3.1.53 \(\int e^{-3 \coth ^{-1}(a x)} \, dx\) [53]

Optimal. Leaf size=60 \[ \frac {4 \sqrt {1-\frac {1}{a^2 x^2}}}{a+\frac {1}{x}}+\sqrt {1-\frac {1}{a^2 x^2}} x-\frac {3 \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a} \]

[Out]

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

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Rubi [A]
time = 0.55, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.875, Rules used = {6303, 6874, 270, 272, 65, 214, 665} \begin {gather*} x \sqrt {1-\frac {1}{a^2 x^2}}+\frac {4 \sqrt {1-\frac {1}{a^2 x^2}}}{a+\frac {1}{x}}-\frac {3 \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(-3*ArcCoth[a*x]),x]

[Out]

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

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 214

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

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*
c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 665

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^m*((a + c*x^2)^(p + 1)/
(2*c*d*(p + 1))), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p
+ 2, 0]

Rule 6303

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.)), x_Symbol] :> -Subst[Int[(1 + x/a)^((n + 1)/2)/(x^2*(1 - x/a)^((n - 1)/2)*Sq
rt[1 - x^2/a^2]), x], x, 1/x] /; FreeQ[a, x] && IntegerQ[(n - 1)/2]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 54, normalized size = 0.90 \begin {gather*} \frac {\sqrt {1-\frac {1}{a^2 x^2}} x (5+a x)}{1+a x}-\frac {3 \log \left (a \left (1+\sqrt {1-\frac {1}{a^2 x^2}}\right ) x\right )}{a} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(-3*ArcCoth[a*x]),x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(247\) vs. \(2(54)=108\).
time = 0.09, size = 248, normalized size = 4.13

method	result	size
risch	\(\frac {\left (a x +1\right ) \sqrt {\frac {a x -1}{a x +1}}}{a}+\frac {\left (-\frac {3 \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}-1}\right )}{\sqrt {a^{2}}}+\frac {4 \sqrt {a^{2} \left (x +\frac {1}{a}\right )^{2}-2 a \left (x +\frac {1}{a}\right )}}{a^{2} \left (x +\frac {1}{a}\right )}\right ) \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}}{a x -1}\)	\(127\)
default	\(-\frac {\left (-3 \sqrt {\left (a x +1\right ) \left (a x -1\right )}\, \sqrt {a^{2}}\, a^{2} x^{2}+3 \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}}{\sqrt {a^{2}}}\right ) a^{3} x^{2}+2 \left (\left (a x +1\right ) \left (a x -1\right )\right )^{\frac {3}{2}} \sqrt {a^{2}}-6 \sqrt {\left (a x +1\right ) \left (a x -1\right )}\, \sqrt {a^{2}}\, a x +6 \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}}{\sqrt {a^{2}}}\right ) a^{2} x -3 \sqrt {a^{2}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}+3 a \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}}{\sqrt {a^{2}}}\right )\right ) \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{a \sqrt {a^{2}}\, \left (a x -1\right ) \sqrt {\left (a x +1\right ) \left (a x -1\right )}}\)	\(248\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a*x-1)/(a*x+1))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-(-3*((a*x+1)*(a*x-1))^(1/2)*(a^2)^(1/2)*a^2*x^2+3*ln((a^2*x+(a^2)^(1/2)*((a*x+1)*(a*x-1))^(1/2))/(a^2)^(1/2))
*a^3*x^2+2*((a*x+1)*(a*x-1))^(3/2)*(a^2)^(1/2)-6*((a*x+1)*(a*x-1))^(1/2)*(a^2)^(1/2)*a*x+6*ln((a^2*x+(a^2)^(1/
2)*((a*x+1)*(a*x-1))^(1/2))/(a^2)^(1/2))*a^2*x-3*(a^2)^(1/2)*((a*x+1)*(a*x-1))^(1/2)+3*a*ln((a^2*x+(a^2)^(1/2)
*((a*x+1)*(a*x-1))^(1/2))/(a^2)^(1/2)))/a*((a*x-1)/(a*x+1))^(3/2)/(a^2)^(1/2)/(a*x-1)/((a*x+1)*(a*x-1))^(1/2)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 111 vs. \(2 (54) = 108\).
time = 0.25, size = 111, normalized size = 1.85 \begin {gather*} -a {\left (\frac {2 \, \sqrt {\frac {a x - 1}{a x + 1}}}{\frac {{\left (a x - 1\right )} a^{2}}{a x + 1} - a^{2}} + \frac {3 \, \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{2}} - \frac {3 \, \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{2}} - \frac {4 \, \sqrt {\frac {a x - 1}{a x + 1}}}{a^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.33, size = 66, normalized size = 1.10 \begin {gather*} \frac {{\left (a x + 5\right )} \sqrt {\frac {a x - 1}{a x + 1}} - 3 \, \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) + 3 \, \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a*x + 5)*sqrt((a*x - 1)/(a*x + 1)) - 3*log(sqrt((a*x - 1)/(a*x + 1)) + 1) + 3*log(sqrt((a*x - 1)/(a*x + 1))
- 1))/a

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef

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Mupad [B]
time = 0.04, size = 78, normalized size = 1.30 \begin {gather*} \frac {2\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{a-\frac {a\,\left (a\,x-1\right )}{a\,x+1}}+\frac {4\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{a}-\frac {6\,\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a*x - 1)/(a*x + 1))^(3/2),x)

[Out]

(2*((a*x - 1)/(a*x + 1))^(1/2))/(a - (a*(a*x - 1))/(a*x + 1)) + (4*((a*x - 1)/(a*x + 1))^(1/2))/a - (6*atanh((
(a*x - 1)/(a*x + 1))^(1/2)))/a

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