### 3.162 $$\int \frac{e^{\coth ^{-1}(a x)}}{c-a c x} \, dx$$

Optimal. Leaf size=51 $\frac{2 \left (a+\frac{1}{x}\right )}{a^2 c \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{\tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{a c}$

[Out]

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

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Rubi [A]  time = 0.206206, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 16, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.438, Rules used = {6175, 6178, 852, 1805, 266, 63, 208} $\frac{2 \left (a+\frac{1}{x}\right )}{a^2 c \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{\tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{a c}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^ArcCoth[a*x]/(c - a*c*x),x]

[Out]

(2*(a + x^(-1)))/(a^2*c*Sqrt[1 - 1/(a^2*x^2)]) - 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 852

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[d^(2*m)/a
^m, Int[((f + g*x)^n*(a + c*x^2)^(m + p))/(d - e*x)^m, x], x] /; FreeQ[{a, c, d, e, f, g, n, p}, x] && NeQ[e*f
- d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[f, 0] && ILtQ[m, -1] &&  !(IGtQ[n, 0] && ILtQ[m +
n, 0] &&  !GtQ[p, 1])

Rule 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

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

Mathematica [A]  time = 0.0521113, size = 60, normalized size = 1.18 $\frac{2 a x \sqrt{1-\frac{1}{a^2 x^2}}+(1-a x) \log \left (a x \left (\sqrt{1-\frac{1}{a^2 x^2}}+1\right )\right )}{a c (a x-1)}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.132, size = 247, normalized size = 4.8 \begin{align*} -{\frac{1}{ac \left ( ax-1 \right ) } \left ( \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 ){x}^{2}{a}^{3}+\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }{x}^{2}{a}^{2}-2\,\ln \left ({\frac{{a}^{2}x+\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}{\sqrt{{a}^{2}}}} \right ) x{a}^{2}- \left ( \left ( ax-1 \right ) \left ( ax+1 \right ) \right ) ^{{\frac{3}{2}}}\sqrt{{a}^{2}}-2\,\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }xa+a\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 ) +\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) } \right ){\frac{1}{\sqrt{{a}^{2}}}}{\frac{1}{\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}}{\frac{1}{\sqrt{{\frac{ax-1}{ax+1}}}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.03232, size = 105, normalized size = 2.06 \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} - \frac{2}{a^{2} c \sqrt{\frac{a x - 1}{a x + 1}}}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [A]  time = 1.61373, size = 205, normalized size = 4.02 \begin{align*} -\frac{{\left (a x - 1\right )} \log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right ) -{\left (a x - 1\right )} \log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right ) - 2 \,{\left (a x + 1\right )} \sqrt{\frac{a x - 1}{a x + 1}}}{a^{2} c x - a c} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.18387, size = 107, normalized size = 2.1 \begin{align*} -a{\left (\frac{\log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right )}{a^{2} c} - \frac{\log \left ({\left | \sqrt{\frac{a x - 1}{a x + 1}} - 1 \right |}\right )}{a^{2} c} - \frac{2}{a^{2} c \sqrt{\frac{a x - 1}{a x + 1}}}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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