3.1.20 \(\int e^{3 \coth ^{-1}(a x)} \, dx\) [20]

Optimal. Leaf size=62 \[ -\frac {4 \sqrt {1-\frac {1}{a^2 x^2}}}{a-\frac {1}{x}}+\sqrt {1-\frac {1}{a^2 x^2}} x+\frac {3 \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a} \]

[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 = 0.59, antiderivative size = 62, 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, 665, 270, 272, 65, 214} \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 {4}{a (a-x) \sqrt {1-\frac {x^2}{a^2}}}+\frac {1}{x^2 \sqrt {1-\frac {x^2}{a^2}}}+\frac {3}{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 = 0.87 \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. \(2(56)=112\).
time = 0.09, size = 248, normalized size = 4.00

method	result	size
risch	\(\frac {a x -1}{a \sqrt {\frac {a x -1}{a x +1}}}+\frac {\left (\frac {3 \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}-1}\right )}{\sqrt {a^{2}}}-\frac {4 \sqrt {a^{2} \left (x -\frac {1}{a}\right )^{2}+2 a \left (x -\frac {1}{a}\right )}}{a^{2} \left (x -\frac {1}{a}\right )}\right ) \sqrt {\left (a x +1\right ) \left (a x -1\right )}}{\left (a x +1\right ) \sqrt {\frac {a x -1}{a x +1}}}\)	\(133\)
default	\(-\frac {-3 \sqrt {\left (a x +1\right ) \left (a x -1\right )}\, \sqrt {a^{2}}\, a^{2} x^{2}-3 \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}}{\sqrt {a^{2}}}\right ) a^{3} x^{2}+2 \left (\left (a x +1\right ) \left (a x -1\right )\right )^{\frac {3}{2}} \sqrt {a^{2}}+6 \sqrt {\left (a x +1\right ) \left (a x -1\right )}\, \sqrt {a^{2}}\, a x +6 \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}}{\sqrt {a^{2}}}\right ) a^{2} x -3 \sqrt {a^{2}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}-3 a \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}}{\sqrt {a^{2}}}\right )}{a \sqrt {a^{2}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}\, \left (a x +1\right ) \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}\)	\(248\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.25, size = 110, normalized size = 1.77 \begin {gather*} -a {\left (\frac {2 \, {\left (\frac {3 \, {\left (a x - 1\right )}}{a x + 1} - 2\right )}}{a^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} - a^{2} \sqrt {\frac {a x - 1}{a x + 1}}} - \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}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.37, size = 92, normalized size = 1.48 \begin {gather*} \frac {3 \, {\left (a x - 1\right )} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 3 \, {\left (a x - 1\right )} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right ) + {\left (a^{2} x^{2} - 4 \, a x - 5\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{a^{2} x - a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\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(1/((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(1/((a*x-1)/(a*x+1))^(3/2),x, algorithm="giac")

[Out]

undef

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Mupad [B]
time = 1.29, size = 59, normalized size = 0.95 \begin {gather*} \frac {2\,a\,x+12\,\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )\,\sqrt {\frac {a\,x-1}{a\,x+1}}-10}{2\,a\,\sqrt {\frac {a\,x-1}{a\,x+1}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a*x - 1)/(a*x + 1))^(3/2),x)

[Out]

(2*a*x + 12*atanh(((a*x - 1)/(a*x + 1))^(1/2))*((a*x - 1)/(a*x + 1))^(1/2) - 10)/(2*a*((a*x - 1)/(a*x + 1))^(1
/2))

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