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

Optimal. Leaf size=51 \[ \frac {(x+1)^{3/2}}{2 (1-x)}+\frac {5 \sqrt {x+1}}{2}-\frac {5 \tanh ^{-1}\left (\frac {\sqrt {x+1}}{\sqrt {2}}\right )}{\sqrt {2}} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6129, 78, 50, 63, 206} \[ \frac {(x+1)^{3/2}}{2 (1-x)}+\frac {5 \sqrt {x+1}}{2}-\frac {5 \tanh ^{-1}\left (\frac {\sqrt {x+1}}{\sqrt {2}}\right )}{\sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[x]*x)/(1 - x)^(3/2),x]

[Out]

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 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 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 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 6129

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

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 40, normalized size = 0.78 \[ \frac {\sqrt {x+1} (2 x-3)}{x-1}-\frac {5 \tanh ^{-1}\left (\frac {\sqrt {x+1}}{\sqrt {2}}\right )}{\sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 + x]*(-3 + 2*x))/(-1 + x) - (5*ArcTanh[Sqrt[1 + x]/Sqrt[2]])/Sqrt[2]

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fricas [B]  time = 0.59, size = 91, normalized size = 1.78 \[ \frac {5 \, \sqrt {2} {\left (x^{2} - 2 \, x + 1\right )} \log \left (-\frac {x^{2} + 2 \, \sqrt {2} \sqrt {-x^{2} + 1} \sqrt {-x + 1} + 2 \, x - 3}{x^{2} - 2 \, x + 1}\right ) - 4 \, \sqrt {-x^{2} + 1} {\left (2 \, x - 3\right )} \sqrt {-x + 1}}{4 \, {\left (x^{2} - 2 \, x + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 49, normalized size = 0.96 \[ \frac {5}{4} \, \sqrt {2} \log \left (\frac {\sqrt {2} - \sqrt {x + 1}}{\sqrt {2} + \sqrt {x + 1}}\right ) + 2 \, \sqrt {x + 1} - \frac {\sqrt {x + 1}}{x - 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/4*sqrt(2)*log((sqrt(2) - sqrt(x + 1))/(sqrt(2) + sqrt(x + 1))) + 2*sqrt(x + 1) - sqrt(x + 1)/(x - 1)

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maple [B]  time = 0.04, size = 78, normalized size = 1.53 \[ \frac {\sqrt {-x^{2}+1}\, \sqrt {1-x}\, \left (5 \sqrt {2}\, \arctanh \left (\frac {\sqrt {1+x}\, \sqrt {2}}{2}\right ) x -5 \arctanh \left (\frac {\sqrt {1+x}\, \sqrt {2}}{2}\right ) \sqrt {2}-4 \sqrt {1+x}\, x +6 \sqrt {1+x}\right )}{2 \left (-1+x \right )^{2} \sqrt {1+x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-x^2+1)^(1/2)*(1-x)^(1/2)*(5*2^(1/2)*arctanh(1/2*(1+x)^(1/2)*2^(1/2))*x-5*arctanh(1/2*(1+x)^(1/2)*2^(1/2)
)*2^(1/2)-4*(1+x)^(1/2)*x+6*(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 \[ \int \frac {{\left (x + 1\right )} x}{\sqrt {-x^{2} + 1} {\left (-x + 1\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x\,\left (x+1\right )}{\sqrt {1-x^2}\,{\left (1-x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((x*(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 \[ \int \frac {x \left (x + 1\right )}{\sqrt {- \left (x - 1\right ) \left (x + 1\right )} \left (1 - x\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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