3.386 \(\int \frac {e^{\tanh ^{-1}(x)}}{(1-x)^{3/2}} \, dx\)

Optimal. Leaf size=37 \[ \frac {\sqrt {x+1}}{1-x}-\frac {\tanh ^{-1}\left (\frac {\sqrt {x+1}}{\sqrt {2}}\right )}{\sqrt {2}} \]

[Out]

-1/2*arctanh(1/2*(1+x)^(1/2)*2^(1/2))*2^(1/2)+(1+x)^(1/2)/(1-x)

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Rubi [A]  time = 0.04, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {6127, 627, 47, 63, 206} \[ \frac {\sqrt {x+1}}{1-x}-\frac {\tanh ^{-1}\left (\frac {\sqrt {x+1}}{\sqrt {2}}\right )}{\sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[x]/(1 - x)^(3/2),x]

[Out]

Sqrt[1 + x]/(1 - x) - ArcTanh[Sqrt[1 + x]/Sqrt[2]]/Sqrt[2]

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])

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 6127

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 36, normalized size = 0.97 \[ -\frac {\sqrt {x+1}}{x-1}-\frac {\tanh ^{-1}\left (\frac {\sqrt {x+1}}{\sqrt {2}}\right )}{\sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[1 + x]/(-1 + x)) - ArcTanh[Sqrt[1 + x]/Sqrt[2]]/Sqrt[2]

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fricas [B]  time = 0.44, size = 85, normalized size = 2.30 \[ \frac {\sqrt {2} {\left (x^{2} - 2 \, x + 1\right )} \log \left (-\frac {x^{2} + 2 \, \sqrt {2} \sqrt {-x^{2} + 1} \sqrt {-x + 1} + 2 \, x - 3}{x^{2} - 2 \, x + 1}\right ) + 4 \, \sqrt {-x^{2} + 1} \sqrt {-x + 1}}{4 \, {\left (x^{2} - 2 \, x + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 42, normalized size = 1.14 \[ \frac {1}{4} \, \sqrt {2} \log \left (\frac {\sqrt {2} - \sqrt {x + 1}}{\sqrt {2} + \sqrt {x + 1}}\right ) - \frac {\sqrt {x + 1}}{x - 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(2)*log((sqrt(2) - sqrt(x + 1))/(sqrt(2) + sqrt(x + 1))) - sqrt(x + 1)/(x - 1)

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maple [B]  time = 0.04, size = 69, normalized size = 1.86 \[ \frac {\sqrt {-x^{2}+1}\, \sqrt {1-x}\, \left (\sqrt {2}\, \arctanh \left (\frac {\sqrt {1+x}\, \sqrt {2}}{2}\right ) x -\arctanh \left (\frac {\sqrt {1+x}\, \sqrt {2}}{2}\right ) \sqrt {2}+2 \sqrt {1+x}\right )}{2 \left (-1+x \right )^{2} \sqrt {1+x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-x^2+1)^(1/2)*(1-x)^(1/2)*(2^(1/2)*arctanh(1/2*(1+x)^(1/2)*2^(1/2))*x-arctanh(1/2*(1+x)^(1/2)*2^(1/2))*2^
(1/2)+2*(1+x)^(1/2))/(-1+x)^2/(1+x)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {x+1}{\sqrt {1-x^2}\,{\left (1-x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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