### 3.201 $$\int \frac{e^{-\coth ^{-1}(a x)}}{c-a c x} \, dx$$

Optimal. Leaf size=23 $-\frac{\tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{a c}$

[Out]

-(ArcTanh[Sqrt[1 - 1/(a^2*x^2)]]/(a*c))

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Rubi [A]  time = 0.100635, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.278, Rules used = {6175, 6178, 266, 63, 208} $-\frac{\tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{a c}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/(E^ArcCoth[a*x]*(c - a*c*x)),x]

[Out]

-(ArcTanh[Sqrt[1 - 1/(a^2*x^2)]]/(a*c))

Rule 6175

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

Rule 6178

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_.), x_Symbol] :> -Dist[c^n, Subst[Int[((c
+ d*x)^(p - n)*(1 - x^2/a^2)^(n/2))/x^(m + 2), x], x, 1/x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[c + a*d, 0] &&
IntegerQ[(n - 1)/2] && IntegerQ[m] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p, n/2 + 1] || LtQ[-5, m, -1]) && In
tegerQ[2*p]

Rule 266

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 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 208

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

Rubi steps

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

Mathematica [A]  time = 0.0383503, size = 34, normalized size = 1.48 $-\frac{\log \left (a x \left (\sqrt{\frac{a^2 x^2-1}{a^2 x^2}}+1\right )\right )}{a c}$

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^ArcCoth[a*x]*(c - a*c*x)),x]

[Out]

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

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Maple [B]  time = 0.181, size = 76, normalized size = 3.3 \begin{align*} -{\frac{ax+1}{c}\sqrt{{\frac{ax-1}{ax+1}}}\ln \left ({ \left ({a}^{2}x+\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) } \right ){\frac{1}{\sqrt{{a}^{2}}}}} \right ){\frac{1}{\sqrt{{a}^{2}}}}{\frac{1}{\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.01962, size = 74, normalized size = 3.22 \begin{align*} -a{\left (\frac{\log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right )}{a^{2} c} - \frac{\log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right )}{a^{2} c}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.54787, size = 111, normalized size = 4.83 \begin{align*} -\frac{\log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right ) - \log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right )}{a c} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(log(sqrt((a*x - 1)/(a*x + 1)) + 1) - log(sqrt((a*x - 1)/(a*x + 1)) - 1))/(a*c)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{\int \frac{\sqrt{\frac{a x}{a x + 1} - \frac{1}{a x + 1}}}{a x - 1}\, dx}{c} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.17058, size = 45, normalized size = 1.96 \begin{align*} \frac{\log \left ({\left | -x{\left | a \right |} + \sqrt{a^{2} x^{2} - 1} \right |}\right ) \mathrm{sgn}\left (a x + 1\right )}{c{\left | a \right |}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(abs(-x*abs(a) + sqrt(a^2*x^2 - 1)))*sgn(a*x + 1)/(c*abs(a))