3.322 \(\int e^{\coth ^{-1}(x)} (1-x)^{3/2} \, dx\)

Optimal. Leaf size=68 \[ \frac{2 \left (\frac{1}{x}+1\right )^{3/2} (1-x)^{3/2} x}{5 \left (1-\frac{1}{x}\right )^{3/2}}-\frac{14 \left (\frac{1}{x}+1\right )^{3/2} (1-x)^{3/2}}{15 \left (1-\frac{1}{x}\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0991177, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {6176, 6181, 78, 37} \[ \frac{2 \left (\frac{1}{x}+1\right )^{3/2} (1-x)^{3/2} x}{5 \left (1-\frac{1}{x}\right )^{3/2}}-\frac{14 \left (\frac{1}{x}+1\right )^{3/2} (1-x)^{3/2}}{15 \left (1-\frac{1}{x}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCoth[x]*(1 - x)^(3/2),x]

[Out]

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

Rule 6176

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 6181

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_), x_Symbol] :> -Dist[c^p*x^m*(1/x)^m, Sub
st[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]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
 n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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]

Rubi steps

\begin{align*} \int e^{\coth ^{-1}(x)} (1-x)^{3/2} \, dx &=\frac{(1-x)^{3/2} \int e^{\coth ^{-1}(x)} \left (1-\frac{1}{x}\right )^{3/2} x^{3/2} \, dx}{\left (1-\frac{1}{x}\right )^{3/2} x^{3/2}}\\ &=-\frac{\left ((1-x)^{3/2} \left (\frac{1}{x}\right )^{3/2}\right ) \operatorname{Subst}\left (\int \frac{(1-x) \sqrt{1+x}}{x^{7/2}} \, dx,x,\frac{1}{x}\right )}{\left (1-\frac{1}{x}\right )^{3/2}}\\ &=\frac{2 \left (1+\frac{1}{x}\right )^{3/2} (1-x)^{3/2} x}{5 \left (1-\frac{1}{x}\right )^{3/2}}+\frac{\left (7 (1-x)^{3/2} \left (\frac{1}{x}\right )^{3/2}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1+x}}{x^{5/2}} \, dx,x,\frac{1}{x}\right )}{5 \left (1-\frac{1}{x}\right )^{3/2}}\\ &=-\frac{14 \left (1+\frac{1}{x}\right )^{3/2} (1-x)^{3/2}}{15 \left (1-\frac{1}{x}\right )^{3/2}}+\frac{2 \left (1+\frac{1}{x}\right )^{3/2} (1-x)^{3/2} x}{5 \left (1-\frac{1}{x}\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0170106, size = 41, normalized size = 0.6 \[ -\frac{2 \sqrt{\frac{1}{x}+1} \sqrt{1-x} (x+1) (3 x-7)}{15 \sqrt{\frac{x-1}{x}}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcCoth[x]*(1 - x)^(3/2),x]

[Out]

(-2*Sqrt[1 + x^(-1)]*Sqrt[1 - x]*(1 + x)*(-7 + 3*x))/(15*Sqrt[(-1 + x)/x])

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Maple [A]  time = 0.058, size = 29, normalized size = 0.4 \begin{align*} -{\frac{ \left ( 2+2\,x \right ) \left ( 3\,x-7 \right ) }{15}\sqrt{1-x}{\frac{1}{\sqrt{{\frac{-1+x}{1+x}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(1+x)*(3*x-7)*(1-x)^(1/2)/((-1+x)/(1+x))^(1/2)

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Maxima [C]  time = 1.05526, size = 23, normalized size = 0.34 \begin{align*} -\frac{1}{15} \,{\left (6 i \, x^{2} - 8 i \, x - 14 i\right )} \sqrt{x + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((-1+x)/(1+x))^(1/2)*(1-x)^(3/2),x, algorithm="maxima")

[Out]

-1/15*(6*I*x^2 - 8*I*x - 14*I)*sqrt(x + 1)

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Fricas [A]  time = 1.62552, size = 101, normalized size = 1.49 \begin{align*} -\frac{2 \,{\left (3 \, x^{3} - x^{2} - 11 \, x - 7\right )} \sqrt{-x + 1} \sqrt{\frac{x - 1}{x + 1}}}{15 \,{\left (x - 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)^(3/2),x, algorithm="fricas")

[Out]

-2/15*(3*x^3 - x^2 - 11*x - 7)*sqrt(-x + 1)*sqrt((x - 1)/(x + 1))/(x - 1)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((-1+x)/(1+x))**(1/2)*(1-x)**(3/2),x)

[Out]

Timed out

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Giac [C]  time = 1.17059, size = 59, normalized size = 0.87 \begin{align*} \frac{1}{15} \,{\left (-16 i \, \sqrt{2} + \frac{2 \,{\left (3 \,{\left (x + 1\right )}^{2} \sqrt{-x - 1} + 10 \,{\left (-x - 1\right )}^{\frac{3}{2}}\right )}}{\mathrm{sgn}\left (-x - 1\right )}\right )} \mathrm{sgn}\left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((-1+x)/(1+x))^(1/2)*(1-x)^(3/2),x, algorithm="giac")

[Out]

1/15*(-16*I*sqrt(2) + 2*(3*(x + 1)^2*sqrt(-x - 1) + 10*(-x - 1)^(3/2))/sgn(-x - 1))*sgn(x)