3.4.81 \(\int \frac {e^{\tanh ^{-1}(x)} x}{\sqrt {1-x}} \, dx\) [381]

Optimal. Leaf size=42 \[ -2 \sqrt {1+x}-\frac {2}{3} (1+x)^{3/2}+2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {1+x}}{\sqrt {2}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 42, 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 = {6264, 81, 52, 65, 212} \begin {gather*} -\frac {2}{3} (x+1)^{3/2}-2 \sqrt {x+1}+2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {x+1}}{\sqrt {2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[x]*x)/Sqrt[1 - x],x]

[Out]

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

Rule 52

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 65

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 81

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 6264

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

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Mathematica [A]
time = 0.01, size = 36, normalized size = 0.86 \begin {gather*} -\frac {2}{3} \sqrt {1+x} (4+x)+2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {1+x}}{\sqrt {2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[1 + x]*(4 + x))/3 + 2*Sqrt[2]*ArcTanh[Sqrt[1 + x]/Sqrt[2]]

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Maple [A]
time = 0.82, size = 61, normalized size = 1.45

method	result	size
default	\(-\frac {2 \sqrt {-x^{2}+1}\, \sqrt {1-x}\, \left (3 \arctanh \left (\frac {\sqrt {1+x}\, \sqrt {2}}{2}\right ) \sqrt {2}-\sqrt {1+x}\, x -4 \sqrt {1+x}\right )}{3 \left (x -1\right ) \sqrt {1+x}}\)	\(61\)
risch	\(\frac {2 \left (x +4\right ) \sqrt {1+x}\, \sqrt {\frac {\left (-x^{2}+1\right ) \left (1-x \right )}{\left (x -1\right )^{2}}}\, \left (x -1\right )}{3 \sqrt {-x^{2}+1}\, \sqrt {1-x}}-\frac {2 \arctanh \left (\frac {\sqrt {1+x}\, \sqrt {2}}{2}\right ) \sqrt {2}\, \sqrt {\frac {\left (-x^{2}+1\right ) \left (1-x \right )}{\left (x -1\right )^{2}}}\, \left (x -1\right )}{\sqrt {-x^{2}+1}\, \sqrt {1-x}}\)	\(106\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 79 vs. \(2 (31) = 62\).
time = 0.34, size = 79, normalized size = 1.88 \begin {gather*} \frac {3 \, \sqrt {2} {\left (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 ) + 2 \, \sqrt {-x^{2} + 1} {\left (x + 4\right )} \sqrt {-x + 1}}{3 \, {\left (x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \left (x + 1\right )}{\sqrt {- \left (x - 1\right ) \left (x + 1\right )} \sqrt {1 - x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.42, size = 44, normalized size = 1.05 \begin {gather*} -\frac {2}{3} \, {\left (x + 1\right )}^{\frac {3}{2}} - \sqrt {2} \log \left (\frac {\sqrt {2} - \sqrt {x + 1}}{\sqrt {2} + \sqrt {x + 1}}\right ) - 2 \, \sqrt {x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x\,\left (x+1\right )}{\sqrt {1-x^2}\,\sqrt {1-x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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