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]

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

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Rubi [A]  time = 0.10, antiderivative size = 68, normalized size of antiderivative = 1.00, 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 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 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 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]

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*}

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Mathematica [A]  time = 0.02, size = 41, normalized size = 0.60 \[ -\frac {2 \sqrt {\frac {1}{x}+1} \sqrt {1-x} (x+1) (3 x-7)}{15 \sqrt {\frac {x-1}{x}}} \]

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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fricas [A]  time = 0.41, size = 40, normalized size = 0.59 \[ -\frac {2 \, {\left (3 \, x^{3} - x^{2} - 11 \, x - 7\right )} \sqrt {-x + 1} \sqrt {\frac {x - 1}{x + 1}}}{15 \, {\left (x - 1\right )}} \]

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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giac [C]  time = 0.16, size = 44, normalized size = 0.65 \[ \frac {1}{15} \, {\left (-16 i \, \sqrt {2} + \frac {2 \, {\left (3 \, {\left (x + 1\right )}^{2} \sqrt {-x - 1} + 10 \, {\left (-x - 1\right )}^{\frac {3}{2}}\right )}}{\mathrm {sgn}\left (-x - 1\right )}\right )} \mathrm {sgn}\relax (x) \]

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)

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maple [A]  time = 0.04, size = 29, normalized size = 0.43 \[ -\frac {2 \left (1+x \right ) \left (3 x -7\right ) \sqrt {1-x}}{15 \sqrt {\frac {-1+x}{1+x}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((-1+x)/(1+x))^(1/2)*(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 = 0.46, size = 17, normalized size = 0.25 \[ -\frac {1}{15} \, {\left (6 i \, x^{2} - 8 i \, x - 14 i\right )} \sqrt {x + 1} \]

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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mupad [B]  time = 1.25, size = 30, normalized size = 0.44 \[ \frac {2\,\left (3\,x-7\right )\,\sqrt {\frac {x-1}{x+1}}\,{\left (x+1\right )}^2}{15\,\sqrt {1-x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (1 - x\right )^{\frac {3}{2}}}{\sqrt {\frac {x - 1}{x + 1}}}\, dx \]

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]

Integral((1 - x)**(3/2)/sqrt((x - 1)/(x + 1)), x)

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