sqrt {1-\frac {1}{x}}}+\frac {2 \left (1+\frac {1}{x}\right )^{3/2} \sqrt {1-x} x^2}{5 \sqrt {1-\frac {1}{x}}}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 41, normalized size = 0.58 \begin {gather*} \frac {2 \sqrt {1+\frac {1}{x}} \sqrt {1-x} (1+x) (-2+3 x)}{15 \sqrt {\frac {-1+x}{x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^ArcCoth[x]*Sqrt[1 - x]*x,x]

[Out]

(2*Sqrt[1 + x^(-1)]*Sqrt[1 - x]*(1 + x)*(-2 + 3*x))/(15*Sqrt[(-1 + x)/x])

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Maple [A]
time = 0.10, size = 29, normalized size = 0.41

method	result	size
gosper	\(\frac {2 \left (1+x \right ) \left (3 x -2\right ) \sqrt {1-x}}{15 \sqrt {\frac {-1+x}{1+x}}}\)	\(29\)
default	\(\frac {2 \left (1+x \right ) \left (3 x -2\right ) \sqrt {1-x}}{15 \sqrt {\frac {-1+x}{1+x}}}\)	\(29\)
risch	\(-\frac {2 \sqrt {\frac {\left (1+x \right ) \left (1-x \right )}{-1+x}}\, \left (-1+x \right ) \left (3 x^{2}+x -2\right )}{15 \sqrt {\frac {-1+x}{1+x}}\, \sqrt {1-x}\, \sqrt {-1-x}}\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(1+x)*(3*x-2)*(1-x)^(1/2)/((-1+x)/(1+x))^(1/2)

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Maxima [C] Result contains complex when optimal does not.
time = 0.26, size = 17, normalized size = 0.24 \begin {gather*} -\frac {2}{15} \, {\left (-3 i \, x^{2} - i \, x + 2 i\right )} \sqrt {x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(-3*I*x^2 - I*x + 2*I)*sqrt(x + 1)

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Fricas [A]
time = 0.34, size = 40, normalized size = 0.56 \begin {gather*} \frac {2 \, {\left (3 \, x^{3} + 4 \, x^{2} - x - 2\right )} \sqrt {-x + 1} \sqrt {\frac {x - 1}{x + 1}}}{15 \, {\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)*x*(1-x)^(1/2),x, algorithm="fricas")

[Out]

2/15*(3*x^3 + 4*x^2 - x - 2)*sqrt(-x + 1)*sqrt((x - 1)/(x + 1))/(x - 1)

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Sympy [A]
time = 12.49, size = 80, normalized size = 1.13 \begin {gather*} \frac {2 \left (1 - x\right )^{\frac {5}{2}}}{5 \sqrt {- \frac {x}{- x - 1} + \frac {1}{- x - 1}}} - \frac {14 \left (1 - x\right )^{\frac {3}{2}}}{15 \sqrt {- \frac {x}{- x - 1} + \frac {1}{- x - 1}}} + \frac {4 \sqrt {1 - x}}{15 \sqrt {- \frac {x}{- x - 1} + \frac {1}{- x - 1}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(1 - x)**(5/2)/(5*sqrt(-x/(-x - 1) + 1/(-x - 1))) - 14*(1 - x)**(3/2)/(15*sqrt(-x/(-x - 1) + 1/(-x - 1))) +
4*sqrt(1 - x)/(15*sqrt(-x/(-x - 1) + 1/(-x - 1)))

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Giac [C] Result contains complex when optimal does not.
time = 0.40, size = 49, normalized size = 0.69 \begin {gather*} -\frac {4}{15} i \, \sqrt {2} \mathrm {sgn}\left (x + 1\right ) + \frac {2 \, {\left (3 \, {\left (x + 1\right )}^{2} \sqrt {-x - 1} + 5 \, {\left (-x - 1\right )}^{\frac {3}{2}} - 2 i \, \sqrt {2}\right )} \mathrm {sgn}\left (x\right )}{15 \, \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)*x*(1-x)^(1/2),x, algorithm="giac")

[Out]

-4/15*I*sqrt(2)*sgn(x + 1) + 2/15*(3*(x + 1)^2*sqrt(-x - 1) + 5*(-x - 1)^(3/2) - 2*I*sqrt(2))*sgn(x)/sgn(x + 1
)

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Mupad [B]
time = 1.27, size = 30, normalized size = 0.42 \begin {gather*} -\frac {2\,\left (3\,x-2\right )\,\sqrt {\frac {x-1}{x+1}}\,{\left (x+1\right )}^2}{15\,\sqrt {1-x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*(3*x - 2)*((x - 1)/(x + 1))^(1/2)*(x + 1)^2)/(15*(1 - x)^(1/2))

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